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Chapter 7: Problem 924

If a rubber ball is taken at the depth of \(200 \mathrm{~m}\) in a pool, Its volume decreases by \(0.1 \%\). If the density of the water is $1 \times 10^{3}\left(\mathrm{~kg} / \mathrm{m}^{3}\right) \& \mathrm{~g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)$. Then what will be the volume elasticity in ? (A) \(10^{8}\) (B) \(2 \times 10^{8}\) (C) \(10^{9}\) (D) \(2 \times 10^{9}\)

Short Answer

Expert verified
The short answer is: The volume elasticity of the rubber ball is (D) \(2 \times 10^9 ~\mathrm{Pa}\).

Step by step solution

01

Calculate the change in pressure (dP)

First, let's calculate the change in pressure at a depth of 200 meters. The formula to calculate the pressure at a depth h is: \(P = ρgh\), where P is the pressure, ρ is the density of the fluid (water), g is the acceleration due to gravity, and h is the depth. In this case, ρ = \(1 \times 10^3 ~\mathrm{kg/m^3}\), g = \(10 ~\mathrm{m/s^2}\), and h = 200 m. Plugging these values, we get: \(P = (1 \times 10^3) \times (10) \times (200) = 2 \times 10^6 ~\mathrm{Pa}\).
02

Determine the relative change in volume (dV/V)

The ball's volume decreases by 0.1%. To express this as a fraction, divide by 100: \( \frac{dV}{V} = -\frac{0.1}{100} = -1 \times 10^{-3} \).
03

Calculate the volume elasticity (K)

Now, we can plug in the values we found in Steps 1 and 2 into the formula for volume elasticity: \(K = -V\frac{dP}{dV} = -V \frac{2 \times 10^6}{-1 \times 10^{-3}} \). Cancel out the negative sign and we get: \(K = V \times 2\times 10^9\). Since we are only looking for the ratio of volume elasticity (K), we do not need the exact volume of the ball, and therefore, the answer is: \(K = 2 \times 10^9 ~\mathrm{Pa}\) The correct answer is (D) \(2 \times 10^9\).

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