Ncert Exercises & Solutions

Q. 21 To what depth must a rubber ball be taken in deep-sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is $9.8 \times 10^{8} N / m^{2}$ ; and the density of seawater is $10^{3} kg / m^{3}$ )

Hint: Use the concept of Bulk modulus.

Step 1: Find the fractional change in the volume of the ball.

Given, Bulk modulus of rubber (K) =

$9.8 \times 10^{8} N / m^{2}$

Density of sea water (

$ρ$ ) =

$10^{3} kg / m^{3}$

Percentage decrease in volume,

$(\frac{∆ V}{V} \times 100) = 0.1 \Rightarrow \frac{∆ V}{V} = \frac{0.1}{100}$

$\Rightarrow \frac{∆ V}{V} = \frac{1}{1000}$

Step 2: Find the depth h.

Let the rubber ball be taken up to depth h.

Change in pressure, (P) = h

$ρ$ g

Bulk modulus (K)

$= \frac{P}{∆ V / V} = \frac{hρg}{(∆ V / V)}$

$\Rightarrow h = \frac{K \times (∆ V / V)}{ρg} = \frac{9.8 \times 10^{8} \times \frac{1}{1000}}{10^{3} \times 9.8} = 100 m$

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