JEE Advanced 2021 Paper 1 · Physics · Calorimetry and Heat Transfer · hard

A small thin sheet of mass 0.10 g and area 1.0×10^-4 m^2 hangs from a long light thread. It is kept vertical by a light frame, so that a horizontal…

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A small thin sheet of mass

0.10\ \mathrm{g}

and area

1.0\times10^{-4}\ \mathrm{m^2}

hangs from a long light thread. It is kept vertical by a light frame, so that a horizontal laser beam of intensity

1.5\times10^{9}\ \mathrm{W\,m^{-2}}

falls normally on it. The sheet absorbs half of the incident light and specularly reflects the other half. Take

c=3\times10^{8}\ \mathrm{m\,s^{-1}}

and

g=10\ \mathrm{m\,s^{-2}}

. In equilibrium, the tension in the thread is

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Answer: (C) $1.25\ \mathrm{mN}$

The radiation force on a surface normal to the beam is

F=(f_a+2f_r)\frac{IA}{c}

where

f_a

and

f_r

are the absorbed and reflected fractions. With

f_a=f_r=0.5

F=1.5\,IA/c = 1.5\times1.5\times10^{9}\times10^{-4}/(3\times10^{8})=7.5\times10^{-4}

N. The thread supports the weight

mg=0.1\times10^{-3}\times10=1.0\times10^{-3}

N vertically and balances the horizontal radiation force, so

T=\sqrt{(mg)^2+F^2}=\sqrt{1.0^2+0.75^2}\ \mathrm{mN}=1.25

mN (the thread makes

\tan^{-1}0.75\approx37^\circ

with the vertical). If all light were absorbed,

F=0.5

mN and

T=1.12

mN; if all were reflected,

F=1.0

mN and

T=1.41

mN;

1.00

mN ignores the radiation force. The correct tension is

1.25

mN.

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