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Standard XIII

Physics

Bulk Modulus

If a rubber b...

Question

If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is $1 \times 10^{3} k g / m^{3}$ and $g = 10 m / s^{2}$ , then the volume elasticity in $n / m^{2}$ will be

$10^{8}$

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$2 \times 10^{8}$

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$10^{9}$

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$2 \times 10^{9}$

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Solution

The correct option is D
$2 \times 10^{9}$

$K = \frac{△ P}{\frac{△ V}{V}} = \frac{h ρ g}{\frac{△ V}{V}} = \frac{200 \times 10^{3} \times 10}{\frac{0.1}{100}} = 2 \times 10^{9}$

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If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is 1×103 kg/m3 and g=10 m/s2, then the volume elasticity in n/m2 will be

The correct option is D 2×109 K=△P△VV=hρg△VV=200×103×100.1100=2×109

If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is $1 \times 10^{3} k g / m^{3}$ and $g = 10 m / s^{2}$ , then the volume elasticity in $n / m^{2}$ will be

The correct option is D
$2 \times 10^{9}$

$K = \frac{△ P}{\frac{△ V}{V}} = \frac{h ρ g}{\frac{△ V}{V}} = \frac{200 \times 10^{3} \times 10}{\frac{0.1}{100}} = 2 \times 10^{9}$