1

A copper cube weighs 0.25 N in air, 0.17N when completely immersed in paraffin oil and 0.15 N when completely immersed in water.The ratio of upthrust in oil to upthrust in water is

A

3 : 5

B

4 : 5

C

7 : 10

D

13 : 10

CORRECT OPTION:
b

Wt. in air = 0.25 N

Wt.in water = 0.15 N

therefore lost in wt. in water = Upthrust in water

= 0.25 - 0.15

= 0.10 N.

Wt. in paraffin oil = 0.17 N.

.. loss in wt.in oil = upthrust in the oil

Ratio upthrust in oil to upthrust in water

= 0.18/0.10 = 8/10 = 4/5 Or 4 : 5

Wt.in water = 0.15 N

therefore lost in wt. in water = Upthrust in water

= 0.25 - 0.15

= 0.10 N.

Wt. in paraffin oil = 0.17 N.

.. loss in wt.in oil = upthrust in the oil

Ratio upthrust in oil to upthrust in water

= 0.18/0.10 = 8/10 = 4/5 Or 4 : 5

2

A wheel and axle is used to raise a load of 500 N by the application of an effort of 250 N. If the radii of the wheel and the axle are 0.4 cm and 0.1 cm respectively, the efficiency of the machine is

A

20 %

B

40 %

C

50 %

D

60 %

CORRECT OPTION:
c

Efficiency = M.A/V.R X 100/1 M.A = Load/Effort = 500/250

= 2

V.R. = radius of wheel/radius of axle = 0.4/0.1 = 4.

therefore Efficiency = 2/4 x 100/1 = 50 %

= 2

V.R. = radius of wheel/radius of axle = 0.4/0.1 = 4.

therefore Efficiency = 2/4 x 100/1 = 50 %

3

The hydrostatic blood pressure difference between the head and the feet of a boy standing straight is 1.65 x 10^{4} Nm^{-2}. Find the height of the boy.

A

0.5 m

B

0.6 m

C

1.5 m

D

2.0 m

CORRECT OPTION:
c

[ Density of blood = 1.1 x 10^{3} kgm^{-3}, g = 10ms^{-2}]

Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity

1.6.5 x 10^{4} = 1.1 x 10^{3} x 10 x h

therefore h = 1.65 x 10^{4}/1.1 x 10^{3} x 10 = 1.5m

Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity

1.6.5 x 10

therefore h = 1.65 x 10

4

A body weighing 80N stands in an elevator that is about to move. The force exerted by floor on the body as the elevator moves upward with an acceleration of 5 ms^{-2} is

[g = 10ms^{-2}]

[g = 10ms

A

40 N

B

80 N

C

120 N

D

160 N

CORRECT OPTION:
d

Let M = mass of elevator

m = mass of the body

g = acceleration of free fall

a = upward acceleration of the elevator (5m/s^{s})

when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg

However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.

Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.

Therefore M x 10 - 8 x 10 = (M + 8) 5

10M - 80 = 5M + 40

10M - 5M = 40 x 80

5M = 120 => M = 120/5 = 24kg.

Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a

= (24 x 8) x5

= 32 x 5

= 160 N

m = mass of the body

g = acceleration of free fall

a = upward acceleration of the elevator (5m/s

when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg

However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.

Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.

Therefore M x 10 - 8 x 10 = (M + 8) 5

10M - 80 = 5M + 40

10M - 5M = 40 x 80

5M = 120 => M = 120/5 = 24kg.

Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a

= (24 x 8) x5

= 32 x 5

= 160 N

5

If the distance between two suspended masses 10kg each is tripled, the gravitational force of attraction between them is reduced by

A

one half

B

one third

C

one quarter

D

one ninth

CORRECT OPTION:
d

Let the distance apart the two masses = m.

From the equation F & M_{1} m_{2}/ϒ^{2}

F_{1} α 10 x 10/ϒ^{2} = F α 100/ ϒ^{2}

F^{2} α 10 X 10/(3ϒ)^{2} => F^{2} α 100/9ϒ^{2}

therefore F^{2}/F^{1} = 100/9ϒ^{2}/100/ϒ^{2} = 100/9ϒ^{2} ϒ^{2}/100 = 1/9

Therefore F_{2} : F _{1} = 1 : 9

From the equation F & M

F

F

therefore F

Therefore F

6

Two force each of 10N acts on a body, one towards the north and the other towards the east.

The magnitude and the direction of the resultant force are

The magnitude and the direction of the resultant force are

A

20N, 45°E

B

20N, 45°W

C

10√2N, 45°W

D

10√2N, 45°E

CORRECT OPTION:
d

From Pythagoras,

R2 = 10^{2 + 102}

= 100 + 100

= 200

therefore R2 = √200 = √2 x 100

= 10√2, in the direction 45°E

R2 = 10

= 100 + 100

= 200

therefore R2 = √200 = √2 x 100

= 10√2, in the direction 45°E

7

A particle in circular motion performs 30 oscillation in 6 seconds. Its angular velocity is

A

5 rad s^{-1}

B

6 rad s^{-1}

C

5π rad s^{-1}

D

10π rad s^{-1}

CORRECT OPTION:
d

Angular velocity w = 2π f.

But f = frequency no. of oscillations per second.

= 30/6 = 5Hz.

therefore W = 2π x 5

= 10π rad. s^{-1}

But f = frequency no. of oscillations per second.

= 30/6 = 5Hz.

therefore W = 2π x 5

= 10π rad. s

8

The effect of a particle in a fluid attaining its terminal velocity is that the

A

acceleration is maximum

B

weight is equal to the retarding force

C

buoyancy force is equal to the to the viscous retardind force

D

buoyancy force is more than the weight of the fluid displaced

CORRECT OPTION:
b

9

A coin place below a rectangular glass block of thickness 9cm and refractive index 1.5 is viewed vertically above the block. The apparent displacement of the coin is

A

3 cm

B

5 cm

C

6 cm

D

8 cm

CORRECT OPTION:
a

Real depth = thickness of the glass block = 9cm.

Refractive index of the glass material = 1.5

But Refractive index = Real depth/Apparent depth

=> 1.5 = 9/Apparent depth

therefore Apparent depth = 9/1.5 = 6cm.

therefore Apparent displacement of the Coin = 9.6

= 3cm

Refractive index of the glass material = 1.5

But Refractive index = Real depth/Apparent depth

=> 1.5 = 9/Apparent depth

therefore Apparent depth = 9/1.5 = 6cm.

therefore Apparent displacement of the Coin = 9.6

= 3cm

10

The radiator of a motor car is cooled by

A

radiation

B

conduction

C

convection

D

radiation and conduction

CORRECT OPTION:
c

The function of the radiator of a motor car is to "collect" heat from the circulating water and radiate some out, but it is itself cooled by the fan adjacent to it. whose function is primarily based on convection

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