Physics MCQs for NEET — Practice Questions with Answers

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A difference of temperature of $25 ^\circ C $ is equivalent to a difference of

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Explanation

$ { \triangle C \over 100 ^\circ } = { \triangle F \over 180 ^\circ } $ $ \therefore {25 \over 100} = { \triangle F \over 180 } $

What is the value of absolute temperature on the Celsius Scale ?

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Explanation

$ T = 273.15 + t^\circ C $ $ 0 = 273.15 + t^\circ C $

The temperature of a substance increases by $ 27 ^\circ $ What is the value of this increase of Kelvin scale ?

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Explanation

equal

At Which temperature the density of water is maximum?

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Explanation

$ 4 ^\circ $

The temperature on celsius scale is $ 25 ^\circ C $ . What is the corresponding temperature on the Fahrenheit Scale?

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Explanation

$ { c \over 5 } = {F -32 \over 9 } $

The temperature of a body on Kelvin Scale is found to be x.K.when it is measured by Fahrenhit thesmometes. it is found to be $x^0F$ , then the value of x is .

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Explanation

$ {F -32 \over 9 } = { K -273 \over 5 } $

A Centigrade and a Fahrenhit thesmometes are dipped in boiling wates-The wates temperature is lowered until the Farenhit thesmometes registered $140 ^\circ $ what is the fall in thrmometers

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Explanation

$ { \triangle Tc \over 100 } = { \triangle TF \over 180 } $ $ { \triangle Tc \over 100 } \triangle TC = 40 ^ \circ C $

A uniform metal rod is used as a bas pendulum. If the room temperature rises by $ 10 ^\circ C $ and the efficient of line as expansion of the metal ofthe rod is, $ 2 \times 10^{-6} 0_c^{-1} $ what will have percentage increase

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Explanation

$ { \triangle T \over T } = { 1 \over 2} 2 \triangle 0 = { 1 \over 2} \times 2 \times 10 ^ {-6} \times 10 = 10 ^ {-5} $ $ \% in increase = { \triangle T \over T } \times 100 = 10 ^ {-5} \times 100 = 1 \times 10 ^ {-3} \% $

A gas expands from 1 litre to 3 litre at atmospheric pressure. The work done by the gas is about

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Explanation

$ W = p \triangle V $

Each molecule of a gas has f degrees of freedom. The radio $ { C_P \over C_V} = \gamma $ for the gas is

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Explanation

$C_p = \left( { f \over 2 } + 1 \right) R, C_v = {f \over 2} R $ $ { C_p \over C_v }= { \left( { f \over 2 } + 1 \right) R \over { f \over 2} R } = { {f \over 2} + 1 \over { f \over 2} } = { f+2 \over f} = 1 + { 2 \over f } $

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