Maths-

General

Easy

Question

# If is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan is–

## The correct answer is:

### Any point P on ellipse is (a cos , b sin )

Equation of the diameter CP is y = x

The normal to ellipse at P is

ax sec – by cosec = a2e2

Slopes of the lines CP and the normal GP are tan andtan

tan = =

=sin cos = sin 2

The greatest value of tan = .1 = .

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### 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

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when source is fixed and observer is moving towards it

when source is moving towards observer at rest

= 900 km/hr

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### 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.

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### 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

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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

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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

\ acceleration of the block is = m/4kx

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### 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

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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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### 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-

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### 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:

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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.