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

Physics-General

- 1032 km/hr
- 800 km/hr
- 1000 km/hr
- 900 km/hr

#### Answer:The correct answer is: 900 km/hrwhen source is fixed and observer is moving towards it

when source is moving towards observer at rest

= 900 km/hr

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### Related Questions to study

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#### The locus of P such that PA^{2} + PB^{2} = 10 where A = (2, 0) and B = (4, 0) is

#### The locus of P such that PA^{2} + PB^{2} = 10 where A = (2, 0) and B = (4, 0) is

maths-General

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#### 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–

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

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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 = .

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 = .

#### 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–

maths-General

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 = .

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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#### If are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

are collinear.

e = = =

e = = =

#### If are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

maths-General

are collinear.

e = = =

e = = =

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#### 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

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#### 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

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

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-

#### 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

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#### 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

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#### 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