JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 24)
A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.
Answer
2
Explanation
Image
Excess pressure inside the smaller soap bubble
$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)
The excess pressure inside equivalent soap bubble
$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)
From (i) & (ii)
$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$
$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$
$$ = {1 \over 6} + {1 \over 3}$$
Req = 2 cm
Excess pressure inside the smaller soap bubble
$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)
The excess pressure inside equivalent soap bubble
$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)
From (i) & (ii)
$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$
$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$
$$ = {1 \over 6} + {1 \over 3}$$
Req = 2 cm
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