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JAMB Physics Past Questions & Answers - Page 248

1,236.

Hot water is added to three times its mass of water at 10^oC and the resultant temperature is 20^oC. What is the initial temperature of the hot water?

100^oC

80^oC

50^oC

40^oC

View answer & explanation

Correct answer is C

Heat loss = heat gained

m_h x c_h x $\theta$ _h = m_c x c_c x $\theta$ _c

m_h x c_h x ( $\theta$ - 20) = m_c x c_c x (20 - 10)

c_h = c_c ; m_c = 3m_h

therefore, $\theta$ = 50^oC

1,237.

As a result of air at the top of a barometer, the height of the mercury column is 73.5cm when it should be 75.0cm; the volume of the space above the mercury is 8.0cm3. Calculate the correct barometer height when the barometer reads 74.0cm and the volume of the space above the mercury is 6.0cm3.

72.0cm

74.5cm

75.1cm

76.0cm

View answer & explanation

Correct answer is D

Using, P1V1 = P2V2

(75 - 73.5) x 8 = (X - 74) x 6

X = 76cm

1,238.

The length of a side of a metallic cube at 20°C is 5.0cm. Given that the linear expansivity of the metal is 4.0 x 10-5k^-1, find the volume of the cube at 120°C .

126.50cm³

126.25cm³

126.00cm³

125.00cm³

View answer & explanation

Correct answer is A

Recall that: Volume = L * L * L → 5 * 5 * 5 = 125

Y = $\frac{V_2 - V_1}{V_1 \times \theta}$

3 x 4 x 10-5 = $\frac{V_2 - 125}{125 \times (120 - 20)}$

V2 = 126.50cm³

1,239.

The distance between the fixed points of a centigrade thermometer is 20cm. What is the temperature when the mercury level is 4.5cm above the lower mark?

22.5^oC

29.0^oC

100.0^oC

View answer & explanation

Correct answer is A

20cm .... 100^oC

4.5cm .... $\theta$

0cm ....0^oC

$\frac{4.5 - 0}{20 - 0}$ = $\frac{0 - 0}{100 - 0}$

$\theta$ = 22.5^oC

1,240.

A cube of side 10cm and mass 0.5kg floats in a liquid with only $\frac{1}{5}$ of its height above the liquid surface. What is the relative density of the liquid?

0.125cm^-1

0.250gcm^-1

0.625cm^-1

2.500cm^-1

View answer & explanation

Correct answer is C

Height below liquid = 1 - $\frac{1}{5}$ = $\frac{4}{5}$

volume of cube = 10³cm³

volume of liquid displaced = $\frac{4}{5}$ x 1000

= 800cm³

density of liquid displaced;

= $\frac{500}{800}$

= 0.625cm^-1