JAMB Physics Past Questions & Answers

1,861.

A progressive wave equation is represented by y = a sin(150 πt - πx/4). if the phase difference of a progressive wave is 45°, the value of x in the equation is

3 cm

4 cm

1 cm

2 cm

View answer & explanation

Correct answer is C

Y = sin π(150πt - πx/4)
comparing the above with the standard wave equation:
Y = A sin π ((2vt)/λ - (2x)/λ)
=> the phase difference x/4 = 2x/λ
∴ 1/4 = 2/λ => λ = 8cm
∴ 2πx/8 = 45°
since π = 180°

45° X 8 = 2 X 180° x

∴ x = 1cm

1,862.

An engine of a car of power 80 kW moves on a rough road with a velocity of 32 ms^-1. The forces required to bring it to rest is

2.56 x 10⁶N

2.50 x 10⁶N

2.80 x 10³N

2.50 x 10³N

View answer & explanation

Correct answer is D

In terms of dimension: power , P, = ML²T³
Force F₁ = MLT^-2; Vel, V, LT^-1
From the above, => P = F x V
i.e, ML²T^-3 = MLT^-2 x LT^-1
ML²T^-3
=> 80,000 = F x 32

∴ F =	80,000
	32

= 2500N
= 2.50 x 10³N

1,863.

A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At the what angle should he fire the bullet to achieve this?

68^o

53^o

30^o

45^o

View answer & explanation

Correct answer is B

Horizontal range = 3 x maxi. height
=> (U²Sin2θ) / g = (3 x U²sin²θ) / 2g

∴sin2θ =	3sin²θ
	2

=> 2sinθcosθ =	3sin²θ
	2

∴ 4sinθcosθ = 3sin²θ
=> 4cosθ = 3sinθ
∴ ^sinθ/_cosθ = ⁴/₃; => tanθ = ⁴/₃ = 1.3333
θ = tan^-1 (1.3333)
= 53^o

1,864.

An object is embedded in a block of ice, 10 cm below the plane surface. if the refractive index of the ice is 1.50, the apparent depth below the surface is?

2.50 cm

6.67 cm

7.50 cm

7.63 cm

View answer & explanation

Correct answer is B

Refraction index, n = (real depth)/apparent depth
1.5 = (10)/X
X = (10)/1.5 = 6.67 cm

1,865.

An object 4 cm high is placed 15 cm from a concave mirror of focal length 5 cm. The size of the image is?

2 cm

3 cm

4 cm

5 cm

View answer & explanation

Correct answer is A

(1)/v + (1)/u = (1)/f
(1)/v + (1)/15 = (1)/5
(1)/v = (1)/5 - (1)/15 = (3 - 1)/15 = (2)/15
∴ v = (15)/2 = 7.5 cm
magnification = (v)/u = (image size)/object size
(7.5)/15 = (X)/4
∴ image size. X = (7.5 x 4)/15 = 2 cm