WAEC Physics Past Questions & Answers - Page 290

1,446.

A metal of mass 1.5 kg was heated from 27oC to 47oC in 4 minutes by a boiling ring of 75 W rating. Calculate the specific heat capacity of the metal, (Neglect heat losses to the surrounding)

A.

2.5 X103J kg-10C-1

B.

6.0 X102J kg-10C-1

C.

2.5 x 102J kg-10C-1

D.

1.4 X102J kg-10C-1

E.

1.0 x 102J kg-10C-1

Correct answer is B

I.Vt = mcθ; C =
I.Vt/
=
75 x 4 x 60/1.5 x 20

= 6.0 X102J kg-10C-1

1,447.

Which of the following instruments is used to measure relative humidity?

A.

Hydrometer

B.

Barometer

C.

Manometer

D.

Hypsometer

E.

Hygrometer

Correct answer is E

No explanation has been provided for this answer.

1,448.

A small circular membrane is 10cm below the surface of a pool of mercury when the barometric height is 76 cm of mercury. If the density of mercury is 13600\(kgm^{-3}\), what is the pressure on the membrane in \(Nm^{-2}\)? \((g =10ms^{-2})\)

A.

1.17 x 107 Nm-2

B.

6.80 x 105Nm-2

C.

1.17x105Nm-2

D.

1.03 x 105Nm-2

E.

1.36 x104 Nm-2

Correct answer is C

Pressure at a depth for fluid with constant density such as mercury is given as 

\(p = p_{0} + \rho hg\)

where \(p_{0}\) = atmospheric pressure.

\(p_{0} = \rho hg =13600 \times \frac{76}{100} \times 10\)

= \(103,360 Nm^{-2}\)

\(p = 103,360 + (13600 \times \frac{10}{100} \times 10)\)

\(p = 103,360 + 13,600\)

\(p = 116,960 Nm^{-2} \approxeq 1.17 \times 10^{5} Nm^{-2}\)

 

1,449.

Which of the following best explains why a person suffers a more severe burn when his skin is exposed to steam than when boiling water pours on his skin?

A.

Steam is at a higher temperature than boiling water

B.

Steam possesses greater heat energy per unit mass than boiling water

C.

Steam spreads more easily over a wider area of the skin man boiling water

D.

Steam penetrates more deeply into the skin than boiling water

E.

the specific latent heat of vaporization is released in changing from boiling water to steam

Correct answer is B

No explanation has been provided for this answer.

1,450.

A gas which obeys Charles’ law exactly has a volume of 283cm3 at 10oC. What is its volume at 30oC?

A.

142cm3

B.

293cm3

C.

303cm3

D.

566cm3

E.

849cm3

Correct answer is C

Using Charles law:

V1/T1 = V2/T2;

(283 * 303) / 283 

V2= 303