2010 - JAMB Physics Past Questions and Answers - page 5
41
A cell of e.m.f 12V and internal resistance 4Ω is connected to an external resistor of resistance 2Ω. Find the current flow.
A
5 A
B
4 A
C
3 A
D
2 A
42
A
75.0 cm
B
66.7 cm
C
33.3 cm
D
25.0 cm
correct option: c
1/R = 1/8 + 1/8 = 2/8
∴ R = 4Ω
l/l1 = R/R1
(x)/(100-x) = 4/8
4(100-x) = 8x
400-4x = 8x
400 = 8x + 4x = 12x
x = 400/12 = 33.3cm
Users' Answers & Comments∴ R = 4Ω
l/l1 = R/R1
(x)/(100-x) = 4/8
4(100-x) = 8x
400-4x = 8x
400 = 8x + 4x = 12x
x = 400/12 = 33.3cm
43
Three 4Ω resistors connected in parallel have a potential difference of 16 V applied across them. What is the total current in the circuit?
A
14 A
B
12 A
C
10 A
D
8 A
correct option: b
1/R = 1/4 + 1/4 + 1/4 = 3/4
R = 4/3
I = V/R
= 12 A
Users' Answers & CommentsR = 4/3
I = V/R
| = | 16 |
| 4/3 |
| = | 16 x 3 |
| _4 |
= 12 A
44
A
N1,920.00
B
N192.00
C
N19.20
D
N1.92
correct option: b
Energy consumed = power x time
= 200W x 24hr = 4800wh
= 4.8 kwh
But 1 kwh cost N40
∴ 4.8kwh costs 4.8 x 40
= N192
Users' Answers & Comments= 200W x 24hr = 4800wh
= 4.8 kwh
But 1 kwh cost N40
∴ 4.8kwh costs 4.8 x 40
= N192
45
A
south and north
B
north and south
C
north and north
D
south and south
correct option: c
IN magnetisation by stroking, the last touched has a pole opposite to that of the magnet used. Therefore, the poles T and V are N and N respectively
Users' Answers & Comments46
In Faradays law of electrolysis, a graph of the mass deposited against the quantity of electricity is plotted. The slope of the graph gives
A
the energy released
B
the electrochemical equivalent
C
the current flowing
D
the charge released
correct option: b
In Faradays law of electrolysis
M = It = Q
Where M is the mass of substance deposit . Q is the quantity of charge. A graph of m against Q is a straight line passing through the origin with slope equal to the electrochemical equivalent.Z
Users' Answers & CommentsM = It = Q
Where M is the mass of substance deposit . Q is the quantity of charge. A graph of m against Q is a straight line passing through the origin with slope equal to the electrochemical equivalent.Z
47
Rationalize \( \frac{2\sqrt{3}} + \sqrt{5} \sqrt{5} - \sqrt{3} \)
A
3√15 - 11
B
\(\frac{3\sqrt {15} - 11}{2}\)
C
3√15 + 11
D
\(\frac{3\sqrt {15} + 11}{2}\)
correct option: d
\(\frac{2\sqrt{3}} + \sqrt{5}}{\sqrt{5} - \sqrt{3}\) = \(\frac{\sqrt{5}} + 2\sqrt{3}}{\sqrt{5} - \sqrt{3}\) x \(\frac{\sqrt{5}} + \sqrt{3}}{\sqrt{5} + \sqrt{3}\)
\(\frac{5 + \sqrt{15} + 2\sqrt{15} + 2 \times 3}{(\sqrt{5} - \sqrt{3})^2\)
= \(\frac{5 + 3\sqrt{15} +6} {5 - 3)\)
= \(\frac{11 + 3\sqrt{15}}{2}\)
Users' Answers & Comments\(\frac{5 + \sqrt{15} + 2\sqrt{15} + 2 \times 3}{(\sqrt{5} - \sqrt{3})^2\)
= \(\frac{5 + 3\sqrt{15} +6} {5 - 3)\)
= \(\frac{11 + 3\sqrt{15}}{2}\)
48
An object is weighted at different locations on the earth. What will be the right observation?
A
Both the mass and weight are constant
B
Both the mass and weight vary
C
the mass is constant while the weight varies
D
the weight is constant while the mass varies
correct option: c
The weight of a body is the resultant of earth's gravitational pull on it. The earth is not a perfect sphere, it is more flattened at the poles and bulges at the equator. Hence an object at the pole is closer to the earth's center than the one at the equator. The weight of a body at the poles should be more than that of when weighed at the equator.
Thus weight varies from place to place while the mass of a body being the quantity of matter it contains remains the same at different points on the earth.
Users' Answers & CommentsThus weight varies from place to place while the mass of a body being the quantity of matter it contains remains the same at different points on the earth.
49
The radioisotope \( ^{235}_{92}U \) decays by emitting two alpha particles, three beta particles and a gamma ray. What is the mass and atomic number of the resulting element?
A
215 and 88
B
91 and 227
C
227 and 91
D
92 and 238
correct option: c
Users' Answers & Comments