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Standard IX

Physics

Factors Affecting Pressure

A rock weighs...

Question

A rock weighs $1400 N$ in air , has apparent weight of $900 N$ when submerged in fresh water $(998 K g / m^{3})$ . The volume ( in $m^{3}$ ) of the rock is $(g = 9.8 m / s^{2})$ :

0.14

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0.60

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0.90

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0.05

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Solution

The correct option is D $0.05$
Loss of weight $= 1400 - 900 = 500 N$
Let
V=volume of rock
$ρ$ =density of water $= 998 k g / m^{3}$
Loss of weight=Weight of displaced liquid

$500 = ρ V g$

$V = \frac{500}{998 \times 9.8}$

$V = 0.05 m^{3}$

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A rock weighs 1400 N in air , has apparent weight of 900 N when submerged in fresh water (998 Kg/m3). The volume ( in m3) of the rock is (g=9.8 m/s2):

The correct option is D 0.05Loss of weight =1400−900=500NLetV=volume of rockρ=density of water=998kg/m3Loss of weight=Weight of displaced liquid500=ρVgV=500998×9.8V=0.05m3

A rock weighs $1400 N$ in air , has apparent weight of $900 N$ when submerged in fresh water $(998 K g / m^{3})$ . The volume ( in $m^{3}$ ) of the rock is $(g = 9.8 m / s^{2})$ :

The correct option is D $0.05$
Loss of weight $= 1400 - 900 = 500 N$
Let
V=volume of rock
$ρ$ =density of water $= 998 k g / m^{3}$
Loss of weight=Weight of displaced liquid

$500 = ρ V g$

$V = \frac{500}{998 \times 9.8}$

$V = 0.05 m^{3}$