The volume of an air bubble is doubled as it | vedclass

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Question

(a)

$2 P_0=P_0+ho g h$

$\Rightarrow P_0=ho g h$

$\Rightarrow P_0=75 \,cm 	ext { mercury } \quad 	ext { [Atmospheric pressure] }$

$\R

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(a)

$2 P_0=P_0+ho g h$

$\Rightarrow P_0=ho g h$

$\Rightarrow P_0=75 \,cm 	ext { mercury } \quad 	ext { [Atmospheric pressure] }$

$\Rightarrow ho_{	ext {mercury }} 	imes g 	imes \frac{75}{100}=ho_{	ext {water }} 	imes g 	imes h$

$\Rightarrow \frac{ho_m}{ho_w} 	imes \frac{75}{100}=h$

$\Rightarrow \frac{40}{3} 	imes \frac{75}{100}=h \quad\left[\because \frac{ho_m}{ho_w}=\frac{40}{3} 	ext { (given) }ight]$

$\Rightarrow h=10 \,m$

The volume of an air bubble is doubled as it rises from the bottom of lake to its surface. The atmospheric pressure is $75 \,cm$ of mercury. The ratio of density of mercury to that of lake water is $\frac{40}{3}$. The depth of the lake in metre is .........

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