Physics questions and answers

Physics Questions and Answers

If you want to learn more about the nature and properties of matter and energy or you're simply preparing for a Physics exam, these Physics past questions and answers are ideal for you.

3,721.

The instantaneous value of the induced e.m.f as a function of time is ε = εo sin ωt where εo is the peak value of the e.m.f. The instantaneous value of the e.m.f., one quarter of the period is

A.

0.00

B.

εo/4

C.

εo/2

D.

εo

Correct answer is D

Period is the time to complete one cycle of 360o
=> Periodic Time T ≡ 1 cycle of 360o
T/4 = 360/4 = 90o
∴ Instantaneous e.m.f E = Eo sin ωt
= Eo sin 90
But sin 90 = 1: => E = Eo x 1
= Eo

3,723.

Three cells each of e.m.f 1.5 V and internal resistance 2.5Ω are connected as shown in the diagram above. Find the net e.m.f and the internal resistance

A.

1.5V, 7.50Ω

B.

1,5V, 0.83Ω

C.

4.5V, 7.50Ω

D.

4.5V, 0.83Ω

Correct answer is B

Since the cells are connected in parallel the net e.m.f is that of one cell, E.
However the effective internal resistance should be added as parallel resistance.
=> 1/v + 1/25 + 1/25 + 1/25 = 1.2
∴ v = 0.83Ω
∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω

3,724.

The fundamental frequency of a plucked wire under a tension of 400N is 250Hz. When the frequency is changed to 500 Hz at constant length, the tension is

A.

1600N

B.

400N

C.

160N

D.

40N

Correct answer is A

The fundamental frequency is related to tension as
F ∝ √T

=>F= K
√T

F1=F2=>250=500
√T1√T2√400√T2

∴√T2 = 500√400 = 2 x 20
250

∴ T2 = 402 = 1600N

3,725.

The ratio of the coefficient of linear expansion of two metals 1/2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e2 is 1:2, the ratio of the original lengths l1/l2 is

A.

8/3

B.

3/2

C.

2/3

D.

3/8

Correct answer is C

Ratio of their linear expansion = 1/2= 3:4.
When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ

=> e1 = 1l1Δθ = 1l1 = 1
e22l2Δθ2l22

But 1 = 3, ∴3xl1=1
244l22

=> 6l1 = 4l2
l1=4=2OR 2 : 3
l263