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.

176.

The speed of fast-moving neutrons in a nuclear reactor can be reduced by using?

A.

graphite rods

B.

concrete shield

C.

iron rods

D.

boron rods

Correct answer is A

Moderation is the process of the reduction of the initial high speed (high kinetic energy) of the free neutron.

Since energy is conserved, this reduction of the neutron speed takes place by transfer of energy to a material called a moderator.

There are several different types of moderating materials, and each have places where they are used more effectively.

Typically-used moderator materials include heavy water, light water, and graphite.

177.

.An inductor of inductance 10 H is connected across an a.c circuit source of 50 V, 100 Hz. What is the current in the circuit? [\(\pi\) = 3.14]

A.

0.200A

B.

0.070A

C.

0.050A

D.

0.008A

Correct answer is D

Given Data: L = 10, V = 50, F = 100, I = ?

Inductive Reactance [X\(_L\)] = 2\(\pi\)FL → 2 * 3.14 * 100 * 10

X\(_L\) = 6,280

Current[I] = \(\frac{V}{X_L}\) → \(\frac{50}{6280}\)

I = 0.0079 

≈ 0.008A 

178.

A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a 395 \(\Omega\) resistor in series with it. Determine the internal resistance of the galvanometer

A.

2.5\(\Omega\)

B.

5.0\(\Omega\)

C.

8.0\(\Omega\)

D.

10.0\(\Omega\)

Correct answer is B

r = resistance of the galvanometer

l = current through galvanometer = 20mA or 0.02A

V1 = p.d across the galvanometer = I X r = 0.02 X r

: V1 = 0.02r

V2 =  p.d across the  multiplier = 8 - 0.02r


R = resistance of the multiplier = 395Ω

where R = \(\frac{V}{I}\) 

--> 395 = \(\frac{8−0.02r}{0.02}\)

CROSS MULTIPLY

8−0.02r = 395 * 0.02

8−0.02r = 7.9

--> 0.02r = 8 - 7.9

∴ r = \(\frac{0.1}{0.02}\)

 = 5Ω

179.

Which of the following statements about electric potential energy is not correct?

A.

The electric potential energy of a positively charged particle increases when it moves to a point of higher potential

B.

The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential

C.

The electric potential energy of a positively charged particle decreases when it moves to a point of higher potential

D.

The work done in taking a charged particle around a closed path in an electric field is zero.

Correct answer is B

The potential energy for a positive charge increases when it moves against an electric field

And decreases when it moves with the electric field;

The opposite is true for a negative charge. Unless the unit charge crosses a changing magnetic field, its potential at any given point does not depend on the path taken.

180.

Three cells each of emf, 1.0 V, and internal resistance, 2 \(\Omega\), are connected in parallel across a 3\(\Omega\)  resistor. Determine the current in the resistor.

A.

0.90A

B.

0.39A

C.

0.30A

D.

0.01A

Correct answer is C

Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?

3 resistance in parallel = \(\frac{1}{r_1}\) + \(\frac{1}{r_2}\) + \(\frac{1}{r_3}\)

\(\frac{1}{r_T}\)= \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)

\(\frac{1}{r_T}\) = \(\frac{1+1+1}{2}\)

\(\frac{1}{r_T}\) = \(\frac{3}{2}\)

cross multiply

r\(_T\) = \(\frac{2}{3}\) or 0.67

E = I(R+r) 

1 = I(3+0.67)

1 = I(3.67)

1 ÷ 3.67 = I

0.27 = I

: I ≈ 0.30A