Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

Q: Are these Physics questions free to attempt?

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.

Q: Do I need an account to practise Physics MCQs?

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

Q: Are these previous year NEET questions?

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

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

● ● ● ● ●

Newton’s law of cooling, holds good only if the temperature difference between the body and the surroundings is

Less than 10 $° C$

More than 10 $° C$

Less than 100 $° C$

More than 100 $° C$

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

Explanation

(a) I holds good only for small temperature difference between the body and the surroundings i.e. less than 10 $°$ C.

● ● ● ● ●

The temperature of a body falls from 50 $° C$ to 40 $° C$ in 10 minutes. If the temperature of the surroundings is 20 $° C$ Then temperature of the body after another 10 minutes will be

36.6 $° C$

33.3 $° C$

35 $° C$

30 $° C$

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

Explanation

(b) In first case $\frac{50 - 40}{10} = K [\frac{50 + 40}{2} - 20]$ ….(i)
In second case $\frac{40 - θ_{2}}{10} = K [\frac{40 + θ_{2}}{2} - 20]$ ….(ii)
By solving $θ_{2} = 33.3 ° C$ .

● ● ● ● ●

A body takes 5 minutes to cool from 90 $° C$ to 60 $° C$ . If the temperature of the surroundings is 20 $° C$ , the time taken by it to cool from 60 $° C$ to 30 $° C$ will be.

5 min

8 min

11 min

12 min

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

Explanation

(c) $\frac{90 - 60}{5} = K (\frac{90 + 60}{2} - 20) \Rightarrow 6 = k \times 55 \Rightarrow k = \frac{6}{55} And, \frac{60 - 30}{t} = \frac{6}{55} (\frac{60 + 30}{2} - 20) \Rightarrow t = 11 minute$

● ● ● ● ●

An object is cooled from 75°C to 65°C in 2 minutes in a room at 30°C. The time taken to cool another object from 55°C to 45°C in the same room in minutes is -

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

Explanation

(a) According to Newton’s law of cooling
in first case, $\frac{75 - 65}{t} = K [\frac{75 + 65}{2} - 30]$ ...(i)

In second case, $\frac{55 - 45}{t} = K [\frac{55 + 45}{2} - 30]$ ...(ii)

Dividing eq. (i) by (ii) we get

$\frac{5 t}{10} = \frac{40}{20} \Rightarrow t = 4 minutes$

● ● ● ● ●

A cane is taken out from a refrigerator at 0°C. The atmospheric temperature is 25°C. If $t_{1}$ is the time taken to heat from 0°C to 5°C and $t_{2}$ is the time taken from 10°C to 15°C, then

$t_{1} > t_{2}$

$t_{1} < t_{2}$

$t_{1} = t_{2}$

There is no relation

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

Explanation

(b) According to Newton’s law of cooling.

● ● ● ● ●

A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is 36 $° C$ . Then the difference of temperature at the two surfaces of A will be

6 $° C$

12 $° C$

18 $° C$

24 $° C$

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

● ● ● ● ●

Ice starts forming in lake with water at 0 $° C$ and when the atmospheric temperature is -10 $° C$ . If the time taken for 1 cm of ice be 7 hours, then the time taken for the thickness of ice to change from 1 cm to 2 cm is

7 hours

14 hours

Less than 7 hours

More than 7 hours

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

Explanation

(d) $t = \frac{ρL}{2 Kθ} (x_{2}^{2} - x_{1}^{2}) \Rightarrow t \propto (x_{2}^{2} - x_{1}^{2}) \Rightarrow \frac{t}{t'} = \frac{(x_{2}^{2} - x_{1}^{2})}{(x'_{2}^{2} - x'_{1}^{2})} \Rightarrow \frac{9}{t'} = \frac{(1^{2} - 0^{2})}{(2^{2} - 1^{2})} \Rightarrow t = 21 hours$

● ● ● ● ●

A cylinder of radius R made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

$K_{1} + K_{2}$

$\frac{K_{1} K_{2}}{K_{1} + K_{2}}$

$\frac{K_{1} + 3 K_{2}}{4}$

$\frac{3 K_{1} + K_{2}}{4}$

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

● ● ● ● ●

A room is maintained at 20 $° C$ by a heater of resistance 20 ohm connected to 200 volt mains. The temperature is uniform through out the room and heat is transmitted through a glass window of area $1 m^{2}$ and thickness 0.2 cm. What will be the temperature outside? Given that thermal conductivity K for glass is 0.2 cal/m $° C$ and J = 4.2 J/cal

15.24 $° C$

15.00°C

24.15 $° C$

None of the above

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

Explanation

(a) Heat developed by the heater

$H = \frac{V^{2}}{R} . \frac{t}{J} = \frac{{(200)}^{2} \times t}{20 \times 4.2}$

Heat conducted by the glass $H = \frac{0.2 \times 1 \times (20 - θ) t}{0.002} \Rightarrow θ = 15.24 ° C$

● ● ● ● ●

There is formation of layer of snow x cm thick on water, when the temperature of air is $- θ ° C$ (less than freezing point). The thickness of layer increases from x to y in the time t, then the value of t is given by

$\frac{(x + y) (x - y) ρL}{2 kθ}$

$\frac{(x - y) ρL}{2 kθ}$

$\frac{(x + y) (x - y) ρL}{kθ}$

$\frac{(x - y) ρLk}{2 θ}$

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

Explanation

(a)

$∴$ $If dx thichness forms in time dt with x thickness of ice above it, then; \Rightarrow Heat released due to its formation = Heat gained by the ice \Rightarrow mL = \frac{KAθ}{x} dt \Rightarrow ρAdxL = \frac{KAθ}{x} dt \Rightarrow \int_{x_{1}}^{x_{2}} ρxdxL = \int_{0}^{t} Kθdt \Rightarrow t = \frac{ρL}{2 Kθ} ({x_{2}}^{2} - {x_{1}}^{2}) t = \frac{ρL}{2 kθ} (x^{2} - y^{2}) = \frac{ρL (x + y) (x - y)}{2 Kθ}$

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.