A ball falling in a lake of depth 200 m shows 0.1 % decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water =1000 kg / m 3 )

Question

A ball falling in a lake of depth 200 m shows 0.1 % decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water =1000 kg / m 3 ) (A)  19.6× 108N/m2  (B)  19.6× 10-10N/m2  (C)  19.6× 1010N/m2  (D)  19.6× 10-8N/m2

Tardigrade Admin · Accepted Answer

When strain is small, the ratio of normal stress to the volume-strain is called the bulk modulus of the material of the body.

K = \frac{p}{Δ V / V}

where

p

is pressure

(= \frac{F}{A})

, and

\frac{Δ V}{V}

the change in volume
Given,

Δ V = 0.1%, V = 0.001

volt,

p = h ρ g

p = h ρ g p = 200 \times 1000 \times 9.8

= 1.96 \times 1 0^{6} N / m^{2}

∴ K = \frac{1.96 \times 1 0 ^{6}}{0.001}

= 19.6 \times 1 0^{8} N / m^{2}

Q. A ball falling in a lake of depth $200 m$ shows $0.1%$ decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water $= 1000 k g / m^{3}$ )

$19.6 \times 1 0^{8} N / m^{2}$

$19.6 \times 1 0^{- 10} N / m^{2}$

$19.6 \times 1 0^{10} N / m^{2}$

$19.6 \times 1 0^{- 8} N / m^{2}$

Q. A ball falling in a lake of depth 200m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water =1000kg/m3 )

19.6×108N/m2

19.6×10−10N/m2

19.6×1010N/m2

19.6×10−8N/m2

Q. A ball falling in a lake of depth $200 m$ shows $0.1%$ decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water $= 1000 k g / m^{3}$ )

$19.6 \times 1 0^{8} N / m^{2}$

$19.6 \times 1 0^{- 10} N / m^{2}$

$19.6 \times 1 0^{10} N / m^{2}$

$19.6 \times 1 0^{- 8} N / m^{2}$