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The work done in pulling up a block of wood at constant speed weighing \$200 N\$ for length of \$10 m\$ on a smooth plane inclined at an angle of \$30^{\\circ}\$ with the horizontal is

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$\"\"$

\$\\begin{aligned} & W = mg \\sin \\theta \\times s = mg \\sin 30^{\\circ} \\times s \\\\ & W =200 \\times \\frac{1}{2} \\times 10=10^3 J =1 kJ \\end{aligned}\$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=co1V2EJlRHk"}},"videoType":"youtube","videoDetails":{"name":"The work done in pulling up a block of wood weighing \$200 N\$ for length of \$10 m\$ on a smoot....","image":"https://i.ytimg.com/vi/co1V2EJlRHk/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=co1V2EJlRHk","description":"The work done in pulling up a block of wood weighing \$200 N\$ for length of \$10 m\$ on a smooth plane inclined at an angle of \$30^{\\circ}\$ with the horizontal is 📲PW App Link - https://bit.ly/YTA","duration":"3:55","publishedAt":"2024-07-18T09:41:45Z","_id":"68afd47048ef5f8c860f5437","createdAt":"2025-08-28T04:00:48.405Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"e3curcc4qfkuy2wwedimid3ys","qbgId":"8jl5sjjlr1w70hq5wqre6tsla","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950efffd"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":43.09,"score":1.51,"timeTaken":139,"options":[{"optionId":"6adb4867d589b45d950efffe","markedPercentile":7.19},{"optionId":"6adb4867d589b45d950efffc","markedPercentile":9.96},{"optionId":"6adb4867d589b45d950efffd","markedPercentile":43.09},{"optionId":"6adb4867d589b45d950efffb","markedPercentile":4.52}]}},{"question":{"_id":"6adb4867d589b45d950f0009","type":"Single","questionNumber":2,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

In uniform circular motion,

"},"options":[{"_id":"6adb4867d589b45d950f0036","texts":{"en":"

Both velocity and acceleration are constant,

"}},{"_id":"6adb4867d589b45d950f0037","texts":{"en":"

Velocity changes and acceleration is constant,

"}},{"_id":"6adb4867d589b45d950f0038","texts":{"en":"

Velocity is constant and acceleration changes,

"}},{"_id":"6adb4867d589b45d950f0039","texts":{"en":"

Both, velocity and acceleration change.

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Uniform Circular Motion (Expression for Centripetal Acceleration, Angular Displacement, Angular Speed, Time Period, And Frequency)","_id":"khj89ed9k46gtw7o3y46sgmed"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"68x4gamsqqc7o9nof51ndd35y","solutions":["6adb4867d589b45d950f0039"],"solutionDescription":[{"_id":"6adb4867d589b45d950f000a","texts":{"en":"

Both, velocity and acceleration change.

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=Xl1zsPseRl0"}},"videoType":"youtube","videoDetails":{"name":"In uniform circular motion,\n(A) Both velocity and acceleration are constant,\n(B) Velocity change...","image":"https://i.ytimg.com/vi/Xl1zsPseRl0/mqdefault.jpg","embedCode":"https://www.youtube.com/embed/Xl1zsPseRl0","description":"In uniform circular motion, (A) Both velocity and acceleration are constant, (B) Velocity changes and acceleration is constant, (C) Velocity is constant and acceleration changes, (D) Both, velocity an","duration":"3:56","publishedAt":"2022-11-16T15:57:59Z","_id":"669d139adabbeb0018a24362","createdAt":"2024-07-21T13:56:42.315Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5apsjaa0ck0iz7srib1p3jv0z","qbgTopicId":"khj89ed9k46gtw7o3y46sgmed","qbgId":"12a09f1tjxkhxpejvrrp4s6g0","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0039"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":37.06,"score":1.06,"timeTaken":45,"options":[{"optionId":"6adb4867d589b45d950f0038","markedPercentile":7.01},{"optionId":"6adb4867d589b45d950f0036","markedPercentile":12.47},{"optionId":"6adb4867d589b45d950f0039","markedPercentile":37.06},{"optionId":"6adb4867d589b45d950f0037","markedPercentile":22.57}]}},{"question":{"_id":"6adb4867d589b45d950efff8","type":"Single","questionNumber":3,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If atmosphere exerts only frictional force on a moving block which is given as \$ \\overrightarrow{\\mathrm{F}}=-\\mathrm{k} \\overrightarrow{\\mathrm{v}} \$ then what is the maximum possible rate of heat loss if a ball of mass \$ \\mathrm{m} \$ is dropped from sufficient height under gravity:

"},"options":[{"_id":"6adb4867d589b45d950f0022","texts":{"en":"

\$ \\frac{\\mathrm{m}^{2} \\mathrm{~g}^{2}}{\\mathrm{k}} \$

"}},{"_id":"6adb4867d589b45d950f0023","texts":{"en":"

\$ \\frac{\\mathrm{mg}}{\\mathrm{k}} \$

"}},{"_id":"6adb4867d589b45d950f0024","texts":{"en":"

\$ \\frac{\\mathrm{k}}{\\mathrm{m}^{2} \\mathrm{~g}^{2}} \$

"}},{"_id":"6adb4867d589b45d950f0025","texts":{"en":"

\$ \\frac{m^{4} g^{4}}{k^{3}} \$

"}}],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Power","_id":"j3nppv2meqbk2bwn0qsrgf61o"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"2iqy456cyqb4c47v2ucyfiuaw","solutions":["6adb4867d589b45d950f0022"],"solutionDescription":[{"_id":"6adb4867d589b45d950efff9","texts":{"en":"

ma = mg – kv (v is max when a = 0)

V_max = $\frac{mg}{k}$

P_loss = $\vec{F} . \vec{v} = - {kv}^{2}$

$=- \frac{m^{2} g^{2}}{k}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=D2KY7Qc_M0w"}},"videoType":"youtube","videoDetails":{"name":"If atmosphere exerts only frictional force on a moving block which ...","image":"https://i.ytimg.com/vi/D2KY7Qc_M0w/mqdefault.jpg","embedCode":"https://www.youtube.com/embed/D2KY7Qc_M0w","description":"If atmosphere exerts only frictional force on a moving block which is given as \$ \\overrightarrow{\\mathrm{F}}=-\\mathrm{k} \\overrightarrow{\\mathrm{v}} \$ then \$ \\mathrm{P} \$ what is the maximum possi","duration":"4:53","publishedAt":"2023-02-13T10:11:14Z","_id":"66a891352a58800018e726e6","createdAt":"2024-07-30T07:07:33.887Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"j3nppv2meqbk2bwn0qsrgf61o","qbgId":"d0fcakqq2pw38llugnbu48mgd","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0022"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":18.76,"score":0.65,"timeTaken":84,"options":[{"optionId":"6adb4867d589b45d950f0024","markedPercentile":3.23},{"optionId":"6adb4867d589b45d950f0025","markedPercentile":1.07},{"optionId":"6adb4867d589b45d950f0022","markedPercentile":18.76},{"optionId":"6adb4867d589b45d950f0023","markedPercentile":5.94}]}},{"question":{"_id":"6adb4867d589b45d950efff5","type":"Single","questionNumber":4,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A stone is tied to a massless string of length 2.5 m and whirled in a vertical circle. At the instant the string is horizontal, the tension in the string is exactly equal to the weight of the stone. The speed of the stone at this instant is: (Take $g = 10 {m/s}^{2}$ )

"},"options":[{"_id":"6adb4867d589b45d950f001d","texts":{"en":"

5 m/s

"}},{"_id":"6adb4867d589b45d950f001e","texts":{"en":"

10 m/s

"}},{"_id":"6adb4867d589b45d950f001f","texts":{"en":"

15 m/s

"}},{"_id":"6adb4867d589b45d950f0020","texts":{"en":"

25 m/s

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Vertical Circular Motion","_id":"vcu9zds4bq3wvxp1d4sc59eey"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"kx5astfaeicx2n0dfub455r36","solutions":["6adb4867d589b45d950f001d"],"solutionDescription":[{"_id":"6adb4867d589b45d950efff6","texts":{"en":"

At the horizontal position, the tension provides the centripetal force. $T = \frac{m v^{2}}{L}$

Given $T = m g$ (Weight):

$\Rightarrow m g = \frac{m v^{2}}{L} \Rightarrow v^{2} = g L$

$v = \sqrt{10 \times 2.5} = \sqrt{25} = 5 m/s$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/embed/75VOY1p33b0"}},"videoType":"youtube","videoDetails":{"name":"Phase 1 Class 11th NRTS14 English Physics....","image":"https://i.ytimg.com/vi/75VOY1p33b0/mqdefault.jpg","embedCode":"https://www.youtube.com/embed/75VOY1p33b0","description":"Class 11th NRTS14 English Physics VIDEO SOLUTION 🎓Disha Mentorship - https://disha.live 📘More Questions: https://pwbooks-app.onelink.me/y2BH/hpkmntlw 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐P","duration":"1:19","publishedAt":"2026-03-21T14:18:16Z","_id":"69fec2a220f7ba65c52cabc4","createdAt":"2026-05-09T05:14:10.839Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5apsjaa0ck0iz7srib1p3jv0z","qbgTopicId":"vcu9zds4bq3wvxp1d4sc59eey","qbgId":"4i958g9g7kcjfzj9kbw3ouu00","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f001d"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":41.1,"score":1.51,"timeTaken":114,"options":[{"optionId":"6adb4867d589b45d950f001e","markedPercentile":5.46},{"optionId":"6adb4867d589b45d950f001d","markedPercentile":41.1},{"optionId":"6adb4867d589b45d950f001f","markedPercentile":3.7},{"optionId":"6adb4867d589b45d950f0020","markedPercentile":3.86}]}},{"question":{"_id":"6adb4867d589b45d950f0003","type":"Single","questionNumber":5,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A body of mass 2 kg is moving under the influence of a central force whose potential energy is given by U = 2r³J. If the body is moving in a circular orbit of 5 m, its energy will be:

"},"options":[{"_id":"6adb4867d589b45d950f002c","texts":{"en":"

625 J

"}},{"_id":"6adb4867d589b45d950f002d","texts":{"en":"

250 J

"}},{"_id":"6adb4867d589b45d950f002e","texts":{"en":"

500 J

"}},{"_id":"6adb4867d589b45d950f002f","texts":{"en":"

125 J

"}}],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Energy","_id":"62f4rf88qwu8ewxgp5dvl9y39"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","solutions":["6adb4867d589b45d950f002c"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0004","texts":{"en":"

$F = - \frac{d U}{d r} = - 6 r^{2}$

$Now \frac{m v^{2}}{r} = 6 r^{2}$

$∴ \frac{1}{2} m v^{2} = 3 r^{3}$

E = K + U = 3r³ + 2r³ = 5r³

At r = 5m, E = 625 J

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/481eeb31-24e9-4ec3-a45f-053652bdd152/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6683bef83bbca60018aaac1e","id":"6683bef83bbca60018aaac1e","name":"1.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/481eeb31-24e9-4ec3-a45f-053652bdd152/master.mpd","duration":"00:02:35","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-07-02T08:49:00.961Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"62f4rf88qwu8ewxgp5dvl9y39","qbgId":"0ktr9rjqcrxnhwxcuk710f4r0","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f002c"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":22.8,"score":0.69,"timeTaken":97,"options":[{"optionId":"6adb4867d589b45d950f002d","markedPercentile":14},{"optionId":"6adb4867d589b45d950f002f","markedPercentile":2.84},{"optionId":"6adb4867d589b45d950f002c","markedPercentile":22.8},{"optionId":"6adb4867d589b45d950f002e","markedPercentile":5.12}]}},{"question":{"_id":"6adb4867d589b45d950f0000","type":"Single","questionNumber":6,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If a stone is thrown up vertically and return to ground, its potential energy is maximum

"},"options":[{"_id":"6adb4867d589b45d950f0027","texts":{"en":"

During the upward journey

"}},{"_id":"6adb4867d589b45d950f0028","texts":{"en":"

At the maximum height

"}},{"_id":"6adb4867d589b45d950f0029","texts":{"en":"

During the return journey

"}},{"_id":"6adb4867d589b45d950f002a","texts":{"en":"

At the bottom

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"The Concept of Potential Energy","_id":"jtt2wisyxm146cgl2q7s54qe5"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"tkel1cr5vgscyxu33r7aop4cx","solutions":["6adb4867d589b45d950f0028"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0001","texts":{"en":"

Potential energy \$ =m g h \$

Potential energy is maximum when \$ h \$ is maximum

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/d846c7be-fd70-4f2a-8e81-8728d4e9266e/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"634559fce33baa001808f628","id":"634559fce33baa001808f628","name":"DPPP11ND1C6Q117","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/d846c7be-fd70-4f2a-8e81-8728d4e9266e/master.mpd","duration":"00:01:46","status":"Ready","types":["DASH","HLS"],"createdAt":"2022-10-11T11:56:44.812Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"jtt2wisyxm146cgl2q7s54qe5","qbgId":"2bzfe89gdlt2mdc83glwn0xzq","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0028"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":60.72,"score":2.32,"timeTaken":29,"options":[{"optionId":"6adb4867d589b45d950f0028","markedPercentile":60.72},{"optionId":"6adb4867d589b45d950f0027","markedPercentile":2.57},{"optionId":"6adb4867d589b45d950f0029","markedPercentile":4.68},{"optionId":"6adb4867d589b45d950f002a","markedPercentile":3.82}]}},{"question":{"_id":"6adb4867d589b45d950effef","type":"Single","questionNumber":7,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A pendulum bob has a speed \$ 3 \\mathrm{~m} / \\mathrm{s} \$ while passing through its lowest position, length of the pendulum is \$ 0.5 \\mathrm{~m} \$ then its speed when it makes an angle of \$ 60^{\\circ} \$ with the vertical is :- \$ \\left(\\mathrm{g}=10 \\mathrm{~m} / \\mathrm{s}^{2}\\right) \$

"},"options":[{"_id":"6adb4867d589b45d950f0011","texts":{"en":"

\$ 2 \\mathrm{~m} / \\mathrm{s} \$

"}},{"_id":"6adb4867d589b45d950f0012","texts":{"en":"

\$ 1 \\mathrm{~m} / \\mathrm{s} \$

"}},{"_id":"6adb4867d589b45d950f0013","texts":{"en":"

\$ 4 \\mathrm{~m} / \\mathrm{s} \$

"}},{"_id":"6adb4867d589b45d950f0014","texts":{"en":"

\$ 3 \\mathrm{~m} / \\mathrm{s} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Motion in a Vertical Circle","_id":"zmqbeskxldg8wbl68vopabhby"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"54zuucbhs66yumv9xw90rl0vu","solutions":["6adb4867d589b45d950f0011"],"solutionDescription":[{"_id":"6adb4867d589b45d950efff0","texts":{"en":"

At the Lowest Position:

Speed $v_0 = 3$  m/s.

Kinetic Energy: $\\displaystyle K_0 = \\tfrac{1}{2}m(3^2)=4.5\\,m$ .

Taking the lowest point as zero potential, total energy: $E_0=4.5\\,m$ .

At $60^\\circ$ with the Vertical:

For a pendulum of length $L=0.5$  m, the vertical rise is

$\\Delta h = L - L\\cos60^\\circ = 0.5 - 0.5 \\times 0.5 = 0.25\\,\\text{m}.$

Gain in potential energy: $\\Delta U = mg\\Delta h = m \\times 10 \\times 0.25 = 2.5\\,m$ .

Energy Conservation:
Total energy remains constant, so at $60^\\circ$ the kinetic energy is

$K = E_0 - \\Delta U = 4.5\\,m - 2.5\\,m = 2\\,m.$

Equate kinetic energy to $\\tfrac{1}{2}m\\,v^2$ :

$\\tfrac{1}{2}m\\,v^2 = 2\\,m \\quad\\Longrightarrow\\quad v^2=4 \\quad\\Longrightarrow\\quad v=2\\,\\text{m/s}.$

Thus, the speed at $60^\\circ$ from the vertical is 2 m/s.

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/50286406-8566-4ad7-a3d5-8f59ecb00d67/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"66ac8e744546ff001865560e","id":"66ac8e744546ff001865560e","name":"Q2.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/50286406-8566-4ad7-a3d5-8f59ecb00d67/master.mpd","duration":"00:03:02","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-08-02T07:44:56.887Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"zmqbeskxldg8wbl68vopabhby","qbgId":"easr2a1gk72mn7bft2ovc7wvr","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0011"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":39.73,"score":1.46,"timeTaken":123,"options":[{"optionId":"6adb4867d589b45d950f0014","markedPercentile":2.31},{"optionId":"6adb4867d589b45d950f0028","markedPercentile":60.72},{"optionId":"6adb4867d589b45d950f0011","markedPercentile":39.73},{"optionId":"6adb4867d589b45d950f0013","markedPercentile":5.46},{"optionId":"6adb4867d589b45d950f0012","markedPercentile":5.09}]}},{"question":{"_id":"6adb4867d589b45d950effec","type":"Single","questionNumber":8,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A car runs at constant speed on a circular track of radius \$ 100 \\mathrm{~m} \$ taking \$ 62.8 \\mathrm{~s} \$ on each lap. What is the average speed and average velocity on each complete lap?

"},"options":[{"_id":"6adb4867d589b45d950f000c","texts":{"en":"

Velocity \$ 10 \\mathrm{~m} / \\mathrm{s} \$ speed \$ 10 \\mathrm{~m} / \\mathrm{s} \$

"}},{"_id":"6adb4867d589b45d950f000d","texts":{"en":"

Velocity zero, speed \$ 10 \\mathrm{~m} / \\mathrm{s} \$

"}},{"_id":"6adb4867d589b45d950f000e","texts":{"en":"

Velocity zero, speed zero

"}},{"_id":"6adb4867d589b45d950f000f","texts":{"en":"

Velocity \$ 10 \\mathrm{~m} / \\mathrm{s} \$, speed zero

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Uniform Circular Motion (Expression for Centripetal Acceleration, Angular Displacement, Angular Speed, Time Period, And Frequency)","_id":"khj89ed9k46gtw7o3y46sgmed"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"68x4gamsqqc7o9nof51ndd35y","solutions":["6adb4867d589b45d950f000d"],"solutionDescription":[{"_id":"6adb4867d589b45d950effed","texts":{"en":"

Given:
Radius $r = 100 \\, \\text{m}$ ,
Time for one lap $t = 62.8 \\, \\text{s}$

Circumference $= 2\\pi r = 2 \\times \\pi \\times 100 = 628 \\, \\text{m}$

Average speed:

$\\text{Average speed} = \\frac{\\text{Total distance}}{\\text{Total time}} = \\frac{628}{62.8} = 10 \\, \\text{m/s}$

Average velocity:
Since displacement over a full lap is zero (returns to starting point),

$\\text{Average velocity} = 0$

In a circular path, though speed is constant, average velocity over a complete lap is zero due to zero net displacement.

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/ac4b0554-9add-4e13-82c2-d09fe5b362cb/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"686037f178d1e566c209274b","id":"686037f178d1e566c209274b","name":"lm8gqfyc7oxzycpo0tfi7h800","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/ac4b0554-9add-4e13-82c2-d09fe5b362cb/master.mpd","description":"

youtube_video/lm8gqfyc7oxzycpo0tfi7h800.mp4

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A body slides from an initial height of 10 m, what should be the radius ( $R$ ) of the circular loop so that the body just completes the loop.

$\"\"$

"},"options":[{"_id":"6adb4867d589b45d950f0018","texts":{"en":"

2 m

"}},{"_id":"6adb4867d589b45d950f0019","texts":{"en":"

3 m

"}},{"_id":"6adb4867d589b45d950f001a","texts":{"en":"

4 m

"}},{"_id":"6adb4867d589b45d950f001b","texts":{"en":"

5 m

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Motion in a Vertical Circle","_id":"zmqbeskxldg8wbl68vopabhby"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"txov3nd00cd13z9jcanr86jec","solutions":["6adb4867d589b45d950f001a"],"solutionDescription":[{"_id":"6adb4867d589b45d950efff3","texts":{"en":"

\$mgh =\\frac{1}{2} mv ^2 \\quad[\$ at the lowest point \$v =\\sqrt{5 gR }]\$
\\[
\\Rightarrow h =\\frac{5 R }{2}
\\]

$\"\"$

\$\\begin{aligned} & \\therefore R=\\frac{2}{5} h \\\\ & R=4 m\\end{aligned}\$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=_4s_L-Fd0lY"}},"videoType":"youtube","videoDetails":{"name":"A body slides from an initial height of 10 m, what should be the radius of the circular loop so ....","image":"https://i.ytimg.com/vi/_4s_L-Fd0lY/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=_4s_L-Fd0lY","description":"A body slides from an initial height of 10 m, what should be the radius of the circular loop so that the body just completes the loop. 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://","duration":"5:29","publishedAt":"2024-08-26T10:34:32Z","_id":"684fd079b74ac2d88b9d9127","createdAt":"2025-06-16T08:06:17.483Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"zmqbeskxldg8wbl68vopabhby","qbgId":"f1g0hlpfiywrf826gkmljv0qo","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f001a"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":29.43,"score":0.98,"timeTaken":83,"options":[{"optionId":"6adb4867d589b45d950f001a","markedPercentile":29.43},{"optionId":"6adb4867d589b45d950f001b","markedPercentile":9.98},{"optionId":"6adb4867d589b45d950f0018","markedPercentile":5.3},{"optionId":"6adb4867d589b45d950f0019","markedPercentile":4.22}]}},{"question":{"_id":"6adb4867d589b45d950f0006","type":"Single","questionNumber":10,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

Two particles of equal masses are revolving in circular paths of radii r₁ and r₂ respectively with the same speed. The ratio of their centripetal forces is

"},"options":[{"_id":"6adb4867d589b45d950f0031","texts":{"en":"

$\frac{r_{2}}{r_{1}}$

"}},{"_id":"6adb4867d589b45d950f0032","texts":{"en":"

$\sqrt{\frac{r_{2}}{r_{1}}}$

"}},{"_id":"6adb4867d589b45d950f0033","texts":{"en":"

${(\frac{r_{1}}{r_{2}})}^{2}$

"}},{"_id":"6adb4867d589b45d950f0034","texts":{"en":"

${(\frac{r_{2}}{r_{1}})}^{2}$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Uniform Circular Motion (Expression for Centripetal Acceleration, Angular Displacement, Angular Speed, Time Period, And Frequency)","_id":"khj89ed9k46gtw7o3y46sgmed"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"mu8mgvpphqgnutnhta295n1b9","solutions":["6adb4867d589b45d950f0031"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0007","texts":{"en":"

$m_{1} = m_{2}$

$v_{1} = v_{2}$

$F_{1} = \frac{m_{1} v_{1}^{2}}{r_{1}}, F_{2} = \frac{m_{2} v_{2}^{2}}{r_{2}}$

$∴ \frac{F_{1}}{F_{2}} = \frac{r_{2}}{r_{1}}$

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A string is used to pull a block of mass $m$ vertically up by a distance $h$ at a constant acceleration $\frac{g}{3}$ .The work done by the tension in the string is

"},"options":[{"_id":"6adb4867d589b45d950f008d","texts":{"en":"

\$ \\frac{2}{3} m g h \$

"}},{"_id":"6adb4867d589b45d950f008e","texts":{"en":"

\$ \\frac{-m g h}{3} \$

"}},{"_id":"6adb4867d589b45d950f008f","texts":{"en":"

\$ m g h \$

"}},{"_id":"6adb4867d589b45d950f0090","texts":{"en":"

\$ \\frac{4}{3} m g h \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Calculation of Work","_id":"e3curcc4qfkuy2wwedimid3ys"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","solutions":["6adb4867d589b45d950f0090"],"solutionDescription":[{"_id":"6adb4867d589b45d950f006a","texts":{"en":"

\$\\begin{aligned} & T-m g=m a \\\\ & T=m(g+a) \\\\ & =\\frac{4}{3} m g \\\\ & \\text { Work }(\\mathrm{w})=T . h \\\\ & =\\frac{4}{3} m g h\\end{aligned}\$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=q6TeC5IxHJc"}},"videoType":"youtube","videoDetails":{"name":"A string is used to pull a block of mass \$ m \$ vertically up by a...","image":"https://i.ytimg.com/vi/q6TeC5IxHJc/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=q6TeC5IxHJc","description":"A string is used to pull a block of mass \$ m \$ vertically up by a distance \$ h \$ at a constant acceleration \$ \\frac{g}{3} \$. PIV The work done by the tension in the string is (1) \\( \\frac{2}{3} ","duration":"2:46","_id":"6357eccd45b15f0011061b32","createdAt":"2022-10-25T14:03:57.265Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"e3curcc4qfkuy2wwedimid3ys","qbgId":"suvned30ibt0daoqe4fqhg53u","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0090"],"markedSolutionText":"","score":4,"timeTaken":3},"averageResult":{"correctPercentile":34.89,"score":1.14,"timeTaken":83,"options":[{"optionId":"6adb4867d589b45d950f008e","markedPercentile":14.06},{"optionId":"6adb4867d589b45d950f008f","markedPercentile":4.46},{"optionId":"6adb4867d589b45d950f008d","markedPercentile":6.54},{"optionId":"6adb4867d589b45d950f0090","markedPercentile":34.89}]}},{"question":{"_id":"6adb4867d589b45d950f0057","type":"Single","questionNumber":12,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The kinetic energy of a body of mass \$ 2 \\mathrm{~kg} \$ and momentum of \$ 2 \\mathrm{Ns} \$ is

"},"options":[{"_id":"6adb4867d589b45d950f006f","texts":{"en":"

\$ 1 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f0070","texts":{"en":"

\$ 2 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f0071","texts":{"en":"

\$ 3 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f0072","texts":{"en":"

\$ 4 \\mathrm{~J} \$

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Energy","_id":"62f4rf88qwu8ewxgp5dvl9y39"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"mbc6ij8551io26ii0a18zs5lh","solutions":["6adb4867d589b45d950f006f"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0058","texts":{"en":"

We are given:

\n
Mass $m = 2 \\, \\text{kg}$
\n
\n
Momentum $p = 2 \\, \\text{Ns}$
\n

We need to find the kinetic energy (KE) using the formula:

$KE = \\frac{p^2}{2m}$

Step-by-step:

$KE = \\frac{(2)^2}{2 \\times 2} = \\frac{4}{4} = 1 \\, \\text{J}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=LBgF-SpfmgQ"}},"videoType":"youtube","videoDetails":{"name":"The kinetic energy of a body of mass \$ 2 \\mathrm{~kg} \$ and momen...","image":"https://i.ytimg.com/vi/LBgF-SpfmgQ/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=LBgF-SpfmgQ","description":"The kinetic energy of a body of mass \$ 2 \\mathrm{~kg} \$ and momentum of \$ 2 \\mathrm{Ns} \$ is (A) \$ 1 \\mathrm{~J} \$ (B) \$ 2 \\mathrm{~J} \$ (C) \$ 3 \\mathrm{~J} \$ (D) \$ 4 \\mathrm{~J} \$ \$ P \$","duration":"1:37","_id":"635d265308c8900018df8979","createdAt":"2022-10-29T13:10:43.704Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"62f4rf88qwu8ewxgp5dvl9y39","qbgId":"fssjn7vhcu8h3rjjetcfbv9eo","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f006f"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":58.07,"score":2.22,"timeTaken":36,"options":[{"optionId":"6adb4867d589b45d950f0070","markedPercentile":4.93},{"optionId":"6adb4867d589b45d950f006f","markedPercentile":58.07},{"optionId":"6adb4867d589b45d950f0071","markedPercentile":2.09},{"optionId":"6adb4867d589b45d950f0072","markedPercentile":3.13}]}},{"question":{"_id":"6adb4867d589b45d950f006c","type":"Single","questionNumber":13,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $\sqrt{3 g r}$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

"},"options":[{"_id":"6adb4867d589b45d950f0092","texts":{"en":"

4 : 3

"}},{"_id":"6adb4867d589b45d950f0093","texts":{"en":"

5 : 4

"}},{"_id":"6adb4867d589b45d950f0094","texts":{"en":"

1 : 4

"}},{"_id":"6adb4867d589b45d950f0095","texts":{"en":"

3 : 2

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Vertical Circular Motion","_id":"vcu9zds4bq3wvxp1d4sc59eey"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"kx5astfaeicx2n0dfub455r36","solutions":["6adb4867d589b45d950f0094"],"solutionDescription":[{"_id":"6adb4867d589b45d950f006d","texts":{"en":"

Tension at the highest point

$T_{top} = \frac{m v^{2}}{r} - m g = 2 m g (∴ v_{top} = \sqrt{3 g r})$

Tension at the lowest point

$T_{bottom} = 2 mg + 6 mg = 8 mg$

$∴ \frac{T_{top}}{T_{bottom}} = \frac{2 mg}{8 mg} = \frac{1}{4}$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/0154f514-4884-46cf-a4df-b79d3a142a69/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"676fc230d300931e92d33897","id":"676fc230d300931e92d33897","name":"15 (2).mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/0154f514-4884-46cf-a4df-b79d3a142a69/master.mpd","duration":"00:06:10","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-12-28T09:17:42.468Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5apsjaa0ck0iz7srib1p3jv0z","qbgTopicId":"vcu9zds4bq3wvxp1d4sc59eey","qbgId":"9wjfyk8frvcvjew08vu5lbt1b","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0094"],"markedSolutionText":"","score":4,"timeTaken":1},"averageResult":{"correctPercentile":30.19,"score":1.09,"timeTaken":102,"options":[{"optionId":"6adb4867d589b45d950f0092","markedPercentile":4.22},{"optionId":"6adb4867d589b45d950f0094","markedPercentile":30.19},{"optionId":"6adb4867d589b45d950f0095","markedPercentile":3.98},{"optionId":"6adb4867d589b45d950f0093","markedPercentile":3.44}]}},{"question":{"_id":"6adb4867d589b45d950f0066","type":"Single","questionNumber":14,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A force acts on a 3.0 gm particle in such a way that the position of the particle as a function of time is given by x = 3t – 4t² + t³, where x is in metres and t is in seconds. The work done during the first 4 second is

"},"options":[{"_id":"6adb4867d589b45d950f0088","texts":{"en":"

450 mJ

"}},{"_id":"6adb4867d589b45d950f0089","texts":{"en":"

528 mJ

"}},{"_id":"6adb4867d589b45d950f008a","texts":{"en":"

490 mJ

"}},{"_id":"6adb4867d589b45d950f008b","texts":{"en":"

570 mJ

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"The Work-Energy Theorem","_id":"inks5zjayuix5hobkc7tk53aq"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"cvlkwahvz8llro3c2qia7at5s","solutions":["6adb4867d589b45d950f0089"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0067","texts":{"en":"

x = 3t – 4t² + t³

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

At t = 0, v₁ = 3 m/s

At t = 4s, v₂ = 19 m/s

Work done = $\frac{1}{2} m (v_{2}^{2} - v_{1}^{2})$

$= \frac{1}{2} \times 3 \times 10^{- 3} ({(19)}^{2} - {(3)}^{2})$

= 528 × 10^–3 J

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/831a83b0-1fd1-4e60-a60f-f2550237c0a6/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"67664313cd7ffc536360717d","id":"67664313cd7ffc536360717d","name":"11.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/831a83b0-1fd1-4e60-a60f-f2550237c0a6/master.mpd","duration":"00:02:16","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-12-21T04:24:55.021Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"inks5zjayuix5hobkc7tk53aq","qbgId":"vyk575txozg7ai9e3km2bcgcb","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0089"],"markedSolutionText":"","score":4,"timeTaken":6},"averageResult":{"correctPercentile":24.22,"score":0.82,"timeTaken":167,"options":[{"optionId":"6adb4867d589b45d950f0089","markedPercentile":24.22},{"optionId":"6adb4867d589b45d950f008b","markedPercentile":8.74},{"optionId":"6adb4867d589b45d950f0088","markedPercentile":2.83},{"optionId":"6adb4867d589b45d950f008a","markedPercentile":3.17}]}},{"question":{"_id":"6adb4867d589b45d950f0060","type":"Single","questionNumber":15,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A 10 kg block placed on a rough horizontal floor is being pulley by a constant force of 50 N. Coefficient of kinetic friction between the block and the floor is 0.4. The total work done by all forces acting on the block over a displacement of
5 m is: (g = 10 m/s²)

$\"\"$

"},"options":[{"_id":"6adb4867d589b45d950f007e","texts":{"en":"

50 J

"}},{"_id":"6adb4867d589b45d950f007f","texts":{"en":"

250 J

"}},{"_id":"6adb4867d589b45d950f0080","texts":{"en":"

200 J

"}},{"_id":"6adb4867d589b45d950f0081","texts":{"en":"

450 J

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Work Done by Friction on System","_id":"32bzuqt8layj5z7tj3j0kcddd"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"1bzkg74hte0u2gw3woq0z7u0l","solutions":["6adb4867d589b45d950f007e"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0061","texts":{"en":"

$\"\"$

Total work done = (50 – 40) × 5 = 50 J

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=bgiOj1Q62Uc"}},"videoType":"youtube","videoDetails":{"name":"A 10 kg block placed on a rough horizontal floor is being pulley by a constant force of 50 N. Co....","image":"https://i.ytimg.com/vi/bgiOj1Q62Uc/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=bgiOj1Q62Uc","description":"A 10 kg block placed on a rough horizontal floor is being pulley by a constant force of 50 N. Coefficient of kinetic friction between the block and the floor is 0.4. The total work done by all forces ","duration":"0:54","publishedAt":"2025-01-17T17:23:22Z","_id":"6adb762cbc3d3b3dfbaeeb12","createdAt":"2026-10-23T14:58:52.006Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"32bzuqt8layj5z7tj3j0kcddd","qbgId":"enhm6ufamsil4br7dsmedqapb","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f007e"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":43.1,"score":1.57,"timeTaken":89,"options":[{"optionId":"6adb4867d589b45d950f007f","markedPercentile":6.97},{"optionId":"6adb4867d589b45d950f0081","markedPercentile":3.39},{"optionId":"6adb4867d589b45d950f0088","markedPercentile":2.83},{"optionId":"6adb4867d589b45d950f0080","markedPercentile":5.37},{"optionId":"6adb4867d589b45d950f007e","markedPercentile":43.1}]}},{"question":{"_id":"6adc6bacea69ba4651d6ebbe","type":"Single","questionNumber":16,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","imageIds":{"en":{"_id":"6adc6baaadc9f324fbec1556","name":"QUES_ENG_3gz3w18u7u6sqrw9se4k3igwy.png","baseUrl":"https://static.pw.live/","key":"5b09189f7285894d9130ccd0/5e33309c-4ae3-4af4-9d35-58ae1504e1ef.png"}},"options":[{"_id":"6adc6bacea69ba4651d6ebc1","texts":{"en":"A"}},{"_id":"6adc6bacea69ba4651d6ebc2","texts":{"en":"B"}},{"_id":"6adc6bacea69ba4651d6ebc3","texts":{"en":"C"}},{"_id":"6adc6bacea69ba4651d6ebc4","texts":{"en":"D"}}],"difficultyLevel":1,"pyqYears":[],"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","solutions":["6adc6bacea69ba4651d6ebc3"],"solutionDescription":[{"_id":"6adc6bacea69ba4651d6ebbf","imageIds":{"en":{"_id":"6adc6baa9152a7cd880bfaef","name":"SOLU_ENG_3gz3w18u7u6sqrw9se4k3igwy.png","baseUrl":"https://static.pw.live/","key":"5b09189f7285894d9130ccd0/7cb03698-1479-45c1-8812-c7daf8893790.png"}},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=qlXtIBH7l18"}},"videoType":"youtube","videoDetails":{"name":"A motor car is travelling at \$ 60 \\mathrm{~m} / \\mathrm{s} \$ on a circular road of radius \$ 1....","image":"https://i.ytimg.com/vi/qlXtIBH7l18/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=qlXtIBH7l18","description":"A motor car is travelling at \\( 60 \\mathrm{~m} / \\mathrm{s} \$ on a circular road of radius \$ 1200 \\mathrm{~m} \$. It is increasing its speed at \$ \\mathrm{P} \$ the rate of \\( 4 \\mathrm{~m} / \\mathr","duration":"1:34","publishedAt":"2024-01-24T07:53:44Z","_id":"65f322e11f117d00183838d5","createdAt":"2024-03-14T16:16:33.145Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5apsjaa0ck0iz7srib1p3jv0z","qbgId":"3gz3w18u7u6sqrw9se4k3igwy","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adc6bacea69ba4651d6ebc3"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":39.09,"score":1.45,"timeTaken":70,"options":[{"optionId":"6adc6bacea69ba4651d6ebc4","markedPercentile":1.49},{"optionId":"6adb4867d589b45d950f007f","markedPercentile":6.97},{"optionId":"6adc6bacea69ba4651d6ebc3","markedPercentile":39.09},{"optionId":"6adc6bacea69ba4651d6ebc2","markedPercentile":4.09},{"optionId":"6adc6bacea69ba4651d6ebc1","markedPercentile":5.51}]}},{"question":{"_id":"6adb4867d589b45d950f005a","type":"Single","questionNumber":17,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

In a gravitational force field a particle is taken from $A$ to $B$ along different paths as shown in figure. Then

$\"\"$

"},"options":[{"_id":"6adb4867d589b45d950f0074","texts":{"en":"

Work done along path I will be maximum

"}},{"_id":"6adb4867d589b45d950f0075","texts":{"en":"

Work done along path III will be minimum

"}},{"_id":"6adb4867d589b45d950f0076","texts":{"en":"

Work done along path IV will be minimum

"}},{"_id":"6adb4867d589b45d950f0077","texts":{"en":"

Work done along all the paths will be the same

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"The Work-Energy Theorem","_id":"inks5zjayuix5hobkc7tk53aq"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"cvlkwahvz8llro3c2qia7at5s","solutions":["6adb4867d589b45d950f0077"],"solutionDescription":[{"_id":"6adb4867d589b45d950f005b","texts":{"en":"

Work done along all the paths will be the same

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=a4GXCoW_uTQ"}},"videoType":"youtube","videoDetails":{"name":"In a gravitational force field a particle is taken from \$ A \$ to \$ B \$ along different paths...","image":"https://i.ytimg.com/vi/a4GXCoW_uTQ/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=a4GXCoW_uTQ","description":"In a gravitational force field a particle is taken from \$ A \$ to \$ B \$ along different paths as shown in figure. Then [MP PET 2009] (a) Work done along path I will be maximum (b) Work done along p","duration":"0:56","_id":"637a3587464a4c0011873d65","createdAt":"2022-11-20T14:11:19.133Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"inks5zjayuix5hobkc7tk53aq","qbgId":"h7z1pztvqd9rs81mgabamm75g","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0077"],"markedSolutionText":"","score":4,"timeTaken":1},"averageResult":{"correctPercentile":51.15,"score":1.93,"timeTaken":30,"options":[{"optionId":"6adb4867d589b45d950f0077","markedPercentile":51.15},{"optionId":"6adb4867d589b45d950f0074","markedPercentile":4},{"optionId":"6adb4867d589b45d950f0076","markedPercentile":2.39},{"optionId":"6adb4867d589b45d950f0075","markedPercentile":5.13}]}},{"question":{"_id":"6adb4867d589b45d950f005d","type":"Single","questionNumber":18,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

When a long spring is stretched by \$2\\ cm\$, its potential energy is \$U\$. if the spring is stretched by \$10\\ cm\$, the potential energy in it will be?

"},"options":[{"_id":"6adb4867d589b45d950f0079","texts":{"en":"

\$\\frac{U}{5}\$

"}},{"_id":"6adb4867d589b45d950f007a","texts":{"en":"

\$\\frac{U}{25}\$

"}},{"_id":"6adb4867d589b45d950f007b","texts":{"en":"

\$5U\$

"}},{"_id":"6adb4867d589b45d950f007c","texts":{"en":"

\$25U\$

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"The Concept of Potential Energy","_id":"jtt2wisyxm146cgl2q7s54qe5"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"z1b2lu3h30lbejg97gudjvw5x","solutions":["6adb4867d589b45d950f007c"],"solutionDescription":[{"_id":"6adb4867d589b45d950f005e","texts":{"en":"

We know that, in a spring,

\\[
\\text{Potential energy, } U = \\frac{1}{2} K X^{2}
\\]

Now, when stretched in spring \$ = 2 \\, \\text{cm} \$,

\\[
U = \\frac{1}{2} K (2)^{2} \\quad \\text{…….(1)} \\quad \\text{(Given)}
\\]

\\[
U_{1} = \\frac{1}{2} K (10)^{2} \\quad \\text{…….(2)}
\\]

From \$ \\text{eqn (2) / eqn (1)} \$, we get

\\[
U_{1} = 25U
\\]

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A particle moves in a circle of radius \$ 25 \\mathrm{~cm} \$ at two revolutions per second. The acceleration of the particle in \$ \\mathrm{m} / \\mathrm{s}^{2} \$ is

"},"options":[{"_id":"6adb4867d589b45d950f0083","texts":{"en":"

\$ \\pi^{2} \$

"}},{"_id":"6adb4867d589b45d950f0084","texts":{"en":"

\$ 8 \\pi^{2} \$

"}},{"_id":"6adb4867d589b45d950f0085","texts":{"en":"

\$ 4 \\pi^{2} \$

"}},{"_id":"6adb4867d589b45d950f0086","texts":{"en":"

\$ 2 \\pi^{2} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Uniform Circular Motion (Expression for Centripetal Acceleration, Angular Displacement, Angular Speed, Time Period, And Frequency)","_id":"khj89ed9k46gtw7o3y46sgmed"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"mu8mgvpphqgnutnhta295n1b9","solutions":["6adb4867d589b45d950f0085"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0064","texts":{"en":"

Two revolutions per second ⇒ angular speed $\\omega = 2\\pi \\times 2 = 4\\pi$  rad/s.

Radius $r = 25\\,\\text{cm} = 0.25\\,\\text{m}$ .

Centripetal acceleration:

$a = \\omega^2\\,r = (4\\pi)^2 \\times 0.25 = 16\\pi^2 \\times 0.25 = 4\\pi^2.$

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youtube_video/87qifw8zj9lkyf5t0qzn2eb3b.mp4

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A system of wedge and block as shown in figure, is released with the spring in its natural length. All surfaces are frictionless. Maximum elongation in the spring will be:

$\"\"$

"},"options":[{"_id":"6adb4867d589b45d950f00a2","texts":{"en":"

$\frac{2 m g \sin θ}{K}$

"}},{"_id":"6adb4867d589b45d950f00a3","texts":{"en":"

$\frac{m g \sin θ}{K}$

"}},{"_id":"6adb4867d589b45d950f00a4","texts":{"en":"

$\frac{4 m g \sin θ}{K}$

"}},{"_id":"6adb4867d589b45d950f00a5","texts":{"en":"

$\frac{m g \sin θ}{2 K}$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"The Conservation of Energy","_id":"smxq8xg1ktgam0255dip8hpj6"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"338alakve50rlc35n1tnk9eb3","solutions":["6adb4867d589b45d950f00a2"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00a0","texts":{"en":"

Let x be the elongation in the spring.

Increase in $P E$ of spring = Decrease in $P E$ of block

$∴ \frac{1}{2} k x^{2} = m g x \sin θ$

$\Rightarrow x = \frac{2 m g \sin θ}{k}$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/cc9d808a-e3e8-4275-aa7c-fad3c6d8d73b/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"67738432d68f488708ff0607","id":"67738432d68f488708ff0607","name":"5.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/cc9d808a-e3e8-4275-aa7c-fad3c6d8d73b/master.mpd","duration":"00:01:36","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-12-31T05:42:13.806Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"smxq8xg1ktgam0255dip8hpj6","qbgId":"slwlcg7bypkxhdv2ksrnkkgbk","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f00a2"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":36.88,"score":1.3,"timeTaken":81,"options":[{"optionId":"6adb4867d589b45d950f00a5","markedPercentile":1.49},{"optionId":"6adb4867d589b45d950f00a4","markedPercentile":2.04},{"optionId":"6adb4867d589b45d950f00a2","markedPercentile":36.88},{"optionId":"6adb4867d589b45d950f00a3","markedPercentile":13.76},{"optionId":"6adb4867d589b45d950f0085","markedPercentile":39.36}]}},{"question":{"_id":"6adb4867d589b45d950f00cb","type":"Numeric","questionNumber":21,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A small particle moves to position $(5 \hat{i} - 2 \hat{j} + \hat{k}) m$ from its initial position $(2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ m under the action of force $(5 \hat{i} + 2 \hat{j} + 7 \hat{k}) N$ . The value of work done will be _______J.

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Calculation of Work","_id":"e3curcc4qfkuy2wwedimid3ys"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"nm5uslry85ijhzuz2e8qzs2ih","solutions":[],"solutionText":"40","solutionDescription":[{"_id":"6adb4867d589b45d950f00cc","texts":{"en":"

Work done = Force. Displacement

$W = \vec{F} . \vec{r}$

Here $\vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1}$

${\vec{r}}_{2} = 5 \hat{i} - 2 \hat{j} + \hat{k}$

${\vec{r}}_{1} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$

$\vec{r} = (5 - 2) \hat{i} + (- 2 - 3) \hat{j} + (1 - (- 4)) \hat{k}$

$\vec{r} = 3 \hat{i} - 5 \hat{j} + 5 \hat{k}$

$Workdone = \vec{F} \cdot \vec{r} = (5 \hat{i} + 2 \hat{j} + 7 \hat{k}) \cdot (3 \hat{i} - 5 \hat{j} + 5 \hat{k})$

= (15 – 10 + 35)J

Workdone = 40 Joule

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/6dae699c-337f-4e46-bc13-6d5190f43f76/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"67cec6cda380fb7aeacd26f0","id":"67cec6cda380fb7aeacd26f0","name":"PTSMT8Q025.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/6dae699c-337f-4e46-bc13-6d5190f43f76/master.mpd","duration":"00:02:00","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-03-10T11:02:41.975Z","drmProtected":false}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"e3curcc4qfkuy2wwedimid3ys","qbgId":"03zew7ae58sou9exys7h6oomj","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"40","score":4,"timeTaken":4},"averageResult":{"correctPercentile":38.45,"score":1.39,"timeTaken":79,"options":[]}},{"question":{"_id":"6adb4867d589b45d950f00c8","type":"Numeric","questionNumber":22,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A block of mass m is placed at origin (0, 0) and a force, $F = (2 x + 2) N$ , is applied on the block in $x$ direction so that it reaches at some other point, $P$ (5, 0). The work done by the force is ____ $J$ .

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Calculation of Work","_id":"e3curcc4qfkuy2wwedimid3ys"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"7k38h4sctijer67kyxxz2lcwx","solutions":[],"solutionText":"35","solutionDescription":[{"_id":"6adb4867d589b45d950f00c9","texts":{"en":"

Initial position of the block, x_i = 0 m

Final position of the block, x_f = 5 m

The total work done by the variable force is defined as, $W_{t o t a l} = \int_{x_{i}}^{x_{f}} F (x) d x$

Hence,

$W_{t o t a l} = \int_{0}^{5} (2 x + 2) d x$

$\Rightarrow W_{t o t a l} = {[x^{2} + 2 x]}_{0}^{5}$

$\Rightarrow W_{t o t a l} = [{(5)}^{2} + (2 \times 5)] - [0^{2} + (2 \times 0)]$

$\Rightarrow W_{t o t a l} = 35 J$

The work done by the force is 35 J.

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=dLMkPDCoZ2I"}},"videoType":"youtube","videoDetails":{"name":"A block of mass m is placed at origin (0, 0) and a force, F=(2x+2)N, is applied on the block in ....","image":"https://i.ytimg.com/vi/dLMkPDCoZ2I/mqdefault.jpg","embedCode":"https://www.youtube.com//embed/dLMkPDCoZ2I","description":"A block of mass m is placed at origin (0, 0) and a force, F=(2x+2)N, is applied on the block in x direction so that it reaches at some other point, P (5, 0). The work done by the force is ____ J. 📲PW","duration":"1:18","publishedAt":"2024-09-18T18:56:31Z","_id":"6932e694e03f3dede258d7e2","createdAt":"2025-12-05T14:05:08.260Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"e3curcc4qfkuy2wwedimid3ys","qbgId":"yoi0c90caegmvkdy26iekvnz7","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"35","score":4,"timeTaken":3},"averageResult":{"correctPercentile":33.51,"score":1.18,"timeTaken":51,"options":[]}},{"question":{"_id":"6adb4867d589b45d950f00bf","type":"Numeric","questionNumber":23,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rads⁻¹) is x. Find x/6

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Dynamics of Circular Motion","_id":"193qf6nqvpw3vxlk3guut08vo"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"uqfa2xfsh10asfvdhox9c4iuq","solutions":[],"solutionText":"6","solutionDescription":[{"_id":"6adb4867d589b45d950f00c0","texts":{"en":"

T cos θ component will cancel mg.

$\"\"$

T sin θ component will provide necessary centripetal force to the ball towards centre C.

$\Rightarrow T \sin θ = m r ω^{2} = m (l \sin θ) ω^{2}$

$\Rightarrow T = m l ω^{2}$

$\Rightarrow ω = \sqrt{\frac{T}{m l}}$

$\Rightarrow ω_{\max} = \sqrt{\frac{T_{\max}}{m l}}$

$\Rightarrow ω_{\max} = \sqrt{\frac{324}{0.5 \times 0.5}}$

$\Rightarrow ω_{\max} = 36 {rads}^{- 1}$

So, x =36

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A body of mass \$ 0.5 \\mathrm{~kg} \$ travels in a straight line with velocity \$ v=a x^{3 / 2} \$ where \$ a=5 \\mathrm{~m}^{-1 / 2} s^{-1} \$. What is the work done by the net force during its displacement from \$ x=0 \$ to \$ x=2 \\mathrm{~m} \$ ?

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Calculation of Work","_id":"e3curcc4qfkuy2wwedimid3ys"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5m9tbebxdyzr4sfxutg5f8dk4","subTopicId":"kszi0yzgpo36jatcqhcp9p5jg","solutions":[],"solutionText":"50","solutionDescription":[{"_id":"6adb4867d589b45d950f00c6","texts":{"en":"

m = 0.5 kg

$u = a x^{\frac{3}{2}}, a = 5 m^{- \frac{1}{2}} s^{- 1}$

initial velocity at x = 0, $v_{1} = a {(0)}^{\frac{3}{2}}$ = 0

final velocity at $x = 2, v_{2} = a 2^{\frac{3}{2}} = 5 \times {(2)}^{\frac{3}{2}}$

work done = Increase in KE = $\frac{1}{2} m (v_{2}^{2} - v_{1}^{2})$ = $\frac{1}{2} 0.5 [{(5 \times 2^{3 / 2})}^{2} - 0] = 50 J$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=GJy46wYZjAc"}},"videoType":"youtube","videoDetails":{"name":"A body of mass \$ 0.5 \\mathrm{~kg} \$ travels in a straight line wi...","image":"https://i.ytimg.com/vi/GJy46wYZjAc/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=GJy46wYZjAc","description":"A body of mass \$ 0.5 \\mathrm{~kg} \$ travels in a straight line with velocity \$ v=a x^{3 / 2} \$ where \$ a=5 \\mathrm{~m}^{-1 / 2} \\varsigma^{-1} \$. What is the work done by the net force during it","duration":"2:50","_id":"63722b81463ed40a5181eb0b","createdAt":"2022-11-14T11:50:25.762Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5m9tbebxdyzr4sfxutg5f8dk4","qbgTopicId":"e3curcc4qfkuy2wwedimid3ys","qbgId":"97543g7hdu0q55u5n3nsh7s12","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"50","score":4,"timeTaken":2},"averageResult":{"correctPercentile":16.48,"score":0.55,"timeTaken":73,"options":[]}},{"question":{"_id":"6adb4867d589b45d950f00d1","type":"Numeric","questionNumber":25,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A particle moves in a circle of radius \$ 1.0 \\mathrm{~cm} \$ at a speed given by \$ \\mathrm{v}=2.0 \\mathrm{t} \$ where \$ \\mathrm{v} \$ is in \$ \\mathrm{cm} / \\mathrm{s} \$ and \$ \\mathrm{t} \$ in seconds.

Find the radial acceleration in cm/s² of the particle at \$ \\mathrm{t}=1 \\mathrm{~s} \$.

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Uniform Circular Motion (Expression for Centripetal Acceleration, Angular Displacement, Angular Speed, Time Period, And Frequency)","_id":"khj89ed9k46gtw7o3y46sgmed"},"sectionId":"6adb4674c235743130e8f044","subjectId":"iydv16vvyph4i0hajzqbqxk2t","chapterId":"5apsjaa0ck0iz7srib1p3jv0z","subTopicId":"mu8mgvpphqgnutnhta295n1b9","solutions":[],"solutionText":"4","solutionDescription":[{"_id":"6adb4867d589b45d950f00d2","texts":{"en":"

$a = \frac{v^{2}}{R} = \frac{4 t^{2}}{R} = 4 cm / s^{2}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=8W2G0numLq4"}},"videoType":"youtube","videoDetails":{"name":"A particle moves in a circle of radius \$ 1.0 \\mathrm{~cm} \$ at a ...","image":"https://i.ytimg.com/vi/8W2G0numLq4/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=8W2G0numLq4","description":"A particle moves in a circle of radius \$ 1.0 \\mathrm{~cm} \$ at a speed given by \$ \\mathrm{v}=2.0 \\mathrm{t} \$ where \$ \\mathrm{v} \$ is in \$ \\mathrm{cm} / \\mathrm{s} \$ and \$ \\mathrm{t} \$ in se","duration":"1:58","_id":"6353a05077dda40011de616e","createdAt":"2022-10-22T07:48:32.373Z"}}],"qbgSubjectId":"iydv16vvyph4i0hajzqbqxk2t","qbgChapterId":"5apsjaa0ck0iz7srib1p3jv0z","qbgTopicId":"khj89ed9k46gtw7o3y46sgmed","qbgId":"u8tii9sqpt0ov0xg4dh9os09l","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"4","score":4,"timeTaken":3},"averageResult":{"correctPercentile":32.06,"score":1.19,"timeTaken":45,"options":[]}},{"question":{"_id":"6adb4867d589b45d950f00d7","type":"Single","questionNumber":26,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

One mole of oxygen is allowed to expand isothermally and reversibly from \$ 5 \\mathrm{~m}^{3} \$ to \$ 10 \\mathrm{~m}^{3} \$ at \$ 300 \\mathrm{~K} \$. Calculate work done in expansion of gas.

"},"options":[{"_id":"6adb4867d589b45d950f00f1","texts":{"en":"

\$ -172.89 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f00f2","texts":{"en":"

\$ 1728.98 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f00f3","texts":{"en":"

\$ -1728.98 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f00f4","texts":{"en":"

\$ 172.89 \\mathrm{~J} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Pressure-Volume Work","_id":"vwz3ifjrjgj7mo07jmvxpnd1i"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"h5ritvxt9rio477smh23ule4f","solutions":["6adb4867d589b45d950f00f3"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00d8","texts":{"en":"

1. Work done for reversible isothermal process defined as W =-2.303nRT log( $\frac{V 2}{V 1}$ ).

2. W=-2.303 x 1 x 300 x 8.314 x log( $\frac{10}{5}$ )

W=-2.303 x 1 x 300 x 8.314 x log2 (log2=0.301)

W=-1728.98 J

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8eee035e-5171-4fd9-9526-60fd91a56f8a/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6860323b56136891187175fe","id":"6860323b56136891187175fe","name":"af2812whf132rsvzapduzhmmf","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8eee035e-5171-4fd9-9526-60fd91a56f8a/master.mpd","description":"

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Calculate the work done during the compression of 2 mol of an ideal gas from a volume of 1 m³ to 10 dm³ at 300 K against a pressure of 100 kPa.

"},"options":[{"_id":"6adb4867d589b45d950f00df","texts":{"en":"

+99 kJ

"}},{"_id":"6adb4867d589b45d950f00e0","texts":{"en":"

-99 kJ

"}},{"_id":"6adb4867d589b45d950f00e1","texts":{"en":"

+22.98 kJ

"}},{"_id":"6adb4867d589b45d950f00e2","texts":{"en":"

-22.98 kJ

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Pressure-Volume Work","_id":"vwz3ifjrjgj7mo07jmvxpnd1i"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4867d589b45d950f00df"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00c3","texts":{"en":"

W = -PΔV = -P (V₂ – V₁)

= -100 (10 – 1000 L)

= -100 [-990]

= +99 kJ

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/33d7dabb-1eb2-4609-83b7-cd828d4ee054/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"67dafd258383ee3d52d9f229","id":"67dafd258383ee3d52d9f229","name":"14.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/33d7dabb-1eb2-4609-83b7-cd828d4ee054/master.mpd","duration":"00:02:06","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-03-19T17:21:45.497Z","drmProtected":false}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"vwz3ifjrjgj7mo07jmvxpnd1i","qbgId":"ltl13z0rwz523n2dvc590o05c","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f00df"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":25,"score":0.86,"timeTaken":78,"options":[{"optionId":"6adb4867d589b45d950f00e1","markedPercentile":6.18},{"optionId":"6adb4867d589b45d950f00e2","markedPercentile":2.06},{"optionId":"6adb4867d589b45d950f00e0","markedPercentile":5.92},{"optionId":"6adb4867d589b45d950f00df","markedPercentile":25}]}},{"question":{"_id":"6adb4867d589b45d950f00bc","type":"Single","questionNumber":28,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

A gas expands adiabatically at constant pressure such that \$ \\mathrm{TV}^{1 / 2} \$ The value of \$ \\left(\\mathrm{C}_{\\mathrm{p}} / \\mathrm{C}_{\\mathrm{V}}\\right) \$ of the gas will be:

"},"options":[{"_id":"6adb4867d589b45d950f00da","texts":{"en":"

1.30

"}},{"_id":"6adb4867d589b45d950f00db","texts":{"en":"

1.50

"}},{"_id":"6adb4867d589b45d950f00dc","texts":{"en":"

1.70

"}},{"_id":"6adb4867d589b45d950f00dd","texts":{"en":"

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Heat Capacity","_id":"v2fivjnlwqy5ucu444oyg7ea5"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"nspm6rldgnhgazlzzod4p5hln","solutions":["6adb4867d589b45d950f00db"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00bd","texts":{"en":"

Adiabatic relation: $TV^{\\gamma - 1} = \\text{constant}$
Given: $TV^{1/2} = \\text{constant}$

$\\Rightarrow \\gamma - 1 = \\frac{1}{2} \\Rightarrow \\gamma = \\frac{3}{2} = 1.5$

So, $\\frac{C_p}{C_v} = \\gamma = 1.5$

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The standard enthalpies of formation of \$ \\mathrm{CO}_{2}(\\mathrm{~g}) \$,

\$ \\mathrm{H}_{2} \\mathrm{O}(l) \$ and glucose(s) at \$ 25^{\\circ} \\mathrm{C} \$ are \$ -400 \\mathrm{~kJ} / \\mathrm{mol},-300 \$ \$ \\mathrm{kJ} / \\mathrm{mol} \$ and \$ -1300 \\mathrm{~kJ} / \\mathrm{mol} \$, respectively. The standard enthalpy of combustion per gram of glucose at \$ 25^{\\circ} \\mathrm{C} \$ is

"},"options":[{"_id":"6adb4867d589b45d950f00e9","texts":{"en":"

\$ +2900 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f00ea","texts":{"en":"

\$ -2900 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f00eb","texts":{"en":"

\$ -16.11 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f00ec","texts":{"en":"

\$ +16.11 \\mathrm{~kJ} \$

"}}],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Enthalpies for Different Types of Reactions","_id":"1gxmt29movp1ajf2a3h1i64ay"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"joybev5wcvlapeiy311cz6um5","solutions":["6adb4867d589b45d950f00eb"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00d5","texts":{"en":"

$\begin{array}{c} C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 {CO}_{2} + 6 H_{2} O \\ △_{c} H = 6 (- 400) + 6 (- 300) - (1300) \\ = - 2400 - 1800 + 1300 \\ = - 2900 kJ \\ △_{c} H per gram = \frac{- 2900}{180} \\ = - 16.11 kJ \end{array}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=nX9kXEM1z4Q"}},"videoType":"youtube","videoDetails":{"name":"The standard enthalpies of formation of \$ \\mathrm{CO}_{2}(\\mathrm{~g}) \$,\n\$ \\mathrm{H}_{2} \\m....","image":"https://i.ytimg.com/vi/nX9kXEM1z4Q/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=nX9kXEM1z4Q","description":"The standard enthalpies of formation of \\( \\mathrm{CO}_{2}(\\mathrm{~g}) \$, \$ \\mathrm{P} \$ \$ \\mathrm{H}_{2} \\mathrm{O}(l) \$ and glucose(s) at \$ 25^{\\circ} \\mathrm{C} \$ are \\( -400 \\mathrm{~kJ} /","duration":"5:30","publishedAt":"2024-01-24T09:06:30Z","_id":"65d8b80944bf9c0018ee6f17","createdAt":"2024-02-23T15:21:45.652Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"1gxmt29movp1ajf2a3h1i64ay","qbgId":"ks9efgr4fzy8byq68uhzhwvu7","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f00eb"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":18.04,"score":0.52,"timeTaken":80,"options":[{"optionId":"6adb4867d589b45d950f00ea","markedPercentile":10.46},{"optionId":"6adb4867d589b45d950f00e9","markedPercentile":6.55},{"optionId":"6adb4867d589b45d950f00eb","markedPercentile":18.04},{"optionId":"6adb4867d589b45d950f00ec","markedPercentile":3.1}]}},{"question":{"_id":"6adb4867d589b45d950f00ce","type":"Single","questionNumber":30,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

H₂O(s) $\to_{}^{}$ H₂O(l) This phase transition is carried out at constant temp and pressure then work done during the process:

"},"options":[{"_id":"6adb4867d589b45d950f00e4","texts":{"en":"

W < 0

"}},{"_id":"6adb4867d589b45d950f00e5","texts":{"en":"

W > 0

"}},{"_id":"6adb4867d589b45d950f00e6","texts":{"en":"

W = 0

"}},{"_id":"6adb4867d589b45d950f00e7","texts":{"en":"

can’t determined

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Pressure-Volume Work","_id":"vwz3ifjrjgj7mo07jmvxpnd1i"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4867d589b45d950f00e5"],"solutionDescription":[{"_id":"6adb4867d589b45d950f00cf","texts":{"en":"

Reason: Ice → water at constant P: volume decreases (ice is less dense). Hence $Δ V < 0$ and $w = - P Δ V > 0$ — work is done on the system.

[NCERT Ed. 2025-26, 11^th Page No. 140]

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Standard entropy of \$ \\mathrm{N}_{2}, \\mathrm{H}_{2} \$ and \$ \\mathrm{NH}_{3} \$ are 60,40 and \$ 50 \\mathrm{JK}^{-1} \\mathrm{~mol}^{-1} \$ respectively.

For the reaction \$ 1 / 2 \\mathrm{~N}_{2}+ \$ \$ 3 / 2 \\mathrm{H}_{2} \\rightleftharpoons \\mathrm{NH}_{3}, \\Delta \\mathrm{H}=-30 \\mathrm{~kJ} \$ to be at equilibrium, the temperature should be:

"},"imageIds":{},"options":[{"_id":"6adb4867d589b45d950f014a","texts":{"en":"

\$ 500 \\mathrm{~K} \$

"}},{"_id":"6adb4867d589b45d950f014b","texts":{"en":"

\$ 750 \\mathrm{~K} \$

"}},{"_id":"6adb4867d589b45d950f014c","texts":{"en":"

\$ 1000 \\mathrm{~K} \$

"}},{"_id":"6adb4867d589b45d950f014d","texts":{"en":"

\$ 1250 \\mathrm{~K} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Gibbs Energy Change and Equilibrium","_id":"r2kddjmvtfrzzgxfscia8qyxc"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"juvquiu9zf7d7ouhr9a5dyqfg","solutions":["6adb4867d589b45d950f014b"],"solutionDescription":[{"_id":"6adc82f0ece37a9249ea9d33","texts":{"en":"

At equilibrium:

$ΔG = 0 ⇒ ΔH = TΔS ⇒ T = \\frac{ΔH}{ΔS}$

Given:

\n
ΔH = –30 kJ = –30,000 J
\n
\n
ΔS = S(NH₃) – [½S(N₂) + 3/2 S(H₂)]
= 50 – [0.5×60 + 1.5×40] = 50 – (30 + 60) = –40 J/K
\n

Now,

$T = \\frac{–30000}{–40} = 750 \\text{ K}$

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5 mol of ideal gas expands isothermally and irreversibly from a pressure of 10 atm to \$ 1 \\mathrm{~atm} \$ against constant external pressure of 1 atm work at \$ 300 \\mathrm{~K} \$ will be:

"},"options":[{"_id":"6adb4867d589b45d950f0136","texts":{"en":"

\$ -15.921 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f0137","texts":{"en":"

\$ -11.224 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f0138","texts":{"en":"

\$ -110.83 \\mathrm{~kJ} \$

"}},{"_id":"6adb4867d589b45d950f0139","texts":{"en":"

None of these

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Pressure-Volume Work","_id":"vwz3ifjrjgj7mo07jmvxpnd1i"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"h5ritvxt9rio477smh23ule4f","solutions":["6adb4867d589b45d950f0137"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0116","texts":{"en":"

$w = - P_{ext} ΔV = - 1 \times (\frac{nRT}{P_{2}} - \frac{nRT}{P_{1}}) = - 5 \times 0.0821 \times 300 \times (1 - 0.1) = - 110.83 L.atm$

$= -110.83 \\times 101.3 = -11.224\\, \\text{kJ}$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/1cb4daa8-5341-4377-897f-0d2d7e661c45/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6860323a822d9e8f35f5ddcd","id":"6860323a822d9e8f35f5ddcd","name":"hp3wfsviwa7o8bmw2i5k59h3i","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/1cb4daa8-5341-4377-897f-0d2d7e661c45/master.mpd","description":"

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For which of the following process, $Δ$ S is negative?

"},"options":[{"_id":"6adb4867d589b45d950f0140","texts":{"en":"

H₂(g) $\to$ 2H(g)

"}},{"_id":"6adb4867d589b45d950f0141","texts":{"en":"

N₂(g) (1 atm) $\to$ N₂(g) (8 atm)

"}},{"_id":"6adb4867d589b45d950f0142","texts":{"en":"

2SO₃(g) $\to$ 2SO₂(g) + O₂(g)

"}},{"_id":"6adb4867d589b45d950f0143","texts":{"en":"

C(Diamond) $\to$ C(Graphite)

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"1st Law of Thermodynamics","_id":"igfmhd3djsapsr17eiloesngc"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"qpecsegaskveddo5ui53oetpy","solutions":["6adb4867d589b45d950f0141"],"solutionDescription":[{"_id":"6adb4867d589b45d950f011c","texts":{"en":"

At high pressure, entropy will decrease and $Δ$ S will be negative.

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=HUdZiVFJMcA"}},"videoType":"youtube","videoDetails":{"name":"For which of the following process, ΔS is negative?....","image":"https://i.ytimg.com/vi/HUdZiVFJMcA/mqdefault.jpg","embedCode":"https://www.youtube.com/embed/HUdZiVFJMcA","description":"For which of the following process, ΔS is negative? 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live","duration":"1:40","publishedAt":"2025-04-27T23:03:09Z","_id":"691ed341069dd9ae63660a7a","createdAt":"2025-11-20T08:37:21.390Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"igfmhd3djsapsr17eiloesngc","qbgId":"znyf7xgw1iz0r9izp1yojleg9","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0141"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":23.97,"score":0.77,"timeTaken":36,"options":[{"optionId":"6adb4867d589b45d950f0142","markedPercentile":5.66},{"optionId":"6adb4867d589b45d950f0141","markedPercentile":23.97},{"optionId":"6adb4867d589b45d950f0143","markedPercentile":10.56},{"optionId":"6adb4867d589b45d950f0140","markedPercentile":2.74}]}},{"question":{"_id":"6adb4867d589b45d950f0118","type":"Single","questionNumber":34,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

What is the change in internal energy when a gas contracts from 377 ml to 177 ml under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J heat? [Take: (1 L atm) = 100 J]

"},"options":[{"_id":"6adb4867d589b45d950f013b","texts":{"en":"

–24 J

"}},{"_id":"6adb4867d589b45d950f013c","texts":{"en":"

– 84 J

"}},{"_id":"6adb4867d589b45d950f013d","texts":{"en":"

– 164 J

"}},{"_id":"6adb4867d589b45d950f013e","texts":{"en":"

–248 J

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"1st Law of Thermodynamics","_id":"igfmhd3djsapsr17eiloesngc"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"qpecsegaskveddo5ui53oetpy","solutions":["6adb4867d589b45d950f013c"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0119","texts":{"en":"

$∆ H = ∆ U + P ∆ v$

$- 124 = ∆ U + (\frac{1520}{760}) \times (177 - 377) \times 10^{- 3}$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/852180df-ac30-4d18-8b84-fb88fcafc704/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6880d9ca9ab16ca40416ada7","id":"6880d9ca9ab16ca40416ada7","name":"Q15.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/852180df-ac30-4d18-8b84-fb88fcafc704/master.mpd","duration":"00:02:45","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-07-23T12:47:09.184Z","drmProtected":false}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"igfmhd3djsapsr17eiloesngc","qbgId":"r4845jjm5hg8h0zfnv9hgh1i4","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f013c"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":20.65,"score":0.74,"timeTaken":50,"options":[{"optionId":"6adb4867d589b45d950f013e","markedPercentile":1.28},{"optionId":"6adb4867d589b45d950f013d","markedPercentile":4.82},{"optionId":"6adb4867d589b45d950f013c","markedPercentile":20.65},{"optionId":"6adb4867d589b45d950f013b","markedPercentile":2.33}]}},{"question":{"_id":"6adb4867d589b45d950f011e","type":"Single","questionNumber":35,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

Two moles of an ideal gas expand spontaneously into vacuum. The work is

"},"options":[{"_id":"6adb4867d589b45d950f0145","texts":{"en":"

Zero

"}},{"_id":"6adb4867d589b45d950f0146","texts":{"en":"

\$ 2 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f0147","texts":{"en":"

\$ 4 \\mathrm{~J} \$

"}},{"_id":"6adb4867d589b45d950f0148","texts":{"en":"

\$ 8 \\mathrm{~J} \$

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Free Expansion","_id":"v9ouj01pzyz40c2kt4mkkwahs"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"x8xjuqz1grujybraoqzd0ozjf","solutions":["6adb4867d589b45d950f0145"],"solutionDescription":[{"_id":"6adb4867d589b45d950f011f","texts":{"en":"

$W = - P_{ext} Δ$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=7k1l6LnFL1g"}},"videoType":"youtube","videoDetails":{"name":", Two moles of an ideal gas expand spontaneously into vacuum. The work is :-\n(1) Zero\n(2) 2 J\n(3...","image":"https://i.ytimg.com/vi/7k1l6LnFL1g/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=7k1l6LnFL1g","description":", Two moles of an ideal gas expand spontaneously into vacuum. The work is :- (1) Zero (2) 2 J (3) 4 J (4) 8 J, , 📲PW App Link - https://bit.ly/PW_APP 🌐PW Website - https://bit.ly/PW_APP 📌 PHYS","duration":"6:54","publishedAt":"2022-08-22T09:48:16Z","_id":"63c678b543a12f027e0d9ba0","createdAt":"2023-01-17T10:30:13.121Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"v9ouj01pzyz40c2kt4mkkwahs","qbgId":"d2zekn6v443pg8c4pi0eusv2q","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0145"],"markedSolutionText":"","score":4,"timeTaken":4},"averageResult":{"correctPercentile":45.56,"score":1.76,"timeTaken":18,"options":[{"optionId":"6adb4867d589b45d950f0145","markedPercentile":45.56},{"optionId":"6adb4867d589b45d950f0148","markedPercentile":1.05},{"optionId":"6adb4867d589b45d950f0146","markedPercentile":2.52},{"optionId":"6adb4867d589b45d950f0147","markedPercentile":2.55}]}},{"question":{"_id":"6adb4867d589b45d950f0112","type":"Single","questionNumber":36,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If the bond energies of H–H, Br–Br and H–Br are x, y and z kJ mol^–1 respectively, then enthalpy change of the reaction H₂(g) + Br₂(g) $\to$ 2HBr(g) is:

"},"options":[{"_id":"6adb4867d589b45d950f0131","texts":{"en":"

2z – x – y kJ mol^–1

"}},{"_id":"6adb4867d589b45d950f0132","texts":{"en":"

x + y – 2z kJ mol^–1

"}},{"_id":"6adb4867d589b45d950f0133","texts":{"en":"

y – x + 2z kJ mol^–1

"}},{"_id":"6adb4867d589b45d950f0134","texts":{"en":"

x + y + 2z kJ mol^–1

"}}],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Enthalpies for Different Types of Reactions","_id":"1gxmt29movp1ajf2a3h1i64ay"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4867d589b45d950f0132"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0113","texts":{"en":"

$H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g)$

$ΔH =$ [Bond energy of H–H + Bond energy of Br–Br]

– [2 × bond energy of H–Br]

$= x + y - 2 z kJ {mol}^{- 1}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://youtu.be/pQWyxWZf8dg"}},"videoType":"youtube","videoDetails":{"name":null,"image":null,"embedCode":"https://youtu.be/pQWyxWZf8dg","duration":"00:00","error":"Invalid video url"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"1gxmt29movp1ajf2a3h1i64ay","qbgId":"6i7k88yiqmvhvrzek2mdddf7j","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4867d589b45d950f0132"],"markedSolutionText":"","score":4,"timeTaken":1},"averageResult":{"correctPercentile":30.25,"score":1.06,"timeTaken":30,"options":[{"optionId":"6adb4867d589b45d950f0131","markedPercentile":10.45},{"optionId":"6adb4867d589b45d950f0133","markedPercentile":2.6},{"optionId":"6adb4867d589b45d950f0134","markedPercentile":1.58},{"optionId":"6adb4867d589b45d950f0132","markedPercentile":30.25}]}},{"question":{"_id":"6adb4867d589b45d950f0124","type":"Single","questionNumber":37,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

When a gas absorbs 50 J of heat and does 20 J of work, what is the change in internal energy (ΔU)?

"},"options":[{"_id":"6adb4867d589b45d950f0156","texts":{"en":"

30 J

"}},{"_id":"6adb4867d589b45d950f0157","texts":{"en":"

70 J

"}},{"_id":"6adb4867d589b45d950f0158","texts":{"en":"

20 J

"}},{"_id":"6adb4867d589b45d950f0159","texts":{"en":"

50 J

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Pressure-Volume Work","_id":"vwz3ifjrjgj7mo07jmvxpnd1i"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4867d589b45d950f0156"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0125","texts":{"en":"

According to the first law of thermodynamics, the change in internal energy (ΔU) is given by:

ΔU = Q + W

where:

Q is the heat absorbed by the system (positive).

W is the work done on the system.

Given:
Q = +50 J (heat absorbed)

The system does 20 J of work, so W = - 20J

Substituting the values:

ΔU = 50J + (- 20J) = 30J

The internal energy of the system increases by 30 J.

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A child bought a balloon which became very small in size the next day. Which is correct statement about balloon?

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It is isolated system

"}},{"_id":"6adb4867d589b45d950f012d","texts":{"en":"

It is an open system

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It is a closed system

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It exchanges only energy with the surrounding

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Due to leakage of gas i.e. mass transfer from

system to surrounding.

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Five moles of an ideal gas is expanded isothermally from 5 dm³ to 5 m³ at 300 K. Which of the following is incorrect about the gas?

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No heat is absorbed or rejected by the gas.

"}},{"_id":"6adb4867d589b45d950f0128","texts":{"en":"

There is no change in internal energy of the gas.

"}},{"_id":"6adb4867d589b45d950f0129","texts":{"en":"

There is no change in enthalpy of the gas.

"}},{"_id":"6adb4867d589b45d950f012a","texts":{"en":"

Pressure of the gas will decrease by 1000 times.

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Thermodynamic Processes","_id":"oprz3gpyyysqcaflomith22cf"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"14ggycvwdw16pd421icarfh5j","solutions":["6adb4867d589b45d950f0127"],"solutionDescription":[{"_id":"6adb4867d589b45d950f0110","texts":{"en":"

As the system is doing work but its internal energy is not changing, the system must absorb heat.

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The entropy change when two moles of ideal monoatomic gas is heated from \$ 200^{\\circ} \\mathrm{C} \$ to \$ 300^{\\circ} \\mathrm{C} \$ reversibly and isochorically?

"},"options":[{"_id":"6adc475760094e1ec91c617d","texts":{"en":"

\$ \\frac{3}{2} \\mathrm{R} \\ell \\mathrm{n}\\left(\\frac{300}{200}\\right) \$

"}},{"_id":"6adc475760094e1ec91c617e","texts":{"en":"

\$ \\frac{5}{2} \\mathrm{R} \\ell \\mathrm{n}\\left(\\frac{573}{273}\\right) \$

"}},{"_id":"6adc475760094e1ec91c617f","texts":{"en":"

\$ 3 R \\ln \\left(\\frac{573}{473}\\right) \$

"}},{"_id":"6adc475760094e1ec91c6180","texts":{"en":"

\$ \\frac{3}{2} \\mathrm{R} \\ell \\mathrm{n}\\left(\\frac{573}{473}\\right) \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Spontaneity: Second Law of Thermodynamics","_id":"6g22rlvgsbwexnbj3xmcbmv2l"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"k6mxkd2c3xm08e4qtmk2rmkkv","solutions":["6adc475760094e1ec91c617f"],"solutionDescription":[{"_id":"6adc475760094e1ec91c617b","texts":{"en":"

\n
Convert: 200°C = 473 K, 300°C = 573 K
\n
\n
For monoatomic gas: $C_v = \\frac{3}{2}R$ , and $n = 2$
\n

$Δ S = n C_{v} \ln (\frac{T_{2}}{T_{1}}) = 2 \times \frac{3}{2} R \times \ln (\frac{573}{473}) = 3 R \ln (\frac{573}{473})$

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The standard heats of formation in kcal mol^–1 of NO_2(g)and N₂O_4(g) are 8.0 and 2.0 respectively. The heat of dimerization of NO₂ in kcal is _________ .

2NO_2(g) $⇌$ N₂O_4(g)

"},"options":[{"_id":"6adb4868d589b45d950f018e","texts":{"en":"

10.0

"}},{"_id":"6adb4868d589b45d950f018f","texts":{"en":"

–6.0

"}},{"_id":"6adb4868d589b45d950f0190","texts":{"en":"

–12.0

"}},{"_id":"6adb4868d589b45d950f0191","texts":{"en":"

–14.0

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Enthalpy Change","_id":"rb4yqymmyauvnfo7nkc2rxzoa"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"qiy6uf760q4vg0pnk14lwvd1u","solutions":["6adb4868d589b45d950f0191"],"solutionDescription":[{"_id":"6adb4868d589b45d950f017b","texts":{"en":"

Considering the forward reaction,

ΔH° = Δ_fH°_(products)– Δ_fH°_(reactants)

ΔH° = Δ_fH°(N₂O₄)– 2Δ_fH°(NO₂)

= 2 – 2(8) = – 14 kcal

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Which of the following is true for an adiabatic process?

"},"options":[{"_id":"6adb4868d589b45d950f0189","texts":{"en":"

ΔH = 0

"}},{"_id":"6adb4868d589b45d950f018a","texts":{"en":"

w = 0

"}},{"_id":"6adb4868d589b45d950f018b","texts":{"en":"

q = 0

"}},{"_id":"6adb4868d589b45d950f018c","texts":{"en":"

ΔV = 0

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Thermodynamic Processes","_id":"oprz3gpyyysqcaflomith22cf"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4868d589b45d950f018b"],"solutionDescription":[{"_id":"6adb4868d589b45d950f017e","texts":{"en":"

For an adiabatic process

$q = 0$

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Out of internal energy (I), boiling point (II), pH (III) and standard E.M.F. of the cell (IV) intensive properties are:

"},"options":[{"_id":"6adb4868d589b45d950f0198","texts":{"en":"

I, II

"}},{"_id":"6adb4868d589b45d950f0199","texts":{"en":"

II, III, IV

"}},{"_id":"6adb4868d589b45d950f019a","texts":{"en":"

I, III, IV

"}},{"_id":"6adb4868d589b45d950f019b","texts":{"en":"

All of these

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Intensive Properties","_id":"mpz92x423sdagl0ol361ponuw"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","solutions":["6adb4868d589b45d950f0199"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0175","texts":{"en":"

Boiling point, pH and standard E.M.F. of the cell are intensive properties.

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=D7WN-u6hRp4"}},"videoType":"youtube","videoDetails":{"name":"Out of internal energy (I), boiling point (II), pH (III) and standard E.M.F. of the cell (IV) in....","image":"https://i.ytimg.com/vi/D7WN-u6hRp4/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=D7WN-u6hRp4","description":"Out of internal energy (I), boiling point (II), pH (III) and standard E.M.F. of the cell (IV) intensive properties are: 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live","duration":"1:16","publishedAt":"2025-01-17T16:36:28Z","_id":"6900841f375af404a8ef2470","createdAt":"2025-10-28T08:51:43.018Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"mpz92x423sdagl0ol361ponuw","qbgId":"h25ctody17o7aoj638yqjq28l","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0199"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":36.97,"score":1.34,"timeTaken":33,"options":[{"optionId":"6adb4868d589b45d950f019b","markedPercentile":6.33},{"optionId":"6adb4868d589b45d950f0199","markedPercentile":36.97},{"optionId":"6adb4868d589b45d950f0198","markedPercentile":2.55},{"optionId":"6adb4868d589b45d950f019a","markedPercentile":4.75}]}},{"question":{"_id":"6adb4868d589b45d950f01a0","type":"Single","questionNumber":44,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The entropy change in the fusion of one mole of a solid melting at 27^°C (latent heat of fusion is 2930 J mol^–1) is:

"},"options":[{"_id":"6adb4868d589b45d950f01a3","texts":{"en":"

${9.77 J K}^{–1} {mol}^{–1}$

"}},{"_id":"6adb4868d589b45d950f01a4","texts":{"en":"

${10.73 J K}^{–1} {mol}^{–1}$

"}},{"_id":"6adb4868d589b45d950f01a5","texts":{"en":"

${2930 J K}^{–1} {mol}^{–1}$

"}},{"_id":"6adb4868d589b45d950f01a6","texts":{"en":"

${108.5 J K}^{–1} {mol}^{–1}$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Spontaneity: Second Law of Thermodynamics","_id":"6g22rlvgsbwexnbj3xmcbmv2l"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"k7ao9coyjhadvxejuhk9igfhl","solutions":["6adb4868d589b45d950f01a3"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01a1","texts":{"en":"

${ΔS}_{f} = \frac{{ΔH}_{f}}{T_{f}}$

${ΔS}_{f} = \frac{{2930 J mol}^{–1}}{300 K} {= 9.77 JK}^{–1} {mol}^{–1}$

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One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy is:

"},"options":[{"_id":"6adb4868d589b45d950f0193","texts":{"en":"

0.1 R

"}},{"_id":"6adb4868d589b45d950f0194","texts":{"en":"

2.303 R

"}},{"_id":"6adb4868d589b45d950f0195","texts":{"en":"

10.0 R

"}},{"_id":"6adb4868d589b45d950f0196","texts":{"en":"

100.0 R

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Spontaneity: Second Law of Thermodynamics","_id":"6g22rlvgsbwexnbj3xmcbmv2l"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"k6mxkd2c3xm08e4qtmk2rmkkv","solutions":["6adb4868d589b45d950f0194"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0178","texts":{"en":"

We know,

ΔS = q_rev / T

For a reversible isothermal expansion of an ideal gas:

q_rev = −W_rev = 2.303 nRT log(V₂/V₁)

Given:

n = 1

V₂/V₁ = 10

Therefore,

q_rev = 2.303 RT log 10

Since log 10 = 1,

q_rev = 2.303 RT

Hence,

ΔS = q_rev / T

ΔS = 2.303 RT / T

ΔS = 2.303 R

Therefore,

ΔS = 2.303 R J K⁻¹ mol⁻¹.

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Using the following average bond energies:

C ≡ N = 891 kJ/mol

C—H = 412 kJ/mol

N—H = 391 kJ/mol

H—H = 436 kJ/mol

C—C = 348 kJ/mol

Estimate the magnitude of enthalpy change (ΔH) for the following reaction in kJ/mol: (Nearest Integer)

CH₃CN(g) + 3H₂(g) → CH₃CH₃(g) + NH₃(g)

"},"options":[],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Enthalpies for Different Types of Reactions","_id":"1gxmt29movp1ajf2a3h1i64ay"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"nrl40su4cw5n77xgsqoywyt5u","solutions":[],"solutionText":"210","solutionDescription":[{"_id":"6adb4868d589b45d950f0172","texts":{"en":"

In CH₃CN, assume structure:

3 × C—H = 3 × 412 = 1236

1 × C—C = 348

1 × C≡N = 891

In 3H₂:

3 × H—H = 3 × 436 = 1308

Total energy required to break bonds (approx.):

= 1236 (C—H) + 348 (C—C) + 891 (C≡N) + 1308 (H—H) = 3783 kJ/mol

In CH₃CH₃ (ethane):

6 × C—H = 6 × 412 = 2472

1 × C—C = 348

In NH₃:

3 × N—H = 3 × 391 = 1173

Total energy released in bond formation:

= 2472 + 348 + 1173 = 3993 kJ/mol

ΔH = Bond energy of reactants − Bond energy of products = 3783 −3993 = −210 kJ/mol

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/b2907690-4d3d-44aa-b330-87c8b66313fd/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"683ab06acee2fdfbebedd94b","id":"683ab06acee2fdfbebedd94b","name":"Row 14 (48).mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/b2907690-4d3d-44aa-b330-87c8b66313fd/master.mpd","duration":"00:04:30","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-05-31T07:31:58.210Z","drmProtected":false}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"1gxmt29movp1ajf2a3h1i64ay","qbgId":"oyjno3llv8gcsi4l2kblup8qp","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"210","score":4,"timeTaken":4},"averageResult":{"correctPercentile":13.69,"score":0.4,"timeTaken":71,"options":[{"optionId":"6adb4868d589b45d950f0195","markedPercentile":4.19}]}},{"question":{"_id":"6adb4868d589b45d950f0180","type":"Numeric","questionNumber":47,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The molar heat capacity for an ideal gas at constant pressure is 20.785 J K^–1 mol^–1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is ______ [Nearest integer]
(Given: R = 8.314 J K^–1 mol^–1)

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Heat Capacity","_id":"v2fivjnlwqy5ucu444oyg7ea5"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"k16frdh0zaflhjlqg158zkczn","solutions":[],"solutionText":"2","solutionDescription":[{"_id":"6adb4868d589b45d950f0181","texts":{"en":"

$C_{m, p} - C_{m, v} = R$

$\Rightarrow 20.785 - C_{m, v} = 8.314 \Rightarrow C_{m, v} \Rightarrow 12.471 J K^{- 1} {mol}^{- 1}$

$△ U = {nC}_{m, v} △ T \Rightarrow n = \frac{5000}{12.471 \times (500 - 300)} \Rightarrow n = 2$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/fa5dafab-61ee-4328-8c43-aba65e374f41/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"66a673127dd5b6001892190e","id":"66a673127dd5b6001892190e","name":"05.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/fa5dafab-61ee-4328-8c43-aba65e374f41/master.mpd","duration":"00:03:05","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-07-28T16:34:30.842Z","drmProtected":false}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"v2fivjnlwqy5ucu444oyg7ea5","qbgId":"zv57mt77e61znkbeu27purch5","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"2","score":4,"timeTaken":3},"averageResult":{"correctPercentile":14.55,"score":0.5,"timeTaken":47,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f0186","type":"Numeric","questionNumber":48,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

How many of the following are state function?

(i) Internal energy

(ii) Heat

(iii) Enthalpy

(iv) Entropy

(v) Pressure

(vi) Temperature

(vii) Volume

(viii) Work

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"State of a System","_id":"s8uqun7gm4psp4lg5ct8w4yjd"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"lom9ldklug6bbg3f5vvsgos13","solutions":[],"solutionText":"6","solutionDescription":[{"_id":"6adb4868d589b45d950f0187","texts":{"en":"

(i), (iii), (iv), (v), (vi) and (vii) are state function.

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/1ec10356-cace-41c9-a9ca-d67eec4d741b/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6679600da8302a0018104b3e","id":"6679600da8302a0018104b3e","name":"26.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/1ec10356-cace-41c9-a9ca-d67eec4d741b/master.mpd","duration":"00:02:43","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-06-24T12:01:21.182Z","drmProtected":false}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"s8uqun7gm4psp4lg5ct8w4yjd","qbgId":"boldhxfpczhu5hg0yucgh9upw","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"6","score":4,"timeTaken":2},"averageResult":{"correctPercentile":33.32,"score":1.22,"timeTaken":32,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f019d","type":"Numeric","questionNumber":49,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

While performing a thermodynamics experiment, a student made the following observations,

\$ \\mathrm{HCl}+\\mathrm{NaOH} \\longrightarrow \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}, \\Delta \\mathrm{H}=-57.3 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \$

\$ \\mathrm{CH}_{3} \\mathrm{COOH}+\\mathrm{NaOH} \\rightarrow \\mathrm{CH}_{3} \\mathrm{COONa}+\\mathrm{H}_{2} \\mathrm{O} \$

\$ \\Delta \\mathrm{H}=-55.3 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \$.

The enthalpy of ionization of \$ \\mathrm{CH}_{3} \\mathrm{COOH} \$ as calculated by the student is............ \$ \\mathrm{kJ} \\mathrm{mol}^{-1} \$.

"},"options":[],"difficultyLevel":2,"pyqYears":["2022"],"topicId":{"name":"Enthalpies for Different Types of Reactions","_id":"1gxmt29movp1ajf2a3h1i64ay"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"c6mumitmkcocw9nw680oxdaag","solutions":[],"solutionText":"2","solutionDescription":[{"_id":"6adb4868d589b45d950f019e","texts":{"en":"

Given:

ΔH (HCl + NaOH) = −57.3 kJ/mol (strong acid)
ΔH (CH₃COOH + NaOH) = −55.3 kJ/mol

The difference is due to ionization of CH₃COOH.

So, enthalpy of ionization =

ΔH = −55.3 − (−57.3) = +2.0 kJ/mol

Answer: +2.0 kJ/mol.

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=jJ7ZrFpPInM"}},"videoType":"youtube","videoDetails":{"name":"While performing a thermodynamics experiment, a student made the following observations,\n\\( \\mat...","image":"https://i.ytimg.com/vi/jJ7ZrFpPInM/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=jJ7ZrFpPInM","description":"While performing a thermodynamics experiment, a student made the following observations, \\( \\mathrm{HCl}+\\mathrm{NaOH} \\longrightarrow \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}, \\Delta \\mathrm{H}=-57.3 ","duration":"10:43","publishedAt":"2023-07-08T11:44:09Z","_id":"64abccec70e0ca001852d0a8","createdAt":"2023-07-10T09:18:36.395Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"1gxmt29movp1ajf2a3h1i64ay","qbgId":"elnoih7ypcy6vjxrizhch0our","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"2","score":4,"timeTaken":2},"averageResult":{"correctPercentile":19.5,"score":0.74,"timeTaken":28,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f0183","type":"Numeric","questionNumber":50,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If \$1.0 \\mathrm{kcal}\$ of heat is added to \$1.2 \\mathrm{~L}\$ of \$\\mathrm{O}_2\$ in a cylinder of constant pressure of \$1 \\mathrm{~atm}\$, the volume increases to \$1.5 \\mathrm{~L}\$. Calculate \$\\Delta \\mathrm{U}\$ (in kcal) of the process. (round off to nearest integer)

"},"options":[],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"1st Law of Thermodynamics","_id":"igfmhd3djsapsr17eiloesngc"},"sectionId":"6adb467ea26438bfde20019d","subjectId":"7rgg2rb3rbirmrkpguofa7sor","chapterId":"7kx7aknod2psthzyz0hxpqllr","subTopicId":"qpecsegaskveddo5ui53oetpy","solutions":[],"solutionText":"1","solutionDescription":[{"_id":"6adb4868d589b45d950f0184","texts":{"en":"

\$\\begin{aligned} & \\mathrm{w}=-\\mathrm{P}_{\\mathrm{ext}}\\left(\\mathrm{V}_2-\\mathrm{V}_1\\right)=-1(1.5-1.2)=-0.3 \\mathrm{~L} \\text { atm } \\\\ & =-7.235 \\mathrm{cal} \\\\ & \\Delta \\mathrm{U}=\\mathrm{q}+\\mathrm{w} \\\\ & =1-0.007235=0.993 \\mathrm{kcal}=1 \\mathrm{kcal}\\end{aligned}\$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=HNZh7tOtBpE"}},"videoType":"youtube","videoDetails":{"name":"NEET Class 12 Chemistry Coordination Compounds....","image":"https://i.ytimg.com/vi/HNZh7tOtBpE/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=HNZh7tOtBpE","description":"🎓Disha Mentorship - https://disha.live 📘More Questions: https://pwbooks-app.onelink.me/y2BH/hpkmntlw 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.p","duration":"3:10","publishedAt":"2025-10-07T08:26:30Z","_id":"6adb762cf53ba71e97290a44","createdAt":"2026-10-23T14:58:52.002Z"}}],"qbgSubjectId":"7rgg2rb3rbirmrkpguofa7sor","qbgChapterId":"7kx7aknod2psthzyz0hxpqllr","qbgTopicId":"igfmhd3djsapsr17eiloesngc","qbgId":"uyzwz8gljt15qdxsfzixw2mbt","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"1","score":4,"timeTaken":3},"averageResult":{"correctPercentile":14.78,"score":0.53,"timeTaken":42,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f01cd","type":"Single","questionNumber":51,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The domain of the function

$f (x) = \sqrt{x^{2} - 5 x + 6} + \sqrt{2 x + 8 - x^{2}}, is$

"},"options":[{"_id":"6adb4868d589b45d950f01f3","texts":{"en":"

$\"[\"2, 3\"]\"$

"}},{"_id":"6adb4868d589b45d950f01f4","texts":{"en":"

$\"[\"- 2, 4\"]\"$

"}},{"_id":"6adb4868d589b45d950f01f5","texts":{"en":"

$\"[\"- 2, 2\"]\" \cup \"[\"3, 4\"]\"$

"}},{"_id":"6adb4868d589b45d950f01f6","texts":{"en":"

$\"[\"- 2, 1\"]\" \cup \"[\"2, 4\"]\"$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Domain and Range of Functions","_id":"ekw95veq7nq6pz9c7x9v5z82a"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ca33u7zvbbnt64edqggwp2a1x","subTopicId":"yyaa80c1sdjkhdkquj1bvmvtx","solutions":["6adb4868d589b45d950f01f5"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01ce","texts":{"en":"

We are going to factorize

$x^{2} - 5 x + 6 \geq 0$

$(x - 2) (x - 3) \geq 0$

$\Rightarrow x \in (- \infty, 2] \cup [3, \infty)$ …(1)

$2 x + 8 - x^{2} \geq 0$

$\Rightarrow x^{2} - 2 x - 8 \leq 0$

$\Rightarrow x^{2} - 4 x + 2 x - 8 \leq 0$

$\Rightarrow x (x - 4) + 2 (x - 4) \leq 0$

$\Rightarrow (x + 2) (x - 4) \leq 0$

$x \in \Rightarrow \"[\"- 2, 4\"]\" . . . . . (2)$

By (1) and (2)

Domain = $\"[\"- 2, 2\"]\" \cup \"[\"3, 4\"]\"$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/07a867ff-b601-4b75-9cd5-5e50c82571a9/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6894817a213668c76681bdf3","id":"6894817a213668c76681bdf3","name":"Q2.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/07a867ff-b601-4b75-9cd5-5e50c82571a9/master.mpd","duration":"00:01:40","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-08-07T10:35:41.729Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ca33u7zvbbnt64edqggwp2a1x","qbgTopicId":"ekw95veq7nq6pz9c7x9v5z82a","qbgId":"g1kk2juhzajw6s8t7uv7oxk40","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f01f5"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":42.58,"score":1.61,"timeTaken":87,"options":[{"optionId":"6adb4868d589b45d950f01f6","markedPercentile":1.07},{"optionId":"6adb4868d589b45d950f01f5","markedPercentile":42.58},{"optionId":"6adb4868d589b45d950f01f4","markedPercentile":4.22},{"optionId":"6adb4868d589b45d950f01f3","markedPercentile":3.59}]}},{"question":{"_id":"6adb4868d589b45d950f01c7","type":"Single","questionNumber":52,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

Number greater than 7000 and divisible by 5 that can be formed using only the digit 3, 5, 7, 8 and 9, no digit being repeated, is

"},"options":[{"_id":"6adb4868d589b45d950f01e9","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01ea","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01eb","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01ec","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Number of Permutations under Certain Restrictions","_id":"22tgb7s4ufqzfut9sva1i5mbh"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"wnwygz1jd0pd7jgbj8rbbst19","solutions":["6adb4868d589b45d950f01ec"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01c8","texts":{"en":"

Number of four digit numbers = 3 $\cdot$ 3 $\cdot$ 2 $\cdot$ 1 = 18

Number of five digit numbers =4 $\cdot$ 3 $\cdot$ 2 $\cdot$ 1 $\cdot$ 1 = 24

∴ Total = 42

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/9130bb49-1e56-4213-8abe-b400a378de11/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b41ee2dc45d30340adc1","id":"6ad1b41ee2dc45d30340adc1","name":"2.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/9130bb49-1e56-4213-8abe-b400a378de11/master.mpd","duration":"00:02:09","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:20:32.422Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"22tgb7s4ufqzfut9sva1i5mbh","qbgId":"8gzn2y2gkcfee8wqwhjz2ijmy","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f01ec"],"markedSolutionText":"","score":4,"timeTaken":3},"averageResult":{"correctPercentile":34.2,"score":1.2,"timeTaken":101,"options":[{"optionId":"6adb4868d589b45d950f01ec","markedPercentile":34.2},{"optionId":"6adb4868d589b45d950f01eb","markedPercentile":6.42},{"optionId":"6adb4868d589b45d950f01e9","markedPercentile":2.15},{"optionId":"6adb4868d589b45d950f01ea","markedPercentile":8.13}]}},{"question":{"_id":"6adb4868d589b45d950f01d0","type":"Single","questionNumber":53,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

\$ { }^{26} C_{0}+{ }^{26} C_{1}+{ }^{26} C_{2}+\\ldots .+{ }^{26} C_{13} \$ is equal to :

"},"options":[{"_id":"6adb4868d589b45d950f01f8","texts":{"en":"

\$ 2^{25}-\\frac{1}{2} \\cdot{ }^{26} C_{13} \$

"}},{"_id":"6adb4868d589b45d950f01f9","texts":{"en":"

\$ 2^{25}+\\frac{1}{2} \\cdot{ }^{26} C_{13} \$

"}},{"_id":"6adb4868d589b45d950f01fa","texts":{"en":"

\$ 2^{13} \$

"}},{"_id":"6adb4868d589b45d950f01fb","texts":{"en":"

\$ 2^{26}+\\frac{1}{2} \\cdot{ }^{26} C_{13} \$

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Properties of Binomial Coefficients","_id":"1qhlpp7yb3i0r5z5l71zteydt"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"xoookkg0vv7mgdwl7o16ikxp5","solutions":["6adb4868d589b45d950f01f9"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01d1","texts":{"en":"

$\sum_{r = 0}^{26} {}^{26}C_{r} = 2^{26} 2 ({}^{26}C_{0} + {}^{26}C_{1} + \dots + {}^{26}C_{12}) + {}^{26}C_{13} = 2^{26} Let S = {}^{26}C_{0} + {}^{26}C_{1} + \dots + {}^{26}C_{13} 2 (S - {}^{26}C_{13}) + {}^{26}C_{13} = 2^{26} 2 S - {}^{26}C_{13} = 2^{26} 2 S = 2^{26} + {}^{26}C_{13} S = 2^{25} + \frac{1}{2} \cdot {}^{26}C_{13}$

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The smallest value of \$ x \$ satisfying the inequality \$ { }^{10} C_{x-1}>2 \\cdot{ }^{10} C_{x} \$ is

"},"options":[{"_id":"6adb4868d589b45d950f0214","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0215","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0216","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0217","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Combinations","_id":"lcy4jd33l7nogd53ovi2hcysz"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"8ysx1qt2o5r66c5t0lzdizgdt","solutions":["6adb4868d589b45d950f0217"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01dd","texts":{"en":"

${}^{10}{C_{x - 1}} > 2 \cdot {}^{10}{C_{x}} \frac{10!}{(x - 1)! (11 - x)!} > 2 \cdot \frac{10!}{x! (10 - x)!} \frac{1}{11 - x} > \frac{2}{x} x > 22 - 2 x 3 x > 22 x > 7.33$

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The number of irrational terms in the expansion of \$ (\\sqrt[8]{5}+\\sqrt[6]{2})^{100} \$ is

"},"options":[{"_id":"6adb4868d589b45d950f0207","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0208","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0209","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f020a","texts":{"en":"

"}}],"difficultyLevel":3,"pyqYears":[],"topicId":{"name":"Problems Based on Direct Expansion","_id":"3kz6zpvy6ducdesr6w71knq28"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"mpw5snq21sdy8qdxq9wezhhae","solutions":["6adb4868d589b45d950f0207"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01da","texts":{"en":"

General Term:

\"T_{r+1}

Total terms = $\"100$

For rational terms:

\n
$\"r$
\n
\n
$\"\\frac{100-r}{8}$
\n

Valid $\"r\"$ : $\"12,$

Number of rational terms = 4

\"\\text{Number

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/80f41c95-a5b2-4894-a231-e9144311b37f/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b48d35875ac75b86250a","id":"6ad1b48d35875ac75b86250a","name":"5.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/80f41c95-a5b2-4894-a231-e9144311b37f/master.mpd","duration":"00:02:37","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:22:23.142Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"bkp5mpgm6wa4lyroz395ykf21","qbgTopicId":"3kz6zpvy6ducdesr6w71knq28","qbgId":"d81apm24kl9bfftn65eqb2yyu","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0207"],"markedSolutionText":"","score":4,"timeTaken":3},"averageResult":{"correctPercentile":22.65,"score":0.72,"timeTaken":93,"options":[{"optionId":"6adb4868d589b45d950f020a","markedPercentile":3.16},{"optionId":"6adb4868d589b45d950f0209","markedPercentile":10.09},{"optionId":"6adb4868d589b45d950f0207","markedPercentile":22.65},{"optionId":"6adb4868d589b45d950f0208","markedPercentile":5.82}]}},{"question":{"_id":"6adb4868d589b45d950f01c1","type":"Single","questionNumber":56,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If in a chess toumament each contestant plays once against each of the others and in all 45 games are played, then the number of participants, is

"},"options":[{"_id":"6adb4868d589b45d950f01df","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01e0","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01e1","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01e2","texts":{"en":"

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Combinations","_id":"lcy4jd33l7nogd53ovi2hcysz"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"fxqy0r8zsviquujt4zwf3qefe","solutions":["6adb4868d589b45d950f01e0"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01c2","texts":{"en":"

$(\frac{n}{2}) = 45 \frac{n (n - 1)}{2} = 45 n (n - 1) = 90 n = 10$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8aa50a30-aac4-435f-abac-4dda566efcb7/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b4a835875ac75b86251f","id":"6ad1b4a835875ac75b86251f","name":"6.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8aa50a30-aac4-435f-abac-4dda566efcb7/master.mpd","duration":"00:02:25","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:22:50.733Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"lcy4jd33l7nogd53ovi2hcysz","qbgId":"a82pk5da50ijz3gs0d9d7pln3","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f01e0"],"markedSolutionText":"","score":4,"timeTaken":4},"averageResult":{"correctPercentile":27.83,"score":1.01,"timeTaken":58,"options":[{"optionId":"6adb4868d589b45d950f01e1","markedPercentile":3.64},{"optionId":"6adb4868d589b45d950f0208","markedPercentile":5.82},{"optionId":"6adb4868d589b45d950f01df","markedPercentile":5.25},{"optionId":"6adb4868d589b45d950f01e2","markedPercentile":1.23},{"optionId":"6adb4868d589b45d950f01e0","markedPercentile":27.83}]}},{"question":{"_id":"6adb4868d589b45d950f01c4","type":"Single","questionNumber":57,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

Which term has greatest coefficient in the expansion of ${(1 + \frac{2 x}{3})}^{15} ?$

"},"options":[{"_id":"6adb4868d589b45d950f01e4","texts":{"en":"

\$ 5^{\\text {th }} \$

"}},{"_id":"6adb4868d589b45d950f01e5","texts":{"en":"

\$ 6^{\\text {th }} \$

"}},{"_id":"6adb4868d589b45d950f01e6","texts":{"en":"

\$ 7^{\\text {th }} \$

"}},{"_id":"6adb4868d589b45d950f01e7","texts":{"en":"

\$ 8^{\\text {th }} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Numerically Greatest Term in the Binomial Expansion","_id":"puncod72djy5acxzjuw4hjvp3"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"fcan1mndmoqyrt8zhd4ouuxu4","solutions":["6adb4868d589b45d950f01e6"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01c5","texts":{"en":"

$T_{r + 1} = (\frac{15}{r}) {(\frac{2}{3})}^{r} \frac{T_{r + 1}}{T_{r}} \geq 1 \Rightarrow \frac{16 - r}{r} \cdot \frac{2}{3} \geq 1 \Rightarrow 32 - 2 r \geq 3 r \Rightarrow r \leq 6.4 r = 6 \Rightarrow 7th term (T_{7})$

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Letters of the word \"HISTORY\" are arranged in all possible ways to form 7-letter words without repetition. These words are now arranged in alphabetical order as in a dictionary; the rank of \"HISTORY\" in this dictionary is

"},"options":[{"_id":"6adb4868d589b45d950f01ee","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01ef","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01f0","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01f1","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Rank of a Word in Dictionary","_id":"fbp3kcnsriy84uxq4z3w4vlzx"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"behqq30ke1ghmppumbfnkl7rp","solutions":["6adb4868d589b45d950f01ef"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01cb","texts":{"en":"

\\[
H , I , O , R , S , T , Y
\\]
(i) \$HIO \\square \\square \\square \\square=4\$ !
(ii) \$\\operatorname{HIR} \\square \\square \\square \\square=4\$ !
(iii) HISO \$\\square \\square=3\$ !
(iv) \$\\operatorname{HIsR} \\square \\square \\square=3\$ !
(v) H I S TO R Y \$=1\$
Rank \$=4 ! \\times 2+3 ! \\times 2+1=61\$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/e7c9f67f-5d2f-4946-a0d7-651080491fc4/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b4e9481e470728fa481e","id":"6ad1b4e9481e470728fa481e","name":"8.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/e7c9f67f-5d2f-4946-a0d7-651080491fc4/master.mpd","duration":"00:03:10","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:23:55.611Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"fbp3kcnsriy84uxq4z3w4vlzx","qbgId":"gejn7l0gwisyre7awms3uxq2d","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f01ef"],"markedSolutionText":"","score":4,"timeTaken":3},"averageResult":{"correctPercentile":35.52,"score":1.3,"timeTaken":93,"options":[{"optionId":"6adb4868d589b45d950f01f1","markedPercentile":1.9},{"optionId":"6adb4868d589b45d950f01ee","markedPercentile":6.48},{"optionId":"6adb4868d589b45d950f01f0","markedPercentile":3.42},{"optionId":"6adb4868d589b45d950f01ef","markedPercentile":35.52}]}},{"question":{"_id":"6adb4868d589b45d950f01d6","type":"Single","questionNumber":59,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

In the expansion of \$\\left(1+3 x+2 x^2\\right)^6\$ the coefficient of \$x^{11}\$ is

"},"options":[{"_id":"6adb4868d589b45d950f0202","texts":{"en":"

144

"}},{"_id":"6adb4868d589b45d950f0203","texts":{"en":"

288

"}},{"_id":"6adb4868d589b45d950f0204","texts":{"en":"

216

"}},{"_id":"6adb4868d589b45d950f0205","texts":{"en":"

576

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"General Term in the Expansion of (A + X)^N","_id":"hds7c20n2396c6hcjqxb5407w"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"iq2d3anffux5pjvb6mnitfdv7","solutions":["6adb4868d589b45d950f0205"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01d7","texts":{"en":"

$(1 + 3 x + 2 x^{2})^{6} = (1 + x)^{6} (1 + 2 x)^{6} Term in expansion = (\frac{6}{r}) x^{r} \cdot (\frac{6}{k}) (2 x)^{k} = (\frac{6}{r}) (\frac{6}{k}) 2^{k} x^{r + k} For x^{11}, require r + k = 11 (0 \leq r, k \leq 6) Possible pairs (r, k) = (5, 6), (6, 5) Coefficient = (\frac{6}{5}) (\frac{6}{6}) 2^{6} + (\frac{6}{6}) (\frac{6}{5}) 2^{5} = (6) (1) (64) + (1) (6) (32) = 384 + 192 = 576$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/2fa0f7ba-b62e-463b-a652-cde0f1e29c67/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b509a2b238202fc0ab26","id":"6ad1b509a2b238202fc0ab26","name":"9.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/2fa0f7ba-b62e-463b-a652-cde0f1e29c67/master.mpd","duration":"00:04:44","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:24:28.158Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"bkp5mpgm6wa4lyroz395ykf21","qbgTopicId":"hds7c20n2396c6hcjqxb5407w","qbgId":"xga2thttubz63r49u5l3g17lh","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0205"],"markedSolutionText":"","score":4,"timeTaken":1},"averageResult":{"correctPercentile":28.71,"score":1.06,"timeTaken":86,"options":[{"optionId":"6adb4868d589b45d950f0204","markedPercentile":3.62},{"optionId":"6adb4868d589b45d950f0203","markedPercentile":3.78},{"optionId":"6adb4868d589b45d950f0202","markedPercentile":1.9},{"optionId":"6adb4868d589b45d950f0205","markedPercentile":28.71}]}},{"question":{"_id":"6adb4868d589b45d950f01d3","type":"Single","questionNumber":60,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The number of permutation of all the letters \$ A A A A B B B C \$ in which the $A$ 's appear together in a block of $4$ letters or the $B$ 's appear together in a block of $3$ letters is :

"},"options":[{"_id":"6adb4868d589b45d950f01fd","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01fe","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f01ff","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0200","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Permutations","_id":"u4t8kav5nevzillmhqjfvmr4z"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"qbq433zv9zw2eohb3icl4kozr","solutions":["6adb4868d589b45d950f01fd"],"solutionDescription":[{"_id":"6adb4868d589b45d950f01d4","texts":{"en":"

Using Principle of Inclusion-Exclusion:

$\"n(A\\text{'s$

$\"n(B\\text{'s$

$\"n(A\\text{'s$

\"\\text{Total}

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/4e3fb0f2-03e7-487c-b5fc-f22d1176ec24/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b528752784a7cdcdf47d","id":"6ad1b528752784a7cdcdf47d","name":"10.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/4e3fb0f2-03e7-487c-b5fc-f22d1176ec24/master.mpd","duration":"00:02:41","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:24:58.304Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"u4t8kav5nevzillmhqjfvmr4z","qbgId":"wuzeeusm5931h2tplste2f8bb","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f01fd"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":17.02,"score":0.54,"timeTaken":69,"options":[{"optionId":"6adb4868d589b45d950f0200","markedPercentile":1.42},{"optionId":"6adb4868d589b45d950f0204","markedPercentile":3.62},{"optionId":"6adb4868d589b45d950f01ff","markedPercentile":3.85},{"optionId":"6adb4868d589b45d950f01fe","markedPercentile":8.77},{"optionId":"6adb4868d589b45d950f01fd","markedPercentile":17.02}]}},{"question":{"_id":"6adb4868d589b45d950f024a","type":"Single","questionNumber":61,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The fractional part of $\frac{2^{4 n}}{15}$ is ( $n \in N$ ).

"},"options":[{"_id":"6adb4868d589b45d950f0281","texts":{"en":"

$\frac{1}{15}$

"}},{"_id":"6adb4868d589b45d950f0282","texts":{"en":"

$\frac{2}{15}$

"}},{"_id":"6adb4868d589b45d950f0283","texts":{"en":"

$\frac{4}{15}$

"}},{"_id":"6adb4868d589b45d950f0284","texts":{"en":"

None of these

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Expansion of (A + X)^N","_id":"gw1wyc5nsnwgpglxwap7yow58"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"g52m2bc3m54g857sy0oqzq13q","solutions":["6adb4868d589b45d950f0281"],"solutionDescription":[{"_id":"6adb4868d589b45d950f024b","texts":{"en":"

$\frac{2^{4 n}}{15} = \frac{{(15 + 1)}^{n}}{15}$

= Integer + $\frac{1}{15}$

Hence, the fractional part of $\frac{2^{4 n}}{15}$ is $\frac{1}{15}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=LonLP6UIUvQ"}},"videoType":"youtube","videoDetails":{"name":"Q4.mp4","image":"https://i.ytimg.com/vi/LonLP6UIUvQ/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=LonLP6UIUvQ","description":"Auto-uploaded from Drive file https://drive.google.com/file/d/1HIl-lxbFtSgMfyxZnhPna59z--5i8_WN/view?usp=drivesdk","duration":"0:33","publishedAt":"2026-07-09T12:56:52Z","_id":"6ac886b83298e8da1b789566","createdAt":"2026-10-09T06:16:24.155Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"bkp5mpgm6wa4lyroz395ykf21","qbgTopicId":"gw1wyc5nsnwgpglxwap7yow58","qbgId":"8cbc2di9rwlob0da8olx8hhan","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0281"],"markedSolutionText":"","score":4,"timeTaken":5},"averageResult":{"correctPercentile":26.46,"score":0.95,"timeTaken":26,"options":[{"optionId":"6adb4868d589b45d950f0282","markedPercentile":3.87},{"optionId":"6adb4868d589b45d950f0281","markedPercentile":26.46},{"optionId":"6adb4868d589b45d950f0284","markedPercentile":2.98},{"optionId":"6adb4868d589b45d950f0283","markedPercentile":3.52}]}},{"question":{"_id":"6adb4868d589b45d950f0232","type":"Single","questionNumber":62,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The number of negative integral solutions of the equation \$ x_{1}+x_{2}+x_{3}+x_{4}+15=0 \$ is

"},"options":[{"_id":"6adb4868d589b45d950f0261","texts":{"en":"

364

"}},{"_id":"6adb4868d589b45d950f0262","texts":{"en":"

\$ { }^{15} \\mathrm{C}_{3} \$

"}},{"_id":"6adb4868d589b45d950f0263","texts":{"en":"

\$ { }^{14} \\mathrm{C}_{4} \$

"}},{"_id":"6adb4868d589b45d950f0264","texts":{"en":"

none of these

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Number of Integral Solution of an Equation","_id":"utugb7fffl8kcvlvlqlli0w7p"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"dsvie409ylns8kqo34jij8fxh","solutions":["6adb4868d589b45d950f0261"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0233","texts":{"en":"

$x_{1} + x_{2} + x_{3} + x_{4} + 15 = 0 x_{1} + x_{2} + x_{3} + x_{4} = - 15 y_{1} + y_{2} + y_{3} + y_{4} = 15 n u m b e r o f s o l = (\frac{14}{3}) = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 14 \times 13 \times 2 = 364$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=OVEgZcKjagc"}},"videoType":"youtube","videoDetails":{"name":"Q16.mp4","image":"https://i.ytimg.com/vi/OVEgZcKjagc/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=OVEgZcKjagc","description":"Auto-uploaded from Drive file https://drive.google.com/file/d/1Bq1n38S_M1kW0IxGkohGvjdWabcfp9by/view?usp=drivesdk","duration":"1:12","publishedAt":"2026-07-09T12:57:39Z","_id":"6adb762c9241157d57d3587f","createdAt":"2026-10-23T14:58:52.010Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"utugb7fffl8kcvlvlqlli0w7p","qbgId":"gyn7t02v364vkhxlv0xokjrzs","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0261"],"markedSolutionText":"","score":4,"timeTaken":5},"averageResult":{"correctPercentile":15.65,"score":0.5,"timeTaken":34,"options":[{"optionId":"6adb4868d589b45d950f0261","markedPercentile":15.65},{"optionId":"6adb4868d589b45d950f0263","markedPercentile":4.38},{"optionId":"6adb4868d589b45d950f0282","markedPercentile":3.87},{"optionId":"6adb4868d589b45d950f0262","markedPercentile":5.22},{"optionId":"6adb4868d589b45d950f0264","markedPercentile":2.68}]}},{"question":{"_id":"6adb4868d589b45d950f0247","type":"Single","questionNumber":63,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

\$ \\sum_{r=0}^{n-1} \\frac{{ }^{n} C_{r}}{{ }^{n} C_{r}+{ }^{n} C_{r+1}} \$ equals

"},"options":[{"_id":"6adb4868d589b45d950f0275","texts":{"en":"

\$ \\frac{n}{2} \$

"}},{"_id":"6adb4868d589b45d950f0276","texts":{"en":"

\$ \\frac{n+1}{2} \$

"}},{"_id":"6adb4868d589b45d950f0277","texts":{"en":"

\$ \\frac{n(n+1)}{2} \$

"}},{"_id":"6adb4868d589b45d950f0278","texts":{"en":"

\$ \\frac{n(n-1)}{2(n+1)} \$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Properties of Binomial Coefficients","_id":"1qhlpp7yb3i0r5z5l71zteydt"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"xoookkg0vv7mgdwl7o16ikxp5","solutions":["6adb4868d589b45d950f0275"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0248","texts":{"en":"

\\[ \\begin{gathered} \\sum_{r=0}^{n-1} \\frac{{ }^{n} C_{r}}{{ }^{n} C_{r}+{ }^{n} C_{r+1}}=\\sum_{r=0}^{n-1} \\frac{1}{1+\\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}} \\\\ =\\sum_{r=0}^{n-1} \\frac{1}{1+\\left(\\frac{n !}{r !(n-r) !}\\right)} \\\\ =\\sum_{r=0}^{n-1} \\frac{1}{1+\\frac{n-r}{r+1}}=\\sum_{r=0}^{n-1} \\frac{r+1}{n+1} \\end{gathered} \\]

$= \frac{n}{2}$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=IMzLRk0ft8I"}},"videoType":"youtube","videoDetails":{"name":"Q11.mp4","image":"https://i.ytimg.com/vi/IMzLRk0ft8I/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=IMzLRk0ft8I","description":"Auto-uploaded from Drive file https://drive.google.com/file/d/1d9MJCgzeEFcYrmOtcuVooxrhR2NH54Rm/view?usp=drivesdk","duration":"1:06","publishedAt":"2026-07-09T12:57:20Z","_id":"6ac886b8806780b7368087a3","createdAt":"2026-10-09T06:16:24.158Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"bkp5mpgm6wa4lyroz395ykf21","qbgTopicId":"1qhlpp7yb3i0r5z5l71zteydt","qbgId":"srtztro140swkcxlhf6veqnsw","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0275"],"markedSolutionText":"","score":4,"timeTaken":4},"averageResult":{"correctPercentile":15.76,"score":0.49,"timeTaken":61,"options":[{"optionId":"6adb4868d589b45d950f0278","markedPercentile":5.22},{"optionId":"6adb4868d589b45d950f0276","markedPercentile":2.85},{"optionId":"6adb4868d589b45d950f0277","markedPercentile":5.8},{"optionId":"6adb4868d589b45d950f0275","markedPercentile":15.76}]}},{"question":{"_id":"6adb4868d589b45d950f0241","type":"Single","questionNumber":64,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The value of $f (x) = sgn (x^{2} - 2 x + 3)$ is

"},"options":[{"_id":"6adb4868d589b45d950f026b","texts":{"en":"

{2}

"}},{"_id":"6adb4868d589b45d950f026c","texts":{"en":"

{1}

"}},{"_id":"6adb4868d589b45d950f026d","texts":{"en":"

{1, 2}

"}},{"_id":"6adb4868d589b45d950f026e","texts":{"en":"

None of these

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Functions","_id":"39324dn1mmkigf6ibx6io3k7x"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ca33u7zvbbnt64edqggwp2a1x","subTopicId":"ld7uwtsm12u5dhmitjtzfuvlm","solutions":["6adb4868d589b45d950f026c"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0242","texts":{"en":"

$x^{2} - 2 x + 3 = {(x - 1)}^{2} + 2 > 0$ , $\forall x \in R$

or $f (x) = sgn (x^{2} - 2 x + 3) = 1$

Hence, the value f (x) is {1}.

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/4ca162bd-debc-45f7-9c90-0fe4886db9e0/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"68873efa80832fe37f6cbdf2","id":"68873efa80832fe37f6cbdf2","name":"2025-07-28 14-33-45.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/4ca162bd-debc-45f7-9c90-0fe4886db9e0/master.mpd","duration":"00:01:44","status":"Ready","types":["DASH","HLS"],"createdAt":"2025-07-28T09:12:29.396Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ca33u7zvbbnt64edqggwp2a1x","qbgTopicId":"39324dn1mmkigf6ibx6io3k7x","qbgId":"ty1zvl2c20g959r1v4m3l9u65","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f026c"],"markedSolutionText":"","score":4,"timeTaken":4},"averageResult":{"correctPercentile":24.81,"score":0.88,"timeTaken":30,"options":[{"optionId":"6adb4868d589b45d950f026b","markedPercentile":2.11},{"optionId":"6adb4868d589b45d950f026e","markedPercentile":3.98},{"optionId":"6adb4868d589b45d950f026c","markedPercentile":24.81},{"optionId":"6adb4868d589b45d950f026d","markedPercentile":5.06}]}},{"question":{"_id":"6adb4868d589b45d950f023b","type":"Single","questionNumber":65,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If the sum of the coefficients in the expansion of \$ \\left(x^{2}+\\frac{2}{x^{3}}\\right)^{n} \$ is 243 , the term independent of \$ x \$ is equal to

"},"options":[{"_id":"6adb4868d589b45d950f0252","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0253","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0254","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0255","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Properties of Binomial Coefficients","_id":"1qhlpp7yb3i0r5z5l71zteydt"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"xoookkg0vv7mgdwl7o16ikxp5","solutions":["6adb4868d589b45d950f0252"],"solutionDescription":[{"_id":"6adb4868d589b45d950f023c","texts":{"en":"

${(1^{2} + \frac{2}{1^{3}})}^{n} = 243 (1 + 2)^{n} = 243 3^{n} = 3^{5} \Rightarrow n = 5 T_{r + 1} = (\frac{5}{r}) (x^{2})^{5 - r} {(\frac{2}{x^{3}})}^{r} T_{r + 1} = (\frac{5}{r}) \cdot 2^{r} \cdot x^{10 - 5 r} 10 - 5 r = 0 \Rightarrow r = 2 Term = (\frac{5}{2}) \cdot 2^{2} = 40$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=ESeYqNcuEyA"}},"videoType":"youtube","videoDetails":{"name":"Q13.mp4","image":"https://i.ytimg.com/vi/ESeYqNcuEyA/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=ESeYqNcuEyA","description":"Auto-uploaded from Drive file https://drive.google.com/file/d/12C3YDUVM_rFR1rSWOoci_-umI4y2cWC0/view?usp=drivesdk","duration":"1:15","publishedAt":"2026-07-09T12:57:27Z","_id":"6aa2f69159f6d94b1b3f4974","createdAt":"2026-09-10T18:27:29.142Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"bkp5mpgm6wa4lyroz395ykf21","qbgTopicId":"1qhlpp7yb3i0r5z5l71zteydt","qbgId":"raj0hyt12dvyeql1e19fm1dsj","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0252"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":23.82,"score":0.87,"timeTaken":58,"options":[{"optionId":"6adb4868d589b45d950f0254","markedPercentile":3.25},{"optionId":"6adb4868d589b45d950f0255","markedPercentile":1.55},{"optionId":"6adb4868d589b45d950f0253","markedPercentile":3.76},{"optionId":"6adb4868d589b45d950f0252","markedPercentile":23.82}]}},{"question":{"_id":"6adb4868d589b45d950f0244","type":"Single","questionNumber":66,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

There are 15 stones placed in line numbered 1 to 15. In how many ways a person can mark 5 of these stones so that there are an odd number of stones between any two stones marked-

"},"options":[{"_id":"6adb4868d589b45d950f0270","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0271","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f0272","texts":{"en":"

252

"}},{"_id":"6adb4868d589b45d950f0273","texts":{"en":"

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Special Uses of Ncr","_id":"48o4hwobwbu93epkv6gqzr3af"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"24wcczadxivpjhh2k9zfkzrud","solutions":["6adb4868d589b45d950f0270"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0245","texts":{"en":"

If we mark any odd number or even number stones, there are always odd number of stones in between them
So, ${}^{8}C_{5} + {}^{7}C_{5} = 77$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=Vdg4ZTmvuS8"}},"videoType":"youtube","videoDetails":{"name":"Q3.mp4","image":"https://i.ytimg.com/vi/Vdg4ZTmvuS8/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=Vdg4ZTmvuS8","description":"Auto-uploaded from Drive file https://drive.google.com/file/d/1uMRKkaOJ8m-vqLu17QLVIIqmx8Q8ekoY/view?usp=drivesdk","duration":"2:33","publishedAt":"2026-07-09T12:56:49Z","_id":"6ac886b8e3eaba165fbc737b","createdAt":"2026-10-09T06:16:24.157Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"48o4hwobwbu93epkv6gqzr3af","qbgId":"2vf2s1lxty1l9mzseoknk8mb0","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0270"],"markedSolutionText":"","score":4,"timeTaken":4},"averageResult":{"correctPercentile":12.42,"score":0.4,"timeTaken":45,"options":[{"optionId":"6adb4868d589b45d950f0272","markedPercentile":3.82},{"optionId":"6adb4868d589b45d950f0273","markedPercentile":1.72},{"optionId":"6adb4868d589b45d950f0271","markedPercentile":4.35},{"optionId":"6adb4868d589b45d950f0270","markedPercentile":12.42}]}},{"question":{"_id":"6adb4868d589b45d950f022f","type":"Single","questionNumber":67,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If the coefficients of second, third and fourth terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P., then

"},"options":[{"_id":"6adb4868d589b45d950f0266","texts":{"en":"

$2 n^{2} + 9 n + 7 = 0$

"}},{"_id":"6adb4868d589b45d950f0267","texts":{"en":"

$2 n^{2} - 9 n + 7 = 0$

"}},{"_id":"6adb4868d589b45d950f0268","texts":{"en":"

$2 n^{2} - 9 n - 7 = 0$

"}},{"_id":"6adb4868d589b45d950f0269","texts":{"en":"

$2 n^{2} + 9 n - 7 = 0$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Problems Based on General Term, Particular Terms and Middle Term in a Given Binomial Expansion","_id":"x06cb1kz25gq0jpn0wyuq2w49"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"xx8aubb92s2u9xe2ut22ggb4p","solutions":["6adb4868d589b45d950f0267"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0230","texts":{"en":"

It is given that $^{2 n} C_{1},^{2 n} C_{2}$ and

$^{2 n} C_{3}$ are A.P.

$∴ 2 \cdot^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$

$2 \cdot \frac{(2 n)!}{(2 n - 2)! 2!} = \frac{(2 n)!}{(2 n - 1)!} + \frac{(2 n)!}{(2 n - 3)! 3!}$

$\Rightarrow 2 \frac{(2 n) (2 n - 1)}{2} = 2 n + \frac{(2 n) (2 n - 1) (2 n - 2)}{3!}$

$\Rightarrow 6 (2 n - 1) = 6 + (2 n - 1) (2 n - 2)$

$\Rightarrow 12 n - 6 = 6 + 4 n^{2} - 6 n + 2$

$\Rightarrow 4 n^{2} - 18 n + 14 = 0 \Rightarrow 2 n^{2} - 9 n + 7 = 0$

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If 9 members of a committee sit at a round table so that the CEO and the CFO always sit together, then the number of arrangements is

"},"options":[{"_id":"6adb4868d589b45d950f025c","texts":{"en":"

$2 (7!)$

"}},{"_id":"6adb4868d589b45d950f025d","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f025e","texts":{"en":"

"}},{"_id":"6adb4868d589b45d950f025f","texts":{"en":"

$2 (8!)$

"}}],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Circular Permutation","_id":"00vxxe18moqq0tmt6erqpt94n"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"dgqjxwdr9ju7mmekm12t5q1lh","solutions":["6adb4868d589b45d950f025c"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0236","texts":{"en":"

members sit at a round table; CEO and CFO must sit together
Treat CEO & CFO as one unit → total “people” = 8

Circular arrangements = (8 – 1)! = 7! = 5040

CEO & CFO can swap $\to$ 2!=2ways
Total arrangements = 2 (7!)

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Let $[t]$ denote the greatest integer $\leq t$ . Then the equation in $x, [x]^{2} + 2 [x + 2] - 7 = 0$ has:

"},"options":[{"_id":"6adb4868d589b45d950f0257","texts":{"en":"

Infinitely many solutions

"}},{"_id":"6adb4868d589b45d950f0258","texts":{"en":"

Exactly two solutions

"}},{"_id":"6adb4868d589b45d950f0259","texts":{"en":"

No integral solution

"}},{"_id":"6adb4868d589b45d950f025a","texts":{"en":"

Exactly four integral solutions

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Classification of Functions","_id":"3xzrre5lwku0vhxj13138xr3o"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ca33u7zvbbnt64edqggwp2a1x","subTopicId":"rhlo5auy7yiglbtekk2z6emnk","solutions":["6adb4868d589b45d950f0257"],"solutionDescription":[{"_id":"6adb4868d589b45d950f0239","texts":{"en":"

$[x]^{2} + 2 [x + 2] - 7 = 0 \Rightarrow [x]^{2} + 2 [x] + 4 - 7 = 0 [x] = 1, - 3 \Rightarrow x \in [1, 2) \cup [- 3, - 2)$

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=fQjSq7SfYh0"}},"videoType":"youtube","videoDetails":{"name":"JEE Mains Class: 11 Maths Relations and Functions....","image":"https://i.ytimg.com/vi/fQjSq7SfYh0/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=fQjSq7SfYh0","description":"More Questions: https://pwbooks-app.onelink.me/y2BH/hpkmntlw 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live","duration":"3:20","publishedAt":"2025-08-19T07:00:51Z","_id":"68efda5be29867953982202e","createdAt":"2025-10-15T17:31:07.719Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ca33u7zvbbnt64edqggwp2a1x","qbgTopicId":"3xzrre5lwku0vhxj13138xr3o","qbgId":"x8vydhq52z3zpwgq918yvzgu5","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f0257"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":15.78,"score":0.48,"timeTaken":34,"options":[{"optionId":"6adb4868d589b45d950f0258","markedPercentile":10.15},{"optionId":"6adb4868d589b45d950f025a","markedPercentile":1.82},{"optionId":"6adb4868d589b45d950f0257","markedPercentile":15.78},{"optionId":"6adb4868d589b45d950f0259","markedPercentile":3.63}]}},{"question":{"_id":"6adb4868d589b45d950f023e","type":"Single","questionNumber":70,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The number of elements in the set

$\"{\"(a, b) : 2 a^{2} + 3 b^{2} = 35, a, b \in Z\"}\"$ , where $Z$ is the set of all integers, is equal to

"},"options":[{"_id":"6adb4868d589b45d950f024d","texts":{"en":"

\$2\$

"}},{"_id":"6adb4868d589b45d950f024e","texts":{"en":"

\$4\$

"}},{"_id":"6adb4868d589b45d950f024f","texts":{"en":"

\$8\$

"}},{"_id":"6adb4868d589b45d950f0250","texts":{"en":"

\$12\$

"}}],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Relations","_id":"mdcsvw4exa9ru80o9hlzk8pn6"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ca33u7zvbbnt64edqggwp2a1x","subTopicId":"2jqfp7k6nearzyz6zp32acw1r","solutions":["6adb4868d589b45d950f024f"],"solutionDescription":[{"_id":"6adb4868d589b45d950f023f","texts":{"en":"

$2 a^{2} + 3 b^{2} = 35$

If $a = 0,$ \$ b \\notin Z\$

$a = \pm 1, b^{2} = 11 \Rightarrow$ \$ b \\notin Z\$

$a = \pm 2, b^{2} = 9 \Rightarrow b = 3 or - 3$

$a = \pm 3, b^{2} = \frac{17}{3} \Rightarrow$ \$ b \\notin Z\$

$a = \pm 4, b^{2} = 1 \Rightarrow b = \pm 1$

Required set

\$=\\{(2,3),(2,-3),(-2,3),(-2,-3),(4,1),(4,-1),(-4,1),(-4,-1)\\}\$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8322296f-5a12-472e-8645-82886969b109/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"676d38030988d53bcd8fc060","id":"676d38030988d53bcd8fc060","name":"Q17.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/8322296f-5a12-472e-8645-82886969b109/master.mpd","duration":"00:02:00","status":"Ready","types":["DASH","HLS"],"createdAt":"2024-12-26T11:03:35.240Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ca33u7zvbbnt64edqggwp2a1x","qbgTopicId":"mdcsvw4exa9ru80o9hlzk8pn6","qbgId":"ybk9ibuce85tujbpim4a03roq","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":["6adb4868d589b45d950f024f"],"markedSolutionText":"","score":4,"timeTaken":2},"averageResult":{"correctPercentile":16.63,"score":0.49,"timeTaken":50,"options":[{"optionId":"6adb4868d589b45d950f0250","markedPercentile":1.28},{"optionId":"6adb4868d589b45d950f024e","markedPercentile":7.82},{"optionId":"6adb4868d589b45d950f024d","markedPercentile":7.96},{"optionId":"6adb4868d589b45d950f024f","markedPercentile":16.63}]}},{"question":{"_id":"6adb4868d589b45d950f0296","type":"Numeric","questionNumber":71,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

In how many ways can four persons \$ A, B, C \$ and \$ D \$ be seated at a circular table?

"},"options":[],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Circular Permutation","_id":"00vxxe18moqq0tmt6erqpt94n"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"dgqjxwdr9ju7mmekm12t5q1lh","solutions":[],"solutionText":"6","solutionDescription":[{"_id":"6adb4868d589b45d950f0297","texts":{"en":"

The number of ways to arrange n persons around a circular table is $\"(n$ .

Here, $\"n$ :

\"\\text{Number

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/7ee172ea-dffd-41ae-b979-d4450effd537/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b84a35875ac75b862598","id":"6ad1b84a35875ac75b862598","name":"21_slide_01_solution.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/7ee172ea-dffd-41ae-b979-d4450effd537/master.mpd","duration":"00:01:16","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:38:20.081Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"00vxxe18moqq0tmt6erqpt94n","qbgId":"jad6xm5nm79nw2zzcm6q4k369","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"6","score":4,"timeTaken":5},"averageResult":{"correctPercentile":30.15,"score":1.09,"timeTaken":24,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f0299","type":"Numeric","questionNumber":72,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

How many terms are there in the expansion of \$ (\\sqrt{x}+\\sqrt{y})^{10}+(\\sqrt{x}-\\sqrt{y})^{10} \$.

"},"options":[],"difficultyLevel":1,"pyqYears":[],"topicId":{"name":"Properties of Binomial Coefficients","_id":"1qhlpp7yb3i0r5z5l71zteydt"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"bkp5mpgm6wa4lyroz395ykf21","subTopicId":"19w96onomyozcmrnix818dfhc","solutions":[],"solutionText":"6","solutionDescription":[{"_id":"6adb4868d589b45d950f029a","texts":{"en":"

For the expansion of $\"(a$ where $\"n\"$ is even, the odd-numbered terms remain while the even-numbered terms cancel out.

\n
Formula: Number of terms = $\"\\frac{n}{2}$
\n
\n
Here, $\"n$ (which is even).
\n
\n
Number of terms = $\"\\frac{10}{2}$
\n

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Number of positive divisors of 10! is

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Divisors","_id":"6feo1y2po46nmw579ybchzjba"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"cunsmwzhzl5sj0eumck3lg44l","solutions":[],"solutionText":"270","solutionDescription":[{"_id":"6adb4868d589b45d950f029d","texts":{"en":"

Exponent of 2 is

$\"[\"\frac{10}{2}\"]\" + \"[\"\frac{10}{2^{2}}\"]\" + \"[\"\frac{10}{2^{3}}\"]\" = 5 + 2 + 1 = 8$

Exponent of 3 is

$\"[\"\frac{10}{3}\"]\" + \"[\"\frac{10}{3^{2}}\"]\" = 3 + 1 = 4$

Exponent of 5 is

$\"[\"\frac{10}{5}\"]\" = 2$

Exponent of 7 is

$\"[\"\frac{10}{7}\"]\" = 1$

The number of positive divisors of 10! is

(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 270.

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/5d52bbc6-ed4c-49fb-829c-c2458eee09ef/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b572e2dc45d30340ae05","id":"6ad1b572e2dc45d30340ae05","name":"23.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/5d52bbc6-ed4c-49fb-829c-c2458eee09ef/master.mpd","duration":"00:03:34","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:26:11.887Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"6feo1y2po46nmw579ybchzjba","qbgId":"s16duhmavu9q80l1fv5xvqj3x","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"270","score":4,"timeTaken":4},"averageResult":{"correctPercentile":18.33,"score":0.63,"timeTaken":36,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f029f","type":"Numeric","questionNumber":74,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

If $A = {1, 2}$ and the number of relations on the power set of A is $2^{k}$ then value of $k$ is

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Relations","_id":"mdcsvw4exa9ru80o9hlzk8pn6"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ca33u7zvbbnt64edqggwp2a1x","subTopicId":"a4xxrd8k86xcq5lml2uypact0","solutions":[],"solutionText":"16","solutionDescription":[{"_id":"6adb4868d589b45d950f02a0","texts":{"en":"

$n (P (A)) = 4$

number of relations $= 2^{4 \times 4}$

$\Rightarrow k = 16$

"},"videos":{"en":{"videoType":"penpencilvdo","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/c51291a1-15ab-4f63-b672-35ff838c7053/master.mpd"}},"videoType":"penpencilvdo","videoDetails":{"_id":"6ad1b591e2dc45d30340ae2e","id":"6ad1b591e2dc45d30340ae2e","name":"24.mp4","image":"","videoUrl":"https://d1d34p8vz63oiq.cloudfront.net/c51291a1-15ab-4f63-b672-35ff838c7053/master.mpd","duration":"00:02:36","status":"Ready","types":["DASH","HLS"],"createdAt":"2026-10-16T05:26:43.242Z","drmProtected":false}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ca33u7zvbbnt64edqggwp2a1x","qbgTopicId":"mdcsvw4exa9ru80o9hlzk8pn6","qbgId":"qwksucqcwdyxsae8p8k0qdffy","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"16","score":4,"timeTaken":3},"averageResult":{"correctPercentile":13.72,"score":0.33,"timeTaken":28,"options":[]}},{"question":{"_id":"6adb4868d589b45d950f0293","type":"Numeric","questionNumber":75,"positiveMarks":4,"positiveMarksStr":"4","negativeMarks":1,"negativeMarksStr":"1","texts":{"en":"

The number of words that can be formed by rearranging the letters of the word ' $COVID$ ' such that no letter is in its own place and the word should not start with a vowel is ________.

"},"options":[],"difficultyLevel":2,"pyqYears":[],"topicId":{"name":"Dearrangement","_id":"pysjdxto0ugmuyscb0uxag5oh"},"sectionId":"6adb4687354567414b32a2da","subjectId":"19c8p297y48dy01537cxzinys","chapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","subTopicId":"rueu4f0yg9accdjrx9axkdckd","solutions":[],"solutionText":"22","solutionDescription":[{"_id":"6adb4868d589b45d950f0294","texts":{"en":"

Total derangements of 5 letters = $D_{5} = 44$

Derangements starting with a vowel (O or l)

$= 2 \times (D_{4} \times D_{3})$

= 2 × (9 + 2) = 22

Required number of words = 44 $-$ 22 = 22

"},"videos":{"en":{"videoType":"youtube","videoUrl":"https://www.youtube.com/watch?v=afBjURlcLXU"}},"videoType":"youtube","videoDetails":{"name":"Sachin Jakhar Sir....","image":"https://i.ytimg.com/vi/afBjURlcLXU/mqdefault.jpg","embedCode":"https://www.youtube.com/watch?v=afBjURlcLXU","description":"Permutation & Combinations 🎓Disha Mentorship - https://disha.live 📘More Questions: https://pwbooks-app.onelink.me/y2BH/hpkmntlw 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://ww","duration":"2:13","publishedAt":"2025-12-16T11:18:55Z","_id":"6947de0c1521e063936b51c7","createdAt":"2025-12-21T11:46:20.213Z"}}],"qbgSubjectId":"19c8p297y48dy01537cxzinys","qbgChapterId":"ql4ps0uyfsjqsiyuoztf3ltzn","qbgTopicId":"pysjdxto0ugmuyscb0uxag5oh","qbgId":"lul5u3a9vwp34pgz033oviw91","exam":""},"yourResult":{"isUnderReview":false,"status":"UnAttempted","markedSolutions":[],"markedSolutionText":"","score":0,"scoreStr":"0","timeTaken":0},"topperResult":{"status":"CORRECT","markedSolutions":[],"markedSolutionText":"22","score":4,"timeTaken":4},"averageResult":{"correctPercentile":10.59,"score":0.26,"timeTaken":69,"options":[]}}],"languageCodes":[{"language":"English","isSelected":true,"code":"en"}],"test":{"_id":"6adb45c21e402a4666203855","name":"Practice Test 06","type":"MockAndRegular","template":"NTA"},"isLiveAttempt":true,"comprehensionData":{"comprehensions":[],"comprehensionRanges":[]},"bookmarkEnabled":true}}