Question

A certain pressure ' $P$ ' is applied to $1$ litre of water and $2$ litre of a liquid separately Water gets compressed to $0.01 \%$ whereas the liquid gets compressed to $0.03 \%$ The ratio of Bulk modulus of water to that of the liquid is $\frac{3}{x}$ The value of $x$ is ________

Answer 1

Correct Answer: 1

To solve this problem, we need to understand the concept of Bulk modulus, which is defined as the ratio of pressure change to the relative change in volume. The formula for Bulk modulus \( B \) is:

\( B = \frac{-\Delta P}{\Delta V/V} \)

Given that pressure \( P \) is applied and results in volume compression, we can consider:

1. Initial volume of water \( V = 1 \, \text{litre} \).
2. Volume change for water: \( \Delta V/V = 0.01\% = 0.0001 \).
3. Bulk modulus of water, \( B_1 = \frac{P}{0.0001} \).

For the liquid:

1. Initial volume of liquid \( V = 2 \, \text{litres} \) (the initial volume doesn't affect the ratio of compression).
2. Volume change for the liquid: \( \Delta V/V = 0.03\% = 0.0003 \).
3. Bulk modulus of the liquid, \( B_2 = \frac{P}{0.0003} \).

We are given the ratio \( \frac{B_1}{B_2} = \frac{3}{x} \). Substituting the expressions for \( B_1 \) and \( B_2 \), we have:

\( \frac{\frac{P}{0.0001}}{\frac{P}{0.0003}} = \frac{3}{x} \)

Upon simplifying:

\( \frac{0.0003}{0.0001} = \frac{3}{x} \)

\( 3 = \frac{3}{x} \)

Thus, \( x = 1 \).

This result fits the given range, confirming the correct computation.