Are these NEET questions free to attempt?

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

Do I need an account to practise NEET MCQs?

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

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.

NEET NEET Questions — Free MCQ Practice

● ● ● ● ●

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

A.

6 $° C$

B.

12 $° C$

C.

18 $° C$

D.

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

A.

7 hours

B.

14 hours

C.

Less than 7 hours

D.

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

A.

$K_{1} + K_{2}$

B.

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

C.

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

D.

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

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

A.

15.24 $° C$

B.

15.00°C

C.

24.15 $° C$

D.

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

A.

$\frac{(x + y) (x - y) ρL}{2 kθ}$

B.

$\frac{(x - y) ρL}{2 kθ}$

C.

$\frac{(x + y) (x - y) ρL}{kθ}$

D.

$\frac{(x - y) ρLk}{2 θ}$

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

Explanation

(a)

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

● ● ● ● ●

Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle ABC , right angled at B. The points A and B are maintained at temperatures T and $\sqrt{2} T$ respectively. In the steady state the temperature of the point C is $T_{C}$ . Assuming that only heat conduction takes place, $\frac{T_{C}}{T}$ is equal to -

A.

$\frac{1}{(\sqrt{2} + 1)}$

B.

$\frac{3}{(\sqrt{2} + 1)}$

C.

$\frac{1}{2 (\sqrt{2} - 1)}$

D.

$\frac{1}{\sqrt{3} (\sqrt{2} - 1)}$

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

● ● ● ● ●

A sphere, a cube and a thin circular plate, all made of the same material and having the same mass are initially heated to a temperature of 1000°C. Which one of these will cool first ?

A.

Plate

B.

Sphere

C.

Cube

D.

None of these

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

Explanation

(a) Rate of cooling $\frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc} \Rightarrow \frac{∆ θ}{t} \propto A$ . Since area of plate is largest so it will cool fastest.

● ● ● ● ●

Two identical conducting rods are first connected independently to two vessels, one containing water at 100 $° C$ and the other containing ice at 0 $° C$ . In the second case, the rods are joined end to end and connected to the same vessels. Let $q_{1}$ and $q_{2}$ g / s be the rate of melting of ice in two cases respectively. The ratio of is $q_{1}$ / $q_{2}$

A.

$\frac{1}{2}$

B.

$\frac{2}{1}$

C.

$\frac{4}{1}$

D.

$\frac{1}{4}$

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

● ● ● ● ●

A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature 120 $° C$ , then -

A.

Both the cube and the sphere cool down at the same rate

B.

The cube cools down faster than the sphere

C.

The sphere cools down faster than the cube

D.

Whichever is having more mass will cool down faster

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

Explanation

(b) Rate of cooling of a body

$R = \frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc} \Rightarrow R \propto \frac{A}{m} \propto \frac{Area}{Volume}$

$\Rightarrow$ For the same surface area. $R \propto \frac{1}{Volume}$

$∵$ Volume of cube < Volume of sphere

$\Rightarrow R_{Cube} > R_{Sphere}$ i.e. cube cools down with faster rate.

● ● ● ● ●

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength $λ_{B}$ corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 $μm$ . If the temperature of A is 5802 K -

A.

The temperature of B is 1964 K

B.

$λ_{B}$ =1.5 $μm$

C.

The temperature of B is 11604 K

D.

The temperature of B is 2901 K

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

Explanation

(b) According to Stefan’s law

$E = {eAσT}^{4} \Rightarrow E_{1} = e_{1} {AσT}_{1}^{4} and E_{2} = e_{2} {AσT}_{2}^{4} ∵ E_{1} = E_{2} ∴ e_{1} T_{1}^{4} = e_{2} T_{2}^{4} \Rightarrow T_{2} = {(\frac{e_{1}}{e_{2}} T_{1}^{4})}^{1 / 4} = {(\frac{1}{81} \times {(5802)}^{4})}^{1 / 4} \Rightarrow T_{B} = 1934 K$

And, from Wein's law $λ_{A} \times T_{A} = λ_{B} \times T_{B}$

$\Rightarrow \frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}} \Rightarrow \frac{λ_{B} - λ_{A}}{λ_{B}} = \frac{T_{A} - T_{B}}{T_{A}} \Rightarrow \frac{1}{λ_{B}} = \frac{5802 - 1934}{5802} = \frac{3968}{5802} \Rightarrow λ_{B} = 1.5 μm$

NEET Practice Questions (MCQs) with Answers & Solutions

Frequently Asked Questions