JAMB Physics Past Questions & Answers - Page 22

106.

The relationship between the coefficient of linear expansion \(\alpha\) and volumetric expansion (\(\gamma\)) is-----------------

A.

\(\gamma\) = \(\alpha ^{-3}\)

B.

\(\gamma\) = \(\alpha\)

C.

3\(\alpha\)

D.

\(\gamma\) = \(\alpha ^{3}\)

Correct answer is C

 volumetric expansion (γ) = 3x Linear expansion

107.

A car starts from rest and covers a distance of 40 m in 10 s. Calculate the magnitude of its acceleration

A.

3.20 ms\(^{-2}\)

B.

0.25 ms\(^{-2}\)

C.

0.80 ms\(^{-2}\)

D.

4.00 ms\(^{-2}\)

Correct answer is C

S = ut + \(\frac{1}{2}\)at\(^2\)

40 = 0 x 10 + \(\frac{1}{2}\)a x 10\(^2\)

50a = 40

a = \(\frac{40}{50}\)

= 0.8m/s\(^2\)

108.

Tyres are treaded to?

A.

increase weight of tyres

B.

increase friction

C.

increase its longevity

D.

look good

Correct answer is B

Tire treads provide your tires the ability to grip the road safely enhancing proper traction.

Helps your vehicle to be able to accelerate smoothly and as well to be able to brake more quickly.

109.

A bar magnet is placed near and lying along the axis of a solenoid connected to a galvanometer. The pointer of the galvanometer shows no deflection when?

A.

the magnet is moved towards the stationary solenoid

B.

there is no relative motion

C.

the magnet is moved away from the stationary solenoid

D.

the solenoid is moved away from the stationary magnet

Correct answer is B

If both the coil and the magnet are stationary, then there is no change in the magnetic flux and hence no current is induced in the coil.

110.

An object 40 cm high is 30cm from the pin hole camera. If the height of the image formed is 20 cm. What is the distance of the image from the pin height?

A.

15 cm

B.

70 cm

C.

40 cm

D.

40 cm

Correct answer is A

M → \(\frac{Hi}{Ho}\) = \(\frac{Di}{Do}\)

where M is the magnification, Hi is the height of the image, Ho is the height of the object, Di is the distance from the lens to the in-focus projected image, and Do is the distance from the object to the lens.

But Di = \(\frac{Hi * Do}{Ho}\) → \(\frac{20 * 30}{40}\)

Image distance(Di) = 15cm