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.

356.

A rectangular piece of iron measuring 4 cm by 3 cm at 20\(^o\)C is heated until its temperature increases by 100 C. Calculate the new area of the metal. [Linear expansivity of iron is 1.2x 10\(^{-5}\) K\(^{-1}\)]

A.

12.0144 cm\(^3\)

B.

12.0346 cm\(^3\)

C.

12.0288 cm\(^3\)

D.

12.0173 cm\(^3\)

Correct answer is C

Linear expansivity (\(\alpha\)) 

= \(\frac{A_2 - A_1}{A_1\theta}\)

Original Area(A\(_1\)) = 4 x 3 = 12

New Area (A\(_2\)) = ? 

Temp. rise (\(\theta\)) = 100\(^o\)C 

A\(_2\) - 12 = 1200 x 2 x 1.2 x 10\(^{-5}\) 

A\(_2\) - 12 = 2880 x 10\(^-5\)

A\(_2\) - 12 = 0.0288

A\(_2\) = 12.0288 cm\(^2\)  

357.

A ray of light traveling from a rectangular glass block of refractive index 1.5 into air strikes the block at an angle of incidence of 30. Calculate its angle of refraction.

A.

48.6°

B.

19.5°

C.

20.0

D.

45.0

Correct answer is A

Given that refractive index = 1.5, angle of incidence = 30°, angle of refraction =?

using snell's law = refractive index = \(\frac{sini}{sinr}\)

=\(\frac{1}{n } = \frac{sin30}{sinr}\)

= \(sinr = 1.5\times sin 30\)

r =  \(sin^10.75\)

= 48.6° 

this is because ray is traveling from a more denser medium to a less dense medium

358.

The anomalous expansion of water occurs in the range 

A.

0 °C to 100 °C

B.

0 °C to 4 °C

C.

4 °C to 100 C

D.

-4 °C to 0 °C

Correct answer is B

No explanation has been provided for this answer.

359.

Which property of a wave remains constant when the wave travels from one medium into another?
 

A.

Amplitude

B.

Wavelength

C.

Velocity

D.

Frequency

Correct answer is D

Frequency remains unchanged as waves travel from one medium to another because it is characterized by the source of the wave.

360.

A device consumes 100 W of power when connected to a 120 V source. Calculate its resistance

A.

1.2\(\Omega\)

B.

12,000 \(\Omega\)

C.

20.0\(\Omega\)

D.

144\(\Omega\)

Correct answer is D

P = \(\frac{V^2}{R}\) 

R = \(\frac{120^2}{100}\)

= 144.0\(\Omega\)