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Rotational Motion — JEE Previous Year Questions

Every Rotational Motion question asked in JEE Main and JEE Advanced across the last 186 papers — 266 questions, each with its correct answer. Free to read, no account needed.

Questions

266

Papers it appeared in

172/186

Appearance rate

92%

All 266 Rotational Motion questions

Most recent papers first.

Q1·PhysicsSingle correctJEE Advanced 2026
Consider a large disk of radius RRR and two smaller disks, each of radius r=R/50r = R/50r=R/50, lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation Δθ\Delta\thetaΔθ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities ω\omegaω and 2ω2\omega2ω while the large disk is held stationary. The time τ\tauτ at which the smaller disks are again in contact is: [Use sin⁡(Δθ)=Δθ\sin(\Delta\theta) = \Delta\thetasin(Δθ)=Δθ and ignore gravity.]
  1. (A)τ=51×(2π−451)/ω\tau = 51 \times \left(2\pi - \frac{4}{51}\right)/\omegaτ=51×(2π−514​)/ω
  2. (B)τ=51×(2π−251)/3ω\tau = 51 \times \left(2\pi - \frac{2}{51}\right)/3\omegaτ=51×(2π−512​)/3ω
  3. (C)τ=51×(2π−451)/3ω\tau = 51 \times \left(2\pi - \frac{4}{51}\right)/3\omegaτ=51×(2π−514​)/3ω
  4. (D)τ=51×(2π−251)/ω\tau = 51 \times \left(2\pi - \frac{2}{51}\right)/\omegaτ=51×(2π−512​)/ω

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsNumericalJEE Advanced 2026
Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point CCC such that it can rotate freely around CCC in the XYXYXY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative xxx direction in the XYXYXY plane with speed 100 ms−1^{-1}−1, hits the circumference of the disk at a point PPP. After collision the particle moves along negative yyy direction at a speed of 90 ms−1^{-1}−1. [Given: the acceleration due to gravity (g)=−10j^(g) = -10\hat{j}(g)=−10j^​ ms−2^{-2}−2] Question: After the collision the disk starts to rotate around point CCC in the XYXYXY plane. The maximum change in the height (in m) of its center OOO is:

Correct answer: 0.15

Step-by-step solution →
Q3·PhysicsSingle correctJEE Advanced 2026
List-I shows four planar structures made of uniform solid rods each of mass mmm and length lll. In the List-II the possible moment of inertia of these structures about an axis OCO′OCO'OCO′, which lies in the plane of the structures, are given. Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-IList-II
P.see figure1.54ml2\frac{5}{4}ml^245​ml2
Q.see figure2.16ml2\frac{1}{6}ml^261​ml2
R.see figure3.112ml2\frac{1}{12}ml^2121​ml2
S.see figure4.23ml2\frac{2}{3}ml^232​ml2
5.13ml2\frac{1}{3}ml^231​ml2
  1. (A)P→ 5, Q→ 1, R→ 4, S→ 2
  2. (B)P→ 1, Q→ 3, R→ 4, S→ 2
  3. (C)P→ 5, Q→ 3, R→ 2, S→ 1
  4. (D)P→ 5, Q→ 4, R→ 2, S→ 1

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsNumericalJEE Advanced 2026
Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point CCC such that it can rotate freely around CCC in the XYXYXY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative xxx direction in the XYXYXY plane with speed 100 ms−1^{-1}−1, hits the circumference of the disk at a point PPP. After collision the particle moves along negative yyy direction at a speed of 90 ms−1^{-1}−1. [Given: the acceleration due to gravity (g)=−10j^(g) = -10\hat{j}(g)=−10j^​ ms−2^{-2}−2] Question: Amount of energy loss (in J) in the collision is:

Correct answer: 17.47

Step-by-step solution →
Q5·PhysicsSingle correctJEE Advanced 2026
A solid cylinder of radius RRR rolls without slipping with a center of mass speed v0=gR3v_0 = \sqrt{\frac{gR}{3}}v0​=3gR​​ on a horizontal surface with a vertical edge, as shown in the figure. Here, ggg is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:
  1. (A)0
  2. (B)5gR7\sqrt{\frac{5gR}{7}}75gR​​
  3. (C)gR15\sqrt{\frac{gR}{15}}15gR​​
  4. (D)3gR7\sqrt{\frac{3gR}{7}}73gR​​

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
A solid cylinder having radius RRR and length LLL is slipping on a rough horizontal plane. At time t=0t = 0t=0 the cylinder has a translational velocity v0=49v_0 = 49v0​=49 m/s, perpendicular to its axis and a rotational velocity v0/4Rv_0/4Rv0​/4R about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction μK=0.25\mu_K = 0.25μK​=0.25 and g=9.8g = 9.8g=9.8 m/s2^22)
  1. (A)15
  2. (B)5
  3. (C)10
  4. (D)7.5

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is ________.
  1. (A)(34,1.25)\left(\frac{\sqrt{3}}{4}, 1.25\right)(43​​,1.25)
  2. (B)(34,1.0)\left(\frac{\sqrt{3}}{4}, 1.0\right)(43​​,1.0)
  3. (C)(0, 0)
  4. (D)(1.25, 0)

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
Two identical bodies AAA and BBB of equal masses have initial velocities v⃗1=4i^\vec{v}_{1} = 4\hat{i}v1​=4i^ m/s and v⃗2=4j^\vec{v}_{2} = 4\hat{j}v2​=4j^​ m/s respectively. The body AAA has acceleration a⃗1=6i^+6j^\vec{a}_{1} = 6\hat{i} + 6\hat{j}a1​=6i^+6j^​ m/s2^{2}2 while the acceleration of the other body BBB is zero. The centre of mass of the two bodies moves in __________ path.
  1. (A)circular
  2. (B)parabolic
  3. (C)straight line
  4. (D)elliptical

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
A solid sphere (A)(A)(A) of mass 5m5m5m and a spherical shell (B)(B)(B) of mass mmm, both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of AAA and BBB, they start rolling without slipping with an acceleration of aAa_{A}aA​ and aBa_{B}aB​, respectively. The ratio of aAa_{A}aA​ and aBa_{B}aB​ is __________.
  1. (A)5 : 21
  2. (B)6 : 10
  3. (C)21 : 25
  4. (D)1 : 5

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ________ and ________ respectively.
  1. (A)0.128π Nm, 100
  2. (B)0.0128π Nm, 50
  3. (C)0.128π Nm, 50
  4. (D)0.0128π Nm, 100

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
An object of uniform density rolls up the curved path with the initial velocity v₀ as shown in the figure. If the maximum height attained by an object is 7v0210g\dfrac{7v_{0}^{2}}{10g}10g7v02​​ (g = acceleration due to gravity), the object is a ________.
  1. (A)solid cylinder
  2. (B)ring
  3. (C)disc
  4. (D)solid sphere

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle θ₁ and in the next 2 s it rotates through an angle θ₂. The ratio θ2θ1\dfrac{\theta_{2}}{\theta_{1}}θ1​θ2​​ is ________.
  1. (A)6
  2. (B)3
  3. (C)4
  4. (D)13\dfrac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
A solid sphere of mass MMM and radius RRR is divided into two unequal parts. The smaller part having mass M/8M/8M/8 is converted into a sphere of radius rrr and the larger part is converted into a circular disc of thickness ttt and radius 2R2R2R. If I1I_1I1​ is moment of inertia of a sphere having radius rrr about an axis through its centre and I2I_2I2​ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia I2/I1=I_2/I_1 =I2​/I1​= _____.
  1. (A)35
  2. (B)70
  3. (C)140
  4. (D)210

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
A particle is rotating in a circular path and at any instant its motion can be described as θ=5t440−t33\theta = \frac{5t^{4}}{40} - \frac{t^{3}}{3}θ=405t4​−3t3​. The angular acceleration of the particle after 10 seconds is ______ rad/s2^{2}2.
  1. (A)150
  2. (B)120
  3. (C)130
  4. (D)170

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsNumericalJEE Main 2026
Moment of inertia about an axis ABABAB for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass of 10 kg and radius RRR about an axis parallel to ABABAB axis with separation of 3 m as shown in figure below. The value of RRR is given as α2\sqrt{\frac{\alpha}{2}}2α​​. The value of α\alphaα is ________.

Correct answer: 15

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2026
Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is ______ m. (Take g=10g = 10g=10 m/s2^{2}2)
  1. (A)3.33
  2. (B)3.12
  3. (C)2.22
  4. (D)1.42

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2026
The position of an object having mass 0.1 kg as a function of time ttt is given as r⃗=(10t2i^+5t3j^)\vec{r} = \left(10t^{2}\hat{i} + 5t^{3}\hat{j}\right)r=(10t2i^+5t3j^​) m. At t=1t = 1t=1 s, which of the following statements are correct? A. The linear momentum p⃗=(2i^+1.5j^)\vec{p} = \left(2\hat{i} + 1.5\hat{j}\right)p​=(2i^+1.5j^​) kg·m/s. B. The force acting on the object F⃗=(2i^+3j^)\vec{F} = \left(2\hat{i} + 3\hat{j}\right)F=(2i^+3j^​) N. C. The angular momentum of the object about its origin L⃗=15k^\vec{L} = 15\hat{k}L=15k^ J·s. D. The torque acting on the object about its origin τ⃗=20k^\vec{\tau} = 20\hat{k}τ=20k^ N·m. Choose the correct answer from the options given below :
  1. (A)A, B and C only
  2. (B)B, C and D only
  3. (C)A, C and D only
  4. (D)A, B and D only

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2026
When the position vector r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}r=xi^+yj^​+zk^ changes sign as −r⃗-\vec{r}−r, which one of the following vector will not flip under sign change ?
  1. (A)Linear momentum
  2. (B)Velocity
  3. (C)Acceleration
  4. (D)Angular momentum

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsNumericalJEE Main 2026
A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is _______ J. (g = 10 m/s²)

Correct answer: 30

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2026
Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure. If the mass of each disc is 600 gm and applied torque between two discs is 43×10543 \times 10^{5}43×105 dyne cm, the angular acceleration of the discs about the given axis ABABAB is _______ rad/s2^{2}2.
  1. (A)22
  2. (B)11
  3. (C)100
  4. (D)27

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumericalJEE Main 2026
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n\sqrt{n}n​ cm. The value of nnn is

Correct answer: 265

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2026
Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley is ________ kg.m2^{2}2. (g=9.8(g = 9.8(g=9.8 m/s2)^{2})2)
  1. (A)9.5×10−39.5 \times 10^{-3}9.5×10−3
  2. (B)4.75×10−34.75 \times 10^{-3}4.75×10−3
  3. (C)1.86×10−21.86 \times 10^{-2}1.86×10−2
  4. (D)8.3×10−38.3 \times 10^{-3}8.3×10−3

Correct answer: (A)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2026
A thin uniform rod (X) of mass M and length L is pivoted at a height (L3)\left(\frac{L}{3}\right)(3L​) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is________. (g = gravitational acceleration)
  1. (A)32gL\sqrt{\frac{3}{2}\frac{g}{L}}23​Lg​​
  2. (B)32gL\frac{3}{\sqrt{2}}\sqrt{\frac{g}{L}}2​3​Lg​​
  3. (C)12gL\frac{1}{\sqrt{2}}\sqrt{\frac{g}{L}}2​1​Lg​​
  4. (D)3gL\sqrt{\frac{3g}{L}}L3g​​

Correct answer: (D)

Step-by-step solution →
Q24·PhysicsNumericalJEE Main 2026
A uniform solid cylinder of length LLL and radius RRR has moment of inertia about its axis equal to I1I_1I1​. A small co-centric cylinder of length L/2L/2L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to I2I_2I2​. The ratio I1/I2I_1/I_2I1​/I2​ is _____.

Correct answer: 162

Step-by-step solution →
Q25·PhysicsNumericalJEE Main 2026
Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x256\frac{x}{256}256x​ Mr2r^{2}r2. The value of x is __________.

Correct answer: 109

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2026
Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is :
  1. (A)32Lmd2\frac{3}{2}\frac{L}{md^2}23​md2L​
  2. (B)2Lmd2\frac{2L}{md^2}md22L​
  3. (C)43Lmd2\frac{4}{3}\frac{L}{md^2}34​md2L​
  4. (D)2L5md2\frac{2L}{5md^2}5md22L​

Correct answer: (A)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Main 2026
The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R < L) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M) :
  1. (A)38MR2+712ML2\frac{3}{8}MR^2 + \frac{7}{12}ML^283​MR2+127​ML2
  2. (B)34MR2+16ML2\frac{3}{4}MR^2 + \frac{1}{6}ML^243​MR2+61​ML2
  3. (C)34MR2+712ML2\frac{3}{4}MR^2 + \frac{7}{12}ML^243​MR2+127​ML2
  4. (D)38MR2+16ML2\frac{3}{8}MR^2 + \frac{1}{6}ML^283​MR2+61​ML2

Correct answer: (D)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2026
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω. The ratio of v and ω is :
  1. (A)33
  2. (B)2882\sqrt{88}288​
  3. (C)66
  4. (D)32

Correct answer: (A)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2026
A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is__________kg.m2^{2}2.
  1. (A)0.36
  2. (B)0.72
  3. (C)0.18
  4. (D)0.63

Correct answer: (D)

Step-by-step solution →
Q30·PhysicsNumericalJEE Main 2026
A circular disc has radius R1R_1R1​ and thickness T1T_1T1​. Another circular disc made of the same material has radius R2R_2R2​ and thickness T2T_2T2​. If the moment of inertia of both discs are same and R1R2=2\frac{R_1}{R_2} = 2R2​R1​​=2 then T1T2=1α\frac{T_1}{T_2} = \frac{1}{\alpha}T2​T1​​=α1​. The value of α is ____.

Correct answer: 16

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2026
Given below are two statements : Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles. Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are true
  2. (B)Statement I is true but Statement II is false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q32·PhysicsNumericalJEE Main 2026
Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1r = 0.1r=0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ________ m/s. (Assume string does not slip and g =10= 10=10 m/s2^22)

Correct answer: 2

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2026
A uniform rod of mass m and length lll suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ________. (g acceleration due to gravity)
  1. (A)mg/2mg/2mg/2
  2. (B)mg/4mg/4mg/4
  3. (C)mg/3mg/3mg/3
  4. (D)mgmgmg

Correct answer: (B)

Step-by-step solution →
Q34·PhysicsNumericalJEE Main 2026
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x2\frac{x}{2}2x​ ML2^{2}2 kg m2^{2}2. The value of x is ___________.

Correct answer: 17

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2026
The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is____ (assume no slipping of string on pulley.)
  1. (A)(M−m)g[(136)M+m]\frac{(M-m)g}{\left[\left(\frac{13}{6}\right)M + m\right]}[(613​)M+m](M−m)g​
  2. (B)(M−m)gM+m\frac{(M-m)g}{M+m}M+m(M−m)g​
  3. (C)(M−m)g[(83)M+m]\frac{(M-m)g}{\left[\left(\frac{8}{3}\right)M + m\right]}[(38​)M+m](M−m)g​
  4. (D)(M−m)g2M+m\frac{(M-m)g}{2M+m}2M+m(M−m)g​

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2026
Two cars A and B each of mass 10310^3103 kg are moving on parallel tracks separated by a distance of 10 m, in same direction with speeds 72 km/h and 36 km/h. The magnitude of angular momentum of car A with respect to car B is ___ J.s.
  1. (A)3.6×1053.6 \times 10^53.6×105
  2. (B)10510^5105
  3. (C)3×1053 \times 10^53×105
  4. (D)2×1052 \times 10^52×105

Correct answer: (B)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2025
A rod of linear mass density 'λ\lambdaλ' and length 'LLL' is bent to form a ring of radius 'RRR'. Moment of inertia of ring about any of its diameter is:
  1. (A)λL316π2\dfrac{\lambda L^3}{16\pi^2}16π2λL3​
  2. (B)λL312\dfrac{\lambda L^3}{12}12λL3​
  3. (C)λL34π2\dfrac{\lambda L^3}{4\pi^2}4π2λL3​
  4. (D)λL38π2\dfrac{\lambda L^3}{8\pi^2}8π2λL3​

Correct answer: (D)

Step-by-step solution →
Q38·PhysicsIntegerJEE Main 2025
A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π25\pi25π Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ___ m.

Correct answer: 40

Step-by-step solution →
Q39·PhysicsIntegerJEE Main 2025
A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in the figure. The moment of inertia of the given system about PQ is x15I\dfrac{x}{15}I15x​I, where III is the moment of inertia of the disc about its diameter. The value of xxx is ______.

Correct answer: 199

Step-by-step solution →
Q40·PhysicsIntegerJEE Main 2025
Let M and R be the mass and radius of a disc. A small disc of radius R3\dfrac{R}{3}3R​ is removed from the bigger disc as shown in the figure. The moment of inertia of the remaining part of the bigger disc about an axis AB passing through the centre O and perpendicular to the plane of the disc is 4xMR2\dfrac{4}{x}MR^2x4​MR2. The value of x is __________.

Correct answer: 9

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2025
A rod of length 5L is bent right angle keeping one side length as 2L (as shown in the figure). The position of the centre of mass of the system: (Consider L = 10 cm)
  1. (A)2i^+3j^2\hat{i}+3\hat{j}2i^+3j^​
  2. (B)3i^+7j^3\hat{i}+7\hat{j}3i^+7j^​
  3. (C)5i^+8j^5\hat{i}+8\hat{j}5i^+8j^​
  4. (D)4i^+9j^4\hat{i}+9\hat{j}4i^+9j^​

Correct answer: (D)

Step-by-step solution →
Q42·PhysicsIntegerJEE Main 2025
A solid sphere with uniform density and radius R is rotating with constant angular velocity ω1\omega_1ω1​ about its diameter. After some time during the rotation its mass starts losing at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius becomes R/2R/2R/2 is xω1x\omega_1xω1​. The value of x is ______.

Correct answer: 32

Step-by-step solution →
Q43·PhysicsIntegerJEE Main 2025
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is x5\sqrt{\dfrac{x}{5}}5x​​ where x=x=x= ______.

Correct answer: 4

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2025
Which of the following are correct expression for torque acting on a body? A. τ⃗=r⃗×L⃗\vec{\tau}=\vec{r}\times\vec{L}τ=r×L B. τ⃗=ddt(r⃗×p⃗)\vec{\tau}=\dfrac{d}{dt}(\vec{r}\times\vec{p})τ=dtd​(r×p​) C. τ⃗=r⃗×dp⃗dt\vec{\tau}=\vec{r}\times\dfrac{d\vec{p}}{dt}τ=r×dtdp​​ D. τ⃗=Iα⃗\vec{\tau}=I\vec{\alpha}τ=Iα E. τ⃗=r⃗×F⃗\vec{\tau}=\vec{r}\times\vec{F}τ=r×F (r⃗\vec{r}r = position vector; p⃗\vec{p}p​ = linear momentum; L⃗\vec{L}L = angular momentum; α⃗\vec{\alpha}α = angular acceleration; III = moment of inertia; F⃗\vec{F}F = force; ttt = time). Choose the correct answer from the options given below:
  1. (A)B, D and E Only
  2. (B)C and D Only
  3. (C)B, C, D and E Only
  4. (D)A, B, D and E Only

Correct answer: (C)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2025
A wheel is rolling on a plane surface. The speed of a particle at the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of the wheel, will be:
  1. (A)424\sqrt242​ m/s
  2. (B)8 m/s
  3. (C)4 m/s
  4. (D)828\sqrt282​ m/s

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2025
A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, the acceleration of the center of the sphere is:
  1. (A)3.5 m/s23.5\,\text{m/s}^23.5m/s2
  2. (B)0.35 m/s20.35\,\text{m/s}^20.35m/s2
  3. (C)2.5 m/s22.5\,\text{m/s}^22.5m/s2
  4. (D)0.25 m/s20.25\,\text{m/s}^20.25m/s2

Correct answer: (A)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2025
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be:
  1. (A)20 rad/s
  2. (B)30 rad/s
  3. (C)10 rad/s
  4. (D)0 rad/s

Correct answer: (A)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2025
A square Lamina OABC of length 10 cm is pivoted at ‘O’. Forces act at the lamina as shown in figure. If the lamina remains stationary, then the magnitude of FFF is:
  1. (A)20 N
  2. (B)0 (zero)
  3. (C)10 N
  4. (D)10210\sqrt{2}102​ N

Correct answer: (C)

Step-by-step solution →
Q49·PhysicsIntegerJEE Main 2025
A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by a force of 10 N as shown in the figure. The established torque produces an angular acceleration of 2 rad/s2rad/s^2rad/s2. Moment of inertia of the wheel is ______ kg m2kg\,m^2kgm2. (Acceleration due to gravity =10 m/s2=10\,m/s^2=10m/s2)

Correct answer: 1

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2025
The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is:
  1. (A)12Mr2\dfrac{1}{2}Mr^221​Mr2
  2. (B)38Mr2\dfrac{3}{8}Mr^283​Mr2
  3. (C)32Mr2\dfrac{3}{2}Mr^223​Mr2
  4. (D)2Mr22Mr^22Mr2

Correct answer: (B)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2025
Moment of inertia of a rod of mass ‘M’ and length ‘L’ about an axis passing through its centre and normal to its length is ‘α\alphaα’. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of this cross about an axis passing through its centre and normal to the plane containing the cross is:
  1. (A)α\alphaα
  2. (B)α4\dfrac{\alpha}{4}4α​
  3. (C)α8\dfrac{\alpha}{8}8α​
  4. (D)α2\dfrac{\alpha}{2}2α​

Correct answer: (B)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2025
Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0, they are given an initial velocity V⃗A=V0 AC^\vec{V}_A=V_0\,\hat{AC}VA​=V0​AC^, V⃗B=V0 BA^\vec{V}_B=V_0\,\hat{BA}VB​=V0​BA^ and V⃗C=V0 CB^\vec{V}_C=V_0\,\hat{CB}VC​=V0​CB^. Here, AC^,CB^\hat{AC}, \hat{CB}AC^,CB^ and BA^\hat{BA}BA^ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:
  1. (A)12amV0\dfrac{1}{2}a m V_021​amV0​
  2. (B)3amV03 a m V_03amV0​
  3. (C)32amV0\dfrac{\sqrt{3}}{2}a m V_023​​amV0​
  4. (D)32amV0\dfrac{3}{2}a m V_023​amV0​

Correct answer: (C)

Step-by-step solution →
Q53·PhysicsIntegerJEE Main 2025
The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force F⃗=i^−j^+k^\vec{F}=\hat{i}-\hat{j}+\hat{k}F=i^−j^​+k^ acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is ______ Nm.

Correct answer: 2

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2025
The centre of mass of a thin rectangular plate with sides of length a and b, whose mass per unit area (σ)(\sigma)(σ) varies as σ=σ0xab\sigma=\frac{\sigma_0 x}{ab}σ=abσ0​x​ (where σ0\sigma_0σ0​ is a constant), would be______
  1. (A)(23a,b2)\left(\frac{2}{3}a, \frac{b}{2}\right)(32​a,2b​)
  2. (B)(23a,23b)\left(\frac{2}{3}a, \frac{2}{3}b\right)(32​a,32​b)
  3. (C)(a2,b2)\left(\frac{a}{2}, \frac{b}{2}\right)(2a​,2b​)
  4. (D)(13a,b2)\left(\frac{1}{3}a, \frac{b}{2}\right)(31​a,2b​)

Correct answer: (A)

Step-by-step solution →
Q55·PhysicsIntegerJEE Main 2025
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n = _______. Consider all the bodies have equal masses.

Correct answer: 4

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2025
A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is:
  1. (A)300 g
  2. (B)190 g
  3. (C)200 g
  4. (D)290 g

Correct answer: (B)

Step-by-step solution →
Q57·PhysicsIntegerJEE Main 2025
Two iron solid discs of negligible thickness have radii R1R_1R1​ and R2R_2R2​ and moment of inertia I1I_1I1​ and I2I_2I2​, respectively. For R2=2R1R_2=2R_1R2​=2R1​, the ratio of I1I_1I1​ and I2I_2I2​ would be 1x\frac{1}{x}x1​, where x = _______.

Correct answer: 16

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2025
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:
  1. (A)25\dfrac{2}{5}52​
  2. (B)52\dfrac{5}{2}25​
  3. (C)34\dfrac{3}{4}43​
  4. (D)43\dfrac{4}{3}34​

Correct answer: (B)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2025
An object of mass ‘m’ is projected from origin in a vertical x-y plane at an angle 45° with the x-axis with an initial velocity v0v_0v0​. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, is [g is acceleration due to gravity]
  1. (A)mv0322 g\dfrac{mv_0^3}{2\sqrt2\,g}22​gmv03​​ along negative z-axis
  2. (B)mv0322 g\dfrac{mv_0^3}{2\sqrt2\,g}22​gmv03​​ along positive z-axis
  3. (C)mv0342 g\dfrac{mv_0^3}{4\sqrt2\,g}42​gmv03​​ along positive z-axis
  4. (D)mv0342 g\dfrac{mv_0^3}{4\sqrt2\,g}42​gmv03​​ along negative z-axis

Correct answer: (D)

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2025
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t1t_1t1​ and t2t_2t2​, respectively, then
  1. (A)t1<t2t_1<t_2t1​<t2​
  2. (B)t1=t2t_1=t_2t1​=t2​
  3. (C)t1=2t2t_1=2t_2t1​=2t2​
  4. (D)t1>t2t_1>t_2t1​>t2​

Correct answer: (A)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2025
A uniform solid cylinder of mass ‘m’ and radius ‘r’ rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-
  1. (A)12g\dfrac{1}{\sqrt2}g2​1​g
  2. (B)132g\dfrac{1}{3\sqrt2}g32​1​g
  3. (C)2g3\dfrac{\sqrt2 g}{3}32​g​
  4. (D)2g\sqrt2 g2​g

Correct answer: (C)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2025
A circular disk of radius R meter and mass M kg is rotating about the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t2−8t\theta(t)=5t^2-8tθ(t)=5t2−8t, where θ(t)\theta(t)θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t=2t=2t=2s?
  1. (A)60 MR260\,MR^260MR2
  2. (B)72 MR272\,MR^272MR2
  3. (C)108 MR2108\,MR^2108MR2
  4. (D)8 MR28\,MR^28MR2

Correct answer: (A)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2025
A solid sphere of mass ‘m’ and radius ‘r’ is allowed to roll without slipping from the highest point of an inclined plane of length ‘L’ and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is v1v_1v1​. If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2v_2v2​. The ratio of v12:v22v_1^2:v_2^2v12​:v22​ is
  1. (A)1:21:\sqrt21:2​
  2. (B)1:31:31:3
  3. (C)1:21:21:2
  4. (D)1:31:\sqrt31:3​

Correct answer: (A)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2025
Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be-
  1. (A)2.0 cm
  2. (B)0.5 cm
  3. (C)1.5 cm
  4. (D)1.0 cm

Correct answer: (D)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2025
The torque due to the force (2i^+j^+2k^)\left(2\hat{i}+\hat{j}+2\hat{k}\right)(2i^+j^​+2k^) about the origin, acting on a particle whose position vector is (i^+j^+k^)\left(\hat{i}+\hat{j}+\hat{k}\right)(i^+j^​+k^), would be
  1. (A)i^−j^+k^\hat{i}-\hat{j}+\hat{k}i^−j^​+k^
  2. (B)i^+k^\hat{i}+\hat{k}i^+k^
  3. (C)i^−k^\hat{i}-\hat{k}i^−k^
  4. (D)j^−k^\hat{j}-\hat{k}j^​−k^

Correct answer: (C)

Step-by-step solution →
Q66·PhysicsIntegerJEE Main 2025
The position vectors of two 1 kg particles A and B are given by r_A = (α₁t i + α₂t j + α₃t k) m and r_B = (β₁ i + β₂t² j + β₃t k) m, where (α₁ = 1, α₂ = 3 m/s, α₃ = 2 m/s) and (β₁ = 1 m, β₂ = ... , β₃ = ... ) are constants, with t in seconds. At t = 1 s the magnitude of the angular momentum of particle B with respect to the position of particle A is √L kg m²/s. The value of L is ______.

Correct answer: 90

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2025
A uniform circular disc of radius R and mass M is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure (its boundary passes through the centre). The moment of inertia of the remaining part about the same axis is:
  1. (A)7/32 MR²
  2. (B)9/32 MR²
  3. (C)17/32 MR²
  4. (D)13/32 MR²

Correct answer: (D)

Step-by-step solution →
Q68·PhysicsIntegerJEE Main 2025
A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is FαM\sqrt{\frac{F}{\alpha M}}αMF​​ in SI unit. The value of α\alphaα is ______.

Correct answer: 1

Step-by-step solution →
Q69·PhysicsIntegerJEE Advanced 2024
A disc of mass MMM and radius RRR is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass MMM and radius R/2R/2R/2 is fixed to the motor's thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed ω\omegaω. If the angular speed at which the large disc rotates is ω/n\omega/nω/n, then the value of nnn is ______.

Correct answer: 12

Step-by-step solution →
Q70·PhysicsIntegerJEE Advanced 2024
A thin uniform rod of length LLL and certain mass is kept on a frictionless horizontal table with a massless string of length LLL fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point OOO. If a horizontal impulse PPP is imparted to the rod at a distance x=L/nx = L/nx=L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point OOO, with the rod remaining aligned with the string. In such a case, the value of nnn is ______.

Correct answer: 18

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2024
A heavy iron bar, of weight WWW is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle θ\thetaθ with the horizontal. The weight experienced by the person is:
  1. (A)W2\frac{W}{2}2W​
  2. (B)WWW
  3. (C)Wcos⁡θW\cos\thetaWcosθ
  4. (D)Wsin⁡θW\sin\thetaWsinθ

Correct answer: (A)

Step-by-step solution →
Q72·PhysicsNumericalJEE Main 2024
A string is wrapped around the rim of a wheel of moment of inertia 0.40 kgm20.40\ \mathrm{kgm^{2}}0.40 kgm2 and radius 10 cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 N. The angular velocity of the wheel after 10 s is xxx rad/s, where xxx is ______.

Correct answer: 100

Step-by-step solution →
Q73·PhysicsNumericalJEE Main 2024
A circular disc reaches from top to bottom of an inclined plane of length lll. When it slips down the plane, it takes time t1t_{1}t1​. When it rolls down the plane then it takes (α2)1/2t1\left(\dfrac{\alpha}{2}\right)^{1/2}t_{1}(2α​)1/2t1​ s, where α\alphaα is _______.

Correct answer: 3

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2024
A thin circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity ω\omegaω. If another disc of same dimensions but of mass M2\dfrac{M}{2}2M​ is placed gently on the first disc co-axially, then the new angular velocity of the system is:
  1. (A)4ω5\dfrac{4\omega}{5}54ω​
  2. (B)5ω4\dfrac{5\omega}{4}45ω​
  3. (C)2ω3\dfrac{2\omega}{3}32ω​
  4. (D)3ω2\dfrac{3\omega}{2}23ω​

Correct answer: (C)

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2024
A uniform thin metal plate of mass 10 kg10\,kg10kg with dimensions as shown. The ratio of xxx and yyy coordinates of center of mass of plate is n9\dfrac{n}{9}9n​. The value of nnn is ___

Correct answer: 15

Step-by-step solution →
Q76·PhysicsNumericalJEE Main 2024
A circular table is rotating with an angular velocity of ω\omegaω rad/s about its axis. There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of 1 m on the groove. All the surfaces are smooth. If the radius of the table is 3 m, the radial velocity of the ball w.r.t. the table at the time the ball leaves the table is x2x\sqrt{2}x2​ m/s, where the value of x is ________ .

Correct answer: 2

Step-by-step solution →
Q77·PhysicsNumericalJEE Main 2024
Three balls of masses 2kg, 4kg and 6kg respectively are arranged at centre of the edges of an equilateral triangle of side 2 m. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ______ kg m2^{2}2.

Correct answer: 4

Step-by-step solution →
Q78·PhysicsNumericalJEE Main 2024
If the radius of earth is reduced to three-fourth of its present value without change in its mass, then the value of the duration of the day of earth will be _______ hours 30 minutes.

Correct answer: 13

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2024
The ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for the moment of inertia about their diameter axis AB as shown in the figure, is 8x\sqrt{\dfrac{8}{x}}x8​​. The value of xxx is:
  1. (A)34
  2. (B)17
  3. (C)67
  4. (D)51

Correct answer: (C)

Step-by-step solution →
Q80·PhysicsNumericalJEE Main 2024
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of its rotational kinetic energy to its total kinetic energy is x5\dfrac{x}{5}5x​. The value of xxx is __________.

Correct answer: 2

Step-by-step solution →
Q81·PhysicsNumericalJEE Main 2024
In a system two particles of masses m1=3m_1=3m1​=3 kg and m2=2m_2=2m2​=2 kg are placed at certain distance from each other. The particle of mass m1m_1m1​ is moved towards the center of mass of the system through a distance 2 cm. In order to keep the center of mass of the system at the original position, the particle of mass m2m_2m2​ should move towards the center of mass by the distance ______ cm.

Correct answer: 3

Step-by-step solution →
Q82·PhysicsNumericalJEE Main 2024
A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed vvv. The sphere and the cylinder reaches upto maximum heights h1h_1h1​ and h2h_2h2​, respectively, above the initial level. The ratio h1:h2h_1:h_2h1​:h2​ is n10\dfrac{n}{10}10n​. The value of nnn is ___

Correct answer: 7

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2024
A disc of radius R and mass M is rolling horizontally without slipping with speed vvv. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:
  1. (A)v2g\dfrac{v^2}{g}gv2​
  2. (B)34v2g\dfrac{3}{4}\dfrac{v^2}{g}43​gv2​
  3. (C)12v2g\dfrac{1}{2}\dfrac{v^2}{g}21​gv2​
  4. (D)23v2g\dfrac{2}{3}\dfrac{v^2}{g}32​gv2​

Correct answer: (C)

Step-by-step solution →
Q84·PhysicsNumericalJEE Main 2024
A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B. The time taken by the rod to turn through at right angles will be πx\dfrac{\pi}{x}xπ​ s, where x=x=x= __________.

Correct answer: 4

Step-by-step solution →
Q85·PhysicsNumericalJEE Main 2024
The identical spheres each of mass 2M2M2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m4\,m4m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is 42x\dfrac{4\sqrt2}{x}x42​​, where the value of xxx is ___

Correct answer: 3

Step-by-step solution →
Q86·PhysicsNumericalJEE Main 2024
A body of mass 'mmm' is projected with a speed 'uuu' making an angle of 45∘45^\circ45∘ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 2 mu3Xg\dfrac{\sqrt{2}\,mu^3}{Xg}Xg2​mu3​. The value of 'XXX' is ______.

Correct answer: 8

Step-by-step solution →
Q87·PhysicsNumericalJEE Main 2024
Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x20\dfrac{x}{20}20x​ kg m2^22, where the value of xxx is ______.

Correct answer: 53

Step-by-step solution →
Q88·PhysicsNumericalJEE Main 2024
A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is ______ J.

Correct answer: 6

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2024
A particle of mass mmm is projected with a velocity uuu making an angle of 30∘30^\circ30∘ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height hhh is:
  1. (A)3 mu316g\dfrac{\sqrt3\,mu^3}{16g}16g3​mu3​
  2. (B)3 mu32g\dfrac{\sqrt3\,mu^3}{2g}2g3​mu3​
  3. (C)mu32 g\dfrac{mu^3}{\sqrt2\,g}2​gmu3​
  4. (D)zero

Correct answer: (A)

Step-by-step solution →
Q90·PhysicsNumericalJEE Main 2024
Two discs of moment of inertia I1=4I_1=4I1​=4 kg m2^22 and I2=2I_2=2I2​=2 kg m2^22 about their central axes & normal to their planes, rotating with angular speeds 10 rad/s & 4 rad/s respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is ______ J.

Correct answer: 24

Step-by-step solution →
Q91·PhysicsNumericalJEE Main 2024
Consider a Disc of mass 5 kg5\,kg5kg, radius 2 m2\,m2m, rotating with angular velocity of 10 rad/s10\,rad/s10rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that the discs continue to rotate together without slipping is ___ JJJ.

Correct answer: 250

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2024
A cylinder is rolling down on an inclined plane of inclination 60∘60^\circ60∘. It’s acceleration during rolling down will be x3\dfrac{x}{\sqrt{3}}3​x​ m/s2^22, where x=x=x= ______ . (use g=10g=10g=10 m/s2^22).

Correct answer: 10

Step-by-step solution →
Q93·PhysicsNumericalJEE Main 2024
A body of mass 5 kg is moving with a uniform speed 323\sqrt{2}32​ ms−1^{-1}−1 in X–Y plane along the line y=x+4y=x+4y=x+4. The angular momentum of the particle about the origin will be ___ kg m2^{2}2s−1^{-1}−1.

Correct answer: 60

Step-by-step solution →
Q94·PhysicsNumericalJEE Main 2024
A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7x\dfrac{7}{x}x7​ where x is ________.

Correct answer: 7.00

Step-by-step solution →
Q95·PhysicsNumericalJEE Main 2024
Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is __________ kgm2^22.

Correct answer: 16

Step-by-step solution →
Q96·PhysicsIntegerJEE Advanced 2023
A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xy -plane. The coin is tossed vertically up (+z direction) by applying an impulse of π2×10−2\sqrt{\frac{\pi}{2}}\times10^{-2}2π​​×10−2 N-s at a distance 2/3 cm from its center. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is ________. [Given: The acceleration due to gravity g = 10 m s−2^{-2}−2]

Correct answer: 30

Step-by-step solution →
Q97·PhysicsMultiple correctJEE Advanced 2023
An annular disk of mass M, inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction μ\muμ, as shown in the figure. At some time, an impulse J0x^J_0\hat{x}J0​x^ is applied at a height h above the center of the disk. If h=hmh = h_mh=hm​ then the disk rolls without slipping along the x-axis. Which of the following statement(s) is(are) correct?
  1. (A)For μ≠0\mu \neq 0μ=0 and a→0a \rightarrow 0a→0, hm=b/2h_m = b/2hm​=b/2
  2. (B)For μ≠0\mu \neq 0μ=0 and a→ba \rightarrow ba→b, hm=bh_m = bhm​=b
  3. (C)For h=hmh = h_mh=hm​, the initial angular velocity does not depend on the inner radius a.
  4. (D)For μ=0\mu = 0μ=0 and h=0h = 0h=0, the wheel always slides without rolling.

Correct answer: (A), (B), (C), (D)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Advanced 2023
A bar of mass M = 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m = 0.10 kg is moving on the same horizontal surface with 5.00 ms−1ms^{-1}ms−1 speed on a path perpendicular to the bar. It hits the bar at a distance L/2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity ω\omegaω. Which of the following statement is correct?
  1. (A)ω=6.98\omega = 6.98ω=6.98 rad s−1s^{-1}s−1 and v=4.30 ms−1v = 4.30\ ms^{-1}v=4.30 ms−1
  2. (B)ω=3.75\omega = 3.75ω=3.75 rad s−1s^{-1}s−1 and v=4.30 ms−1v = 4.30\ ms^{-1}v=4.30 ms−1
  3. (C)ω=3.75\omega = 3.75ω=3.75 rad s−1s^{-1}s−1 and v=10.0 ms−1v = 10.0\ ms^{-1}v=10.0 ms−1
  4. (D)ω=6.80\omega = 6.80ω=6.80 rad s−1s^{-1}s−1 and v=4.10 ms−1v = 4.10\ ms^{-1}v=4.10 ms−1

Correct answer: (A)

Step-by-step solution →
Q99·PhysicsNumericalJEE Main 2023
A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (ksph:kcylk_{sph} : k_{cyl}ksph​:kcyl​) is 2:x2 : \sqrt{x}2:x​, then value of x is __________.

Correct answer: 5

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2023
A disc is rolling without slipping on a surface. The radius of the disc is RRR. At t=0t=0t=0, the top most point on the disc is AAA as shown in figure. When the disc completes half of its rotation, the displacement of point AAA from its initial position is:
  1. (A)Rπ2+4R\sqrt{\pi^{2}+4}Rπ2+4​
  2. (B)Rπ2+1R\sqrt{\pi^{2}+1}Rπ2+1​
  3. (C)2R2R2R
  4. (D)2R1+4π22R\sqrt{1+4\pi^{2}}2R1+4π2​

Correct answer: (A)

Step-by-step solution →
Q101·PhysicsNumericalJEE Main 2023
A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be ____ rad s−2^{-2}−2.

Correct answer: 15

Step-by-step solution →
Q102·PhysicsNumericalJEE Main 2023
A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π:22\pi:22π:22 then, the value of its angular speed will be _____ rad/s.

Correct answer: 4

Step-by-step solution →
Q103·PhysicsNumericalJEE Main 2023
For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is x5\dfrac{x}{5}5x​. The value of x is _________.

Correct answer: 2

Step-by-step solution →
Q104·PhysicsNumericalJEE Main 2023
A circular plate is rotating in a horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity ω\omegaω. A person sits at the centre having two dumbbells in his hands. When he stretches out his hands the moment of inertia of the system becomes triple. If E is the initial kinetic energy of the system, then the final kinetic energy will be Ex\dfrac{E}{x}xE​. The value of x is ____.

Correct answer: 3

Step-by-step solution →
Q105·PhysicsNumericalJEE Main 2023
A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s−1\text{s}^{-1}s−1. If the moment of inertia of the sphere about its tangent is x×10−2x \times 10^{-2}x×10−2 times its angular momentum about the diameter. Then the value of x will be ______.

Correct answer: 35

Step-by-step solution →
Q106·PhysicsNumericalJEE Main 2023
A force of −Pk^-P\hat{k}−Pk^ acts on the origin of the coordinate system. The torque about the point (2,−3)(2,-3)(2,−3) is P(ai^+bj^)P(a\hat{i}+b\hat{j})P(ai^+bj^​). If the ratio ab=x2\tfrac{a}{b}=\tfrac{x}{2}ba​=2x​, then the value of x is _______ .

Correct answer: 3

Step-by-step solution →
Q107·PhysicsAssertion & ReasonJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: An electric fan continues to rotate for some time after the current is switched off. Reason R: Fan continues to rotate due to inertia of motion. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)A is correct but R is not correct
  2. (B)Both A and R are correct and R is the correct explanation of A
  3. (C)A is not correct but R is correct
  4. (D)Both A and R are correct but R is NOT the correct explanation of A

Correct answer: (B)

Step-by-step solution →
Q108·PhysicsNumericalJEE Main 2023
If the earth suddenly shrinks to 164\dfrac{1}{64}641​th of its original volume with its mass remaining the same, the period of rotation of earth becomes 24x\dfrac{24}{x}x24​ h. The value of x is _________.

Correct answer: 16

Step-by-step solution →
Q109·PhysicsNumericalJEE Main 2023
The moment of inertia of semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is 1xMR2\frac{1}{x}MR^{2}x1​MR2, where RRR is the radius and MMM is the mass of semicircular ring. The value of xxx will be

Correct answer: 1

Step-by-step solution →
Q110·PhysicsNumericalJEE Main 2023
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m/s (as shown in the figure). The maximum height with respect to the initial position covered by it will be ____ cm.

Correct answer: 75

Step-by-step solution →
Q111·PhysicsNumericalJEE Main 2023
Two identical solid spheres each of mass 2 kg2\,kg2kg and radius 10 cm10\,cm10cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm40\,cm40cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is _____ ×10−3 kg m2\times10^{-3}\,kg\,m^{2}×10−3kgm2.

Correct answer: 176

Step-by-step solution →
Q112·PhysicsNumericalJEE Main 2023
A ring and a solid sphere rotating about an axis passing through their centers have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is 2x\sqrt{\dfrac{2}{x}}x2​​. The value of xxx is

Correct answer: 5

Step-by-step solution →
Q113·PhysicsNumericalJEE Main 2023
Moment of inertia of a disc of mass MMM and radius 'RRR' about any of its diameter is MR24\dfrac{MR^2}{4}4MR2​. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be x2MR2\dfrac{x}{2}MR^22x​MR2. The value of xxx is _________.

Correct answer: 3

Step-by-step solution →
Q114·PhysicsNumericalJEE Main 2023
A solid cylinder is released from rest from the top of an inclined plane of inclination 30∘30^\circ30∘ and length 606060 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is _______ ms−1^{-1}−1 (given g=10g=10g=10 ms−2^{-2}−2).

Correct answer: 2

Step-by-step solution →
Q115·PhysicsNumericalJEE Main 2023
A solid sphere of mass 111 kg rolls without slipping on a plane surface. Its kinetic energy is 7×10−37 \times 10^{-3}7×10−3 J. The speed of the centre of mass of the sphere is __________ cms−1\text{cms}^{-1}cms−1

Correct answer: 10

Step-by-step solution →
Q116·PhysicsNumericalJEE Main 2023
Two discs of same mass and different radii are made of different materials such that their thicknesses are 111 cm and 0.50.50.5 cm respectively. The densities of materials are in the ratio 3:53:53:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x6\dfrac{x}{6}6x​. The value of xxx is _________.

Correct answer: 5

Step-by-step solution →
Q117·PhysicsNumericalJEE Main 2023
A uniform disc of mass 0.5 kg and radius rrr is projected with velocity 18 m/s at t=0t=0t=0 s on a rough horizontal surface. It starts with a purely sliding motion at t=0t=0t=0 s and after 2 s it acquires a purely rolling motion. The total kinetic energy of the disc after 2 s will be _______ J. (Given coefficient of friction is 0.3 and g=10g=10g=10 m/s2^22)

Correct answer: 54

Step-by-step solution →
Q118·PhysicsNumericalJEE Main 2023
A thin uniform rod of length 222 m, cross sectional area 'AAA' and density 'ddd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω\omegaω. If value of ω\omegaω in terms of its rotational kinetic energy EEE is αEAd\sqrt{\dfrac{\alpha E}{Ad}}AdαE​​ then value of α\alphaα is

Correct answer: 3

Step-by-step solution →
Q119·PhysicsNumericalJEE Main 2023
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ________ ms−1^{-1}−1.

Correct answer: 40

Step-by-step solution →
Q120·PhysicsNumericalJEE Main 2023
A particle of mass 100 g100\,g100g is projected at time t=0t=0t=0 with a speed 20 ms−120\,ms^{-1}20ms−1 at an angle 45∘45^\circ45∘ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 st=2\,st=2s is found to be K kg m2/s\sqrt K\,kg\,m^2/sK​kgm2/s. The value of KKK is _____. (Take g=10 ms−2g=10\,ms^{-2}g=10ms−2)

Correct answer: 800

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2023
An object of mass 8 kg hanging from one end of a uniform rod CD of mass 2 kg and length 1 m, pivoted at its end C on a vertical wall, is supported by a cable AB as shown in the figure such that the system is in equilibrium. The tension in the cable is: (Take g=10g=10g=10 m/s2^22)
  1. (A)909090 N
  2. (B)303030 N
  3. (C)300300300 N
  4. (D)240240240 N

Correct answer: (C)

Step-by-step solution →
Q122·PhysicsNumericalJEE Main 2023
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be x7\frac{x}{7}7x​. The value of xxx is ______.

Correct answer: 5

Step-by-step solution →
Q123·PhysicsNumericalJEE Main 2023
ICMI_{CM}ICM​ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of the disc. IABI_{AB}IAB​ is its moment of inertia about an axis AB perpendicular to the plane and parallel to axis CM at a distance 23R\dfrac{2}{3}R32​R from the center, where R is the radius of the disc. The ratio of IABI_{AB}IAB​ and ICMI_{CM}ICM​ is x:9x:9x:9. The value of xxx is _______.

Correct answer: 17

Step-by-step solution →
Q124·PhysicsNumericalJEE Main 2023
A uniform solid cylinder with radius RRR and length LLL has moment of inertia I1I_1I1​ about the axis of the cylinder. A concentric solid cylinder of radius R′=R2R'=\dfrac{R}{2}R′=2R​ and length L′=L2L'=\dfrac{L}{2}L′=2L​ is carved out of the original cylinder. If I2I_2I2​ is the moment of inertia of the carved out portion of the cylinder then I1I2=\dfrac{I_1}{I_2}=I2​I1​​= _________ (Both I1I_1I1​ and I2I_2I2​ are about the axis of the cylinder)

Correct answer: 32

Step-by-step solution →
Q125·PhysicsNumericalJEE Main 2023
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be x\sqrt{x}x​ cm. The value of xxx is _______ .

Correct answer: 110

Step-by-step solution →
Q126·PhysicsNumericalJEE Advanced 2022
A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ=30∘\theta=30^\circθ=30∘ from the horizontal. Two forces of magnitude 1N each, parallel to the incline, act on the sphere, both at distance r = 0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ________ ms−2^{-2}−2. (Take g = 10 ms−2^{-2}−2.)

Correct answer: 2.85 or 2.86

Step-by-step solution →
Q127·PhysicsIntegerJEE Advanced 2022
A particle of mass 1 kg is subjected to a force which depends on the position as F⃗=−k(xi^+yj^)\vec{F} = -k(x\hat{i} + y\hat{j})F=−k(xi^+yj^​) kg ms−2^{-2}−2 with k = 1 kg s−2^{-2}−2. At time t = 0, the particle's position r⃗=(12i^+2j^)\vec{r} = \left(\frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}\right)r=(2​1​i^+2​j^​) m and its velocity v⃗=(−2i^+2j^+2πk^)\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}\right)v=(−2​i^+2​j^​+π2​k^) ms−1^{-1}−1, Let vx_xx​ and vy_yy​ denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z = 0.5 m, the value of (x vy_yy​ − y vx_xx​) is ______ m2^{2}2 s−1^{-1}−1.

Correct answer: 3

Step-by-step solution →
Q128·PhysicsSingle correctJEE Advanced 2022
A flat surface of a thin uniform disk A of radius R is glued to a horizontal table. Another thin uniform disk B of mass M and with the same radius R rolls without slipping on the circumference of A , as shown in the figure. A flat surface of B also lies on the plane of the table. The center of mass of B has fixed angular speed ω\omegaω about the vertical axis passing through the center of A . The angular momentum of B is nMωR2nM\omega R^2nMωR2 with respect to the center of A . Which of the following is the value of n ?
  1. (A)2
  2. (B)5
  3. (C)7/2
  4. (D)9/2

Correct answer: (B)

Step-by-step solution →
Q129·PhysicsNumericalJEE Advanced 2022
At time t=0, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of α=23 rad s−2\alpha = \frac{2}{3}\ rad\ s^{-2}α=32​ rad s−2 . A small stone is stuck to the disk. At t = 0, it is at the contact point of the disk and the plane. Later, at time t=πt = \sqrt{\pi}t=π​ s, the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is 12+x10\frac{1}{2} + \frac{x}{10}21​+10x​. The value of x is______ [Take g= 10ms−2^{-2}−2.]

Correct answer: 0.52

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Q130·PhysicsSingle correctJEE Main 2022
The torque of a force 5i^+3j^−7k^5\hat{i} + 3\hat{j} - 7\hat{k}5i^+3j^​−7k^ about the origin is τ. If the force acts on a particle whose position vector is 2i^+2j^+k^2\hat{i} + 2\hat{j} + \hat{k}2i^+2j^​+k^, then the value of τ will be :
  1. (A)11i^+19j^−4k^11\hat{i} + 19\hat{j} - 4\hat{k}11i^+19j^​−4k^
  2. (B)−11i^+9j^−16k^-11\hat{i} + 9\hat{j} - 16\hat{k}−11i^+9j^​−16k^
  3. (C)−17i^+19j^−4k^-17\hat{i} + 19\hat{j} - 4\hat{k}−17i^+19j^​−4k^
  4. (D)17i^+9j^+16k^17\hat{i} + 9\hat{j} + 16\hat{k}17i^+9j^​+16k^

Correct answer: (C)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2022
Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}i^+2j^​+k^ and −3i^−2j^+k^-3\hat{i} - 2\hat{j} + \hat{k}−3i^−2j^​+k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :
  1. (A)i^−2j^+k^\hat{i} - 2\hat{j} + \hat{k}i^−2j^​+k^
  2. (B)−3i^−2j^+k^-3\hat{i} - 2\hat{j} + \hat{k}−3i^−2j^​+k^
  3. (C)−2i^+2k^-2\hat{i} + 2\hat{k}−2i^+2k^
  4. (D)−2i^−j^+2k^-2\hat{i} - \hat{j} + 2\hat{k}−2i^−j^​+2k^

Correct answer: (A)

Step-by-step solution →
Q132·PhysicsNumericalJEE Main 2022
Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is x4\frac{x}{4}4x​Ma2^{2}2. Then, the value of x will be ________.

Correct answer: 3

Step-by-step solution →
Q133·PhysicsNumericalJEE Main 2022
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ=ρ0(1−x2L2)\rho = \rho_0\left(1 - \dfrac{x^2}{L^2}\right)ρ=ρ0​(1−L2x2​) kg/m and length L (in meter) is 3Lα\dfrac{3L}{\alpha}α3L​ m. The value of α\alphaα is ......... (where x is the distance form end A)

Correct answer: 8

Step-by-step solution →
Q134·PhysicsNumericalJEE Main 2022
A pulley of radius 1.5 m is rotated about its axis by a force F=(12t−3t2)F = (12t - 3t^{2})F=(12t−3t2) N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m2\mathrm{m}^{2}m2, the number of rotations made by the pulley before its direction of motion is reversed, will be Kπ\frac{K}{\pi}πK​. The value of K is ____ .

Correct answer: 18

Step-by-step solution →
Q135·PhysicsNumericalJEE Main 2022
A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of 4 ms−1^{-1}−1, is…….cm. (take g = 10 ms−2^{-2}−2)

Correct answer: 120

Step-by-step solution →
Q136·PhysicsNumericalJEE Main 2022
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be________ m. Given, the length of the rod is 10310\sqrt{3}103​ m.

Correct answer: 5

Step-by-step solution →
Q137·PhysicsNumericalJEE Main 2022
A disc of mass 1 kg and radius R is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be 4x3R4\sqrt{\frac{x}{3R}}43Rx​​ rad s−1^{-1}−1 where x = ______ (g = 10ms−2^{-2}−2)

Correct answer: 5

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2022
A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :
  1. (A)53\sqrt{\frac{5}{3}}35​​
  2. (B)45\sqrt{\frac{4}{5}}54​​
  3. (C)35\sqrt{\frac{3}{5}}53​​
  4. (D)1415\sqrt{\frac{14}{15}}1514​​

Correct answer: (D)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2022
A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is a3R2ω\frac{a}{3}R^2\omega3a​R2ω. The value of a will be :
  1. (A)2
  2. (B)3
  3. (C)5
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q140·PhysicsNumericalJEE Main 2022
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1I_1I1​. The same rod is bent into a ring and its moment of inertia about a diameter is I2I_2I2​. If I1I2\frac{I_1}{I_2}I2​I1​​ is xπ23\frac{x\pi^2}{3}3xπ2​, then the value of x will be________.

Correct answer: 8

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Q141·PhysicsSingle correctJEE Main 2022
A ball is spun with angular acceleration α=6t2−2t\alpha = 6t^2 - 2tα=6t2−2t where t is in second and α is in rads−2^{-2}−2. At t = 0, the ball has angular velocity of 10 rads−1^{-1}−1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:
  1. (A)32t4−t2+10t\frac{3}{2} t^4 - t^2 + 10t23​t4−t2+10t
  2. (B)t42−t33+10t+4\frac{t^4}{2} - \frac{t^3}{3} + 10t + 42t4​−3t3​+10t+4
  3. (C)2t43−t36+10t+12\frac{2t^4}{3} - \frac{t^3}{6} + 10t + 1232t4​−6t3​+10t+12
  4. (D)2t4−t32+5t+42t^4 - \frac{t^3}{2} + 5t + 42t4−2t3​+5t+4

Correct answer: (B)

Step-by-step solution →
Q142·PhysicsNumericalJEE Main 2022
The position vector of 1 kg object is r⃗=(3i^−j^)\vec{r} = \left(3\hat{i} - \hat{j}\right)r=(3i^−j^​) m and its velocity v⃗=(3j^+k)\vec{v} = \left(3\hat{j} + k\right)v=(3j^​+k) ms−1^{-1}−1. The magnitude of its angular momentum is x\sqrt{x}x​ Nm where x is ______ .

Correct answer: 91

Step-by-step solution →
Q143·PhysicsNumericalJEE Main 2022
A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ________N. (Take g = 10 ms−2^{-2}−2)

Correct answer: 10

Step-by-step solution →
Q144·PhysicsSingle correctJEE Main 2022
A 34\sqrt{34}34​ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ff_ff​ and Fw_ww​ are the reaction forces of the floor and the wall, then ratio of Fw_ww​/Ff_ff​ will be: (Use g = 10 m/s2^22)
  1. (A)6110\frac{6}{\sqrt{110}}110​6​
  2. (B)3113\frac{3}{\sqrt{113}}113​3​
  3. (C)3109\frac{3}{\sqrt{109}}109​3​
  4. (D)2109\frac{2}{\sqrt{109}}109​2​

Correct answer: (C)

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2022
Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :
  1. (A)4 cm towards the 10 kg block
  2. (B)2 cm away from the 10 kg block
  3. (C)2 cm towards the 10 kg block
  4. (D)4 cm away from the 10 kg block

Correct answer: (C)

Step-by-step solution →
Q146·PhysicsNumericalJEE Main 2022
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 12xgh\dfrac{1}{2}\sqrt{xgh}21​xgh​ m/s. The value of x is _______ .

Correct answer: 3

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2022
A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads−1^{-1}−1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads−1^{-1}−1).
  1. (A)M(M+m)\frac{M}{\left(M+m\right)}(M+m)M​
  2. (B)(M+2m)2M\frac{\left(M+2m\right)}{2M}2M(M+2m)​
  3. (C)2M(M+2m)\frac{2M}{\left(M+2m\right)}(M+2m)2M​
  4. (D)2(M+2m)M\frac{2\left(M+2m\right)}{M}M2(M+2m)​

Correct answer: (C)

Step-by-step solution →
Q148·PhysicsSingle correctJEE Main 2022
Solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is :-
  1. (A)25\frac{2}{5}52​
  2. (B)27\frac{2}{7}72​
  3. (C)15\frac{1}{5}51​
  4. (D)710\frac{7}{10}107​

Correct answer: (B)

Step-by-step solution →
Q149·PhysicsNumericalJEE Main 2022
Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows: I1I_1I1​= M.I. of solid sphere about its diameter I2I_2I2​ = M.I. of solid cylinder about its axis I3I_3I3​ = M.I. of solid circular disc about its diameter I4I_4I4​ = M.I. of thin circular ring about its diameter If 2(I2+I3)+I4=x.I12(I_2 + I_3) + I_4 = x. I_12(I2​+I3​)+I4​=x.I1​ then the value of x will be ____

Correct answer: 5

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2022
If force F⃗=3i^+4j^−2k^\vec{F}=3\hat{i}+4\hat{j}-2\hat{k}F=3i^+4j^​−2k^ acts on a particle having position vector 2i^+j^+2k^2\hat{i}+\hat{j}+2\hat{k}2i^+j^​+2k^ then, the torque about the origin will be :-
  1. (A)3i^+4j^−2k^3\hat{i}+4\hat{j}-2\hat{k}3i^+4j^​−2k^
  2. (B)−10i^+10j^+5k^-10\hat{i}+10\hat{j}+5\hat{k}−10i^+10j^​+5k^
  3. (C)10i^+5j^−10k^10\hat{i}+5\hat{j}-10\hat{k}10i^+5j^​−10k^
  4. (D)10i^+j^−5k^10\hat{i}+\hat{j}-5\hat{k}10i^+j^​−5k^

Correct answer: (B)

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2022
A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :
  1. (A)7.5 rad
  2. (B)15 rad
  3. (C)20 rad
  4. (D)30 rad

Correct answer: (B)

Step-by-step solution →
Q152·PhysicsNumericalJEE Main 2022
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x×10−2x \times 10^{-2}x×10−2 kg. The value of x is

Correct answer: 6

Step-by-step solution →
Q153·PhysicsNumericalJEE Advanced 2021
A pendulum consists of a bob of mass mmm = 0.1 kg and a massless inextensible string of length LLL = 1.0 m. It is suspended from a fixed point at height HHH = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse PPP = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is JJJ kg-m2^22/s. The kinetic energy of the pendulum just after the lift-off is KKK Joules. The value of JJJ is ____.

Correct answer: 0.18

Step-by-step solution →
Q154·PhysicsMultiple correctJEE Advanced 2021
A horizontal force FFF is applied at the center of mass of a cylindrical object of mass m and radius RRR, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ\muμ. The center of mass of the object has an acceleration aaa. The acceleration due to gravity is ggg. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?
  1. (A)For the same FFF, the value of aaa does not depend on whether the cylinder is solid or hollow
  2. (B)For a solid cylinder, the maximum possible value of aaa is 2μg2\mu g2μg
  3. (C)The magnitude of the frictional force on the object due to the ground is always μmg\mu mgμmg
  4. (D)For a thin-walled hollow cylinder, a=F2ma = \frac{F}{2m}a=2mF​

Correct answer: (B), (D)

Step-by-step solution →
Q155·PhysicsMultiple correctJEE Advanced 2021
One end of a horizontal uniform beam of weight WWW and length LLL is hinged on a vertical wall at point OOO and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point QQQ, at a height LLL above the hinge at point OOO. A block of weight αW\alpha WαW is attached at the point PPP of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of (22)W(2\sqrt{2})W(22​)W. Which of the following statement(s) is(are) correct ?
  1. (A)The vertical component of reaction force at OOO does not depend on α\alphaα
  2. (B)The horizontal component of reaction force at OOO is equal to WWW for α=0.5\alpha = 0.5α=0.5
  3. (C)The tension in the rope is 2W2W2W for α=0.5\alpha = 0.5α=0.5
  4. (D)The rope breaks if α>1.5\alpha > 1.5α>1.5

Correct answer: (A), (B), (D)

Step-by-step solution →
Q156·PhysicsMultiple correctJEE Advanced 2021
A particle of mass MMM = 0.2 kg is initially at rest in the xyxyxy-plane at a point (x = −l-l−l, y = −h-h−h), where lll = 10 m and hhh = 1m. The particle is accelerated at time t = 0 with a constant acceleration a = 10 m/s2^22 along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by L⃗\vec{L}L and τ⃗\vec{\tau}τ, respectively. i^,j^\hat{i}, \hat{j}i^,j^​ and k^\hat{k}k^ are unit vectors along the positive x, y and z-directions, respectively. If k^=i^×j^\hat{k} = \hat{i} \times \hat{j}k^=i^×j^​ then which of the following statement(s) is(are) correct ?
  1. (A)The particle arrives at the point (x=lx = lx=l, y=−hy = -hy=−h) at time t = 2s.
  2. (B)τ⃗=2k^\vec{\tau} = 2\hat{k}τ=2k^ when the particle passes through the point (x=lx = lx=l, y=−hy = -hy=−h)
  3. (C)L⃗=4k^\vec{L} = 4\hat{k}L=4k^ when the particle passes through the point (x=lx = lx=l, y=−hy = -hy=−h)
  4. (D)τ⃗=k^\vec{\tau} = \hat{k}τ=k^ when the particle passes through the point (x=0x = 0x=0, y=−hy = -hy=−h)

Correct answer: (A), (B), (C)

Step-by-step solution →
Q157·PhysicsIntegerJEE Advanced 2021
A thin rod of mass M and length aaa is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a/4a/4a/4 is pivoted on this rod with its center at a distance a/4a/4a/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity Ω\OmegaΩ and the disc rotating about its vertical axis with angular velocity 4Ω4\Omega4Ω. The total angular momentum of the system about the point O is (Ma2Ω48)n\left(\frac{Ma^2\Omega}{48}\right)n(48Ma2Ω​)n . The value of n is____.

Correct answer: 49

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2021
Angular momentum of a single particle moving with constant speed along circular path :
  1. (A)changes in magnitude but remains same in the direction
  2. (B)remains same in magnitude and direction
  3. (C)remains same in magnitude but changes in the direction
  4. (D)is zero

Correct answer: (B)

Step-by-step solution →
Q159·PhysicsSingle correctJEE Main 2021
A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the end of light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint ?
  1. (A)18.7518.7518.75 kgm2^22
  2. (B)1.905×1051.905 \times 10^51.905×105 kgm2^22
  3. (C)19.0519.0519.05 kgm2^22
  4. (D)1.875×1051.875 \times 10^51.875×105 kgm2^22

Correct answer: (C)

Step-by-step solution →
Q160·PhysicsSingle correctJEE Main 2021
Two discs have moments of intertia I1I_{1}I1​ and I2I_{2}I2​ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1\omega_{1}ω1​ and ω2\omega_{2}ω2​ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :
  1. (A)I1I2(I1+I2)(ω1−ω2)2\frac{I_{1}I_{2}}{(I_{1}+I_{2})}\left(\omega_{1}-\omega_{2}\right)^{2}(I1​+I2​)I1​I2​​(ω1​−ω2​)2
  2. (B)(I1−I2)2 ω1ω22(I1+I2)\frac{(I_{1}-I_{2})^{2}\,\omega_{1}\omega_{2}}{2(I_{1}+I_{2})}2(I1​+I2​)(I1​−I2​)2ω1​ω2​​
  3. (C)I1I22(I1+I2)(ω1−ω2)2\frac{I_{1}I_{2}}{2(I_{1}+I_{2})}\left(\omega_{1}-\omega_{2}\right)^{2}2(I1​+I2​)I1​I2​​(ω1​−ω2​)2
  4. (D)(ω1−ω2)22(I1+I2)\frac{(\omega_{1}-\omega_{2})^{2}}{2(I_{1}+I_{2})}2(I1​+I2​)(ω1​−ω2​)2​

Correct answer: (C)

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2021
Moment of inertia of a square plate of side lll about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :
  1. (A)Ml26\frac{Ml^2}{6}6Ml2​
  2. (B)Ml2Ml^2Ml2
  3. (C)Ml212\frac{Ml^2}{12}12Ml2​
  4. (D)23Ml2\frac{2}{3}Ml^232​Ml2

Correct answer: (D)

Step-by-step solution →
Q162·PhysicsSingle correctJEE Main 2021
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2^{2}2.
  1. (A)14.9 kg / m3^{3}3
  2. (B)7.5×1017.5 \times 10^{1}7.5×101 kg / m3^{3}3
  3. (C)7.5×1027.5 \times 10^{2}7.5×102 kg/m3^{3}3
  4. (D)1.49×1021.49 \times 10^{2}1.49×102 kg / m3^{3}3

Correct answer: (D)

Step-by-step solution →
Q163·PhysicsNumericalJEE Main 2021
Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, r2\dfrac{r}{2}2r​ distance from the end A of the handle will be ........ Mr2\mathrm{Mr^2}Mr2.

Correct answer: 52

Step-by-step solution →
Q164·PhysicsNumericalJEE Main 2021
In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias (I1I2)\left(\frac{I_1}{I_2}\right)(I2​I1​​) will be x : 1. The value of x will be __________.

Correct answer: 9

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2021
The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (which is ⊥ to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)
  1. (A)R2:r2R^2 : r^2R2:r2
  2. (B)2r4:R42r^4 : R^42r4:R4
  3. (C)2R4:r42R^4 : r^42R4:r4
  4. (D)2R2:r22R^2 : r^22R2:r2

Correct answer: (C)

Step-by-step solution →
Q166·PhysicsNumericalJEE Main 2021
A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10s is ________ π×10−1\pi \times 10^{-1}π×10−1 Nm.

Correct answer: 4

Step-by-step solution →
Q167·PhysicsNumericalJEE Main 2021
A particle of mass 'm' is moving in time 't' on a trajectory given by r⃗=10αt2i^+5β(t−5)j^\vec{r} = 10\alpha t^2\hat{i} + 5\beta\left(t - 5\right)\hat{j}r=10αt2i^+5β(t−5)j^​ Where α and β are dimensional constants. The angular momentum of the particle become the same as it was for t = 0 at time t = ________seconds.

Correct answer: 10

Step-by-step solution →
Q168·PhysicsSingle correctJEE Main 2021
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Moment of inertia of a circular disc 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be IxI_xIx​, IyI_yIy​ & IzI_zIz​ respectively. The respective radii of gyration about all the three axes will be the same. Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)A is correct but R is not correct .
  2. (B)Both A and B are correct and R is the correct explanation of A .
  3. (C)Both A and R are correct but R is NOT the correct explanation of A.
  4. (D)A is not correct but R is correct.

Correct answer: (D)

Step-by-step solution →
Q169·PhysicsSingle correctJEE Main 2021
Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is :
  1. (A)The ring has the greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.
  2. (B)The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
  3. (C)The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.
  4. (D)All of them will have same velocity.

Correct answer: (B)

Step-by-step solution →
Q170·PhysicsNumericalJEE Main 2021
The centre of a wheel rolling on a plane surface moves with a speed υ0\upsilon_0υ0​. A particle on the rim of the wheel at the same level as the centre will be moving at a speed x υ0\sqrt{x}\,\upsilon_0x​υ0​. Then the value of x is ________.

Correct answer: 2

Step-by-step solution →
Q171·PhysicsNumericalJEE Main 2021
The position of the centre of mass of a uniform semi − circular wire of radius 'R' placed in x − y plane with its centre at the origin and the line joining its ends as x-axis is given by (0,xRπ)\left(0, \frac{xR}{\pi}\right)(0,πxR​). Then, the value of |x| is ________.

Correct answer: 2

Step-by-step solution →
Q172·PhysicsNumericalJEE Main 2021
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses (mM)\left( \frac{m}{M} \right)(Mm​) is 1x\frac{1}{x}x1​. The value of 'x' will be ________.

Correct answer: 4

Step-by-step solution →
Q173·PhysicsNumericalJEE Main 2021
Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is x2\frac{\sqrt{x}}{2}2x​​. Then, the value of x is ___________.

Correct answer: 3

Step-by-step solution →
Q174·PhysicsNumericalJEE Main 2021
A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 sec. The number of rotations made in the process is ____________.

Correct answer: 200

Step-by-step solution →
Q175·PhysicsSingle correctJEE Main 2021
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :
  1. (A)Hollow cylinder
  2. (B)Ring
  3. (C)Solid sphere
  4. (D)Solid cylinder

Correct answer: (D)

Step-by-step solution →
Q176·PhysicsSingle correctJEE Main 2021
Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :
  1. (A)14ML2π2\frac{1}{4}\frac{ML^{2}}{\pi^{2}}41​π2ML2​
  2. (B)25ML2π2\frac{2}{5}\frac{ML^{2}}{\pi^{2}}52​π2ML2​
  3. (C)ML2π2\frac{ML^{2}}{\pi^{2}}π2ML2​
  4. (D)12ML2π2\frac{1}{2}\frac{ML^{2}}{\pi^{2}}21​π2ML2​

Correct answer: (C)

Step-by-step solution →
Q177·PhysicsSingle correctJEE Main 2021
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
  1. (A)ωMM+m\omega\dfrac{M}{M+m}ωM+mM​
  2. (B)ωM+2mM\omega\dfrac{M+2m}{M}ωMM+2m​
  3. (C)ωMM+2m\omega\dfrac{M}{M+2m}ωM+2mM​
  4. (D)ωM−2mM+2m\omega\dfrac{M-2m}{M+2m}ωM+2mM−2m​

Correct answer: (C)

Step-by-step solution →
Q178·PhysicsNumericalJEE Main 2021
The following bodies, (1) a ring (2) a disc (3) a solid cylinder (4) a solid sphere, of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ______. [Mark the body as per their respective numbering given in the question]

Correct answer: 4

Step-by-step solution →
Q179·PhysicsSingle correctJEE Main 2021
A sphere of mass 2kg and radius 0.5 m is rolling with an initial speed of 1 ms−1^{-1}−1 goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ?
  1. (A)0.60 s
  2. (B)0.52 s
  3. (C)0.57 s
  4. (D)0.80 s

Correct answer: (C)

Step-by-step solution →
Q180·PhysicsSingle correctJEE Main 2021
A mass M hangs on a massless rod of length lll which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity ω\omegaω. The angular momentum of M about point A is LAL_ALA​ which lies in the positive z direction and the angular momentum of M about B is LBL_BLB​. The correct statement for this system is :
  1. (A)LAL_ALA​ and LBL_BLB​ are both constant in magnitude and direction
  2. (B)LBL_BLB​ is constant in direction with varying magnitude
  3. (C)LBL_BLB​ is constant, both in magnitude and direction
  4. (D)LAL_ALA​ is constant, both in magnitude and direction

Correct answer: (D)

Step-by-step solution →
Q181·PhysicsNumericalJEE Main 2021
The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position x3aπ\frac{x}{3}\frac{a}{\pi}3x​πa​ , x3aπ\frac{x}{3}\frac{a}{\pi}3x​πa​ where x is ________. (Round off to the Nearest Integer) [a is an area as shown in the figure]

Correct answer: 4

Step-by-step solution →
Q182·PhysicsNumericalJEE Main 2021
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _______. (Assuming the acceleration to be uniform).

Correct answer: 728

Step-by-step solution →
Q183·PhysicsSingle correctJEE Main 2021
A triangular plate is shown. A force F⃗=4i^−3j^\vec{F} = 4\hat{i} - 3\hat{j}F=4i^−3j^​ is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :
  1. (A)−15−203-15 - 20\sqrt{3}−15−203​ , 15−20315 - 20\sqrt{3}15−203​
  2. (B)15+20315 + 20\sqrt{3}15+203​ , 15−20315 - 20\sqrt{3}15−203​
  3. (C)15−20315 - 20\sqrt{3}15−203​ , 15+20315 + 20\sqrt{3}15+203​
  4. (D)−15+203-15 + 20\sqrt{3}−15+203​ , 15+20315 + 20\sqrt{3}15+203​

Correct answer: (A)

Step-by-step solution →
Q184·PhysicsNumericalJEE Main 2021
A force F⃗=4i^+3j^+4k^\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}F=4i^+3j^​+4k^ is applied on an intersection point of x = 2 plane and x-axis. The magnitude of torque of this force about a point (2, 3, 4) is ________. (Round off to the Nearest Integer)

Correct answer: 20

Step-by-step solution →
Q185·PhysicsNumericalJEE Main 2021
Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s−1^{-1}−1. The value of n, to the nearest integer, is _______. [Given : In one complete revolution, the disk rotates by 6.28 rad]

Correct answer: 20

Step-by-step solution →
Q186·PhysicsNumericalJEE Main 2021
A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle θ with the horizontal. The acceleration of the disc will be 2bgsin⁡θ\frac{2}{b}g\sin\thetab2​gsinθ where b is ______. (Round off to the Nearest Integer) (g = acceleration due to gravity) (θ = angle as shown in figure)

Correct answer: 3

Step-by-step solution →
Q187·PhysicsSingle correctJEE Main 2021
Four equal masses, m each are placed at the corners of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :
  1. (A)ml2ml^{2}ml2
  2. (B)2 ml22\,ml^{2}2ml2
  3. (C)3 ml23\,ml^{2}3ml2
  4. (D)3 ml2\sqrt{3}\,ml^{2}3​ml2

Correct answer: (C)

Step-by-step solution →
Q188·PhysicsSingle correctJEE Main 2021
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ‘h’, the square of angular velocity of wheel will be:
  1. (A)2ghI+mr2\frac{2gh}{I+mr^{2}}I+mr22gh​
  2. (B)2gh
  3. (C)2mghI+2mr2\frac{2mgh}{I+2mr^{2}}I+2mr22mgh​
  4. (D)2mghI+mr2\frac{2mgh}{I+mr^{2}}I+mr22mgh​

Correct answer: (D)

Step-by-step solution →
Q189·PhysicsSingle correctJEE Main 2021
Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is :
  1. (A)45\frac{4}{5}54​ma2^{2}2
  2. (B)85\frac{8}{5}58​ma2^{2}2 + mb2^{2}2
  3. (C)45\frac{4}{5}54​ma2^{2}2 + 2mb2^{2}2
  4. (D)85\frac{8}{5}58​ma2^{2}2 + 2mb2^{2}2

Correct answer: (D)

Step-by-step solution →
Q190·PhysicsSingle correctJEE Main 2021
A circular hole of radius (a2)\left(\frac{a}{2}\right)(2a​) is cut out of a circular disc of radius 'a' shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :
  1. (A)1011a\frac{10}{11}a1110​a
  2. (B)23a\frac{2}{3}a32​a
  3. (C)16a\frac{1}{6}a61​a
  4. (D)56a\frac{5}{6}a65​a

Correct answer: (D)

Step-by-step solution →
Q191·PhysicsSingle correctJEE Main 2021
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as – I1I_1I1​ = M.I. of thin circular ring about its diameter, I2I_2I2​ = M.I. of circular disc about an axis perpendicular to disc and going through the centre, I3I_3I3​ = M.I. of solid cylinder about its axis and I4I_4I4​ = M.I. of solid sphere about its diameter. Then :-
  1. (A)I1=I2=I3<I4I_1 = I_2 = I_3 < I_4I1​=I2​=I3​<I4​
  2. (B)I1+I2=I3+52I4I_1 + I_2 = I_3 + \frac{5}{2}I_4I1​+I2​=I3​+25​I4​
  3. (C)I1+I3<I2+I4I_1 + I_3 < I_2 + I_4I1​+I3​<I2​+I4​
  4. (D)I1=I2=I3>I4I_1 = I_2 = I_3 > I_4I1​=I2​=I3​>I4​

Correct answer: (D)

Step-by-step solution →
Q192·PhysicsNumericalJEE Main 2021
A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is _______×10−1\times 10^{-1}×10−1 kg m2^{2}2.

Correct answer: 8

Step-by-step solution →
Q193·PhysicsSingle correctJEE Advanced 2020
A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale)-
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q194·PhysicsSingle correctJEE Advanced 2020
A football of radius R is kept on a hole of radius r (r<R)(r < R)(r<R) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle θ\thetaθ from the horizontal as shown in the figure below. The maximum value of θ\thetaθ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) -
  1. (A)sin⁡θ=rR\sin\theta = \frac{r}{R}sinθ=Rr​
  2. (B)tan⁡θ=rR\tan\theta = \frac{r}{R}tanθ=Rr​
  3. (C)sin⁡θ=r2R\sin\theta = \frac{r}{2R}sinθ=2Rr​
  4. (D)cos⁡θ=r2R\cos\theta = \frac{r}{2R}cosθ=2Rr​

Correct answer: (A)

Step-by-step solution →
Q195·PhysicsMultiple correctJEE Advanced 2020
A rod of mass mmm and length LLL, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed vvv strikes the rod horizontally at a distance xxx from its pivoted end and gets embedded in it. The combined system now rotates with angular speed ω\omegaω about the pivot. The maximum angular speed ωM\omega_MωM​ is achieved for x=xMx = x_Mx=xM​. Then
  1. (A)ω=3vxL2+3x2\omega = \frac{3vx}{L^2 + 3x^2}ω=L2+3x23vx​
  2. (B)ω=12vxL2+12x2\omega = \frac{12vx}{L^2 + 12x^2}ω=L2+12x212vx​
  3. (C)xM=L3x_M = \frac{L}{\sqrt{3}}xM​=3​L​
  4. (D)ωM=v2L3\omega_M = \frac{v}{2L}\sqrt{3}ωM​=2Lv​3​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q196·PhysicsSingle correctJEE Main 2020
Four point masses, each of mass m, are fixed at the corners of a square of side l. The square is rotating with angular frequency ω\omegaω, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is:
  1. (A)ml2ωml^{2}\omegaml2ω
  2. (B)5ml2ω5ml^{2}\omega5ml2ω
  3. (C)3ml2ω3ml^{2}\omega3ml2ω
  4. (D)2ml2ω2ml^{2}\omega2ml2ω

Correct answer: (C)

Step-by-step solution →
Q197·PhysicsSingle correctJEE Main 2020
Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is:
  1. (A)MR22\dfrac{MR^{2}}{2}2MR2​
  2. (B)M(R2+H2)4\dfrac{M\left(R^{2}+H^{2}\right)}{4}4M(R2+H2)​
  3. (C)MH23\dfrac{MH^{2}}{3}3MH2​
  4. (D)MR23\dfrac{MR^{2}}{3}3MR2​

Correct answer: (A)

Step-by-step solution →
Q198·PhysicsSingle correctJEE Main 2020
The liner mass density of a thin rod AB of length L varies from A to B as λ(x)=λ0(1+xL)\lambda(x) = \lambda_{0}\left(1+\frac{x}{L}\right)λ(x)=λ0​(1+Lx​), where x is the distance from A. If M is the mass of the rod hen its moment of inertia about an axis passing through A and perpendicular to the rod is:
  1. (A)512ML2\frac{5}{12}ML^{2}125​ML2
  2. (B)718ML2\frac{7}{18}ML^{2}187​ML2
  3. (C)25ML2\frac{2}{5}ML^{2}52​ML2
  4. (D)37ML2\frac{3}{7}ML^{2}73​ML2

Correct answer: (B)

Step-by-step solution →
Q199·PhysicsNumericalJEE Main 2020
The centre of mass of a solid hemisphere of radius 8 cm is x cm from the centre of the flat surface. Then value of x is ______.

Correct answer: 3.00

Step-by-step solution →
Q200·PhysicsNumericalJEE Main 2020
A force F⃗=(i^+2j^+3k^)\vec{F} = (\hat{i} + 2\hat{j} + 3\hat{k})F=(i^+2j^​+3k^) N acts at a point (4i^+3j^−k^)(4\hat{i} + 3\hat{j} - \hat{k})(4i^+3j^​−k^) m. Then the magnitude of torque about the point (i^+2j^+k^)(\hat{i} + 2\hat{j} + \hat{k})(i^+2j^​+k^) m will be x\sqrt{x}x​ N-m. The value of x is ________.

Correct answer: 195.00

Step-by-step solution →
Q201·PhysicsSingle correctJEE Main 2020
A wheel is rotating freely with an angular speed ω\omegaω on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:
  1. (A)34\frac{3}{4}43​
  2. (B)56\frac{5}{6}65​
  3. (C)14\frac{1}{4}41​
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q202·PhysicsNumericalJEE Main 2020
A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be __________.

Correct answer: 20.00

Step-by-step solution →
Q203·PhysicsSingle correctJEE Main 2020
Consider two uniform discs of the same thickness and different radii R1=RR_1 = RR1​=R and R2=αRR_2 = \alpha RR2​=αR made of the same material. If the ratio of their moments of inertia I1I_1I1​ and I2I_2I2​, respectively, about their axes is I1:I2=1:16I_1 : I_2 = 1 : 16I1​:I2​=1:16 then the value of α is:
  1. (A)4
  2. (B)2\sqrt{2}2​
  3. (C)2
  4. (D)222\sqrt{2}22​

Correct answer: (C)

Step-by-step solution →
Q204·PhysicsSingle correctJEE Main 2020
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O′ (corner point) is:
  1. (A)1/2
  2. (B)2/3
  3. (C)1/4
  4. (D)1/8

Correct answer: (C)

Step-by-step solution →
Q205·PhysicsNumericalJEE Main 2020
ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I0I_0I0​. If part ADE is removed, the moment of inertia of the remaining part about the same axis is NI016\frac{NI_0}{16}16NI0​​ where N is an integer. Value of N is _________.

Correct answer: 11

Step-by-step solution →
Q206·PhysicsNumericalJEE Main 2020
A circular disc of mass M and radius R is rotating about its axis with angular speed ω1\omega_1ω1​. If another stationary disc having radius R2\frac{R}{2}2R​ and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed ω2\omega_2ω2​. The energy lost in the process is p% of the initial energy. Value of p is ________.

Correct answer: 20

Step-by-step solution →
Q207·PhysicsSingle correctJEE Main 2020
A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to its as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length ℓ\ellℓ = 1 m, the value of θ is approximately: (take g = 10 m/s2^{2}2)
  1. (A)49°
  2. (B)55°
  3. (C)69°
  4. (D)63°

Correct answer: (D)

Step-by-step solution →
Q208·PhysicsSingle correctJEE Main 2020
A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) ml212ω2sin⁡θcos⁡θ\dfrac{ml^{2}}{12}\omega^{2}\sin\theta\cos\theta12ml2​ω2sinθcosθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHF_HFH​ and FVF_VFV​ about the CM. The value of θ is then such that:
  1. (A)cos⁡θ=glω2\cos\theta = \dfrac{g}{l\omega^{2}}cosθ=lω2g​
  2. (B)cos⁡θ=2g3lω2\cos\theta = \dfrac{2g}{3l\omega^{2}}cosθ=3lω22g​
  3. (C)cos⁡θ=3g2lω2\cos\theta = \dfrac{3g}{2l\omega^{2}}cosθ=2lω23g​
  4. (D)cos⁡θ=g2lω2\cos\theta = \dfrac{g}{2l\omega^{2}}cosθ=2lω2g​

Correct answer: (C)

Step-by-step solution →
Q209·PhysicsNumericalJEE Main 2020
A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre __________.

Correct answer: 9.00

Step-by-step solution →
Q210·PhysicsNumericalJEE Main 2020
A massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG is N20ma2\frac{N}{20}ma^{2}20N​ma2 where N is an integer. The value of N is __________.

Correct answer: 25.00

Step-by-step solution →
Q211·PhysicsSingle correctJEE Main 2020
Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is I=M(R24+L212)I = M\left(\dfrac{R^{2}}{4} + \dfrac{L^{2}}{12}\right)I=M(4R2​+12L2​). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is
  1. (A)23\dfrac{2}{3}32​
  2. (B)32\sqrt{\dfrac{3}{2}}23​​
  3. (C)32\dfrac{3}{2}23​
  4. (D)23\sqrt{\dfrac{2}{3}}32​​

Correct answer: (B)

Step-by-step solution →
Q212·PhysicsSingle correctJEE Main 2020
Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I0/IAI_{0}/I_{A}I0​/IA​ of moment of inertia I0I_{0}I0​ of the system about an axis passing the centroid and about center of any of the spheres IAI_{A}IA​ and perpendicular to the plane of the triangle is
  1. (A)1513\dfrac{15}{13}1315​
  2. (B)1315\dfrac{13}{15}1513​
  3. (C)2313\dfrac{23}{13}1323​
  4. (D)1323\dfrac{13}{23}2313​

Correct answer: (D)

Step-by-step solution →
Q213·PhysicsNumericalJEE Main 2020
One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30∘30^{\circ}30∘ from the horizontal (see figure). Its angular speed when it hits the table is given as n\sqrt{n}n​ s−1^{-1}−1, where n is an integer. The value of n is ____.

Correct answer: 15

Step-by-step solution →
Q214·PhysicsSingle correctJEE Main 2020
A rod of length L has non-uniform linear mass density given by ρ(x)=a+b(xL)2\rho(x) = a + b\left(\frac{x}{L}\right)^{2}ρ(x)=a+b(Lx​)2, where a and b are constants and 0≤x≤L0 \le x \le L0≤x≤L. The value of x for the centre of mass of the rod is at:
  1. (A)32(a+b2a+b)L\frac{3}{2}\left(\frac{a + b}{2a + b}\right)L23​(2a+ba+b​)L
  2. (B)43(a+b2a+3b)L\frac{4}{3}\left(\frac{a + b}{2a + 3b}\right)L34​(2a+3ba+b​)L
  3. (C)34(2a+b3a+b)L\frac{3}{4}\left(\frac{2a + b}{3a + b}\right)L43​(3a+b2a+b​)L
  4. (D)32(2a+b3a+b)L\frac{3}{2}\left(\frac{2a + b}{3a + b}\right)L23​(3a+b2a+b​)L

Correct answer: (C)

Step-by-step solution →
Q215·PhysicsSingle correctJEE Main 2020
Consider a uniform rod of mass M = 4 m and length ℓ\ellℓ pivoted about its centre. A mass m moving with velocity v making angle θ=π4\theta = \dfrac{\pi}{4}θ=4π​ to the rod's long axis collides with one end of the rod and sticks to it. The angular sped of the rod-mass system just after the collision is
  1. (A)37vℓ\dfrac{3}{7}\dfrac{v}{\ell}73​ℓv​
  2. (B)327vℓ\dfrac{3\sqrt{2}}{7}\dfrac{v}{\ell}732​​ℓv​
  3. (C)47vℓ\dfrac{4}{7}\dfrac{v}{\ell}74​ℓv​
  4. (D)372vℓ\dfrac{3}{7\sqrt{2}}\dfrac{v}{\ell}72​3​ℓv​

Correct answer: (B)

Step-by-step solution →
Q216·PhysicsSingle correctJEE Main 2020
As shown in figure when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e. on the surface of the cavity. R can be determined by the equation:
  1. (A)(R2^{2}2 + R + 1) (2 −-− R) = 1
  2. (B)(R2^{2}2 + R −-− 1) (2 −-− R) = 1
  3. (C)(R2^{2}2 −-− R + 1) (2 −-− R) = 1
  4. (D)(R2^{2}2 −-− R −-− 1) (2 −-− R) = 1

Correct answer: (A)

Step-by-step solution →
Q217·PhysicsSingle correctJEE Main 2020
The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4 kg. (the coordinates of the same are shown in figure) are:
  1. (A)(0.75 m, 0.75 m)
  2. (B)(0.75 m, 1.75 m)
  3. (C)(1.25 m, 1.50 m)
  4. (D)(1 m, 1.75 m)

Correct answer: (B)

Step-by-step solution →
Q218·PhysicsSingle correctJEE Main 2020
A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:
  1. (A)8.75×10−48.75\times 10^{-4}8.75×10−4 J
  2. (B)8.75×10−38.75\times 10^{-3}8.75×10−3 J
  3. (C)6.25×10−46.25\times 10^{-4}6.25×10−4 J
  4. (D)1.13×10−31.13\times 10^{-3}1.13×10−3 J

Correct answer: (A)

Step-by-step solution →
Q219·PhysicsSingle correctJEE Main 2020
As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be
  1. (A)r34ghr\sqrt{\dfrac{3}{4gh}}r4gh3​​
  2. (B)1r2gh3\dfrac{1}{r}\sqrt{\dfrac{2gh}{3}}r1​32gh​​
  3. (C)r32ghr\sqrt{\dfrac{3}{2gh}}r2gh3​​
  4. (D)1r4gh3\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}r1​34gh​​

Correct answer: (D)

Step-by-step solution →
Q220·PhysicsSingle correctJEE Main 2020
Three point particles of masses 10 kg, 1.5 kg and 2.5 kg are placed at there corners of a right angle triangle of sides 4.0 cm, 3.0 ;cm and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:
  1. (A)0.6 cm right and 2.0 cm above 1 kg mass.
  2. (B)2.0 cm right and 0.9 cm above 1 kg mass.
  3. (C)1.5 cm right and 1.2 cm above 1 kg mass.
  4. (D)0.9 cm right and 2.0 cm above 1 kg mass.

Correct answer: (D)

Step-by-step solution →
Q221·PhysicsNumericalJEE Main 2020
Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is μ\muμ = 0.4, the maximum possible value of 100×ba100 \times \dfrac{b}{a}100×ab​ for box not to topple before moving is __________.

Correct answer: 50.00

Step-by-step solution →
Q222·PhysicsSingle correctJEE Main 2020
Mass per unit are of a circular disc of radius a depends on the distance r from its centre as σ(r)=A+Br\sigma(r) = A + Brσ(r)=A+Br. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:
  1. (A)2πa4(A4+B5)2\pi a^{4}\left(\dfrac{A}{4}+\dfrac{B}{5}\right)2πa4(4A​+5B​)
  2. (B)2πa4(A4+aB5)2\pi a^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)2πa4(4A​+5aB​)
  3. (C)2πa4(aA4+B5)2\pi a^{4}\left(\dfrac{aA}{4}+\dfrac{B}{5}\right)2πa4(4aA​+5B​)
  4. (D)πa4(A4+aB5)\pi a^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)πa4(4A​+5aB​)

Correct answer: (B)

Step-by-step solution →
Q223·PhysicsSingle correctJEE Main 2020
The radius of gyration of a uniform rod of length ℓ\ellℓ, about an axis passing through a point ℓ4\dfrac{\ell}{4}4ℓ​ away from the centre of the rod, an perpendicular to it is:
  1. (A)18ℓ\dfrac{1}{8}\ell81​ℓ
  2. (B)748ℓ\sqrt{\dfrac{7}{48}}\ell487​​ℓ
  3. (C)14ℓ\dfrac{1}{4}\ell41​ℓ
  4. (D)38ℓ\sqrt{\dfrac{3}{8}}\ell83​​ℓ

Correct answer: (B)

Step-by-step solution →
Q224·PhysicsMultiple correctJEE Advanced 2019
A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60∘60^\circ60∘ with vertical ? [g is the acceleration due to gravity]
  1. (A)The radial acceleration of the rod's center of mass will be 3g4\frac{3g}{4}43g​
  2. (B)The angular speed of the rod will be 3g2L\sqrt{\frac{3g}{2L}}2L3g​​
  3. (C)The angular acceleration of the rod will be 2gL\frac{2g}{L}L2g​
  4. (D)The normal reaction force from the floor on the rod will be Mg16\frac{Mg}{16}16Mg​

Correct answer: (A), (B), (D)

Step-by-step solution →
Q225·PhysicsSingle correctJEE Main 2019
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :
  1. (A)(37m,712m)\left(\dfrac{\sqrt{3}}{7} m, \dfrac{7}{12} m\right)(73​​m,127​m)
  2. (B)(712m,38m)\left(\dfrac{7}{12} m, \dfrac{\sqrt{3}}{8} m\right)(127​m,83​​m)
  3. (C)(34m,512m)\left(\dfrac{\sqrt{3}}{4} m, \dfrac{5}{12} m\right)(43​​m,125​m)
  4. (D)(712m,34m)\left(\dfrac{7}{12} m, \dfrac{\sqrt{3}}{4} m\right)(127​m,43​​m)

Correct answer: (D)

Step-by-step solution →
Q226·PhysicsSingle correctJEE Main 2019
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as (σ0r)\left(\dfrac{\sigma_0}{r}\right)(rσ0​​), then the radius of gyration of the disc about its axis passing through the centre is :
  1. (A)a+b3\dfrac{a+b}{3}3a+b​
  2. (B)a2+b2+ab3\sqrt{\dfrac{a^{2}+b^{2}+ab}{3}}3a2+b2+ab​​
  3. (C)a+b2\dfrac{a+b}{2}2a+b​
  4. (D)a2+b2+ab2\sqrt{\dfrac{a^{2}+b^{2}+ab}{2}}2a2+b2+ab​​

Correct answer: (B)

Step-by-step solution →
Q227·PhysicsSingle correctJEE Main 2019
A uniform rod of length ℓ\ellℓ is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q228·PhysicsSingle correctJEE Main 2019
A particle of mass m is moving along a trajectory given by x=x0+acos⁡ω1tx = x_{0} + a\cos\omega_{1}tx=x0​+acosω1​t y=y0+bsin⁡ω2ty = y_{0} + b\sin\omega_{2}ty=y0​+bsinω2​t The torque, acing on the particle about the origin, at t = 0 is:
  1. (A)my0aω12k^my_{0}a\omega_{1}^{2}\hat{k}my0​aω12​k^
  2. (B)m(−x0b+y0a)ω12k^m(-x_{0}b + y_{0}a)\omega_{1}^{2}\hat{k}m(−x0​b+y0​a)ω12​k^
  3. (C)−m(−x0bω22−y0aω12)k^-m(-x_{0}b\omega_{2}^{2} - y_{0}a\omega_{1}^{2})\hat{k}−m(−x0​bω22​−y0​aω12​)k^
  4. (D)Zero

Correct answer: (A)

Step-by-step solution →
Q229·PhysicsSingle correctJEE Main 2019
The time dependence of the position of a particle of mass m = 2 is given by r⃗(t)=2ti^−3t2j^\vec{r}(t) = 2t\hat{i} - 3t^{2}\hat{j}r(t)=2ti^−3t2j^​ Its angular momentum with respect to the origin at time t = 2 is .
  1. (A)−48k^-48\hat{k}−48k^
  2. (B)48(i^+j^)48(\hat{i} + \hat{j})48(i^+j^​)
  3. (C)36k^36\hat{k}36k^
  4. (D)−34(k^−i^)-34(\hat{k} - \hat{i})−34(k^−i^)

Correct answer: (A)

Step-by-step solution →
Q230·PhysicsSingle correctJEE Main 2019
Two coaxial discs, having moments of inertia I1_{1}1​ and I12\dfrac{I_{1}}{2}2I1​​, area rotating with respectively angular velocities ω1\omega_{1}ω1​ and ω12\dfrac{\omega_{1}}{2}2ω1​​, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef_{f}f​ and Ei_{i}i​ are the final and initial total energies, then (Ef_{f}f​ - Ei_{i}i​) is:
  1. (A)I1ω126\dfrac{I_{1}\omega_{1}^{2}}{6}6I1​ω12​​
  2. (B)38I1ω12\dfrac{3}{8}I_{1}\omega_{1}^{2}83​I1​ω12​
  3. (C)−I1ω1212-\dfrac{I_{1}\omega_{1}^{2}}{12}−12I1​ω12​​
  4. (D)−I1ω1224-\dfrac{I_{1}\omega_{1}^{2}}{24}−24I1​ω12​​

Correct answer: (D)

Step-by-step solution →
Q231·PhysicsSingle correctJEE Main 2019
A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second is 5 s is close to
  1. (A)2.0×10−52.0 \times 10^{-5}2.0×10−5 Nm
  2. (B)4.0×10−64.0 \times 10^{-6}4.0×10−6 Nm
  3. (C)1.6×10−51.6 \times 10^{-5}1.6×10−5 Nm
  4. (D)7.9×10−67.9 \times 10^{-6}7.9×10−6 Nm

Correct answer: (A)

Step-by-step solution →
Q232·PhysicsSingle correctJEE Main 2019
A thin disc of mass M and radius R has mass per unit area σ(r)=kr2\sigma(r) = kr^{2}σ(r)=kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:
  1. (A)MR22\dfrac{MR^{2}}{2}2MR2​
  2. (B)MR23\dfrac{MR^{2}}{3}3MR2​
  3. (C)MR26\dfrac{MR^{2}}{6}6MR2​
  4. (D)2MR23\dfrac{2MR^{2}}{3}32MR2​

Correct answer: (D)

Step-by-step solution →
Q233·PhysicsSingle correctJEE Main 2019
A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M8\dfrac{7M}{8}87M​ and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1I_{1}I1​ be the moment of inertia of the disc about its axis and I2I_{2}I2​ be the moment of inertia of the new sphere about its axis. The ratio of I1/I2I_{1}/I_{2}I1​/I2​ is given by:
  1. (A)285
  2. (B)185
  3. (C)65
  4. (D)140

Correct answer: (D)

Step-by-step solution →
Q234·PhysicsSingle correctJEE Main 2019
Moment of inertia of a body about a given axis is 1.5 kg m2^{2}2 Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2^{2}2 must be applied about the axis of rotation for a duration of:
  1. (A)2 s
  2. (B)5 s
  3. (C)2.5 s
  4. (D)3 s

Correct answer: (A)

Step-by-step solution →
Q235·PhysicsSingle correctJEE Main 2019
A thin smooth rod of length L and mass M is rotating freely with angular speed ω0\omega_{0}ω0​ about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod will be:-
  1. (A)Mω0M+3m\dfrac{M\omega_{0}}{M + 3m}M+3mMω0​​
  2. (B)Mω0M+m\dfrac{M\omega_{0}}{M + m}M+mMω0​​
  3. (C)Mω0M+2m\dfrac{M\omega_{0}}{M + 2m}M+2mMω0​​
  4. (D)Mω0M+6m\dfrac{M\omega_{0}}{M + 6m}M+6mMω0​​

Correct answer: (D)

Step-by-step solution →
Q236·PhysicsSingle correctJEE Main 2019
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ\thetaθ, where θ\thetaθ is the angle by which it has rotated, is given as kθ2k\theta^2kθ2. If its moment of inertia is I then the angular acceleration of the disc is:
  1. (A)kIθ\dfrac{k}{I}\thetaIk​θ
  2. (B)k2Iθ\dfrac{k}{2I}\theta2Ik​θ
  3. (C)k4Iθ\dfrac{k}{4I}\theta4Ik​θ
  4. (D)2kIθ\dfrac{2k}{I}\thetaI2k​θ

Correct answer: (D)

Step-by-step solution →
Q237·PhysicsSingle correctJEE Main 2019
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsphh_{sph}hsph​ and hcylh_{cyl}hcyl​ on the incline. The radio hsphhcyl\frac{h_{sph}}{h_{cyl}}hcyl​hsph​​ is given by:
  1. (A)1
  2. (B)45\frac{4}{5}54​
  3. (C)25\frac{2}{\sqrt{5}}5​2​
  4. (D)1415\frac{14}{15}1514​

Correct answer: (D)

Step-by-step solution →
Q238·PhysicsSingle correctJEE Main 2019
A thin circular plate of mass M and radius R has its density varying as p(r)=p0rp(r)=p_0 rp(r)=p0​r with P0P_0P0​ as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I=aMR2I=aMR^{2}I=aMR2. The value of the coefficient a is:
  1. (A)85\dfrac{8}{5}58​
  2. (B)12\dfrac{1}{2}21​
  3. (C)35\dfrac{3}{5}53​
  4. (D)32\dfrac{3}{2}23​

Correct answer: (A)

Step-by-step solution →
Q239·PhysicsSingle correctJEE Main 2019
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut off, the coordinates of the centre of mass of the remaining portion will be:
  1. (A)(5a3,5b3)\left(\dfrac{5a}{3}, \dfrac{5b}{3}\right)(35a​,35b​)
  2. (B)(2a3,2b3)\left(\dfrac{2a}{3}, \dfrac{2b}{3}\right)(32a​,32b​)
  3. (C)(3a4,3b4)\left(\dfrac{3a}{4}, \dfrac{3b}{4}\right)(43a​,43b​)
  4. (D)(5a12,5b12)\left(\dfrac{5a}{12}, \dfrac{5b}{12}\right)(125a​,125b​)

Correct answer: (D)

Step-by-step solution →
Q240·PhysicsSingle correctJEE Main 2019
Four particles A, B, C and D with masses mA=mm_A = mmA​=m, mB=2mm_B = 2mmB​=2m, mc=3mm_c = 3mmc​=3m and mD=4mm_D = 4mmD​=4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is:
  1. (A)a5(i^−j^)\dfrac{a}{5}(\hat{i} - \hat{j})5a​(i^−j^​)
  2. (B)Zero
  3. (C)a5(i^+j^)\dfrac{a}{5}(\hat{i} + \hat{j})5a​(i^+j^​)
  4. (D)a(i^+j^)a(\hat{i} + \hat{j})a(i^+j^​)

Correct answer: (A)

Step-by-step solution →
Q241·PhysicsSingle correctJEE Main 2019
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : [Take g = 10 m/s2^{2}2]
  1. (A)2 kg-m2^{2}2/s
  2. (B)8 kg-m2^{2}2/s
  3. (C)6 kg-m2^{2}2/s
  4. (D)3 kg-m2^{2}2/s

Correct answer: (C)

Step-by-step solution →
Q242·PhysicsSingle correctJEE Main 2019
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is:
  1. (A)12 cm
  2. (B)16 cm
  3. (C)14 cm
  4. (D)18 cm

Correct answer: (B)

Step-by-step solution →
Q243·PhysicsSingle correctJEE Main 2019
The position vector of the centre of mass r⃗\vec{r}r cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:
  1. (A)r⃗\vec{r}r cm =138Lx^+58Ly^= \frac{13}{8}L\hat{x} + \frac{5}{8}L\hat{y}=813​Lx^+85​Ly^​
  2. (B)r⃗\vec{r}r cm =58Lx^+138Ly^= \frac{5}{8}L\hat{x} + \frac{13}{8}L\hat{y}=85​Lx^+813​Ly^​
  3. (C)r⃗\vec{r}r cm =38Lx^+118Ly^= \frac{3}{8}L\hat{x} + \frac{11}{8}L\hat{y}=83​Lx^+811​Ly^​
  4. (D)r⃗\vec{r}r cm =118Lx^+38Ly^= \frac{11}{8}L\hat{x} + \frac{3}{8}L\hat{y}=811​Lx^+83​Ly^​

Correct answer: (A)

Step-by-step solution →
Q244·PhysicsSingle correctJEE Main 2019
The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q245·PhysicsSingle correctJEE Main 2019
A circular disc D1_11​ of mass M and radius R has two identical discs D2_22​ and D3_33​ of the same mass M and radius R attached rigidly as its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1_11​ as shown in the figure, will :
  1. (A)MR2^{2}2
  2. (B)3MR2^{2}2
  3. (C)45\frac{4}{5}54​MR2^{2}2
  4. (D)23\frac{2}{3}32​MR2^{2}2

Correct answer: (B)

Step-by-step solution →
Q246·PhysicsSingle correctJEE Main 2019
The equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E and F are the mid points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0_00​. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:
  1. (A)I = 1516\frac{15}{16}1615​I0_00​
  2. (B)I = 34\frac{3}{4}43​I0_00​
  3. (C)I = 916\frac{9}{16}169​I0_00​
  4. (D)I = I04\frac{I_0}{4}4I0​​

Correct answer: (A)

Step-by-step solution →
Q247·PhysicsSingle correctJEE Main 2019
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
  1. (A)20 rad/s220\text{ rad/s}^{2}20 rad/s2
  2. (B)16 rad/s216\text{ rad/s}^{2}16 rad/s2
  3. (C)12 rad/s212\text{ rad/s}^{2}12 rad/s2
  4. (D)10 rad/s210\text{ rad/s}^{2}10 rad/s2

Correct answer: (B)

Step-by-step solution →
Q248·PhysicsSingle correctJEE Main 2019
The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is 1N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians):
  1. (A)π6\frac{\pi}{6}6π​
  2. (B)π3\frac{\pi}{3}3π​
  3. (C)π8\frac{\pi}{8}8π​
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (A)

Step-by-step solution →
Q249·PhysicsSingle correctJEE Main 2019
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is θ\thetaθ, then:
  1. (A)tan⁡θ=123\tan \theta = \dfrac{1}{2\sqrt{3}}tanθ=23​1​
  2. (B)tan⁡θ=12\tan \theta = \dfrac{1}{2}tanθ=21​
  3. (C)tan⁡θ=23\tan \theta = \dfrac{2}{\sqrt{3}}tanθ=3​2​
  4. (D)tan⁡θ=13\tan \theta = \dfrac{1}{3}tanθ=31​

Correct answer: (D)

Step-by-step solution →
Q250·PhysicsSingle correctJEE Main 2019
A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30° fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s−1^{-1}−1) will be (g = 10 ms−2^{-2}−2).
  1. (A)302\sqrt{\dfrac{30}{2}}230​​
  2. (B)30\sqrt{30}30​
  3. (C)202\sqrt{\dfrac{20}{2}}220​​
  4. (D)302\dfrac{\sqrt{30}}{2}230​​

Correct answer: (B)

Step-by-step solution →
Q251·PhysicsMultiple correctJEE Advanced 2018
Consider a body of mass 1.0 kg1.0\ kg1.0 kg at rest at the origin at time t=0t = 0t=0. A force F⃗=(αt i^+β j^)\vec{F} = \left(\alpha t\,\hat{i} + \beta\,\hat{j}\right)F=(αti^+βj^​) is applied on the body, where α=1.0 Ns−1\alpha = 1.0\,Ns^{-1}α=1.0Ns−1 and β=1.0 N\beta = 1.0\,Nβ=1.0N. The torque acting on the body about the origin at time t=1.0 st = 1.0\ st=1.0 s is τ⃗\vec{\tau}τ. Which of the following statements is (are) true?
  1. (A)∣τ⃗∣=13 Nm\left|\vec{\tau}\right| = \frac{1}{3}\,Nm∣τ∣=31​Nm
  2. (B)The torque τ⃗\vec{\tau}τ is in the direction of the unit vector + k^+\ \hat{k}+ k^
  3. (C)The velocity of the body at t=1st = 1st=1s is v⃗=12(i^+2j^)ms−1\vec{v} = \frac{1}{2}\left(\hat{i} + 2\hat{j}\right)ms^{-1}v=21​(i^+2j^​)ms−1
  4. (D)The magnitude of displacement of the body at t=1 st = 1\ st=1 s is 16 m\frac{1}{6}\,m61​m

Correct answer: (A), (C)

Step-by-step solution →
Q252·PhysicsNumericalJEE Advanced 2018
A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60∘60^{\circ}60∘ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2−3)/10 s(2 - \sqrt{3})/\sqrt{10}\ s(2−3​)/10​ s, then the height of the top of the inclined plane, in metres, is __________. Take g=10 ms−2g = 10\ ms^{-2}g=10 ms−2.

Correct answer: 0.75

Step-by-step solution →
Q253·PhysicsSingle correctJEE Advanced 2017
PARAGRAPH 2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω0\omega_{0}ω0​. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ\muμ and the acceleration due to gravity is g. The total kinetic energy of the ring is
  1. (A)Mω02R2M\omega_{0}^{2}R^{2}Mω02​R2
  2. (B)12Mω02(R−r)2\dfrac{1}{2}M\omega_{0}^{2}(R-r)^{2}21​Mω02​(R−r)2
  3. (C)Mω02(R−r)2M\omega_{0}^{2}(R-r)^{2}Mω02​(R−r)2
  4. (D)32Mω02(R−r)2\dfrac{3}{2}M\omega_{0}^{2}(R-r)^{2}23​Mω02​(R−r)2

Correct answer: (C)

Step-by-step solution →
Q254·PhysicsSingle correctJEE Advanced 2017
Consider regular polygons with number of sides n=3,4,5,…n = 3, 4, 5, \dotsn=3,4,5,… as shown in the figure. The center of mass of all the polygons is at height hhh from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ\DeltaΔ. Then Δ\DeltaΔ depends on nnn and hhh as
  1. (A)Δ=hsin⁡2 ⁣(πn)\Delta = h\sin^{2}\!\left(\dfrac{\pi}{n}\right)Δ=hsin2(nπ​)
  2. (B)Δ=h(1cos⁡(πn)−1)\Delta = h\left(\dfrac{1}{\cos\left(\frac{\pi}{n}\right)} - 1\right)Δ=h(cos(nπ​)1​−1)
  3. (C)Δ=hsin⁡ ⁣(2πn)\Delta = h\sin\!\left(\dfrac{2\pi}{n}\right)Δ=hsin(n2π​)
  4. (D)Δ=htan⁡2 ⁣(π2n)\Delta = h\tan^{2}\!\left(\dfrac{\pi}{2n}\right)Δ=htan2(2nπ​)

Correct answer: (B)

Step-by-step solution →
Q255·PhysicsMultiple correctJEE Advanced 2017
A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?
  1. (A)The x component of displacement of the center of mass of the block M is : −mRM+m-\frac{mR}{M+m}−M+mmR​.
  2. (B)The position of the point mass is : x=−2 mRM+mx = -\sqrt{2}\,\frac{mR}{M+m}x=−2​M+mmR​.
  3. (C)The velocity of the point mass m is : v=2gR1+mMv = \sqrt{\frac{2gR}{1+\frac{m}{M}}}v=1+Mm​2gR​​.
  4. (D)The velocity of the block M is: V=−mM2gRV = -\frac{m}{M}\sqrt{2gR}V=−Mm​2gR​.

Correct answer: (A), (C)

Step-by-step solution →
Q256·PhysicsMultiple correctJEE Advanced 2017
A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vortical is θ\thetaθ. Which of the following statements about its motion is/are correct?
  1. (A)The midpoint of the bar will fall vertically downward
  2. (B)The trajectory of the point A is a parabola
  3. (C)Instantaneous torque about the point in contact with the floor is proportional to sin⁡θ\sin\thetasinθ
  4. (D)When the bar makes an angle θ\thetaθ with the vertical, the displacement of its midpoint from the initial position is proportional to (1−cos⁡θ)(1-\cos\theta)(1−cosθ)

Correct answer: (A), (C), (D)

Step-by-step solution →
Q257·PhysicsSingle correctJEE Advanced 2017
PARAGRAPH 2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω0\omega_{0}ω0​. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ\muμ and the acceleration due to gravity is g. The minimum value of ω0\omega_{0}ω0​ below which the ring will drop down is
  1. (A)gμ(R−r)\sqrt{\dfrac{g}{\mu(R-r)}}μ(R−r)g​​
  2. (B)2gμ(R−r)\sqrt{\dfrac{2g}{\mu(R-r)}}μ(R−r)2g​​
  3. (C)3g2μ(R−r)\sqrt{\dfrac{3g}{2\mu(R-r)}}2μ(R−r)3g​​
  4. (D)g2μ(R−r)\sqrt{\dfrac{g}{2\mu(R-r)}}2μ(R−r)g​​

Correct answer: (A)

Step-by-step solution →
Q258·PhysicsMultiple correctJEE Advanced 2017
A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ\tauτ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct?
  1. (A)If the force is applied at point P tangentially then τ\tauτ decreases continuously as the wheel climbs
  2. (B)If the force is applied normal to the circumference at point X then τ\tauτ is constant
  3. (C)If the force is applied normal to the circumference at point P then τ\tauτ is zero
  4. (D)If the force is applied tangentially at point S then τ≠0\tau \neq 0τ=0 but the wheel never climbs the step

Correct answer: (A)

Step-by-step solution →
Q259·PhysicsMultiple correctJEE Advanced 2016
The position vector r⃗\vec{r}r of a particle of mass m is given by the following equation r⃗(t)=αt3i^+βt2j^\vec{r}(t) = \alpha t^{3}\hat{i} + \beta t^{2}\hat{j}r(t)=αt3i^+βt2j^​, where α=10/3 m s−3\alpha = 10/3\ \text{m s}^{-3}α=10/3 m s−3, β=5 m s−2\beta = 5\ \text{m s}^{-2}β=5 m s−2 and m=0.1m = 0.1m=0.1 kg. At t=1t = 1t=1 s, which of the following statement(s) is(are) true about the particle?
  1. (A)The velocity v⃗\vec{v}v is given by v⃗=(10i^+10j^) m s−1\vec{v} = \left(10\hat{i} + 10\hat{j}\right)\ \text{m s}^{-1}v=(10i^+10j^​) m s−1
  2. (B)The angular momentum L⃗\vec{L}L with respect to the origin is given by L⃗=−(5/3)k^ N m s\vec{L} = -(5/3)\hat{k}\ \text{N m s}L=−(5/3)k^ N m s
  3. (C)The force F⃗\vec{F}F is given by F⃗=(i^+2j^) N\vec{F} = (\hat{i} + 2\hat{j})\ \text{N}F=(i^+2j^​) N
  4. (D)The torque τ⃗\vec{\tau}τ with respect to the origin is given by τ⃗=−(20/3)k^ N m\vec{\tau} = -(20/3)\hat{k}\ \text{N m}τ=−(20/3)k^ N m

Correct answer: (A), (B), (D)

Step-by-step solution →
Q260·PhysicsMultiple correctJEE Advanced 2016
Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length ℓ=24 a\ell = \sqrt{24}\,aℓ=24​a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω\omegaω. The angular momentum of the entire assembly about the point 'O' is L⃗\vec{L}L (see the figure). Which of the following statement(s) is(are) true?
  1. (A)The magnitude of angular momentum of the assembly about its center of mass is 17 ma2ω/217\,ma^{2}\omega/217ma2ω/2
  2. (B)The magnitude of the z-component of L⃗\vec{L}L is 55 ma2ω55\,ma^{2}\omega55ma2ω
  3. (C)The magnitude of angular momentum of center of mass of the assembly about the point O is 81 ma2ω81\,ma^{2}\omega81ma2ω
  4. (D)The center of mass of the assembly rotates about the z-axis with an angular speed of ω/5\omega/5ω/5

Correct answer: (D) or (A), (D)

Step-by-step solution →
Q261·PhysicsIntegerJEE Advanced 2015
Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds v1v_{1}v1​ and v2v_{2}v2​, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and v1=3v_{1} = 3v1​=3 m/s, then v2v_{2}v2​ in m/s is (g=10g = 10g=10 m/s2^{2}2)

Correct answer: 7

Step-by-step solution →
Q262·PhysicsIntegerJEE Advanced 2015
The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r)=k(rR)\rho_{A}(r) = k\left(\dfrac{r}{R}\right)ρA​(r)=k(Rr​) and ρB(r)=k(rR)5\rho_{B}(r) = k\left(\dfrac{r}{R}\right)^{5}ρB​(r)=k(Rr​)5, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are IAI_{A}IA​ and IBI_{B}IB​, respectively. If IBIA=n10\dfrac{I_{B}}{I_{A}} = \dfrac{n}{10}IA​IB​​=10n​, the value of n is

Correct answer: 6

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Q263·PhysicsMultiple correctJEE Advanced 2015
A ring of mass M and radius R is rotating with angular speed ω\omegaω about a fixed vertical axis passing through its centre O with two point masses each of mass M8\dfrac{M}{8}8M​ at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is 89ω\dfrac{8}{9}\omega98​ω and one of the masses is at a distance of 35R\dfrac{3}{5}R53​R from O. At this instant the distance of the other mass from O is
  1. (A)23R\dfrac{2}{3}R32​R
  2. (B)13R\dfrac{1}{3}R31​R
  3. (C)35R\dfrac{3}{5}R53​R
  4. (D)45R\dfrac{4}{5}R54​R

Correct answer: (D)

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Q264·PhysicsIntegerJEE Advanced 2014
A horizontal circular platform of radius 0.5 m0.5 \text{ m}0.5 m and mass 0.45 kg0.45 \text{ kg}0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg0.05 \text{ kg}0.05 kg are attached to the platform at a distance 0.25 m0.25 \text{ m}0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms−19 \text{ ms}^{-1}9 ms−1 with respect to the ground. The rotational speed of the platform in rad s−1\text{rad s}^{-1}rad s−1 after the balls leave the platform is

Correct answer: 4

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Q265·PhysicsIntegerJEE Advanced 2014
A uniform circular disc of mass 1.5 kg1.5 \text{ kg}1.5 kg and radius 0.5 m0.5 \text{ m}0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F=0.5 NF = 0.5 \text{ N}F=0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZXYZXYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s−1\text{rad s}^{-1}rad s−1 is

Correct answer: 2

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Q266·PhysicsIntegerJEE Advanced 2013
A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s−1^{-1}−1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s−1^{-1}−1) of the system is

Correct answer: 8

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Rotational Motion — frequently asked

How many questions from Rotational Motion appear in JEE?

Rotational Motion has appeared in 172 of the last 186 JEE Main and JEE Advanced papers — about 92% of them — contributing 266 questions in total across those papers.

Is Rotational Motion an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 92% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Rotational Motion questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201
  • Laws of Motion 196

All 28 Physics chapters →

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