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Q.
In each situation of column-I, a system involving two bodies is given. All strings and pulleys are light and friction is absent everywhere. Initially each body of every system is at rest. Consider the system in all situation of column I from rest till any collision occurs. Then match the statements in column-I with the corresponding results in column-II.
Column – I | Column – II | ||
A) | The block plus wedge system is placed over smooth horizontal surface. After the system is released from rest, the centre of mass of system. | P) | Shifts towards right and also downwards |
B) | The string connecting both the blocks of mass m is horizontal. Left block is placed over smooth horizontal table as shown. After the two block system is released from rest, the centre of mass of system | Q) | Shifts only downwards |
C) | The block and monkey have same mass. The monkey starts climbing up the rope. After the monkey starts climbing up, the centre of mass of monkey + block system. | R) | Shifts only upwards |
D) | Both block of mass m are initially at rest. The left block is given initial velocity downwards. Then, the centre of mass of two block system afterwards. | S) | Does not shift |
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a
A – Q, B – P, C – R, D – S
b
A – P, B – R, C – Q, D – R
c
A – P, B – S, C – Q, D – S
d
A – S, B – R, C – Q, D – P
answer is A.
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Detailed Solution
A) Initial velocity of centre of mass of given system is zero and net external force is in vertical direction. Since there is shift of mass downward, the centre of mass has only downward shift.
B) Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts a force on string which has a horizontal component towards right. Hence centre of mass of system has a rightward shift.
C) Both block and monkey moves up, hence centre of mass of given system shifts vertically upwards.
D) Net external force on given system is zero. Hence centre of mass of given system remains at rest