An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top the volume of the bubble is thrice its initial volume. If the atmospheric pressure is 72 cm of Hg, and mercury is 17 times heavier than kerosene, the depth of the tank must be :

2.16 m

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2.88 m

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12.24 m

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24.48 m

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Solution

The correct option is D 24.48 m
Since by boyle's law
$P_{1} V_{1} = P_{2} V_{2}$
& $U_{2} = 3 V_{1}$
$P_{1} = 3 P_{2}$
& $P_{2} = P_{0}$
so $P_{1} = 3 P_{0}$
& $P_{1} = P_{2} + ϱ g h$
$3 P_{0} = P_{0} + ϱ g h$
$ϱ_{k} g h = 2 P_{0}$
$ϱ k \times h = 2 \times 72 c m \times 17 ϱ k / g$
$h = 2 \times 72 \times 17 c m$
$h = 2448 c m$
$h = 24.48 m$

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An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top the volume of the bubble is thrice its initial volume. If the atmospheric pressure is 72 cm of Hg, and mercury is 17 times heavier than kerosene, the depth of the tank must be :

The correct option is D 24.48 mSince by boyle's law P1V1=P2V2& U2=3V1 P1=3P2& P2=P0so P1=3P0& P1=P2+ϱgh 3P0=P0+ϱgh ϱkgh=2P0 ϱk×h=2×72cm×17ϱk/g h=2×72×17cm h=2448cm h=24.48m