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
Physics-General
- g cos
- g cot
- g sin
- g tan
Answer:The correct answer is: g cot
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physics-
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
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
physics-General
physics-
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 A.
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 A.
physics-General
maths-
The condition that the line 
x + my = n may be a normal to the ellipse 
+ 
= 1 is
The condition that the line x + my = n may be a normal to the ellipse + = 1 is
maths-General
maths-
Let L = 0 is a tangent to ellipse 
+ 
= 1 and S, 
be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from on L = 0 is
Let L = 0 is a tangent to ellipse + = 1 and S, be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from on L = 0 is
maths-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
maths-
The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is
The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is
maths-General
physics-
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft (
) is 1 kHz, then velocity of the aircraft will be
when source is fixed and observer is moving towards it
when source is moving towards observer at rest
= 900 km/hr
when source is moving towards observer at rest
= 900 km/hr
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft ( ) is 1 kHz, then velocity of the aircraft will be
physics-General
when source is fixed and observer is moving towards it
when source is moving towards observer at rest
= 900 km/hr
when source is moving towards observer at rest
= 900 km/hr
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 NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
Hence NAB remains constant and NBC increases with increase in a.
and NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
Hence NAB remains constant and NBC 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 NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
Hence NAB remains constant and NBC increases with increase in a.
and NBC – NAB sin30° = ma or NBC = ma + NAB sin 30° .... (2)
Hence NAB remains constant and NBC 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,
at2 Þ a should be same in both cases, because h and t are same in both cases as given.
at2 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, at2 Þ a should be same in both cases, because h and t are same in both cases as given.
at2 Þ a should be same in both cases, because h and t are same in both cases as given.