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# A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes an angle with horizontal as shown in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero. The magnitude of acceleration of wedge is

- g tan
- g cot
- g sin
- g cos

## The correct answer is: g cot

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### Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:

F - Kx = mb and kx = ma

Hence m (b – a) = F – 2kx

Hence m (b – a) = F – 2kx

### Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:

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F - Kx = mb and kx = ma

Hence m (b – a) = F – 2kx

Hence m (b – a) = F – 2kx

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### Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx

\ acceleration of the block is = m/4kx

\ acceleration of the block is = m/4kx

### Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

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Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx

\ acceleration of the block is = m/4kx

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The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)

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cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g

and T – mg = ma .... (2)

cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g

### In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is

physics-General

The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)

and T – mg = ma .... (2)

cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g

and T – mg = ma .... (2)

cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g

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The free body diagram of cylinder is as shown. Since net acceleration of cylinder is horizontal

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