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JAMB Physics Past Questions & Answers - Page 86

426.

Which of the following instruments helps to maintain the correct humidity and temperature of a building?

A.

A compressor

B.

A refrigerator

C.

Air blower

D.

Air conditioner

View answer & explanation

Correct answer is D

Compressor is an instrument to supply air or other gas at increased pressure
Refrigerator is an appliance to keep things cold
Air blower is used to release air
Air conditioner is used to remove heat and maintain a stable temperature in an occupied space

427.

Three 3ω resistance are connected in parallel what is the equivalent resistance?

A.

1ω

B.

9ω

C.

3ω

D.

0.33ω

View answer & explanation

Correct answer is A

Resistor connected in a parallel have an equivalence given by
$\frac{1}{\text{Reff}}$ = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$

= $\frac{1 + 1 + 1}{3}$ = $\frac{3}{3}$ = 1

Reff = 1ω

428.

A block and tackle pulley arrangement has 6 pulleys in which an effort of 50N supports a load of 200N, calculate the efficiency of the machine

A.

80.0%

B.

76.6%

C.

66.7%

D.

60.7%

View answer & explanation

Correct answer is C

Efficiency = $\frac{ma}{V.R}$ x $\frac{100}{1}$

V.R = number of pulleys = 6

ma = $\frac{\text{Load}}{\text{Effort}}$ = $\frac{200}{50}$ = 4

Efficiency = $\frac{4}{6}$ × 100 = 66.67%

429.

A cell of internal resistance 2 π supplies current to a 6 π resistor. The efficiency of the cell is

A.

12.0%

B.

25.0%

C.

33.3%

D.

75.0%

View answer & explanation

Correct answer is C

Internal resistance determine the maximum current that can be supplied

Efficiency = $\frac{r}{R}$ × 100

= $\frac{2}{6}$ × 100

= 33.3%

430.

Calculate the specific latent heat of vaporization of steam of 1.13 x 106J of heat energy is required to convert 15kg of steam to water.

A.

7.53 x 105 Jkg $^{-1}$

B.

7.53 x 10-2Jkg $^{-1}$

C.

$7.53 \times 10^{4}$ Jkg $^{-1}$

D.

7.53 x 10-3Jkg $^{-1}$

View answer & explanation

Correct answer is C

Quantity of heat = mass(m) × specific latent heat (L)
Q = mL
L = $\frac{Q}{m}$

= $\frac{1.13 \times 10^{6}}{15}$

= $7.53 \times 10^{4}$ Jkg $^{-1}$

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