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1.

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.

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A.
1 : 4
B.
2 : 1
C.
2 : 3
D.
2 : 5
Explanation. Give a better Explanation

Let the forces be F1, and F2.
F1 = m1 a1 = 3 x 2 = 6N

F2 = m2 a2 = 1 x 9 = 9N

F1 : F2 = 6:9 = 2:3

Thus ratio of forces is 2: 3



2.

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

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A.
2 N
B.
20 N
C.
200 N
D.
400 N
Explanation. Give a better Explanation

F = \(\frac{mv ^2}{r}\)

= \(mω^2r\)

(Since V = ωr)

F = 10\(^{-2}\) x 100\(^2\) x 0.2

F =20 N



3.

A car of mass 800kg attains a speed of 25m/s in 20 secs. The power developed in the engine is

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A.
1 . 25 x 104 W
B.
2 . 50 x 104 W
C.
1 . 25 x 106W
D.
2 . 50 x 106 W
Explanation. Give a better Explanation

Power = Force x \(\frac{distance}{time}\) = \(\frac{m(v - u)}{t} \times v\)

\(\frac{800(25 - 0)}{20} \times 25\)

= 2.5 x 10\(^{4}\) watts



4.

A 3000W electric cooker is to be used on a 200V mains circuit. Which of the fuses below can be used safely with the cooker?

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A.
2A
B.
5A
C.
10A
D.
20A
Explanation. Give a better Explanation

The maximum current drawn by the cooker is given by;
I = \(\frac{P}{V}\)
∴ I = \(\frac{\text{3000}}{200}\)
I = 15A
The fuse rating = I \(\times\) 125% (the fuse is rated at 125% of the normal operating current)

Hence, the fuse rating = 15A \(\times\) (\(\frac{125}{100}\)) = 18.75amps

Therefore, the best fuse is 20A.



5.

At a fixed point below a liquid surface, the pressure downwards is P1 and pressure upwards is P2. It can be deduced that

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A.
P1 > P2
B.
P1 P2
C.
P1 < P2
D.
P1 = P2
Explanation. Give a better Explanation

As the temperature increases, the molecules in the gas move faster, impacting the gas's container more frequently and exerting a greater force. This increases the pressure. And similarly As the pressure goes up, the temperature also goes up. So the temperature and the pressure is directly proportional to each other. At a fixed point, temperature is kept constant. Hence, at this point, it can be deduced that \(P_{1} = P_{2}\)



6.

Dry
Day 1 - 30°C | Day 2 - 29°C | Day 3 - 25°C

Wet
Day 1 - 22°C | Day 2 - 22°C | Day 3 - 21°C

The reading above are for three consecutive days from a wet and dry bulb hygrometer. It can be concluded that the relative humidity for the three days

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A.
decreased steadily
B.
remained unchanged
C.
increased steadily
D.
was least on Day 1
Explanation. Give a better Explanation
  1 2 3
Dry 30°C 29°C 25°C
Wet 22°C 22°C

21°C 

Diff

8°C 7°C 4°C

\(Relative Humidity \propto \frac{1}{temp}\)

Hence, the relative humidity increases steadily



7.

One end of a long wire is fixed while a vibrator is attached to the other end. When the vibrator is energized, the types of waves generated in the wire are

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A.
stationary and longitudinal
B.
stationary and transverse
C.
progressive and transverse
D.
progressive and longitudinal
Explanation. Give a better Explanation

The vibrator is fixed at the end of the rope, hence, it is stationary. Also, the wave in this case, as well as the vibration are in the same direction, hence, it is longitudinal.



8.

In a thermonuclear reaction, the total initial mass is 5.02 x \(10^{-{2}{7}}\)kg and the total final mass is 5.01 x \(10^{-{2}{7}}\)kg. The energy released in the process is
[c = 3.0 x \(10^8\) ms-1]

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A.
9.0 x 10-13J
B.
9.0 x 10-12J
C.
9.0 x 10-11J
D.
9.0 x 10-10J
Explanation. Give a better Explanation

In thermonuclear reactions,
Energy = mC2
m = Change in Mass
C = Speed of light
∴ 5.02 x 10-27 - 5.01 x 10-27 x [3 x 108]2
0.01 x 10-27 x 9 x 1016
E = 9.0 x 10-13J



9.
A gas with initial volume of 2 X 10-6m3 is allowed to expand to six times its initial value at constant pressure of 2 x 105Nm-2. The work done is
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A.
2.0J
B.
4. 0 J
C.
12.0 J
D.
2.4 J
Explanation. Give a better Explanation
P = F/A => F = PA = PV/L
W = FD = (PV/L) x L = PV
= 2 x 105
(2 x 10-6 x 6)
= 2 x 105 x 1.2 x 10-5
= 2.4 J


10.
A cell of internal resistance r supplies current to a 6.0 resistor and its efficiency is 75%. Find the value of r.
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A.
4.5
B.
1.0
C.
8.0
D.
2.0
Explanation. Give a better Explanation
Efficiency = power out put/power input * 100/1
= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2
i.e r = 2


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