JAMB Physics Past Questions & Answers - Page 28

136.

Two bodies have masses in the ratio3:1. They experience forces which impart to them, acceleration in the ratio 2:9 respectively. Find the ratio of forces the masses experienced

A.

1:4

B.

2:1

C.

2:3

D.

2:5

Correct answer is C

Given \(\frac{m_{1}}{m_{2}} = \frac{3}{1}\) \(\Rightarrow\) m\(_{1}\) = 3m\(_{2}\)

\(\frac{a_{1}}{a_{2}} = \frac{2}{9}\) = a\(_{1}\) = \(\frac{2a_{2}}{9}\)

\(\frac{F_{1}}{F_{2}}\) = ?

Since F = ma

\(\Rightarrow\)  \(\frac{F_{1}}{F_{2}} = \frac{m_{1}a_{1}}{m_{2}a_{2}}\) = \(\frac{(3 \not m_{2})(2 \not a_{2})}{\not m_{2} a_{2} \not a_{3}}\)

\(\frac{F_{1}}{F_{2}} = \frac{2}{3}\) \(\Rightarrow\) 2:3

137.

The inner diameter of a test tube can be measured accurately using a?

A.

micrometer screw gauge

B.

pair of dividers

C.

meter rule

D.

pair of vernier calipers

Correct answer is D

Vernier calipers are also used to measure the internal and external diameters of a tube

138.

A certain radioactive source emits radiation that was found to be deflected by both magnetic and electric fields. The radiation is?

A.

beta rays

B.

gamma rays

C.

x-rays

D.

ultra-violet rays

Correct answer is A

 Both the electric field and magnetic field affect the β-particles.

139.

If the fraction of the atoms of a radioactive material left after 120 years is 1/64, what is the half-life of the material?

A.

24 years

B.

20 years

C.

10 years

D.

2 years

Correct answer is B

Given amount remaining = \(\frac{1}{64}\)

= \(\frac{1}{64} = \frac{1}{(2)^{6}}\)

\(\therefore\) = \(\frac{120 \text {years}}{6}\) = 20 years

\(\Rightarrow\) T\(\frac{1}{2}\) = 20 years

140.

If the frequency of an emitted x-ray is 1.6 x 10\(^{16}\)Hz, the accelerating potential is?

 [e = 1.6 x 10\(^{-19}\)J, h = 6.63 x 10-\(^{34}\)Js]

A.

6630.0V

B.

663.0V

C.

66.3V

D.

6.6V

Correct answer is C

Given f = 1.6 x10\(^{16}\)H\(_{3}\), h = 6.63 x 10\(^{-39}\)Js

e = 1.6 x 10\(^{-19}\)C, V = ?

V = \(\frac{w}{e} = \frac{hf}{e}\) = \(\frac{1.6 \times 10^{16} \times 6.63 \times 10^{34}}{1.6 \times 10^{-19}}\)

= 66.3V