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

● ● ● ● ●

Hot water cools from 60 $° C$ to 50 $° C$ in first 10 minutes and from 50 $° C$ to 42 $° C$ in next 10 minutes. The temperature of surrounding is :

$5 ° C$

$10 ° C$

$15 ° C$

$20 ° C$

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

Explanation

$(2) (\frac{θ_{1} - θ_{2}}{t}) = C (\frac{θ_{1} + θ_{2}}{2} - θ_{0}) \frac{10}{10} = C [55 - θ_{0}] (1) \frac{8}{10} = C [46 - θ_{0}] (2) \frac{10}{8} = \frac{55 - θ_{0}}{46 - θ_{0}} 460 - 10 θ_{0} = 440 - 8 θ_{0} 20 = 2 θ_{0} θ_{0} = 10 ° C$

● ● ● ● ●

$If α_{c} and α_{f} denote the numerical values of coefficient of linear expansion of a solid, expressed per ° C and per ° F respectively, then$

$α_{c} > α_{f}$

$α_{f} > α_{c}$

$α_{f} = α_{c}$

$α_{f} + α_{c} = 0$

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

Explanation

$(1) α = \frac{∆ L}{L ∆ T} α \propto \frac{1}{∆ T}$

● ● ● ● ●

$A pendulum clock runs fast by 5 \sec . per day at 20 ° C and goes slow by 10 \sec . per day at 35 ° C . It shows correct time at a temperature of$

$27.5 ° C$

$25 ° C$

$30 ° C$

$33 ° C$

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

Explanation

$(2) \frac{1}{2} α (35 - T) \times 86400 = 10 . . . (1) \frac{1}{2} α (T - 20) \times 86400 = 5 . . . (2) Solving (1) and (2) T = 25^{0} C .$

● ● ● ● ●

A constrained steel rod of length l, area of cross-section A Young's modulus Y and coefficient of linear expansion $α$ is heated through $t ° C$ . The work that can be performed by the rod when heated is

(YA $α$ t)(l $α$ t)

$\frac{1}{2}$ (YA $α$ t)(l $α$ t)

$\frac{1}{2}$ (YA $α$ t) $\frac{1}{2}$ (l $α$ t)

2(YA $α$ t)(l $α$ t)

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

Explanation

$W = \frac{1}{2} F ∆ L w h e r e F = Y A α ∆ t a n d ∆ L = α L ∆ t$

● ● ● ● ●

A spherical black body with a radius of 12cm radiates 450-watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:

225

450

1000

1800

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

Explanation

(d)

Radiated power of the black body, P = $δ A T^{4}$

where A=surface area of the body

T=temperature of the body

and, $δ$ =Stefan's constant

when radius of the sphere is halved, new area,

$A^{'} = \frac{A}{4}$

$∴$ power radiated.

$P^{'} = δ (\frac{A}{4}) {(2 T)}^{4} = \frac{16}{4} (δ A T^{4})$

=4P= $4 \times 450 = 1800$ Watts.

● ● ● ● ●

Two identical bodies are made of a material for which the heat capacity increases with temperature.One of these is at $100^{°}$ C,while the other one is at $0^{°}$ C. If the two bodies are brought into contact, then assuming no heat loss,the final common temperature is -

$50^{°} C$

more than $50^{°}$ C

less than $50^{°}$ C but greater than $0^{°}$ C

$0 {}_{0}C$

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

Explanation

(b) Heat lost by 1st body=heat gained by 2nd body. Body at $100^{°}$ C temperature has greater heat capacity than body at $0^{°}$ C so final temperature will be closer to $100^{°}$ C. So $T_{c} > 50^{°} C .$

● ● ● ● ●

A body cools from a temperature 3T to 2T in10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be -

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$

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

Explanation

(b)

$C a s e 1 : T e m p . f a l l s f r o m 3 T t o 2 T \frac{3 T - 2 T}{10} = k (\frac{3 T + 2 T}{2} - T) C a s e 2 : L e t f i n a l t e m p . b e x \frac{2 T - x}{10} = k (\frac{x + 2 T}{2} - T) S o l v i n g b o t h e q u a t i o n s, x = \frac{3 T}{2}$

● ● ● ● ●

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁, at wavelength 500nm is U₂ and that at 1000nm is U₃ . Given Wien's constant $b = 2.88 \times 10^{6} n m - K$ . Which of the following is correct?

U₃=0

U₁>U₂

U₂>U₁

U₁=0

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

Explanation

(c) Given, temperature, T₁=5760K Since, it is given that energy of radiation emitted by the body at wavelength 250 nm in U₁, at wavelength 500 nm is U₂and that at 1000 nm is U₃.

∴ According to Wien’s law, we get λ_mT=b

where, b=Wien’s constant=2.88 x 10⁶ nmK

=> λ_m=b/T
=>λ_m=2.88x10⁶ nmK/5760K
=> λ_m=500 nm

∴ λ_m= wavelength corresponding to maximum energy,so, U₂>U₁.

● ● ● ● ●

A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4x10⁵J/Kg and g=10N/Kg]

544 km

136 km

68 km

34 km

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

Explanation

(b) According to question as conservation of energy, energy gained by the ice during its fall from height h is given by
E=mgh

As given, only one quarter of its energy is absorbed by the ice.
so, mgh/4=mL_I
=> h=mL_I x 4/mg
=L_I x 4/g
=3.4x10⁵x4/10=13.6x10⁴=13600m=136km

● ● ● ● ●

The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be

44.0 J/s

16.8 J/s

8.0 J/s

4.0 J/s

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

Explanation

Here, ΔT₁=110-100=10°C

$\frac{d Q_{1}}{d t}$ =4J/s

ΔT₂=210-200=10°C

$\frac{d Q_{2}}{d t}$ =?

As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. 10°C. So, the same rate of heat will flow in the second case

Hence, $\frac{d Q_{2}}{d t}$ =4 J/s

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.