JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 14)
A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 $$-$$ P0 between its inside and outside would be :
12 cm
2.4 cm
6 cm
4.8 cm
Explanation
Pressure difference inside the inner bubble,
p2 $$-$$ p1 = $${{4T} \over {{r_2}}}$$b . . . . . (1)
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And for outer bubble
p1 $$-$$ p0 = $${{4T} \over {{r_1}}}$$ . . . . . . . (2)
$$\therefore\,\,\,$$ p2 $$-$$ p0 = 4T $$\left( {{1 \over {{r_2}}} + {1 \over {{r_1}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ p2 $$-$$ p0 = $${{4T} \over r}$$
Here r is the radius of the bubble.
$$\therefore\,\,\,$$ $${1 \over r} = {1 \over {{r_2}}} + {1 \over {{r_1}}}$$
$$ \Rightarrow $$$$\,\,\,$$ r = $${{{r_1}{r_2}} \over {{r_1} + {r_2}}}$$
= $${{4 \times 6} \over {4 + 60}}$$
= 2.4 cm
p2 $$-$$ p1 = $${{4T} \over {{r_2}}}$$b . . . . . (1)
And for outer bubble
p1 $$-$$ p0 = $${{4T} \over {{r_1}}}$$ . . . . . . . (2)
$$\therefore\,\,\,$$ p2 $$-$$ p0 = 4T $$\left( {{1 \over {{r_2}}} + {1 \over {{r_1}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ p2 $$-$$ p0 = $${{4T} \over r}$$
Here r is the radius of the bubble.
$$\therefore\,\,\,$$ $${1 \over r} = {1 \over {{r_2}}} + {1 \over {{r_1}}}$$
$$ \Rightarrow $$$$\,\,\,$$ r = $${{{r_1}{r_2}} \over {{r_1} + {r_2}}}$$
= $${{4 \times 6} \over {4 + 60}}$$
= 2.4 cm
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