Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

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

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

$A temperature of 100 ° F (Farenhiet scale) is equal to T K (Kelvin scale) . The value of T is$

310.9

37.8

100

122.4

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

● ● ● ● ●

A black body at temperature 300K radiates heat at the rate E. If its temperature is increased by 600K, the rate of radiation will increase to -

16E

64E

81E

256E

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

Explanation

According to Stefan-Boltzmann law, the rate of radiation emitted by a black body is proportional to the fourth power of its absolute temperature. If the temperature increases from 300K to 900K, the rate of radiation increases by (900/300)^4 = 81 times.

● ● ● ● ●

$A body cools down from 80 ° C to 70 ° C in 5 minutes . the temperature of the body will fall from 70 ° C to 60 ° C in a time$

Less than 5 minutes

equal to 5 minutes

More than 5 minutes

Can't say anything like the temperature of surrounding is not known

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

Explanation

$T_{s} = T A p p l y i n g N e w t o n' s l a w o f c o o l i n g, F o r 1^{s t} c a s e - \frac{10}{5} = - b A (75 - T) . . . . (1) F o r 2^{n d} c a s e - \frac{10}{t} = - b A (65 - T) . . . . (2) f r o m (1) a n d (2) w e c a n s a y t h a t t > 5$

● ● ● ● ●

Heat capacity is equal to the product of:

mass and gas constant

mass and specific heat

latent heat and volume of water

mass and Avogadro number

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

Explanation

Heat capacity = mass x specific heat

● ● ● ● ●

When a block of iron floats in Hg at $0 ° C$ , a fraction $K_{1}$ of its volumen= is submerged, while at temperature of $60 ° C$ a fraction $K_{2}$ is seen to be submersed. If the coefficient of volume expansion of iron is $γ_{F e}$ and that of mercury is $γ_{H g}$ , then the ratio $\frac{K_{1}}{K_{2}}$ can be expressed as:

$\frac{1 + 60 γ_{F e}}{1 + 60 γ_{H g}}$

$\frac{1 - 60 γ_{F e}}{1 + 60 γ_{H g}}$

$\frac{1 + 60 γ_{F e}}{1 - 60 γ_{H g}}$

$\frac{1 + 60 γ_{H e}}{1 - 60 γ_{F e}}$

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

Explanation

$L e t V_{0} i s t h e v o l u m e o f t h e b l o c k . T h e n v o l u m e o f t h e b l o c k a t 60^{0} C - V = V_{0} (1 + γ_{i r o n} T) L e t V_{0}' i s t h e v o l u m e o f b l o c k s u b m e r g e d i n t h e m e r c u r y, t h e n v o l u m e o f t h e b l o c k s u b m e r g e d i n t h e m e r c u r y a t 60^{0} C - V' = V_{0}' (1 + γ_{m e r c u r y} T) (V o l u m e o f t h e b l o c k s u b m e r g e d i n m e r c u r y = v o l u m e o f m e r c u r y d i s p l a c e d) T h e n, \frac{V}{V'} = \frac{V_{0} (1 + γ_{i} T)}{V_{0}' (1 + γ_{m} T)} \Rightarrow \frac{1}{K_{2}} = \frac{1}{K_{1}} \frac{(1 + γ_{i} T)}{(1 + γ_{m} T)} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{(1 + γ_{i} T)}{(1 + γ_{m} T)} = \frac{(1 + 60 γ_{i})}{(1 + 60 γ_{m})}$

● ● ● ● ●

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.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

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.

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.