An air bubble rises from bottom to top of a liquid of density 1.5 g/cm $^{3}$ . If its volume is doubled, the depth of liquids is

689 cm

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398 cm

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760 cm

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152 cm

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Solution

The correct option is A 689 cm
In this case, $(P + ρ g h) V$ $=$ $2 V P$
where, $P$ $=$ atm pressure
$ρ$ $=$ density of the liquid
$h$ $=$ height of the lake
$V$ $=$ volume of the bubble
Solving we get,
$h$ $=$ $\frac{P}{ρ g}$
here, $P$ $=$ 76 X 13.6 X 9.8
$ρ$ $=$ 1.5 $g m / c m^{3}$ $=$ 1500 $k g / m^{3}$
Putting these values we get, $h$ $=$ 689 cm

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An air bubble rises from bottom to top of a liquid of density 1.5 g/cm3. If its volume is doubled, the depth of liquids is

The correct option is A 689 cmIn this case, (P+ρgh)V = 2VPwhere, P = atm pressureρ = density of the liquidh = height of the lakeV = volume of the bubbleSolving we get,h = Pρghere, P = 76 X 13.6 X 9.8ρ = 1.5 gm/cm3 = 1500 kg/m3Putting these values we get, h = 689 cm

An air bubble rises from bottom to top of a liquid of density 1.5 g/cm $^{3}$ . If its volume is doubled, the depth of liquids is