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.

1,401.

The difference observed in solids, liquids and gases can be accounted for by

A.

The different molecules in each of them

B.

The spaces and forces acting between the molecules

C.

Their melting points

D.

Their relative masses

Correct answer is B

No explanation has been provided for this answer.

1,402.

Which of the following gas laws is equivalent to the work done?

A.

Boyle's law

B.

Charles law

C.

Pressure law

D.

van der Waal's law

Correct answer is A

Mathematically Boyle's law = PV = K
Where P = Pressure
and V = Volume


By definition Pressure = \( \frac{F}{A} = \frac{F}{M \times M} \)

Again Workdone = force x displacement

therefore \( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \\

PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x \)

= force x displacement

1,403.

A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?

A.

4.05m

B.

4.08m

C.

5.05m

D.

6.08m

Correct answer is B

from hooke's law F=Ke
\(
\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}

\frac{4}{0.02} = \frac{15}{e_2} \)

e1 = \( \frac{0.02 x 15}{4} = 0.075M \)

The new length of the string is 4 + 0.075
= 4.075
= 4.08

1,404.

A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?

A.

50N

B.

100N

C.

175N

D.

210N

Correct answer is B

Efficiency = \( \frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1} \)

And for incline plane

V.R =\( \frac{1}{Sin\theta} \\
= V.R = \frac{1}{Sin30} =\frac{1}{0.5} = 2 \\

\text{therefore } \frac{60}{120} = \frac{M.A}{2}
=M.A = \frac{120}{100} = 1.2\\

\text{but M.A } = \frac{load}{effort} = \frac{120}{E}\\

\text{therefore} \frac{120}{E} = 1.2 \\

= E =\frac{120}{1.2 } = 100N \\ \)


Therefore Effort up the plane = 100N

1,405.

An alternating current with a frequency of 100HZ has a period of

A.

0.03sec

B.

0.01sec

C.

0.04sec

D.

0.15sec

Correct answer is B

F = 100HZ, period, T = 1/f = 1/100

= 0.01 sec

The period (T) = 0.01sec