Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

The area of a hole of heat furnace is $10^{- 4} m^{2}$ . It radiates $1.58 \times 10^{5}$ calories of heat per hour. If the emissivity of the furnace is 0.80, then its temperature is

1500 K

2000 K

2500 K

3000 K

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\Rightarrow \frac{1.58 \times 10^{5} \times \times 4.2}{60 \times 60} = 5.6 \times 10^{- 8} \times 10^{- 4} \times 0.8 \times T^{4} T \approx 2500 K$

● ● ● ● ●

Two spheres P and Q, of same colour having radii 8 cm and 2 cm are maintained at temperatures 127 $° C$ and 527 $° C$ respectively. The ratio of energy radiated by P and Q is

0.054

0.0034

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\Rightarrow Q \propto {AT}^{4} \propto r^{2} T^{4} (∵ A = 4 {πr}^{2}) \Rightarrow \frac{Q_{P}}{Q_{Q}} = {(\frac{r_{P}}{r_{Q}})}^{2} {(\frac{T_{P}}{T_{Q}})}^{4} = {(\frac{8}{2})}^{2} {\{\frac{(273 + 127)}{(273 + 527)}\}}^{4} = 1$

● ● ● ● ●

A body radiates energy 5W at a temperature of 127 $° C$ . If the temperature is increased to 927 $° C$ , then it radiates energy at the rate of

410 W

81 W

405 W

200 W

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{927 + 273}{127 + 273})}^{4} \Rightarrow P_{1} = 405 W$

● ● ● ● ●

The temperatures of two bodies A and B are respectively 727 $° C$ and 327 $° C$ . The ratio of the rates of heat radiated by them is

727:327

5 : 3

25 : 9

625 : 81

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $Q \propto T^{4} \Rightarrow \frac{H_{A}}{H_{B}} = {(\frac{273 + 727}{273 + 327})}^{4} = {(\frac{10}{6})}^{4} = {(\frac{5}{3})}^{4} = \frac{625}{81}$

● ● ● ● ●

The energy emitted per second by a black body at 27 $° C$ is 10 J. If the temperature of the black body is increased to 327 $° C$ , the energy emitted per second will be

20 J

40 J

80 J

160 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $Q_{Black body} = {AσT}^{4} t \Rightarrow Q \propto T^{4} \Rightarrow Q_{2} = Q_{1} {(\frac{T_{2}}{T_{1}})}^{4} = 10 {(\frac{273 + 327}{273 + 27})}^{4} = 10 {(\frac{600}{300})}^{4} = 160 J$

● ● ● ● ●

The radiant energy from the sun incident normally at the surface of earth is $20 kcal / m^{2} \min$ . What would have been the radiant energy incident normally on the earth, if the sun had a temperature twice of the present one ?

$160 kcal / m^{2} \min$

$40 kcal / m^{2} \min$

$320 kcal / m^{2} \min$

$80 kcal / m^{2} \min$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{E_{2}}{20} = {(\frac{2 T}{T})}^{4} = 16 \Rightarrow E_{2} = 320 kcal / m^{2} \min$

● ● ● ● ●

If the temperature of the sun (black body) is doubled, the rate of energy received on earth will be increased by a factor of

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) Amount of energy radiated ∝ $(Temperature)^{4}$

● ● ● ● ●

The ratio of energy of emitted radiation of a black body at 27 $° C$ and 927 $° C$ is

1 : 4

1 : 16

1 : 64

1 : 256

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $\frac{Q_{1}}{Q_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{273 + 27}{273 + 927})}^{4} = {(\frac{1}{4})}^{4} = \frac{1}{256}$

● ● ● ● ●

Two spherical black bodies of radii $r_{1}$ and $r_{2}$ and with surface temperature $T_{1}$ and $T_{2}$ respectively radiate the same power. Then the ratio of $r_{1}$ and $r_{2}$ will be

${(\frac{T_{2}}{T_{1}})}^{2}$

${(\frac{T_{2}}{T_{1}})}^{4}$

${(\frac{T_{1}}{T_{2}})}^{2}$

${(\frac{T_{1}}{T_{2}})}^{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) For black body, $P = {AεσT}^{4}$ . For same power $A \propto \frac{1}{T^{4}}$

$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{T_{2}}{T_{1}})}^{2}$

● ● ● ● ●

A black body is at a temperature 300 K. It emits energy at a rate, which is proportional to

300

${(300)}^{2}$

${(300)}^{3}$

${(300)}^{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $E \propto T^{4}$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.