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.

1,376.

In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is

A.

0.03A

B.

0.04A

C.

0.05A

D.

0.06A

Correct answer is C

From faraday's law of electrolysis

M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4

\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\
\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)

1,377.

A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2, Calculate the angle through which the ray is minimally deviated in the prism

A.

30.0\(^{\circ} \)

B.

37.2\(^{\circ} \)

C.

42.0\(^{\circ} \)

D.

48.6\(^{\circ} \)

Correct answer is B

For the equilateral glass prism A = 600

\( \frac{3}{2} \text{Sin} \frac{\left( \frac{Dm + 60}{2}\right)}{\left(\frac{60}{2}\right)}= \frac{\left( \text{ Sin }\frac{Dm + 60}{2} \right)}{0.5} \\

\implies Dm = 37.2^{\circ} \)

1,378.

A bead traveling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1 determine the work done by the friction

A.

\( 2 \times 10^{-4}J \)

B.

\( 2 \times 10^{-3}J \)

C.

\( 2 \times 10^{1}J \)

D.

\( 2 \times 10^{2}J \)

Correct answer is B

Work done by friction = Friction force x Displacement from the relation, f = uR

where u = coefficient of friction; R= Normal reaction

F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)

1,379.

If a force of 50N stretches a wire from 20m to 20.01m, what is the amount of force required to stretch the same material from 20m to 20.05m?

A.

250N

B.

200N

C.

100N

D.

50N

Correct answer is A

Original length = 20m
first new length = 20.01m

first increase in length = 20.01m - 20m = 0.01m

Second new length = 20.05m
second increase in length = 20.05 - 20m = 0.05m

From hooke's law = \( \frac{F_1}{e_1} = \frac{F_2}{e_2} \implies \frac{50}{0.01} = \frac{F_2}{0.05} \\

\implies F_2 = \frac{50 \times 0.05}{0.01} = 250N \)

1,380.

A cell whose internal resistance is 0.5Ω delivers a current of 4A to an external resistor,The loss voltage of the cell is

A.

1.250V

B.

2.000V

C.

8.000V

D.

0.125V

Correct answer is B

E = V + v

where E = emf of cell.
V = terminal p.d
v = loss voltage of the cell.

But from ohm's law V=IR ; v = Iv

\( \implies \) lost voltage = V = Ir
= 4 x 0.5
= 2.0v