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.

1,381.

Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4

A.

600

B.

290

C.

990

D.

900

Correct answer is B

From the relation refractive index = Sin \( \left(\frac{A+Dm}{2}\right) \)

Where A = refractive angle = 600, Sin\( \left(\frac{A}{2}\right) \)

\( \implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \\

= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \\

= \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\\

\text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \\

\text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \\

= 44^0 12^1 \\

60^0 + D_m = 88^0 24^1 \\
D_m = 88^0 24^1 - 60^0 \\

= 28^0 24 \\
29^0 \)

1,382.

Am elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. The original length of the string is

A.

12cm

B.

20cm

C.

18cm

D.

15cm

Correct answer is C

For Hooke's law F = Ke

\( \implies F/e = K \\

f1/e1=f2/e2 \\

\text{Let the original length} = t_0 \\

\text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \\
\text{if f1} = 40N \text{f2} = 60N \\
\text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \\
\implies 40(45 - t_0) = 60(36 - t_0) \\
\text{therefore } 1800 - 40t_0 \\

2160 - 60t_0 \\
60t_0 - 40t_0 = 2160 - 1800 \\
20t_0 = 360 \\
t_0 = 18cm \)

1,383.

If two inductors of inductances 3H and 6H are arranged in series, the total inductance is

A.

9.0H

B.

18.0H

C.

0.5H

D.

2.0H

Correct answer is A

l =l1 + l2 = 3 + 6 = 9H

1,384.

A force of 200N acts between two objects at a certain distance apart. The value of the force when the distance is halved is

A.

400N

B.

200N

C.

100N

D.

800N

Correct answer is D

let the masses of the two objects be M1 and M2
and their distance apart = r

Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)

for a new distance of r/2, we have a force of

F2 = \( \frac{GM2M2}{r/2} \\

\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\
F2 \times (r/2)^2 = GM1M1 \\

\text{Since } GM1M1 = GM1M1 \\

\implies F1 \times r^2 = F2 \times (r/2)^2 \\

\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\

F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)

1,385.

The velocity of sound in air will be doubled if it's absolute temperature is

A.

Quadrupled

B.

Constant

C.

Halved

D.

Doubled

Correct answer is A

In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin.

That V \( \propto \sqrt{T} \implies V^2 \propto T. \\

\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2} \\

\text{Thus Let } V_1 = 4m/s \\
T_1 = 10K \\
\text{Therefore } V_2 = 2V_1 = 8m/s \\

\implies \frac{4^2}{10} = \frac{8^2}{T_2} \\

T_2 = \frac{64 \times 10}{16} = 40K\\

T_2 = 4T_1 \)

Thus when the velocity of sound in air is doubled, it's absolute temperature will be quadrupled.