1

The unit of moment of a couple can be expressed in?

A

N m^{-1}

B

N m^{-2}

C

N m

D

N m^{2}

CORRECT OPTION:
c

The unit of a couple is Newton-meter (N M)

2

A gramophone record takes 5s to reach its constant angular velocity of 4\(\pi\) rads^{-1} from rest. Find its constant angular acceleration.

A

0.4\(\pi\) rads^{-2}

B

0.8\(\pi\) rads^{-2}

C

1.3\(\pi\) rads^{-2}

D

20.0\(\pi\) rads^{-2}

CORRECT OPTION:
b

From the radiation:

V = +at, and if it starts from N & T, U = 0

And for angular motion, W = O + at

4R = aχ5

∴ a = (4a)/5 = 0. 8\(\pi\) rads^{-2}

V = +at, and if it starts from N & T, U = 0

And for angular motion, W = O + at

4R = aχ5

∴ a = (4a)/5 = 0. 8\(\pi\) rads

3

A car of mass 1500 kg goes round a circular curve of radius 50m at a speed of 40 ms^{-1}. The magnitude of the centripetal force on the car is ?

A

4.8 x 10^{4} N

B

4.8 x 10^{3 } N

C

1.2 x 10^{3 } N

D

1.2 x 10^{2} N

CORRECT OPTION:
a

Centipetal force of the car = ma = (MV)^{2}/r

= (1500 x 40 x40)/50

= 48000 = 4.8 x 10^{4} N

= (1500 x 40 x40)/50

= 48000 = 4.8 x 10

4

Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them?

[G = 6.67 X 10^{-11} Nm^{2} Kg^{2}]

[G = 6.67 X 10

A

3.50 x 10^{-10}N

B

3.71 x 10^{-8} N

C

3.57 x 10^{-2}N

D

4.00 x 10^{2}N

CORRECT OPTION:
b

The force of attraction between the two spheres is given by F (GM_{1} M_{2})/r^{2} = (6.67X 10^{-11} X5X10)/(0.3)^{2}

= 3.71 X 10^{-8}N.

= 3.71 X 10

5

A machine has a velocity ratio of 4. if it requires 800 N to overcome a load of 1600 N,what is the efficiency of the machine?

A

60%

B

50%

C

40%

D

2%

CORRECT OPTION:
b

Efficiency = (M.A)/V.R X (100)/1

But M.A = (Load)/Effort = (1600)/800 = 2

Efficiency = (2)/4 x (100)/1 = 50%

But M.A = (Load)/Effort = (1600)/800 = 2

Efficiency = (2)/4 x (100)/1 = 50%

6

A reservoir 500 m deep is filled with a fluid of density 850 kgm^{-3}. If the atmospheric pressure is 1.05 x 10^{5} Nm^{-2}, the pressure at the bottom of the reservoir is.

[g = 10ms^{-2}]

[g = 10ms

A

4.72 x 10^{6} Nm^{-2}

B

4.36 x 10^{6} Nm^{-2}

C

4.28 x 10^{6} Nm^{-2}

D

4.25 x 10^{6} Nm^{-2}

CORRECT OPTION:
d

In general, pressure varies with depth in the relation p = lgh

= 850 x 10 x 500

= 4250000

= 4.25 x 10^{6} NM^{-2}

= 850 x 10 x 500

= 4250000

= 4.25 x 10

7

A piece of iron weights 250 N in air and 200 N in a liquid of density 1000 kgm^{-3}. the volume of the iron is?

A

5.0 x 10^{-2} ms^{-2}

B

4.5 x 10^{--2} ms^{-3}

C

2.5 x 10^{-2} ms^{-8}

D

2.0 x 10^{-2} ms^{-3}

CORRECT OPTION:
c

With in the air = 250N

wt. in the liquid = 200N

loss in wt. in the liquid = 250 - 200 = 50N

∴ Wt. the liquid of equal vol. of the iron = 50N

∴If the density of the liquid is 1000 lgm^{-3}

then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m^{3} = 2.5 x 10^{-2} m^{8}.

wt. in the liquid = 200N

loss in wt. in the liquid = 250 - 200 = 50N

∴ Wt. the liquid of equal vol. of the iron = 50N

∴If the density of the liquid is 1000 lgm

then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m

8

A clinical thermometer is different from other mercury in glass thermometers owing to

A

the constriction on its stem

B

the grade of mercury used in it

C

its long stem

D

its wide range of temperatures

CORRECT OPTION:
a

9

I. Use a liquid with a high melting point.

II. use a liquid of high volume expansivity.

III. use a capillary tube of large diameter.

which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?

II. use a liquid of high volume expansivity.

III. use a capillary tube of large diameter.

which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?

A

I only

B

II. only

C

I and II only

D

II and III only

CORRECT OPTION:
b

Using the liquid of high volume expansively II only

10

2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water?

[specific heat capacity of water = 4200 jkg^{-1}k^{-1}]

[specific heat capacity of water = 4200 jkg

A

30°

B

25°

C

15°

D

10°

CORRECT OPTION:
d

Electricity heat energy supplies by the heater = lvt = 3.5 x 200 x (2 x 60) = 84000j.

and this heat is supplied to the 2 kg of the water MC ∆ t = 84000

2 x 4200 x ∆ t = 84000

∴ ∆t = (84000)/8400 = 10°c

∴ increase in temp. of water = 10°c

and this heat is supplied to the 2 kg of the water MC ∆ t = 84000

2 x 4200 x ∆ t = 84000

∴ ∆t = (84000)/8400 = 10°c

∴ increase in temp. of water = 10°c

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