1

A car of mass 800kg attains a speed of 25m/s in 20 secs. The power developed in the engine is

A

1 . 25 x 10^{4} W

B

2 . 50 x 10^{4} W

C

1 . 25 x 10^{6}W

D

2 . 50 x 10^{6} W

CORRECT OPTION:
b

Power = Force x \(\frac{distance}{time}\) = \(\frac{m(v - u)}{t} \times v\)

\(\frac{800(25 - 0)}{20} \times 25\) = 2.5 x 10^{4} watts

\(\frac{800(25 - 0)}{20} \times 25\) = 2.5 x 10

2

When the brakes in a car are applied, the frictional force force on the tyres is

A

is a disadvantage because it is in the direction of the motion of the car

B

is a disadvantage because it is in the opposite direction of the motion of the car

C

is an advantage because it is in the direction of the motion of the car

D

is an advantage because it is in the opposite direction of the motion of the car

CORRECT OPTION:
d

3

If the stress on a wire is 10^{7}NM^{-2} and the wire is stretched from its original length of 10.00 cm to 10.05 cm. The young's modulus of the wire is

A

5 . 0 x 10^{4} Nm^{-2}

B

5 . 0 x 10^{5} Nm^{-2}

C

2 . 0 x 10^{8} Nm^{-2}

D

2 . 0 x 10^{9} Nm^{-2}

CORRECT OPTION:
d

Y = stress/strain, but stress = 10^{7}Nm^{-2} (given).

Strain = e/l_{0}. = (10.05 - 10.00)/10.00 = 0.005

Thus Y = 10^{7}/0.005 = 2 . 0 x 10^{9}Nm^{-2}

Strain = e/l

Thus Y = 10

4

A solid weigh 10 .00 N in air, 6 N when fully immersed in water and 7 . 0 N when fully immersed in a liquid X. Calculate the relative density of the liquid, X.

A

5/3

B

4/3

C

3/4

D

7/10

CORRECT OPTION:
c

R.D = loss in weight in liquid X loss in weight in water.

R.D = (10 - 7)/(10 - 6) = 3/4

R.D = (10 - 7)/(10 - 6) = 3/4

5

When temperature of a liquid increases, its surface tension

A

Decreases

B

Increases

C

Remain Constant

D

Increases then decreases

CORRECT OPTION:
a

6

A gas at a volume of V_{0} in a container at pressure P_{0} is compressed to one-fifth of its volume. What will be its pressure if the magnitude of its original temperature T is constant?

A

P_{0}/5

B

4P_{0}/5

C

P_{0}

D

5P_{0}

CORRECT OPTION:
d

At constant T, P_{1} V_{1} = P_{2} V_{2}.

Let P_{1} = P_{0} and V_{1} = V_{0}, then solve accordingly.

Let P

7

A piece of substance of specific head capacity 450JKg^{-1}K^{-1} falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms^{-2}]

A

2/9°C

B

4/9°C

C

9/4°C

D

9/2°C

CORRECT OPTION:
b

Energy at the height = mgh

This energy is converted to heat energy - mcΔt

∴ From the principle of energy conversion,

mcΔt = mgh

m x 450 x Δt = m x 10 x 20

∴ Δt = m x 10 x 20/(m x 450)

= 200/450

= 4/9°C

This energy is converted to heat energy - mcΔt

∴ From the principle of energy conversion,

mcΔt = mgh

m x 450 x Δt = m x 10 x 20

∴ Δt = m x 10 x 20/(m x 450)

= 200/450

= 4/9°C

8

I. A liquid boils when its saturated vapour pressure is equal to the external pressure

II. Dissolved substances in pure water lead to increase in the boiling point.

III. When the external pressure is increased, the boiling point increases.

IV. Dissolved substances in pure water decrease the boiling point.

Which combination of the above are peculiarities of the boiling point of a liquid?

II. Dissolved substances in pure water lead to increase in the boiling point.

III. When the external pressure is increased, the boiling point increases.

IV. Dissolved substances in pure water decrease the boiling point.

Which combination of the above are peculiarities of the boiling point of a liquid?

A

I, II and III

B

I, II, III and IV

C

I, II and IV

D

II, III and IV

CORRECT OPTION:
a

9

The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is

A

3.0 x 10^{2}Km^{-1}

B

3.0 x 10^{3}Km^{-1}

C

5.0 x 10^{3}Km^{-1}

D

3.0 x 10^{4}Km^{-1}

CORRECT OPTION:
b

Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)

= \(\frac{80 - 20}{0.02}\)

= \(\frac{60}{0.02}\)

= 3000Km^{-1}

= 3.0 x 10^{3}Km^{-1}

= \(\frac{80 - 20}{0.02}\)

= \(\frac{60}{0.02}\)

= 3000Km

= 3.0 x 10

10

Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss;

Specific heat capacity of copper = 400JKg^{-1}K^{-1}

Latent heat of fusion of ice = 3.3 x 10^{5}JKg^{-1}

Specific heat capacity of copper = 400JKg

Latent heat of fusion of ice = 3.3 x 10

A

8/33kg

B

33/80kg

C

80/33kg

D

33/8kg

CORRECT OPTION:
a

Heat given off by the copper = MCt

= 2 x 400 x (100 - 0)

Heat absorbed by the ice = ML

= M x 3.3 x 10^{5}

Heat lost = Heat gained

∴ M x 3.3 x 10^{5} = 2 x 400 x 100

∴ M = (2 x 400 x 100)/3.3 x 10^{5}

= 8/33Kg

= 2 x 400 x (100 - 0)

Heat absorbed by the ice = ML

= M x 3.3 x 10

Heat lost = Heat gained

∴ M x 3.3 x 10

∴ M = (2 x 400 x 100)/3.3 x 10

= 8/33Kg

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