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Physics NEET Questions — Free MCQ Practice

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A capillary tube of radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be

A.

M

B.

2M

C.

M/2

D.

4M

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Explanation

(b)

$Initial mass of liquid in the tube, M = {πR}^{2} H \times ρ When radius is doubled, As height of liquid α \frac{1}{radius} \Rightarrow Final height of liquid = \frac{H}{2} Final mass of liquid = π (2 R)^{2} \frac{H}{2} \times ρ = 2 M$

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There is a horizontal film of soap solution. On it, a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then what will be the tension in the thread ?

A.

${πR}^{2} / T$

B.

${πR}^{2} T$

C.

$2 πRT$

D.

$2 RT$

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A large number of water drops each of radius r combine to have a drop of radius R. If the surface tension is T and the mechanical equivalent of heat is J, then the rise in temperature will be

A.

$\frac{2 T}{rJ}$

B.

$\frac{3 T}{RJ}$

C.

$\frac{3 T}{J} (\frac{1}{r} - \frac{1}{R})$

D.

$\frac{2 T}{J} (\frac{1}{r} - \frac{1}{R})$

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Explanation

$Volume of n small drops = Volume of large drop . n \frac{4}{3} {πr}^{3} = \frac{4}{3} {πR}^{3} {nr}^{3} = R^{3} Change in surface area ∆ A = n 4 {πr}^{2} - 4 {πR}^{2} = 4 π ({nr}^{2} - R^{2}) Reduction in surface energy = T \cdot ∆ A = T 4 π ({nr}^{2} - R^{2}) = T 4 π (\frac{R^{3}}{r} - R^{2}) = T 4 {πR}^{3} [\frac{1}{r} - \frac{1}{R}] Heat \Rightarrow Q = \frac{W}{J} = \frac{T \cdot 4 {πR}^{3}}{J} [\frac{1}{r} - \frac{1}{R}] Q = ms ∆ T \frac{T \cdot 4 {πR}^{3}}{J} [\frac{1}{r} - \frac{1}{R}] = 1 \times \frac{4}{3} {πR}^{3} \times 1 \times ∆ T ∆ T = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

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One kilogram of ice at $0 °$ C is mixed with one kilogram of water at 80 $° C$ . The final temperature of the mixture is (Take: Specific heat of water = 4200 J $k g^{- 1} k^{- 1}$ , Latent heat of ice = 336 kJ $k g^{- 1}$ )

A.

$0 °$ C

B.

$50 ° C$

C.

$40 ° C$

D.

$60 ° C$

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The fastest mode of heat transfer is

A.

Conduction

B.

Radiation

C.

Convection

D.

all are equally fast

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Explanation

In radiation, heat is transferred by electromagnetic waves, which have the fastest speed, so it is the fastest mode of heat transmission.

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

A.

310.9

B.

37.8

C.

100

D.

122.4

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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 -

A.

16E

B.

64E

C.

81E

D.

256E

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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.

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

A.

Less than 5 minutes

B.

equal to 5 minutes

C.

More than 5 minutes

D.

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

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Explanation

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

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Heat capacity is equal to the product of:

A.

mass and gas constant

B.

mass and specific heat

C.

latent heat and volume of water

D.

mass and Avogadro number

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Explanation

Heat capacity = mass x specific heat

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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:

A.

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

B.

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

C.

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

D.

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

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Explanation

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

Physics MCQs for NEET — Practice Questions with Answers

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