Physics questions and answers

Physics Questions and Answers

If you want to learn more about the nature and properties of matter and energy or you're simply preparing for a Physics exam, these Physics past questions and answers are ideal for you.

1,591.

A piece of copper of mass 20g at a temperature of 110°C was dropped into a mixture of ice and water at 0°C. If the final steady temperature of the mixture is 0°C . Calculate the amount of ice that melted [Specific heat capacity of copper = 0.4 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 330Jg\(^{-1}\)]

A.

0.37g

B.

0.60g

C.

2.40g

D.

2.70g

Correct answer is D

Mice = \(\frac{MC}{\theta}\)

= \(\frac{20 \times 0.4 \times 110}{330} 

Heat lost by copper = Heat gained by ice

m\(_{copper}\) c \(\theta\) = m\(_{ice}\) L

20 x 0.4 x (110 - 0) = m\(_{ice}\) x 330

m\(_{ice}\) = \(\frac{20 \times 0.4 \times 110}{330}\)

= \(\frac{8}{3}\)

= 2.67 g

= 2.70g (to 1 d.p)

1,592.

In the formation of sea breeze. wind blows from

A.

sky to land

B.

sea to sky

C.

land to sea

D.

sea to land

Correct answer is D

A sea breeze describes a wind that blows from the ocean (sea) inland towards land.

1,593.

The thermopile is a device for detecting

A.

radioactive radiations

B.

radiant energy

C.

x-rays

D.

the presence of electrons

Correct answer is B

No explanation has been provided for this answer.

1,594.

A metal sheet of area 100cm\(^2\) was heated through 70°C. Calculate its new area if the linear expansivity of the metal is 0.000017K\(^{-1}\).

A.

100.06 cm2

B.

100.12cm2

C.

100.24cm2

D.

100.36cm2

Correct answer is C

Formula: A\(_2\) = A\(_1\)(1 + \(\beta \theta\))

where \(\beta\) = The area expansivity

\(\theta\) = Change in temperature.

\(\beta\) = 2\(\alpha\)

where \(\alpha\) = linear expansivity of the body.

⇒ \(\beta\) = 2 x 0.000017 

= 0.000034 K\(^{-1}\)

\(\therefore\) A\(_2\) = 100 (1 + (0.000034 x 70))

= 100(1 + 0.00238)

= 100(1.00238)

= 100.238 cm\(^2\)

= 100.24 cm\(^2\) (to 2 d.p)

1,595.

A body is pulled through a distance of 500m by a force of 20N. If the power developed is 0.4kW, calculate the time for which the force acts

A.

250.0s

B.

25.0s

C.

2.5s

D.

0.5s

Correct answer is B

Force x Distance = Power x Time

Time = \(\frac{Force \times Distance}{Power}\)

= \(\frac{20 \times 500}{0.4 \times 1000}\)

= 25.0s