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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.
| 201. |
A body of mass `m_(1) = 4kg moves at `5 hat(i) ms^(-1)` and another body of mass `m_(2) = 2kg` moves at `10hat(i)ms^(-1)`. The kinetic energy of center of mass isA. `200/3J`B. `500/3J`C. `400/3J`D. `800/3J` |
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Answer» Correct Answer - C `v_(CM) = (m_(1)(dr_(1))/(dt)+m_(2)(dr_(2))/dt)/(m_(1) + m_(2)) = (4 xx 5hat(i) + 2 xx 10hat(j))/(4+2)` `v_(CM) = (40hat(i))/6 = 20/3hat(i)` The kinetic energy, `K = 1/2mv^(2) = 1/2xx(4+2) xx(20 xx 20)/(3 xx 3)` `=1/2 x 6 xx (20 xx 20)/(3 xx 3) = 400/3 J` |
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| 202. |
A pendulum comsists of a wooden bob of mass `m` and length `l`. A bullet of mass `m_1` is fired towards the pendulum with a speed `v_1` and it emerges from the bob with speed `v_1/3`. The bob just completes motion along a vertical circle. Then `v_1` is A. (a) `m/m_1sqrt(5gl)`B. (b) `(3m)/(2m_1)sqrt(5gl)`C. (c) `2/3(m/m_1)sqrt(5gl)`D. (d) `(m_1/m)sqrt(gl)` |
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Answer» Correct Answer - B Let velocity of m just after collision is v. Then, from conservation of momentum we have, `m_1v_1=mv+m_1v_1/3implies:. v=2/3(m_1v_1)/(m)` Now, for just completing the circle, `v=sqrt(5gl)` `:. 2/3(m_1v_1)/(m)=sqrt(5gl)` `:. v_1=(3m)/(2m_1)sqrt(5gl)` |
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| 203. |
A bob of mass `m` attached with a string of length `l` tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass `m` gently stuck to it. Find the maximum angle from the vertical to which it rotates.A. (a) `cos^-1(2/3)`B. (b) `cos^-1(3/4)`C. (c) `cos^-1(1/4)`D. (d) `60^@` |
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Answer» Correct Answer - B `v_m=sqrt(2gl)` From conservation of linear momentum, `v_(2m)=(v_m)/(2)=(sqrt(2gl))/(2)` Now, `h=(v_(2m))/(2g)=l/4` But `h=l(1-cos theta)` `:. l/4=l(1-costheta)` `:. theta=cos^-1(3/4)` |
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| 204. |
Comprehension # 1 If net force on a system in a particular direction is zero (say in horizontal direction) we can apply: `Sigmam_(R )x_(R )= Sigmam_(L)x_(L), Sigmam_(R )v_(R )= Sigmam_(L)v_(L)` and `Sigmam_(R )a_(R ) = Sigmam_(L)a_(L)` Here R stands for the masses which are moving towards right and L for the masses towards left, x is displacement, v is veloctiy and a the acceleration (all with respect to ground). A small block of mass `m = 1 kg` is placed over a wedge of mass `M = 4 kg` as shown in figure. Mass m is released from rest. All surface are smooth. Origin O is as shown. Final velocity of the wedge is ..........`m//s` :-A. `sqrt3`B. `sqrt2`C. `(1)/(sqrt2)`D. `(1)/(sqrt3)` |
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Answer» Correct Answer - B By appplying conservation of momentum `mv_(1) + Mv_(2) = 0 rArr v_(1) = -4v_(2)` …..(i) By applying conservation of energy `(1)/(2)mv_(1)^(2) + (1)/(2)Mv_(2)^(2) = mgh rArr (v_(1)^(2))/(2) + 2v_(2)^(2) = 20` ……(ii) `v_(2) = sqrt2m//s , v_(1) = 4 sqrt2 m//s`. |
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| 205. |
A block of mass `M` is tied to one end of a massless rope the other end of the rope is in the hands of a man of mass `2M` as shown in the figure. The block and the man are resting on a rough plank of mass `M` as shown in the figure. The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is massless and frictionless. What is the displacement of the plank when the block meets the pulley ( Man does not leave his position on plank during the pull) A. `0.5 m`B. `1m`C. ZeroD. `2//3 m` |
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Answer» Correct Answer - A `Deltabar(x) = (3M.x + M(x + 2))/(4M) = 0 rArr x = (1)/(2)` |
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| 206. |
In an elastic collision in absence of external force, which of the following is /are correctA. The linear momentum is conservedB. The potential energy is conserved in collisionC. The final kinetic energy is less than the initial kinetic energyD. The final kinetic energy is grater then initial kinetic energy |
| Answer» Correct Answer - A | |
| 207. |
Statement-1: The centre of mass of body may lie where there is no mass. Statement-2: Centre of mass of a body is a point, where the whole mass of the body is supposed to be concentrated.A. Both statement-1 and 2 are true and statement-2 is the correct explanation of statement -1.B. Both statement-1 and 2 are true but satatement-2 is not correct explanation of statement -1.C. Statement-1 is true but statement-2 is falseD. Both statement 1 and 2 are false. |
| Answer» Correct Answer - B | |
| 208. |
Motion of Mass Center in Vector Form A 2.0 kg particle has a velocity of Vecv_1=(2.0hati-3.0hatj)m/s, and a 3.0 kg particle has a velocity vecv_2=(1.0hati+6.0hatj)m/s. (a) How fast is the center of mass of the particle system moving? (b) Find velocities of both the particles in centroidal frame. |
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Answer» (a) Velocity of the mass center `vec(v)_(c ) = (m_(1)vec(v)_(1) + m_(2)vec(v)_(2))/(m_(1)+m_(2))` `vec(v)_(c ) = (m_(1)vec(v)_(1)+m_(2)vec(v)_(2))/(m_(1)+m_(2)) rarr vec(v)_(c ) = (2 (2.0hat(i) - 3.0hat(j)) + 3 (1.0hat(i) + 6.0 hat(j)))/(2+3)=(1.4hat(i)+2.4hat(j))m//s` (b) Velocity of the first particle in centrodial frame `vec(v)_(1//c) = vec(v)_(1) - vec(v)_(c ) rarr " " vec(v)_(2//c) = (2.0 hat(i) - 3.0 hat(j) - (1.4hat(j)) = 0.6 (hat(i) + hat(j)) m//s` Velocity of the second particle in centrodial frame `vec(v)_(2//c) = vec(v)_(2) - vec(c ) rarr " " vec(v)_(2//c) = (1.0 hat(j)) - (1.4 hat(i) + 2.4 hat(j)) = - 0.4 hat(i) + 3.6 hat(j) m//s` |
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| 209. |
A bullet ofmass m is fired with a velocity `10m//s` at angle `theta` with horizontal. At the hightest point of its trajectory, it collsides head-on with a bob of mass `3m` suspended by a massless string of length `2//5m` and gets embedded in the bob. After th collision the string moves through an angle of `60^(@)`. The vertical coordinate of the initial positon of the bob w.r.t. the point of firing of the bullet isA. `(9)/(4)m`B. `(9)/(5)m`C. `(24)/(5)m`D. None of these |
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Answer» Correct Answer - (B) `H_(max) = (u^(2)sin^(2) theta)/(2g) = ((100)(9//25))/(20) = (9)/(5)m` |
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| 210. |
A bullet ofmass m is fired with a velocity `10m//s` at angle `theta` with horizontal. At the hightest point of its trajectory, it collsides head-on with a bob of mass `3m` suspended by a massless string of length `2//5m` and gets embedded in the bob. After th collision the string moves through an angle of `60^(@)`. The horizontal coordinate of the intital position of the bob w.r.t. the point of firing of the bullet is nA. `(9)/(5)m`B. `(24)/(5)m`C. `(9)/(4)m`D. None of these |
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Answer» Correct Answer - (B) `(R )/(2) = (2u^(2) sin theta cos theta)/(2g) = ((100)((3)/(5))((4)/(5)))/(10) = (24)/(5) m` |
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| 211. |
In the arrangement shown in Fig. the ball and the block have the same mass `m=1 kg` each, `theta=60^(@)` and length `l = 2.50m`. Coefficient of friction between the block and the floor is `0.5`. When the ball is released from the position shown in Fig. it collides with the block and the block stops after moving a distance `2.50 m`. The velocity of block just after collision isA. `10m//s`B. `5m//s`C. `2.5m//s`D. `3m//s` |
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Answer» Correct Answer - B When ball is released, it moves along a vertical circle with centre at `A`. Kinetic energy of ball just before collision is equal to loss of its potential energy from point of release to the point of collision. Therefore, velocity `v_(1)` of ball, just before collision, is given by `1/2mv_(1)^(2)mg(l-lcostheta)impliesv_(1)=5m//s` After collision, the block starts to move towards right. But it is retarded by force of friction and ultimately it comes to rest. According to law of conservation of energy, Kinetic energy of the block just after collision `=` Work done by it against friction. Therefore, its velocity `v`, just after collision is given by `1/2mv_(2)^(2)=mumgs` where `m=0.54` and `s=2.50 m:.v_(2)=5m//s` Coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))` where `u_(2)=0, u_(1)=5m/s` and `v_(1)=(5-5e)` …………i Applying law of conservation momentum. `"mu"_(1)+"mu"_(2)=mv_(1)+mv_(2)` `5m+(mxx0)=mv_(1)+5m` or `v_(1)=0` substituting in eqn i we get `e=1` |
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| 212. |
In the arrangement shown in Fig. the ball and the block have the same mass `m=1 kg` each, `theta=60^(@)` and length `l = 2.50m`. Coefficient of friction between the block and the floor is `0.5`. When the ball is released from the position shown in Fig. it collides with the block and the block stops after moving a distance `2.50 m`. The coefficient of restitution for collision between the ball and the block isA. `0.5`B. `0.75`C. `1.0`D. `0.3` |
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Answer» Correct Answer - C When ball is released, it moves along a vertical circle with centre at `A`. Kinetic energy of ball just before collision is equal to loss of its potential energy from point of release to the point of collision. Therefore, velocity `v_(1)` of ball, just before collision, is given by `1/2mv_(1)^(2)mg(l-lcostheta)impliesv_(1)=5m//s` After collision, the block starts to move towards right. But it is retarded by force of friction and ultimately it comes to rest. According to law of conservation of energy, Kinetic energy of the block just after collision `=` Work done by it against friction. Therefore, its velocity `v`, just after collision is given by `1/2mv_(2)^(2)=mumgs` where `m=0.54` and `s=2.50 m:.v_(2)=5m//s` Coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))` where `u_(2)=0, u_(1)=5m/s` and `v_(1)=(5-5e)` …………i Applying law of conservation momentum. `"mu"_(1)+"mu"_(2)=mv_(1)+mv_(2)` `5m+(mxx0)=mv_(1)+5m` or `v_(1)=0` substituting in eqn i we get `e=1` |
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| 213. |
A body A of mass M while falling wertically downwards under gravity brakes into two parts, a body B of mass `(1)/(3)` M and a body C of mass `(2)/(3)` M. The center of mass of bodies B and C taken together shifts compared to that of body A towardsA. depaends on height of breakingB. does not shiftC. body CD. body B |
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Answer» Correct Answer - B In order to shift centre of mass, the system must experience an external force, as there is no external force responsible for explosion, hence centre of mass does not shift. |
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| 214. |
A body A of mass M while falling wertically downwards under gravity brakes into two parts, a body B of mass `(1)/(3)` M and a body C of mass `(2)/(3)` M. The center of mass of bodies B and C taken together shifts compared to that of body A towardsA. depends on height of breakingB. does not shiftC. shift towards body CD. shift towards body B |
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Answer» Correct Answer - B Before breaking, the centre of mass of system is moving under gravity. Thus, acceleration of the centre of mass is gravitational acceleration. During breaking, internal force comes into play which are not responsible for the acceleration of the centre of mass. This inducates that, the acceleration of centre of mass remains the same (equal to gravitational acceleration). thus, the center of mass of system continues its original path. |
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| 215. |
No external force: Stationery mass relative to an inertial frame remains at rest A man of mass m is standing at one end of of a plank of mass M. The length of the plank is L and it rests on a frictionless horizontal ground. The man walks to the other end of the plank. Find displacement of the plank and man relative to the ground. |
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Answer» Denoting x-coordinates of the man, mass center of plank and mass center of the man-plank system by `x_(m) x_(p)` and `x_(c )`, we can write the following equation. `(Sigmam_(i))vec(r)_(c ) = Sigma m_(i)vec(r)_(i) rarr" " (m+M)vec(x)_(c ) = m vec(x)_(m)+ Mvec(x)_(p)` Net force on the system relative to the ground is zero. Therefore mass center of the system which is at rest before the man starts walking, remains at rest `(Delta vec(x)_(c ) = 0)` after while the man walks on the plank. `(Delta bar(x)_(c ) = 0) rarr " " m Delta bar(x)_(m) + M Delta vec(x)_(p) = 0` The man walks displacement `(Delta bar(x)_(m))/(p = -L hat(i))` relative to the plank. Denoting displacements of the man and the plank relative to the ground by `Delta vec(x)_(m)` and `Delta vec(x)_(p)`, we can write `Delta vec(x)_(m//p) = Delta vec(x)_(m) - Delta vec(x)_(p) rarr " " Delta bar(x)_(m) - Delta bar(x)_(p) = - L hat(i)` From the above equations (1) and (2), we have `Delta vec(x)_(m) = -(MLhat(i))/(m+M)` The plank moves a distance `(mL)/(m+M)` towards right relative to the ground. `Delta vec(x)_(m) = - (ML hat(i))/(m + M)` The plank moves a distance `(mL)/(m + M)` towards right relative to the ground. |
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| 216. |
Consider the following two statements:A. Kinear momentum of the system remains constant. B. Centre of mass of the system remains at restA. A implies B and B implies AB. A does not imply B and B does not imply A.C. A implies B but B does not imply A.D. B implies A but A does not imply B. |
| Answer» Correct Answer - D | |
| 217. |
A particle of mass m has momentum p. Its kinetic energy will beA. mpB. `p^(2)m`C. `(p^(2)/(m)`D. `p^(2)/(2m)` |
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Answer» Correct Answer - D `K = p^(2)/(2m)` and `p = sqrt(2Km)` are two standard results. |
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| 218. |
Consider the following two statements:A. linear momentum of the system remains constant. B. Centre of mass of the system remains at restA. A implies B and B implies AB. A does not imply B and B does not imply A.C. A implies B but B does not imply AD. B implies A but A does not imply B |
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Answer» Correct Answer - D If CM is at rest, it definitetlyy means momentum of hte system is constant. But momentum of the system is constant, it does not mean center of mass is at rest. |
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| 219. |
A particle is projected from a point of an angle with the horizontal. At any instant t, if p is the linear momentum and E the kinetic energy, then which of the following graph is/are correct?A. B. C. D. |
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Answer» Correct Answer - D `p^(2) = 2Em` or `p^(2) alpha E` i.e., `p^(21)` versus E graph is a straight line passing through origin. |
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| 220. |
Statement -1: If kinetic energy of a system is zero then linear momentum of each of particle of the system must be zero. Statement-2: If linear momentum of a system is zero then kinetic energy of system must be zero. Statement-3: If linear momentum of a particle is zero then kinetic energy is zero. Select the correct sequence.A. TTFB. TFTC. FFTD. TFF |
| Answer» Correct Answer - B | |
| 221. |
The momentum of a moving particle is vectorially given a, `vecp=p_(0)(costhati+sinthatj)`where `t` stands for time. Choose the correct option:A. The applied force is constant.B. The momentum is constant.C. The applied force always remains perpendicular to the momentum.D. The applied force is always parallel to the momentum. |
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Answer» Correct Answer - C `p=p_(0)(costhati+sinthatj)` `vecF=(vec(dp))/(dt)=p_(0)[-sinthati+costhatj]` Both`vecp` and `vecF` are function of time. So none of the is constant. We see that `vecF.vecp=0`, so `vecF_|_^(r)vecp` |
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| 222. |
A bullet of mass `50 g` is fired from below into the bob of mass `450 g` of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of `1.8 m`. find the speed of the bullet. |
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Answer» Let the speed of the bullet be `v`. Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, be `v`. By the principle of conservation of linear momentum. `V=((0.05kg)v)/(0.45g+0.05kg)=v/10` The string becomes loose and the bob will go up with a deceleration of `g=m//s^(2)`. As it comes rest at a height of `1.8 m`, using the equation `v^(2)=u^(2)+2ax`, we get `1.8m=((v/10)^(2))/(2xx10m//s^(2))` or `v=60m//s` |
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| 223. |
The hero of a stunt film fires `50 g` bullets form a machine gun, each at a speed of `1.0 km//s`. If he fires `20` bullets in `4s`, what average force does he exert against the machine gun during this period? |
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Answer» The momentum of each bullet `(0.050kg)(1000m//s)=50kgm//s`. ltbr The gun is imparted this much of momentum by each bullet fired. Thus, the rate of change of momentum of the gun is `50 kg m/s xx20)/4s=250N`. In order to hold the gun, the hero must exert a force of `250 N` against the gun. |
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| 224. |
A man of mass 50 kg starts moving on the earth and acquires speed of 1.8 m/s. with what speed does the earth recoil? Mass of earth =`6xx10^24kg.` |
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Answer» `m_1v_1=m_2v_2` `rarr 50xx1.8=6xx10^24xxv_2` `:. v_2=(50xx1.8)/(6xx10^24)` `15xx10^-24m/sec` `=1.5xx10^-23 m/sec` So, the earth will recoil at a speed of `1.5xx10^-23m/sec`. |
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| 225. |
A uranium 238 nucleus, initially at rest emits an alpha particle with a speed of `1.4xx10^7m/s`. Calculate the recoil speed of the residual nucleus thorium 234. Assume that the mas of a nucleus is proportional to the mass number. |
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Answer» Correct Answer - A::B::D `p_i=p_f` `:. 0=(4)(1.4xx10^7)+(234)v` `:. v=-24xx10^5m//s` Here, negative sign implies that this velocity is opposite to the direction of velocity of alpha particle. |
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| 226. |
Find the ratio of the linear omenta of two particles of masses 1.0 kg and 4.0 kg if their kinetic energies are equal. |
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Answer» Let the mass of the two particle be `m_1,m_2` respectively `m_1=1kgm_2=4kg` `:.` According to question `(1/2)m_1v_2^2=(1/2)m_2v_2^2` `rarr m_1/m_2=v_2^2/v_1^2` `rarr v_2/v_1=sqrt(m_1/m_2)` `rarr v_1/v_2=(sqrt(m_2/m_1)` `Now, (m_1v_1)/(m_2v_2)=m_1/m_2 sqrt(m_2/m_1)` `=sqrt(m_1/m_2)=sqrt(1/4)=1/2` `rarr (m_1v_1)/(m_2v_20=1:2` |
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| 227. |
A small ball of mass `m` collides with as rough wal having coeficient of friction `mu` at an angle `theta` with the normal to the wall. If after collision the ball moves wilth angle `alpha` with the normal to the wall and the coefficient of restitution is `e`, then find the reflected velocity `v` of the ball just after collision. |
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Answer» `mvcosalpha-[m(-ucostheta)]=intNdt` `mvsinalpha-musintheta=-muint Ndt` `and e=(vcosalpha)/(ucostheta)impliesvcosalpha=eucostheta`……….i `or mvsinalpha-musintheta-mu(mvcosalpha+mucostheta)`……….ii `From eqn i and ii v=u/(sinalpha)[sintheta-mucostheta(e+1)]` |
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| 228. |
Which of the following is/are correct?A. If centre of mass of three particles is at rest and it is known that two of them are moving along different lines, then the third particle must also be moving.B. If centre of mass remains at rest, then net work done by the forces acting on the system must be zero.C. If centre of mass remains at rest, then the net external force must be zero.D. If speed of centre of mass is changing, then there must be some net work being done on the system from outside. |
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Answer» Correct Answer - A::C::D As `Sigma vecP=0` and `vecP_(1)=vecP_(2)=!vecP_(3)` cannot be zero. b. For any system for two or more than two particles, `KE` can change due to work done by internal forces while the centre of mss remains at rest c. `veca_(CM)=0impliesSigmavecF_(ext)=0` d. It is possible to have changing speed of `CM` without any work done by ext. forces on the system. when a person accelerates himself on as rough horizontal surface without any flipping between his shoes and ground, work done by friction, normal reaction and weight all are zero but speed of centre of mass changes. |
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| 229. |
A ball of mass `1 kg` drops vertically on to the floor wit a speed of `25 m//s`. It rebounds with an initial velocity of `10 m//s`. What impulse acts on the ball during contact ?A. `35 kg m//s` downwardsB. `35 kg m//s` upwardsC. `30 kg m//s` downwardsD. `30kg m//s` upwards |
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Answer» Correct Answer - B Impulse `= P_(f) - P_(1) = 1 xx 10 - 1 xx (-25) = 35 kg m//s uarr` |
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| 230. |
Find the distance between centre of gravity and centre of mass of a two particle system attached to the ends of a light rod. Each particle has same mass. Length of the rod is R, where R is the radius of earth. A. RB. `R//2`C. zeroD. `R//4` |
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Answer» Correct Answer - B `x_(cm) = (m xx 0 + m xx R)/(m + m) = (R )/(2)` `x_(CG) = (W_(1)(0) + W_(2)(R ))/(W_(1) + W_(2)) = (mgR)/(mg) = R` `:. X_(CG) - x_(cm) = R - (R )/(2) = (R )/(2)` |
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| 231. |
After calling a wall of `3 m` heigh a mass of weight W drops himself to the ground. If his body comes to a complete stop in `0.15 s`. After his feet touch the ground, calculate the average impulsive force in the vertical direction exerted by ground on his feet.A. `5W`B. `5.21W`C. `3W`D. `6W` |
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Answer» Correct Answer - B Velcoity before strike `u = sqrt(2gh)` Impulse `F Deltat = m(v - u)` `rArr F = (m(v - u))/(t) = (w (0 - sqrt(2gh)))/(g xx 0.15) = 5.21 W` |
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| 232. |
A uniform thin rod AB of length L has linear mass density `mu(x) = a + (bx)/(L)`, where x is measured from A. If the CM of the rod lies at a distance of `((7)/(12)L)` from A, then a and b are related as :_A. `2a = b`B. `3a = 2b`C. `a = 2b`D. `a = b` |
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Answer» Correct Answer - A `x_(cm) = (int xdm)/(int dm)` `rArr (7L)/(12) = (int_(0)^(L)x(a + (bx)/(L)) dx)/(int_(0)^(L) (a + (bx)/(L)) dx)` Solving we get `2a = b` |
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| 233. |
A system of particles has its centre of mass at the origin. The x-coordinates of all the particlesA. may be positiveB. may be negativeC. must be non-negativeD. may be non-positive |
| Answer» Correct Answer - D | |
| 234. |
In which of the following cases the centre of mass of a system is certainly not at its centre ?A. A uniformly shaped rod whose density continuously increases from left to rightB. A uniformly shaped rod whose density is constant throughoutC. A uniformly shaped rod whose density decreases from left to right upto the centre and then increasesD. A uniformly shaped rod whose density increases from left to right upto the centre and then decreases |
| Answer» Correct Answer - A | |
| 235. |
In which o the following cases the centre of mass of a rod is certainly not at its centre?A. The linear mass density continuously decreases from left to right.B. The linear mass density continuously increases from left to right.C. The linear mass density decreases from left to fight upto centre and then increases.D. The linear mass density increases from left to right upto cente and then decreases. |
| Answer» Correct Answer - (CD) | |
| 236. |
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity `V m//s` in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is A. `250 m//s`B. `250sqrt(2)m//s`C. `400m//s`D. `500m//s` |
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Answer» Correct Answer - D `R=usqrt((2u)/g` `implies 20=V_(1)sqrt((2xx5)/10)` and `100 =V_(2)sqrt((2xx5)/10)` `implies V_(1)=20 m//s, V_(2)=100 m//s` Applying momentum conservation just before and just after the collision `(0.01)(V)=(0.2)(20)+(0.01)(100)` `V=500 m//s` |
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| 237. |
The system in Fig. is released from rest from the position shown. After blocks have moved distance `H//3`. collar `B` is removed and block `A` and `C` continue to move. The speed of `C` just before it strikes the ground is A. `4/3sqrt(gH)`B. `2(sqrt(gH))/3`C. `sqrt((13gH))/3`D. `2sqrt(2gH)` |
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Answer» Correct Answer - C Initial acceleration of the system before the collar is removed. `a_(1)=(2mg)/(3m)=(2g)/3` velocity at the time when the collar is removed: `v_(1)=sqrt(2a_(1)H/3)` Acceleration after the collar is removed `a_(2)=(mg)/(2m)=g/2` Final velocity `=v_(2)^(2)=v_(1)^(2)+2a_(2)H` `=(2a_(1)H)/3+(2g)/2H=2/3 (2g)/3H+gH` `implies v_(2)=sqrt((13gH)/3` |
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| 238. |
A 2 kg block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of `150 ms^(-1)` is shot into the block and sticks to it. The block then slides 2.7m along the table top and comes to a stop. The force of friction between the block and the table isA. 0.052 NB. 3.63 NC. 2.50 ND. 1.04 N |
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Answer» Correct Answer - A Velocity of block just after collision will be `v = (5 xx 10^(-3) xx 150)/(2+5 xx 10^(-3)` (From conservation of linear momentum) `=0.374 ms^(-1)` Let, F be the force of friction. Then, work done against friction = initial kinetic energy or `F xx 2.7 = 1/2 xx 2.005 xx (0.374)^(2)` or `F=0.052N` |
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| 239. |
Two bocks of masses 10 g and 20 kg are plced on the X-axis. The first mass is moved on the axis by a distance of 2 cm. By what distance should the second mass be moved to keep the position of the centre of mass unchanged? |
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Answer» Two masses `m_1 and m_2` are placed on the X-axis. `m_1=10kg, m_2=20kg` `the first mass is displaced by distance of 2 cm. `:. X_(cm)=(m_1x_1+m_2x_2)/(m_1+m_2)` `=(10xx2+20x_2)/30` `rarr 0=(20+20x_2)/30` `rarr 20+20x_2=0` `rarr 20=-20x_2` `rarr x_2=-1` `:. ` Second mass should be displaced by a distance 1 cm towards left so as to keep the position of centre of mass unchanged. |
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| 240. |
An open water tight railway wagon of mass `5 xx 10(3) kg` coats at an initial velocity `1.2 m//s` without friction on a railway track. Rain drops fall vertically downwards into the wagon. The velocity of the wagon after it has collected`10^(3) kg` of water will be :-A. `0.5 m//s`B. `2 m//s`C. `1 m//s`D. `1.5 m//s` |
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Answer» Correct Answer - C `COLM rArr m_(1)u_(1) + m_(2)u_(2) = (m_(1) + m_(2))v` `5 xx 10^(3) xx 1.2 + 0 = (5 + 1) xx 10^(3) xv rArr v = 1 m//s` |
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| 241. |
Two particles of masses `m_(1),m_(2)` move with initial velocities `u_(1)` and `u_(2)`. On collision, one of the particles get excited to higher level, after absording enegry. If final velocities of particles be `v_(1)` and `v_(2)` then we must haveA. `m_(1)^(2)u_(1) + m_(2)^(2)u_(2)-epsilon =- m_(1)^(2)v_(1) + m_(2)^(2)v_(2)`B. `1/2 m_(1)u_(1)^(2) + 1/2m_(2)u_(2)^(2)=1/2m_(1)v_(1)^(2) + 1/2m_(2)v_(2)^(2) - epsilon`C. `1/2 m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)-epsilon=1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)`D. `1/2m_(1)^(2)u_(1)^(2)+1/2m_(2)^(2)u_(2)^(2) + epsilon=1/2m_(1)^(2)v_(1)^(2)+1/2m_(2)^(2)v_(2)^(2)` |
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Answer» Correct Answer - C Total initial energy =`1/2(m_(1)u_(1)^(2) + 1/2(m_(2)u_(2)^(2)` ltbr Sinc,e after collision one particle absorb energy `epsilon`. `therefore` Total final energy = `1/2(m_(1)v_(1)^(2) + 1/2m_(2)(v_(1)^(2) + epsilon` From conservation of energy `1/2m_(1)u_(1)^(2) + 1/2m_(2)u_(2)^(2) = 1/2(m_(1)v_(1)^(2) + 1/2(m_(2)v_(2)^(2) + epsilon` `rArr 1/2(m_(1)u_(1)^(2) + 1/2(m_(2)u_(2)^(2) -epsilon = 1/2m_(1)v_(1)^(2) + 1/2m_(2)v_(2)^(2)` |
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| 242. |
Two particles of masses `m_(1)` and `m_(2)` and velocities `u_(1)` and `alphau_(1)(alpha!=0)` make an elastic head on collision. If the initial kinetic energies of the two particles are equal and `m_(1)` comes to rest after collision, thenA. `u_(1)/u_(2)=sqrt(2)+1`B. `u_(1)/u_(2)=sqrt(2)-1`C. `m_(2)/m_(1)=3+2sqrt(2)`D. `m_2/m_1=3-2sqrt2` |
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Answer» Correct Answer - A::C::D Since both masses have equal kinetic energy, so `1/2m_(1)u_(1)^(2)=1/2m_(2)(alphau_(1))^(2)` `impliesm_(2)=(m_(1))/(alpha^(2))` By law of conservation of momentum `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)` `m_(1)u_(1)+(m_(1))/(alpha^(2))(alphau_(1))=0+((m_(1))/(alpha^(2)))v_(2)` `v_(2)=alphau_(1)(1+alpha)` Further by law of conservation of energy, we have `1/2m_(1)u_(1)^(2)+1/2((m_(1))/(alpha^(2)))(alphau_(1))^(2)=0+1/2((m_(1))/(alpha^(2)))v_(2)^(2)` `m_(1)u_(1)^(2)=1/2((m_(1))/(alpha^(2))) v_(2)^(2)` `v_(2)=sqrt(2)(alphau_(1))` From Eqs ii and iii we get `(v_(2))/(alphau_(1))=sqrt(2)=1+alphaimpliessqrt(2)-1` `(u_(2))/(u_(1))=sqrt(2)-1` `(u_(1))/(u_(2))=sqrt(2)+1`[on rotationaliting] `(m_(2))/(m_(1))=1/(alpha^(2))=(1/(sqrt(2)-1))^(2)=(sqrt(2)+1)^(2)` `(m_(2))/(m_(1))+3+2sqrt(2)` |
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| 243. |
A space shuttle while travelling at a speed of 4000 km/h with respect to the earth disconnects and ejects a module backward, weighing one sixth of the resudual part. If the shuttle ejects the disconnected module at a speed of 100 km/h with respect to the state of the shuttle before the ejection, find the final velocity of the shuttle. |
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Answer» Suppose the mass of het shuttle including the module is M. The mass of the module will be M/6. The total linear momentum before disconnection `=M(4000 km/h)` The velocity of the ejected module with respect to the earth =it velocilty with respect to the shutle +the velocity of the shuttle with respect to the earth `=-100km/h+4000km/h=3900km/h` If the fiN/Al velocity of te shuttle is V then the total fiN/Al liner momentum `=(5M)/6V+M/6xx3900 km/h` By the principle of conservation of linear mometum, `M(4000 km/h)=(5M)/6V+M/6xx3900km/h` or `V=4020km/h` |
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| 244. |
Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls while in airA. depends on the direction of the motion of the ballsB. depends on the masses of the two ballsC. depends on the speeds of the two ballsD. is equla to g |
| Answer» Correct Answer - D | |
| 245. |
Two balls having masses m and 2m are fastened to two light strings of same length l figure. The other ends of the strings ar fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. a. Find velocities of the balls just after their collision. b. How high will the balls rise after the collision. |
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Answer» Let the velocilty of m reaching at lower end `=v_1` From work energy principle `:.(1/2)xxmv_1^2-(1/2)xxm(0)^2=mgl` `rarr v_1=sqrt(2gl)` similarly velocity of heavy block will be ltbr. `v_2=sqt(2gl)` `:. V_1=v_2=u(say)` Let fiN/A,l velocity of m and 2m are `v_1 and v_2` respectively. According to law of conservation of momenum `rarr mxxu-2mu=mv_1+2mv_2` `rarr v_1+2v_2=-u` `Again v_1-v-2=[(u-v)]=-2u` Substracting `3v_2=u` `rarr v_2=u/3=sqrt(2gl)/3` substituting in ii ltbr. `v_1-v_2=-2u` `rarr v_1=-2u+v_2` `=-2u+(u/3)` `=(-5)/3u=(-5)/3xxsqrt(2gl)` `(sqrt(50gl))/3` b. putting the work energy principle (1/2)xx2mxx(0)^2-(1/2)xx2mxx(v_2)^2` `=-2mxxgxxh` `[hrarr` height reached by heavy ball]` `rarr h=(L/9)` similarly `(1/2)xxmx(0)^2-(1/2)xxmv_1^2=mxxgxxh_2` [height reached by small ball] ltbr. `(1/2)xx(50gL)/9 gxxh_2` `h_2=(0.25L)/9` some `h_2` is more than 2L the velocity at highest point will not be zero. |
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| 246. |
A small bucket of mass M kg is attached to a lone inextensible cord of length L m. The bucket is released from rest when the cord is in a horizontal position. At its lowest position, the bucket scoops up m kg of water and swings up to a height h. The height h in meter is :-A. `((M)/(M + m))^(2) L`B. `((M)/(M + m)) L`C. `((M + m)/(M))^(2) L`D. `((M + m)/(M)) L` |
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Answer» Correct Answer - A At the lowest position COLM : `Msqrt(2gL) = (M + m)v` ……(i) COLM : `(1)/(2) (M + m)v^(2) = (M + m)gh` …..(ii) `rArr v = sqrt(2gh) = (Msqrt2gL)/((M + m)) rArr h = ((M)/(m + M))^(2) L` |
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| 247. |
A girl throws a ball with an initial velocity v at an angle `45^@`. The ball strikes a smooth vertical wall at a horizontal distance d from the girl and after rebound returns to her hands. What is the coefficient of restitution between wall and ball ? `(v^2 gt dg)`A. `v^(2) -gd)`B. `(gd)/(v^(2)-gd)`C. `(gd)/v^(2)`D. `v^(2)/(gd)` |
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Answer» Correct Answer - B `T = (d)/(v/sqrt(2)) + (d)/(ev//sqrt(2)) = (1+1/e)sqrt(d)/v)` or `(2v//sqrt(2))/g = (1+1/e)(sqrt(2d)/v)` or `e = (gd)/(v^(2)-gd)` |
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| 248. |
A smooth sphere is moving on a horizontal surface with velocity vector `2hati+2hatj` immediately before it hits a vertical wall. The wall is parallel to `hatj` and the coefficient of restitution of the sphere and the wall is `e=1/2`. Find the velocity of the sphere after it hits the wall? |
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Answer» Correct Answer - A::B Component parallel to wall will remain unchanged and component perpendicular to wall will become e times but in opposite direction. |
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| 249. |
A sphere A of mass m, travelling with speed v, collides directly with a stationary sphere B. If A is brought to rest and B is given a speed V, find (a) the mass of B (b) the coefficient of restitution between A and B? |
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Answer» Correct Answer - A::B (a) `p_i=p_f` `:. mv=MV` `impliesM=(mv)/(V)` (b) `e=(RVOS)/(RVOA)=V/v` |
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| 250. |
A ball is projected from a given point with velocity u at some angle with the horizontal and after hitting a vertical wall returns to the same point. Show that the distance of the point from the wall must be less than `(eu^2)/((1+e)g)`, where e is the coefficient of restitution. |
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Answer» `(x)/(ucosalpha)+(x)/(eucosalpha)=T=(2usinalpha)/(g)` or `x=(eu^2sin2alpha)/((1+e)g)impliesx_(max)=(eu^2)/((1+e)g)` at `2alpha=90^@` |
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