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 material of the ball

Question

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 material of the ball (A)  19.6 × 108 N/m2  (B)  19.6 × 10-10 N/m2  (C)  19.6 × 1010N/m2  (D)  19.6 × 10-8 N/m2

Tardigrade Admin · Accepted Answer

The bulk modulus B=(Δ p/Δ V/V)=(hρ g/0.1/100) =(200× 103× 9.8/1/1000) =19.6× 108N/m2

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 material of the ball

$19.6 \times 10^{8} N / m^{2}$

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

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

$19.6 \times 10^{- 8} N / m^{2}$

The bulk modulus $B = \frac{Δ p}{Δ V / V} = \frac{h ρ g}{0.1/100}$ $= \frac{200 \times 10 ^{3} \times 9.8}{1/1000}$ $= 19.6 \times 10^{8} N / m^{2}$

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 material of the ball

19.6×108N/m2

19.6×10−10N/m2

19.6×1010N/m2

19.6×10−8N/m2

The bulk modulus B=ΔV/VΔp​=0.1/100hρg​ =1/1000200×103×9.8​ =19.6×108N/m2

$19.6 \times 10^{8} N / m^{2}$

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

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

$19.6 \times 10^{- 8} N / m^{2}$

The bulk modulus $B = \frac{Δ p}{Δ V / V} = \frac{h ρ g}{0.1/100}$ $= \frac{200 \times 10 ^{3} \times 9.8}{1/1000}$ $= 19.6 \times 10^{8} N / m^{2}$