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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig. If the extension of the spring is `x_(0)` at time `t`, then the displacement of the first block at this instant isA. `1/2((Ft^(2))/(2m)+x_(0))`B. `-1/2((Ft^(2))/(2m)+x_(0))`C. `1/2((Ft^(2))/(2m)-x_(0))`D. `(Ft^(2))/(2m)+x_(0)` |
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Answer» Correct Answer - A a. |
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| 52. |
Two blocks each of mass m, connected by an un-stretched spring are kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. (a)Find accelaration of the mass center. (b) Find the displacement of the centre of mass as function of time t. (c) If the extension of the spring is `X_(0)` at an instant t, find the displacements of the two blocks relative to the ground at this instant. |
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Answer» (a) Forces in vertical direction on the system are weights of the blocks and normal reaction from the ground. They balance themselves and have no net resultant. The only external force on the system is the apllied force F in the horizontal direction towards the right. `Sigma vec(F)_(i)= M vec(a)_(c ) rarr" "F=(m+m)a_(c )` `a_ (c ) = (F)/(2m)` towards right (b) The mass center moves with constant acceleration, therefore it displacement in time t is given by equation of constant accelaration motion. `x=ut+1/2at^2rarr`" "`x_c=(Ft^2)/(4m)` (c) Positions `x_(A)` and `x_(B)` of particles A and B forming a system and position `x_(C )`mass center are obtained by following eq. `Mvec(r )_(c ) = Sigmam_(i)vec(r )_(i)` Substituting values we obtain `2mx_(c )= mx_(A) + mx_(B)" "x_(c )= (x_(A) + x_(B))/(2)` Now using result obtained in part(b), we have `x_(A)+x_(B)=(Ft^(2))/(2m)` Extension in the spring at this instant is `x_(0)=x_(B)-x_(A)` From the above two equations, we have `x_(A) = 1/2((Ft^(2))/(2m)-x_(0))` and `x_(B) = 1/2((Ft^(2))/(2m) + x_(0))`. |
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| 53. |
A man of mass m moves with a constant speed on a plank of mass M and length l kept initially at rest on a frictionless horizontal surface, from one end to the other in time t. The speed of the plank relative to grounud while man is moving, isA. `l/t(M/m)`B. `l/t(m/(m+M))`C. `l/t(M/(M +m))`D. None of these |
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Answer» Correct Answer - B `v_(r) = l/t`, Now , `m(v_(r)-v) = Mv` (v= speed of plank) `v=(mv_(r)/(M +m)` = l/t(m/(M +m)` |
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| 54. |
Two objects are at rest on a level frictionless surface. The objects are not connected. A force `F` is applied to one of the objects, which then moves with acceleration a. Mark the correct statement(s).A. The concept of centre of mass cannot be applied be-cause the external force does not act on both the objects.B. The centre of mass moves with acceleration that could be greater than `a`.C. The centre of mass moves with acceleration that must be equal to `a`.D. The centre of mass moves with acceleration that must be less than `a` |
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Answer» Correct Answer - D Let the masses be `m_(1)` and `m_(2)`. Then `F=m_(1)aimpliesa=F/(m_(1))` `a_(CM)=(m_(1)a+m_(2)xx0)/(m_(1)+m_(2))=F/(m_(1)+m_(2))lta` |
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| 55. |
A highly elastic ball moving at a speed of `3 m//s` approaches a wall moving towards it with a speed of `3 m//s`. After the collision. the speed of the ball will be A. `3m//s`B. `6m//s`C. `9m//s`D. zero |
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Answer» Correct Answer - C For elastic collision `e=1`. let speed of the ball be `v` towards left after collision then w.r.t wall, incident velocity `=` reflected velocity. We get `3+3=v-3impliesv=9m//s` |
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| 56. |
A system consists of two particles. At `t=0`, one particle is at the origin, the other, which has a mass of `0.60kg`, is on the y-axis at `y=80m`. At `t=0`, the centre of mass of the system is on the `y-axis` at `y=24m` and has a velocity given by `(6.0m//s^3)t^2hatj`. (a) Find the total mass of the system. (b) Find the acceleration of the centre of mass at any time t. (c) Find the next external force acting on the system at `t=3.0s`. |
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Answer» Correct Answer - A::B::C (a) `y_(CM)=(m_1y_1+m_2y_2)/(m_1+m_2)` `:. 24=((0.6)(80)+(m_2)(0))/("Total mass")` `:. "Total mass" =2kg` (b) `a_(CM)=(d)/(dt)(v_(CM))=(d)/(dt)[6t^2]hatj` `=(12t)hatjm//s^2` (c) `F_(ext)=m_(CM) a_(CM)=m_(CM)(d)/(dt)(v_(CM))` `=m_(CM)(12t)hatj` At `t=3s`, `F_(ext)=(2kg)(12)(3)hatjN` `=(72hatj)N` |
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| 57. |
`x-y` is the vertical plane as shown in figure. A particle of mass `1kg` is at `(10m, 20m)` at time `t=0`. It is released from rest. Another particles of mass `2kg` is at `(20m, 40m)` at the same instant. It is projected with velocity `(10hati+10hatj)m//s`. After `1s`. Find |
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Answer» Correct Answer - A::B::C (a) `a_(COM)=(m_1a_1+m_2a_2)/(m_1+m_2)` `=((1)(-10hatj)+(2)(-10hatj))/(1+2)` `=(-10hatj)m//s^2` (b) `v_(COM)=(m_1v_1+m_2v_2)/(m_1+m_2)` `=(m_1(u_1+a_1t)+m_2(u_2+a_2t))/(m_1+m_2)` `=((1)[0+(-10hatj)(1)]+2[(10hati+10hatj)+(-10hatj)(1)])/(1+2)` `=10/3(2hati-hatj)m//s` (c) `r_(CM)=(m_1r_1+m_2r_2)/(m_1+m_2)` `=(m_1[r_1+ut+1/2at^2]_1+m_2[r_1+ut+1/2at^2]_2)/(m_1+m_2)` `=(1[(10hati+20hatj)+0+1/2(-10hatj)(1)^2]+2[(20hati+40hatj)+(10hati+10hatj)(1)+1/2(-10hatj)(1)^2])/(1+2)` `=(70/3hati+35hatj)m` |
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| 58. |
A uniform circular disc of radius a is taken. A circular portion of radius b has been removed from it as shown in the figure. If the center of hole is at a distance c from the center of the disc, the distance `x_(2)` of the center of mass of the remaining part from the initial center of mass O is given by A. `(pib^(2))/(a^(2) - c^(2))`B. `(cb^(2))/(a^(2)-b^(2))`C. `(pic^(2))/(a^(2)-b^(2))`D. `(ca^(2))/((c^(2)-b^(2)))` |
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Answer» Correct Answer - B Center of mass of whole system was at point O. Hence `x_(2)(pia^(2) - pib^(2)) = c(pib^(2)) rArr x_92) = (cb^(2))/(a^(2)-b^(2))` |
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| 59. |
A bullet is fired from the gun. The gun recoils, the kinetic energy of the recoil shall be-A. KB. more than KC. less than KD. `sqrt(K)` |
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Answer» Correct Answer - C After firing, the momentum of gun and bullet is same. Therefore by the relation, `K = p^92)/(2m)` we have, `K alpha 1/m` (as p is same) Hence, as the mass of gun is greater, its kinetic energy will be loss. |
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| 60. |
A bullet of mass a and velocity b is fired into a large block of mass c. the final velocity of the system isA. `c/(a+b) b`B. `a/(a+c) b`C. `(a+b)/c a`D. `(a+c)/a.b` |
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Answer» Correct Answer - B From momentum conservation `P_(i)=P_(f)` ab=(a+c)v `v=(ab)/(a+c)` |
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| 61. |
A thin rod of length `6 m` is lying along the x-axis with its ends at `x = 0` and `x = 6m`. It linear density (mass/length) varies with x as `kx^(4)`. Find the position of centre of mass of rod in meters.A. B. C. D. |
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Answer» Correct Answer - A `x_(cm) = (intxdm)/(intdm) = (int_(0)^(L) k ((x)/(L))^(n) xdm)/(int_(0)^(L) k ((x)/(L))^(n) dx) = ((n + 1)/(n + 2)) L` For `n = 0, x_(cm) = (L)/(2)` and for `n rarr theta, x_(cm) = L` |
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| 62. |
A block of mass `0.50kg` is moving with a speed of `2.00m//s` on a smooth surface. It strikes another mass of `1 kg` at rest and they move as a single body. The energy loss during the collision isA. `0.16 J`B. `1.00 J`C. `0.67 J`D. `0.34 J` |
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Answer» Correct Answer - C Energy loss `= (1)/(2)(m_(1) m_(2))/((m_(1) + m_(2)))(u_(1) - u_(2))^(2)` `= ((0.5)(1))/(2[0.5 + 1]) (2 - 0)^(2) = (2)/(3) J` |
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| 63. |
A block of mass `0.50kg` is moving with a speed of `2.00m//s` on a smooth surface. It strikes another mass of `1 kg` at rest and they move as a single body. The energy loss during the collision isA. 1.00 JB. 0.67 JC. 0.34 JD. 0.16 J |
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Answer» Correct Answer - B `P_(i)=p_(f)` (momentum conservation) `rArr 0.5xx2=1.5xxv` `rArr v=0.67` energy loss `=1-(1)/(3)=(2)/(3) =0.67 J` |
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| 64. |
A small ball `B` of mass `m` is suspended with light inelastic string of length `L` from a block `A` of same mass in which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle `theta` from equilibrium position and then released. Tension in string when it is vertical, isA. `mg`B. `mg(2-costheta)`C. `mg(3-2costheta)`D. none of these |
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Answer» Correct Answer - D Loss in `PE =` gain in `KE` `mg(L-Lcostheta)=1/2mv_(1)^(2)+1/2mv_(2)^(2)` Applying momentum conservation principle in horizontal direction. `P_(ix)=P_(fx)` `0=mv_(1)-mv_(2)` `:. v_(1)=v_(2)=v` `:. Mg(L-Lcostheta)=mv^(2)` `:. v=sqrt(gl(1-costheta))` `=T-mg=(m(2v^(2)))/l` ` :. T=mg+(4mgl(1-costheta))/l` `:. T=mg+4mg(1-costheta)` |
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| 65. |
For the system shown in the figure, a small block of mass `m` and smooth irregular shaped block of mass `M`, both free to move are placed on a smooth horizontal plane. Theminimum velocity `v` imparted to block so that it will overcome the highest point of `M` is A. `sqrt(2gh)`B. `sqrt(2(1+m/M)gh)`C. `sqrt((2m)/Mgh)`D. `sqrt(2(1+M/m)gh` |
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Answer» Correct Answer - B For the given situation, block should be able to reach just at highest point of `M`, and their relative velocity is to become zero. Let their common velocity of `v_(C)`. `COLMrarr (M+m)v_(C)=mvimplies v_(C)=(mv)/(M+m)` `COME: 1/2 (M+m)v_(C)^(2)+mgh=1/2mv^(2)` Solve to get `v=sqrt(2(1+m/M)gh)` |
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| 66. |
Two indentical balls A and B moving with velocities `u_(A)` and `u_(B)` I the same direction collide. Coefficient of restitution is e. (a) Deduce expression for velocities of the balls after the collision. (b) If collision is perfectly elastic, what do you observe? |
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Answer» Equation expressing momentum conservation is `v_(A) + v_(B) = u_(A) + u_(B)` …(A) Equation of coefficient of restitution is `v_(B) - v_(A) = e u_(A) - e U_(B)m` …(B) (a) From the above two equations, velocities `v_(A)` and `v_(B)` are `v_(A) = ((1-e)/2)u_(A)+((1+e)/2)u_(B)` ...(i) `v_(B) = ((1+e)/2)u_(A) + ((1-e)/2)u_(B)` ...(ii) (b) For perfectly elastic impact `e = 1`, velocities `v_(A)` and `v_(B)` are `v_(A) = u_(B)` ...(iii) `v_(B) = u_(A)` ...(iv) Identical bodies exchange their velocities after perfectly elastic impact. |
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| 67. |
If a ball strickes with a velocity `u_(1)` at the wall which itself is approaching it with a velocity `u_(2)` then find the velocity of the ball after collision with the wall. |
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Answer» As the wall is heavy so after collison it will continue to move with the same velocity `u_(2)`. Relative velocity of separation is equal to relative velocity of approach. Hence, `v_(1)-v_(2)=e(u_(1)-u_(2))` `impliesv_(1)(-u_(2))=-e[u_(1)-(-u_(2))]` `implies v_(1)=-e(u_(1)+u_(2))-u_(2)` In case of perfectly elastic collision `e=1` `v_(1)=-u_(1)-u_(2)-u_(3)=-(u_(1)+2u_(2))` Negative sign indicates backward direction. |
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| 68. |
A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth of its initial velocity. The mass of the other bodyis-A. `0.6 kg`B. `2.4 kg`C. `3 kg`D. `4 kg` |
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Answer» Correct Answer - A COLM : `m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) + m_(2)v_(2)` `1 xx u + 0 = 1 xx (u)/(4) + mv_(2) rArr (3)/(4)u = mv_(a)` ….(i) `e = -((v_(2) - v_(1))/(u_(2) - u_(1))) = - ((v_(2) - u//4)/(0 - u)) = (v_(2) - u//4)/(u)` `v_(2) = (u)/(4) + u = (5u)/(4)` .....(ii) (i) & (ii) `(3)/(4)u = m((5)/(4)u) rArr m = (3)/(5) = 0.6 kg` |
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| 69. |
A straight rod of length L has one of its end at the origin and the other at `X=L`. If the mass per unit length of the rod is given by `Ax` where A is constant, where is its centre of mass?A. `L/3`B. `L/2`C. `(2L)/(3)`D. `(3L)/4` |
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Answer» Correct Answer - C `X_(CM) = (intXdm)/(intdm) = (int_(0)^(L) (Ax).x.dx)/int_(0)^(L)(Ax).dx = (2L)/3` |
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| 70. |
A ballistic pendulum is a device that was used to measure the speeds of bullets before the development of electronic tiring, devices. The device consists of a large block of wood of mass `M`, hanging from two long cords. A bullet of mass `m` is fired into the block. the bullet comes quickly into rest and the block `+` bullet rises to a vertical distance `h` before the pendulum comes momentarily to rest as the ends of the arc. ltbr. In the process. the linear momentum is conserved. In such a collision. some kinetic energy is dissipated as heat: so mechanical energy is not conserved. When there is a loss in mechanical energy, the collision is said to be inelastic. Further when two bodies coalesce, the collision is said to be perfectly inelastic. The energy dissipated as heat in the collision isA. `1/2"mu"^(2)-mgh`B. `1/2"mu"^(2)-(M+m)gh`C. `1/2(M+m)u^(2)-(M+m)gh`D. cannot be estimated |
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Answer» Correct Answer - B Energy dissipated in collision `=` initial energy `-` final energy `=1/2"mu"^(2)-(m+M)gh` |
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| 71. |
A ball of mass `2m` impinges directly on a ball of mass m, which is at rest. If the velocity with which the larger ball impringes be equal to the velocity of the smaller mass after impact then the coefficient of restitution :-A. `(1)/(3)`B. `(3)/(4)`C. `(1)/(2)`D. `(2)/(5)` |
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Answer» Correct Answer - C `COLM rArr 2mu + 0 = 2mv + mu rArr v = (u)/(2)` `e = -((v_(2) - v_(1))/(u_(2) - u_(1))) = - ((u - u//2)/(0 - u)) = (1)/(2) underset(u)underset(rarr)overset(2m)"O" overset(m)"O" rArr underset(v_(LET))underset(rarr)overset(2m)"O" underset(u) underset(rarr)overset(m)"O"` |
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| 72. |
A body of mass `2 kg` is projected upward from the surface of the ground at `t = 0` with a velocity of `20 m//s`. One second later a body B, also of mass `2 kg`, is dropped from a height of `20m`. If they collide elastically, then velocities just after collision are :-A. `v_(A) = 5 m//s` downward, `v_(B) = 5 m//s` upwardB. `v_(A) = 10 m//s` downard, `v_(B) = 5 m//s` upwardC. `v_(A) = 10 m//s` upward, `v_(B) = 10 m//s` downwardD. both move downward with velocity `5 m//s` |
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Answer» Correct Answer - A After `1 s , v_(A) = 20 - 10 xx 1 = 10 m//s` and `v_(B) = 0 + 10 xx 1 = 10 m//s` At the time of collision `V_(A) = V_(B) = 5m//s` after collision, velocity gets enterchanged. |
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| 73. |
A ball of mass `1 kg` strink a heavy platform, elastically, moving upwards with a velocity of `5 m//s`. The speed of the ball just begore the collision is `10 m//s` downwards. Then the impulse imparted by the platform on the ball is :- A. `15 N-s`B. `10 N-s`C. `20 N-s`D. `30 N-s` |
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Answer» Correct Answer - D `e = - ((v_(2) - v_(1))/(u_(2) - u_(1))) rArr 1 = - (5 - v_(1))/(5 - (-10)) = (v_(1) - 5)/(15)` `rArr v_(1) = 20 m//s` `:.` Impulse on ball `= m(vec(v) - vec(u)) = 1 xx [20 -(10)] = 30N-s` |
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| 74. |
Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces to a height `h//4`, after the collision & the second ball to a height `h//6`. The impulse applied by the first & second ball on the floor are `I_(1)` and `I_(2)` respectively. Then :-A. `5I_(1) = 6I_(2)`B. `6I_(1) = 5I_(1)`C. `I_(1) = 2I_(2)`D. `2I_(1) = I_(2)` |
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Answer» Correct Answer - A For the `I^(st)` ball : `(h)/(4) = e_(1)^(2)h` For the `II^(nd)` ball `(h)/(16) = e_(2)^(2)h` Impulse on first ball `= I_(1) = mv_(0) (1 + e_(1)) = (3)/(2)mv_(0)` Impulse on second ball `I_(2) = mv_(0)(1 + e_(2)) = (5)/(4)mv_(0)` `rArr (I_(1))/(I_(2)) = (3//2 mv_(0))/(5//4 mv_(0)) = (6)/(5) rArr 5I_(1) = 6I_(2)` |
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| 75. |
Two balls of same mass are dropped from the same height onto the floor. The first ball bounces upwards from the floor elastically. The second ball stricks to the floor. The first applies an impulse to the floor of `I_(1)` and the second applies an imupulse `I_(2)`. The impulses obey :-A. `I_(2) = 2I_(1)`B. `I_(2) = (I_(1))/(2)`C. `I_(2) = 4I_(1)`D. `I_(2) = (I_(1))/(4)` |
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Answer» Correct Answer - B `m u - I_(1) = -m u :. I_(1) = 2m u` & `m u - I_(2) = 0 rArr I_(2) = m u` (for `II^(rd)` ball) `:. I_(2) = 1//2` |
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| 76. |
Statement I: If a ball projected up obliquely from the ground breaks up into several fragments in its path, the centre of the system of all fragments moves in the same parabolic path compared to initial one till all fragments are in air. Statement II: In the situation of Statement 1, at the instant of breaking, the fragments may be thrown in different directions with different speeds.A. Statement -1 is True, Statement -2 si True , Statement -2 is a correct explanation for Statement -1.B. Statement -1 is True, Statement -2 si True , Statement -2 is not correct explanation for Statement -1.C. Satement-1 is True, Statement-2 is Satement-1 is True, Statement-2 is False.False.D. Satement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - B | |
| 77. |
Statement I: If a ball projected up obliquely from the ground breaks up into several fragments in its path, the centre of the system of all fragments moves in the same parabolic path compared to initial one till all fragments are in air. Statement II: In the situation of Statement 1, at the instant of breaking, the fragments may be thrown in different directions with different speeds.A. Both assertion and reason are true and reason is the correct explanation of assertion.B. Both assertion and reason are true but reason is not the correct explanation of assertion.C. Assertion is true and reason is false.D. Assertion is false and reason is true. |
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Answer» Correct Answer - B The centre of mass of the fragments will continue its parabolic path. After the breakage, fragments may move in different directions. Both statements are correct but Statement II is not the explanation of Statement I. |
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| 78. |
A block of mass m is connect to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring. |
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Answer» Correct Answer - B The centre of mass of the system (two blocks+spring) moves with an acceleration `a=(F)/(m+M)`. Let us solve the problem in a frame of reference fixed to the centre of mass of the system. As this frame is accelerated with respect to the ground, we have to apply a pseudo force `ma` towards left on the block of mass m and `Ma` towards left on the block of mass M. The net external force on m is `F_1=ma=(mF)/(m+M)` (towards left) and the net external force on M is `F_2=F-Ma=F-(MF)/(m+M)=(mF)/(m+M)` (towards right) As the centre of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The extension of the spring will be maximum at this instant. Suppose, the left block is displaced through a distance `x_1` and the right block through a distance `x_2` from the initial positions. The total work done by the external force `F_1` and `F_2` in this period are `W=F_1x_1+F_2x_2=(mF)/(m+M)(x_1+x_2)` This should be equal to the increase in the potential energy of the spring, as there is no change in the kinetic energy. Thus, `(mF)/(m+M)(x_1+x_2)=1/2k(x_1+x_2)^2` or `x_1+x_2=(2mF)/(k(m+M))` This is the maximum extension of the spring. |
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| 79. |
A particle of mass `2kg` is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The force time graph is shown in figure. Find the velocity of the particle at the end of `10s`. |
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Answer» Using impulse =Change in linear momentum (or area under `F-t` graph) We have, `m(v_f-v_i)=Area` or `2(v_f-0)=1/2xx2xx10+2xx10+1/2xx2xx(10+20)+1/2xx4xx20` `=10+20+30+40` or `2v_f=100` `:. v_f=50m//s` |
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| 80. |
A block of mass `2kg` is attached with a spring of spring constant `4000Nm^-1` and the system is kept on smooth horizontal table. The other end of the spring is attached with a wall. Initially spring is stretched by `5cm` from its natural position and the block is at rest. Now suddenly an impulse of `4kg-ms^-1` is given to the block towards the wall. Approximate distance travelled by the block when it comes to rest for a second time (not including the initial one) will be (Take `sqrt(45)=6.70`)A. (a) `30cm`B. (b) `25cm`C. (c) `40cm`D. (d) `20cm` |
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Answer» Correct Answer - B Again using, `E_i=E_f` `:. 1/2xx2xx(2)^2+1/2xx4000xx(0.05)^2` `=1/2xx4000xxx^2` `:. x=0.067m=6.7m`=compression `:. d=5cm+6.7cm+6.7cm+6.7cm` `~~25cm` |
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| 81. |
A block of mass `2kg` is attached with a spring of spring constant `4000Nm^-1` and the system is kept on smooth horizontal table. The other end of the spring is attached with a wall. Initially spring is stretched by `5cm` from its natural position and the block is at rest. Now suddenly an impulse of `4kg-ms^-1` is given to the block towards the wall. Find the velocity of the block when spring acquires its natural lengthA. (a) `5ms^-1`B. (b) `3ms^-1`C. (c) `6ms^-1`D. (d) None of these |
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Answer» Correct Answer - B `Impu lse=m u` `:. u=("Impulse")/(m)=`4/2=2m//s` Now, `E_i=E_f` `:. 1/2xx2xx(2)^2+1/2xx4000xx(0.05)^2` `=1/2xx2xxv^2` Solving we get, `v=3m//s` |
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| 82. |
Find the position of centre of mass of the uniform lamina shown in figure. |
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Answer» Here `A_(1)` = area of complete circle `pia^(2)` `A_(2)=` area of small circle `=pi(a/2)^(2)=(pia^(2))/4` `(x_(1),y_(1))=` coordinates of the centre of mass of the large circle `=(0,0)` `(x_(2),y_(2))=` coordinates of centre of mass of the small circle `=(a/2,0)` Using `x_(CM)=(A_(1)x_(1)-A_(2)x_(2))(A_(1)-A_(2))` we get `x_(CM)=(pia^(2)xx0-(pia^(2))/4xxa/2)/(pia^(2)-(pia^(2))/3)=-a/6` `y_(CM)=`(as `y_(1)` and `y_(2)` both are zero) Therefore coordinates of CM of the lamina shown in figure are `(-a//6, 0)` |
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| 83. |
A time varying force, `F=2t` is acting on a particle of mass `2kg` moving along x-axis. velocity of the particle is `4m//s` along negative x-axis at time `t=0`. Find the velocity of the particle at the end of 4s. |
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Answer» Correct Answer - D Given force is a funciton of time. So, linear impulse can be obtained by integration. Further, this impulse is equal to change in linear momentum. Thus, `J=intFdt=Deltap=p_f-p_i=m(v_f-v_i)` or `v_f=1/mintFdt+v_i=1/2int_0^4(2t)dt-4` `(v_i=-4m//s)` `=1/2[t^2]_10^4-4=1/2[16-0]-4` `=+4m//s` Therefore, the final velocity is `4m//s` along positive x-direction. |
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| 84. |
Find the centre of mass of as uniform `L` -shaped lamina (a thin flast plate) with dimensions as shown in figure. The mass of lamina is `3 kg`. |
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Answer» The plate has uniform density and same thickness everywhere. So its CM will coincide with the centroid. Divide the given plate two parts of area `A_(1)` and `A_(2)` as shown in the figure. We have `A_(1)=2xx1m^(2)` with its centroid `C_(1)(1,1//2)` and `A_(2)=1xx1m^(2)` with its centroid `C_(2)(1//2,3//2)` the centroid of the whole plate can be defined as `barx=(A_(1)x_(1)+A_(2)x_(2))/(A_(1)+A_(2))=(2xx1+1xx1/2)/(2+1)=5/6m` `bary=(A_(1)y_(1)+A_(2)y_(2))/(A_(1)+A_(2))=(2xx1/2+1xx3/2)/(2+1)=5/6m` |
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| 85. |
An equilateral traingular plate of mass `4m` of side a is kept as shown. Consider two cases: (i) a point mass `4m` is placed at the vertex P of the plate (ii) a point mass `m` is placed at the vertex R of the plate. In both cases the x-coordinate of centre of mass remains the same. Then x coordinate of centre of mass of the plate is A. (a) `a/3`B. (b) `a/6`C. (c) `(6a)/(7)`D. (d) `(2a)/(3)` |
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Answer» Correct Answer - B `((4m)X_1+(4m)(a/2))/(8m)=((4m)X_1+(m)(a))/(5m)` Solving this equaiton, we get `X_1=X`-coordinate of COM of plate `=a/6` |
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| 86. |
Consider a rectangular plate of dimensions `axxb`. If this plate is considered to be made up of four rectangles of dimensions `a/2xxb/2` and we now remove one out of four rectangles. Find the position where the centre of mass of the remaining system will lie? |
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Answer» The rectangular plate is shown in the figure of is removed. We can find the `x` and `y` coordinates of the centre of mas of this system, taking origin at the centre of plate. The coordinates of the three remaining rectangles are `(a/4, b/4), ((-a)/4 (+b)/4)` and `((-a)/4, (-b)/4)`. By geometry, masses of these rectangles can be taken as `M/4`. Now `x`-coordinate of the centre of mass: `X_(CM)=(M/4 a/4 M/4 a/4-M/4 a/4)/(3M/4)=a/12` and `y`-coordinate of the centre of mass: `y_(CM)=(M/4b/4+M/4b/4-M/4b/4)/(3M/4)=b/12` |
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| 87. |
Consider a rectangular plate of dimensions `axxb`. If this plate is considered to be made up of four rectangles of dimensions `a/2xxb/2` and we now remove one out of four rectangles. Find the position where the centre of mass of the remaining system will lie? |
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Answer» Correct Answer - A::B `x_(CM)=(A_1x_1-A_2x_2)/(A_1-A_2)` `=((ab)(0)-((ab)/(4))(a/4))/(ab-(ab)/(4))` `=-a/12` `y_(CM)=(A_1y_1-A_2y_2)/(A_1-A_2)` `=((ab)(0)-((ab)/(4))(b/4))/(ab-(ab)/(4))` `=-b/12` |
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| 88. |
The figure shows a smooth curved track terminating in a smooth horizontal part. A spring of spring constant `400(N)//(m)` is attached at one end to a wedge fixed rigidly with the horizontal part . A 40 g mass is released from rest at a height of 5 m on the curved track. The maximum compression of the spring will beA. 9.8 mB. 9.8cmC. 0.98 mD. 0.009 km |
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Answer» Correct Answer - B Applying the conservation of momentum to the system. `mgh = 1/2kx^(2)` Given, m = 0.04 kg, h=5m, k= 400Nm^(-1)` and x = deformation (compression in the spring)` `rArr x=(sqrt(2mgh)/k) = (sqrt(2 xx 0.04 xx 10 xx 5))/400` `=1/10 m = 10cm ~~ 9.8 cm` |
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| 89. |
A block of mass `m` is relesed from rest from a height `h` onto a smooth sledge of mass `M` fitted with an ideal spring of stiffness `k`. |
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Answer» Method 1: a. As no external force in horizontal direction, the linear momentum will be conserved in horizonta direction. Let the velocity of block and wedge be a and `V` respectively up to the case when the block reaches on horizontal surface on the wedge. `mv=MV` ...............i Mechanical energy is also conserved in this case `/_K+/_U=0` `(1/2mv^(2)+1/2MV^(2)-0)+(-mgh)=0` `1/2mv^(2)+1/2M(m/Mv)^(2)=mgh` `v=sqrt((2Mgh)/(M+m))` and `V=sqrt((2m^(2)gh)/(M(M+m)))` Method 2 : Applying work energy theorem from centre of mass frame `W_("gravity")=(/_K)_(CM)` `mgh=1/2((mM)/(m+M))(v+V)^(2)` also, `mv=MVimpliesv=(MV)/m` `impliesmgh=12((mM)/(m+M))V^(2)(M/m1)^(2)` `implies V^(2)=sqrt((2m^(2)gh)/(M(m+M)))` b. Initial linear mometum to of the system is zero and there is no external force on the system., hence at the time of maximum compression of spring the linear momentum of the system should be zero. Hence final kinetic energy should be zero. `/_U+/_K=0implies/_K=K_(f)-K_(i)=0` Hence, `/_U_("gravity")+/_U_("spring")=0` `-mgh+1/2kx^(2)=0impliesx=sqrt((2mgh)/k)` |
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| 90. |
A ball kept in a close box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of massA. of the box remains constantB. of the (box `+` ball) system remains constantC. of the ball remains constantD. of the ball relative to the box remains constant |
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Answer» Correct Answer - B As no external force acts on the ball `+` box system, hence velcilty of the system remains constant. |
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| 91. |
This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible glven as `f(1/2mv^2)` then `f=(m/(M+m))` Statement II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.A. Statement-I is true, Statement-II is ture, Statement-II is a coorect explanation of Statement-IB. Statement-I is true, Statement-II is true, Statement-II is not the correct explanation of Statement-IC. Statement-I is true, Statement-II is false.D. Statement-I is false, Statement-II is true. |
| Answer» Correct Answer - D | |
| 92. |
A moving body of mass m makes a head on elastic collision with another body of mass `2m` which is initially at rest. Find the fraction of kinetic energy lost by the colliding particles after collision. |
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Answer» Substituting the values in above result `(K_i-K_f)/(K_i)=(4m_1m_2)/((m_1+m_2)^2)` `=(4xxmxx2m)/((m+2m)^2)` `=8/9` |
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| 93. |
Assertion: Energy can not be given to a system without giving it momentum. Reason: If kinetic energy is given to a body it means it has acquired momentum.A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Energy can be given in the form of potential energy without giving the momentum. |
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| 94. |
Assertion: During head on collision between two bodies let `Deltap_1` is change in momentum of first body and `Deltap_2` the change in momentum of the other body, then `Deltap_1=Deltap_2`. Reason: Total momentum of the system should remain constant.A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D `Deltap_1+Deltap_2=0` or `Deltap_1=-Deltap_2` |
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| 95. |
The momentum of a system with respect to centre of massA. (a) is zero only if the system is moving uniformlyB. (b) is zero only if no external force acts on the systemC. (c) is always zeroD. (d) can be zero in certain conditions |
| Answer» Correct Answer - C | |
| 96. |
Assertion: A given force applied in turn to a number of different masses may cause the same rate of change in momentum in each but not the same acceleration to all. Reason: `F=(dp)/(dt)` and `a=F/m`A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - A | |
| 97. |
The momentum of a system is definedA. (a) as the product of mass of the system and the velocity of centre of massB. (b) as the vector sum of the momentum of individual particlesC. (c) for bodies undergoing translational, rotational and oscillatory motionD. (d) all the above |
| Answer» Correct Answer - D | |
| 98. |
A steel ball is dropped on a hard surface from a height of `1m` and rebounds to a height of `64cm`. The maximum height attained by the ball after `n^(th)` bounce is (in m)A. (a) `(0.64)^(2n)`B. (b) `(0.8)^(2n)`C. (c) `(0.5)^(2n)`D. (d) `(0.8)^n` |
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Answer» Correct Answer - B `h_n=(e^(2n))hi` Given that `(64cm)=(e^2)(100cm)` `:. e=0.8` Substituting in Eq. (i) we have `h_n=(0.8)^(2n)(1m)` `=(0.8)^(2n)` |
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| 99. |
A loaded `20,000kg` coal wagon is moving on a level track at `6ms^-1`. Suddenly `5000kg` of coal is dropped out of the wagon. The final speed of the wagon isA. (a) `6ms^-1`B. (b) `8ms^-1`C. (c) `4.8ms^-1`D. (d) `4.5ms^-1` |
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Answer» Correct Answer - A Velocity of dropped out mass is also `6m//s`. So its relative velocity is zero and no thrust force will act. Therefore final speed remain `6m//s`. |
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| 100. |
If no external force acts on a systemA. (a) velocity of centre of mass remains constantB. (b) position of centre of mass remains constantC. (c) acceleration of centre of mass remains non-zero and constantD. (d) All of the above |
| Answer» Correct Answer - A | |