# 2002 - JAMB Physics Past Questions & Answers - page 1

1
A copper cube weighs 0.25 N in air, 0.17N when completely immersed in paraffin oil and 0.15 N when completely immersed in water.The ratio of upthrust in oil to upthrust in water is
A
3 : 5
B
4 : 5
C
7 : 10
D
13 : 10
CORRECT OPTION: b
Wt. in air = 0.25 N
Wt.in water = 0.15 N
therefore lost in wt. in water = Upthrust in water
= 0.25 - 0.15
= 0.10 N.
Wt. in paraffin oil = 0.17 N.
.. loss in wt.in oil = upthrust in the oil
Ratio upthrust in oil to upthrust in water
= 0.18/0.10 = 8/10 = 4/5 Or 4 : 5
2
A wheel and axle is used to raise a load of 500 N by the application of an effort of 250 N. If the radii of the wheel and the axle are 0.4 cm and 0.1 cm respectively, the efficiency of the machine is
A
20 %
B
40 %
C
50 %
D
60 %
CORRECT OPTION: c
Efficiency = M.A/V.R X 100/1 M.A = Load/Effort = 500/250
= 2
V.R. = radius of wheel/radius of axle = 0.4/0.1 = 4.
therefore Efficiency = 2/4 x 100/1 = 50 %
3
The hydrostatic blood pressure difference between the head and the feet of a boy standing straight is 1.65 x 104 Nm-2. Find the height of the boy.
A
0.5 m
B
0.6 m
C
1.5 m
D
2.0 m
CORRECT OPTION: c
[ Density of blood = 1.1 x 103 kgm-3, g = 10ms-2]
Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity
1.6.5 x 104 = 1.1 x 103 x 10 x h
therefore h = 1.65 x 104/1.1 x 103 x 10 = 1.5m
4
A body weighing 80N stands in an elevator that is about to move. The force exerted by floor on the body as the elevator moves upward with an acceleration of 5 ms-2 is
[g = 10ms-2]
A
40 N
B
80 N
C
120 N
D
160 N
CORRECT OPTION: d
Let M = mass of elevator
m = mass of the body
g = acceleration of free fall
a = upward acceleration of the elevator (5m/ss)
when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg
However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.
Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.
Therefore M x 10 - 8 x 10 = (M + 8) 5
10M - 80 = 5M + 40
10M - 5M = 40 x 80
5M = 120 => M = 120/5 = 24kg.
Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a
= (24 x 8) x5
= 32 x 5
= 160 N
5
If the distance between two suspended masses 10kg each is tripled, the gravitational force of attraction between them is reduced by
A
one half
B
one third
C
one quarter
D
one ninth
CORRECT OPTION: d
Let the distance apart the two masses = m.
From the equation F & M1 m22
F1 α 10 x 10/ϒ2 = F α 100/ ϒ2
F2 α 10 X 10/(3ϒ)2 => F2 α 100/9ϒ2
therefore F2/F1 = 100/9ϒ2/100/ϒ2 = 100/9ϒ2 ϒ2/100 = 1/9
Therefore F 2 : F 1 = 1 : 9
6
Two force each of 10N acts on a body, one towards the north and the other towards the east.
The magnitude and the direction of the resultant force are
A
20N, 45°E
B
20N, 45°W
C
10√2N, 45°W
D
10√2N, 45°E
CORRECT OPTION: d
From Pythagoras,
R2 = 102 + 102
= 100 + 100
= 200
therefore R2 = √200 = √2 x 100
= 10√2, in the direction 45°E
7
A particle in circular motion performs 30 oscillation in 6 seconds. Its angular velocity is
A
B
C
D
CORRECT OPTION: d
Angular velocity w = 2π f.
But f = frequency no. of oscillations per second.
= 30/6 = 5Hz.
therefore W = 2π x 5
= 10π rad. s-1
8
The effect of a particle in a fluid attaining its terminal velocity is that the
A
acceleration is maximum
B
weight is equal to the retarding force
C
buoyancy force is equal to the to the viscous retardind force
D
buoyancy force is more than the weight of the fluid displaced
CORRECT OPTION: b
9
A coin place below a rectangular glass block of thickness 9cm and refractive index 1.5 is viewed vertically above the block. The apparent displacement of the coin is
A
3 cm
B
5 cm
C
6 cm
D
8 cm
CORRECT OPTION: a
Real depth = thickness of the glass block = 9cm.
Refractive index of the glass material = 1.5
But Refractive index = Real depth/Apparent depth
=> 1.5 = 9/Apparent depth
therefore Apparent depth = 9/1.5 = 6cm.
therefore Apparent displacement of the Coin = 9.6
= 3cm
10
The radiator of a motor car is cooled by
A