1999 - JAMB Physics Past Questions and Answers - page 5
41
A
North - North
B
North - South
C
South - North
D
South - South
correct option: b
Users' Answers & Comments42
A
10.0Ω
B
7.5Ω
C
5.0Ω
D
2.5Ω
correct option: d
From Ohms Law, E = I(R +X)
=> E = I1(R1 + X)
again, E = I2(R2 + X)
=> I1(R1 + X) = I2(R2 + X)
∴ 5(8 + X) = 7(5 + X)
40 + 5X = 35 + 7X
7X - 5X = 40 - 35
2X = 5
X = 5/2
= 2.5Ω
Users' Answers & Comments=> E = I1(R1 + X)
again, E = I2(R2 + X)
=> I1(R1 + X) = I2(R2 + X)
∴ 5(8 + X) = 7(5 + X)
40 + 5X = 35 + 7X
7X - 5X = 40 - 35
2X = 5
X = 5/2
= 2.5Ω
43
A
3.0 x 104J
B
3.0 x 102J
C
2.5 x 10-2J
D
6.0 x 10-3J
correct option: c
Energy stored in a capacitor is given by
E = 1/2CV2
But effective capacitance for the parallel arrangement is given by
C = C1 + C2
= 2 + 3
= 5μF = 5 x 10-6 x (100)2
∴E = 1/2CV2
= 1/2 x 5 x 10-6 x 104
2.9 x 10-2J
Users' Answers & CommentsE = 1/2CV2
But effective capacitance for the parallel arrangement is given by
C = C1 + C2
= 2 + 3
= 5μF = 5 x 10-6 x (100)2
∴E = 1/2CV2
= 1/2 x 5 x 10-6 x 104
2.9 x 10-2J
44
A
R2/R1
B
R1/R2
C
R1+R2/R1
D
R1+R2/R2
correct option: a
Power = IV = I2R = V2/R
In terms of e.m.f E, of a cell we have
IE = I2R = E2/R
∴P1 = E2/R; P2 = E2/R2/E2/R2
= E2/R1 x R2/E2 = R2/R1
∴P1/P2 = R2/R1
Users' Answers & CommentsIn terms of e.m.f E, of a cell we have
IE = I2R = E2/R
∴P1 = E2/R; P2 = E2/R2/E2/R2
= E2/R1 x R2/E2 = R2/R1
∴P1/P2 = R2/R1
45
A
2V
B
10V
C
14V
D
48V
correct option: b
The effective V =
√ VR2 + VL2
=
√ 62 + 82
=
√ 36 + 64
=
√ 100
10V
Users' Answers & Comments√ VR2 + VL2
=
√ 62 + 82
=
√ 36 + 64
=
√ 100
10V
46
The inner diameter of a test tube can be measured accurately using a
A
micrometer screw guage
B
pair of dividers
C
metre rule
D
pair of vernier calipers
correct option: d
Users' Answers & Comments47
Two bodies have masses in the ratio 3:1. 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 option: c
Let the forces be F1, and F2.
F1 = m1 a1 = 3 x 2 = 6N
F2 = m2 a2 = 1 x 9 = 9N
F1F2 = 69 = 23
Thus ratio of forces is 2: 3
Users' Answers & CommentsF1 = m1 a1 = 3 x 2 = 6N
F2 = m2 a2 = 1 x 9 = 9N
F1F2 = 69 = 23
Thus ratio of forces is 2: 3
48
Particles of mass 10-2kg is fixed to the tip of a fan blade which rotates with angular velocity of 100rad-1. If the radius of the blade is 0.2m, the centripetal force is
A
2 N
B
20 N
C
200 N
D
400 N
49
a lead bullet of mass 0.05kg is fired with a velocity of 200ms-1 into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is
A
50 J
B
100 J
C
150 J
D
200 J
correct option: a
From principle of conservation of linear momentum,
(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
Thus V = 10m/s.
K.E = 1/2 (0.05 + 0.95) x 102
K.E = 1/2 (1 x 100) = 50 J
Users' Answers & Comments(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
Thus V = 10m/s.
K.E = 1/2 (0.05 + 0.95) x 102
K.E = 1/2 (1 x 100) = 50 J
50
A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms-1 from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is
(take g = 10ms-2)
(take g = 10ms-2)
A
5 J
B
10 J
C
15 J
D
20 J
correct option: c
Height h = 1/2 (U2/g) = 102/20 = 5m
Energy before hitting ground is = P.E at the height of 15m.
P.E = mgh = 0.1 x 10 x 15 = 15 J
Users' Answers & CommentsEnergy before hitting ground is = P.E at the height of 15m.
P.E = mgh = 0.1 x 10 x 15 = 15 J