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JAMB Physics Past Questions & Answers - Page 34

166.

In the figure, the value of R is

A.

3 $\Omega$

B.

6 $\Omega$

C.

4 $\Omega$

D.

5 $\Omega$

View answer & explanation

Correct answer is B

R $_T$ = $\frac{V}{1} = \frac{12}{2}$ = 6 $\Omega$

R $_T$ = 3 + $\frac{R \times 6}{R + 6}$

R = 6 $\Omega$

167.

In an experiment in which molten naphthalene is allowed to cool, the cooling curve i n the figure was obtained. The temperature 80 $^o$ C is known as the

A.

cooling temperature

B.

boiling point

C.

melting point

D.

vaporization point

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

168.

The fiqure above represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40 $^o$ , then the value of W is

A.

28.0N

B.

48.0N

C.

288.0N

D.

50.0N

View answer & explanation

Correct answer is D

V.E = 6, load = 120N

Efficiency = M.A x $\frac{1}{V.R}$ x 100%

40 = $\frac{120}{effort}$ x $\frac{1}{6}$ x $\frac{100}{1}$

Effort = 50N

169.

Two divers G and H are at depths 20m and 40m, respectively below the water surface in a lake. The pressure on G is p $_1$ while the pressure on H is P $_2$ , If the atmospheric pressure is equivalent to 10m of water, then the value of P $_2$ /P $_1$ is?

A.

0.50

B.

0.60

C.

1.67

D.

2.00

View answer & explanation

Correct answer is C

Atmospheric pressure equivalence of 10m of water
: $\frac{p_2+10}{p_1+10}$

= $\frac{40+10}{20+10$

= $\frac{50}{30}$ = 1.67

170.

Heat is supplied uniformly at the rate of 100W to 1.0 x 10 $^{-2}$ kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5 $^o$ C, then the specific heat capacity of the liquid is

A.

2.0 x 10 $^2$ Jkg $^-1$ K $^-1$

B.

2.0 x 10 $^2$ Jkg $^-1$

C.

4.0 x 10 $^4$ Jkg $^-1$ K $^-1$

D.

4.0 x 10 $^4$ Jkg $^-1$

View answer & explanation

Correct answer is C

Heat energy = Electrical Energy

MC $\theta$ = power x time

5 x 1 x 10 $^{-2}$ x C = 100 x 20

C = $\frac{200}{5 \times 10^2}$

= 4 x 10 $^4$ J/kg/ $^oC$

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