Physics-

General

Easy

Question

# Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a ( ). (g is acceleration due to gravity) Normal reaction (in N) acting on block B is

- (3g + 4a)
- (3g – 4a)
- (4g + 3a)
- (4g – 3a)

## The correct answer is: (3g + 4a)

### Related Questions to study

physics-

### 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

### 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

physics-General

maths-

### The vector which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition

### The vector which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition

maths-General

Maths-

### Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is

### Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is

Maths-General

physics-

### Four identical metal butterflies are hanging from a light string of length at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle and is given by

### Four identical metal butterflies are hanging from a light string of length at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle and is given by

physics-General

physics-

### 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:

physics-General

F - Kx = mb and kx = ma

Hence m (b – a) = F – 2kx

Hence m (b – a) = F – 2kx

physics-

### 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.

physics-General

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

physics-

### 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

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

### 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

maths-

### If S and are two foci of an ellipse += 1 and P a point on it, then SP + P is equal to-

### If S and are two foci of an ellipse += 1 and P a point on it, then SP + P is equal to-

maths-General

physics-

### A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle with the horizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration 'a'. The magnitude of normal reactions exerted by sides AB and BC of carriage on the cylinder be respectively. Neglect friction everywhere. Then as the magnitude of acceleration 'a ' of the carriage is increased, pick up the correct statement:

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

and N

Hence N

and N

_{BC}– N_{AB}sin30° = ma or N_{BC}= ma + N_{AB}sin 30° .... (2)Hence N

_{AB}remains constant and N_{BC}increases with increase in a.### A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle with the horizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration 'a'. The magnitude of normal reactions exerted by sides AB and BC of carriage on the cylinder be respectively. Neglect friction everywhere. Then as the magnitude of acceleration 'a ' of the carriage is increased, pick up the correct statement:

physics-General

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

and N

Hence N

and N

_{BC}– N_{AB}sin30° = ma or N_{BC}= ma + N_{AB}sin 30° .... (2)Hence N

_{AB}remains constant and N_{BC}increases with increase in a.physics-

### In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light.

Since, at

_{2}Þ a should be same in both cases, because h and t are same in both cases as given.### In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light.

physics-General

Since, at

_{2}Þ a should be same in both cases, because h and t are same in both cases as given.physics-

### A rod of length is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the –

At any instant of time the rod makes an angle q with horizontal, the x & y coordinates of centre of rod are

Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.

Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.

### A rod of length is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the –

physics-General

At any instant of time the rod makes an angle q with horizontal, the x & y coordinates of centre of rod are

Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.

Hence the centre C moves along a circle of radius lwith centre at O. \ velocity vector of C is always directed along the tangent drawn at C to the circle of radius l whose centre lies at O.

maths-

### The orthogonal projection of on are unit vectors along three mutually perpendicular directions )

### The orthogonal projection of on are unit vectors along three mutually perpendicular directions )

maths-General

maths-

### If S and are foci and A be one and of minor axis of ellipse += 1, then area of is-

### If S and are foci and A be one and of minor axis of ellipse += 1, then area of is-

maths-General

maths-

### Q is a point on the auxiliary circle corresponding to the point P of the ellipse = 1. If T is the foot of the perpendicular dropped from the focus S. onto the tangent to the auxiliary circle at Q then the SPT is -

### Q is a point on the auxiliary circle corresponding to the point P of the ellipse = 1. If T is the foot of the perpendicular dropped from the focus S. onto the tangent to the auxiliary circle at Q then the SPT is -

maths-General

physics-

### A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ' a ' vertically downward. The tension in the string is equal to :

(Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sinq = ma cos q

Applying Newton’s law along string

Applying Newton’s law along string

### A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ' a ' vertically downward. The tension in the string is equal to :

physics-General

(Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sinq = ma cos q

Applying Newton’s law along string

Applying Newton’s law along string