Physics Questions & Answers - Free Online Practice

2,896.

When 100g of liquids L1 at 78°C was mixed with Xg of liquid L2 at 50°C, the final temperature was 66°C. Given that the specific heat capacity of L2is half that of L1, find x

50g

100g

150g

200g

View answer & explanation

Correct answer is C

Heat lost by L1 = heat gained by L2

cL1 = 2cL2

100 x 2cL2 x (78 - 66)

= xg x cL2 x (66 - 50)

xg = 150g

2,897.

A Motor Tyre is inflated to pressure of 2.0 x 10⁵Nm^-2 when the temperature of air is 27^oC. What will be the pressure in it at 87^oC assuming that the volume of the Tyre does not change?

2.6 x 10⁵Nm^-2

2.4 x 10⁵Nm^-2

2.2 x 10⁵Nm^-2

1.3 x 10⁵Nm^-2

View answer & explanation

Correct answer is B

\(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\)

\(\frac{2 \times 10^5}{27 + 273}\) = \(\frac{p_2}{87 + 273}\)

P₂ = 2.4 x 10⁵Nm^-2

2,898.

The melting point of naphthalene is 78^oC. What is this temperature in kelvin

100k

351k

378k

444k

View answer & explanation

Correct answer is B

T = 78 + 273

= 351k

2,899.

If a solid X floats in liquids P of relative density 2.0 and in liquid Q of relative density 1.5, it can be inferred that the

weight of P displaced is greater than that of Q

weight of P displaced is less than that of Q

volume of P displaced is greater than that of Q

volume of P displaced is less than that of Q

View answer & explanation

Correct answer is A

R.d \(\alpha\) weight of liquid displaced

\(\frac{\text {R.d of P}}{\text {R.d of Q}}\) = \(\frac{\text{weight of p displaced}}{\text{weight of Q displaced}}\)

\(\frac{2}{1.5}\) = \(\frac{\text {weight of p displaced}}{\text {weight of Q displaced}}\)

2,900.

The mass of a specific gravity bottle is 15.2g when it is empty. it is 24.8g when filled with kerosene and 27.2g when filled with distilled water. Calculate the relative density of kerosene

1.25

1.10

0.90

0.80

View answer & explanation

Correct answer is D

Masses of kerosene = 24.8 - 15.2

= 9.6

mass of H₂O = 27.2 - 15.2

= 12

R.d = \(\frac{9.6}{12}\) = 0.8