A certain pressure P is applied to 1 litre of water and 2 litre o… — Solution | JEE Physics

MCQEasyJEE 2023Fluid Pressure & Pascal's Law

A certain pressure $P$ is applied to $1 \, \text{litre}$ of water and $2 \, \text{litre}$ 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 _____.

A
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The same pressure $P$ is applied separately to water and a liquid. Water is compressed by $0.01\%$ and the liquid by $0.03\%$ .

Find: The value of $x$ if

\frac{B_{\text{water}}}{B_{\text{liquid}}} = \frac{3}{x}

Using the bulk modulus relation shown in the solution:

B_{\text{water}} = \frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)}

Since water is compressed by $0.01\%$ ,

B_{\text{water}} = \frac{-\Delta P}{0.01} \times 100

For the liquid compressed by $0.03\%$ ,

B_{\text{liquid}} = \frac{-\Delta P}{0.03} \times 100

Therefore,

\frac{B_{\text{water}}}{B_{\text{liquid}}} = 3

Comparing with

\frac{B_{\text{water}}}{B_{\text{liquid}}} = \frac{3}{x}

we get

x = 1

Therefore, the correct option is A.

Common mistakes

Using the given volumes $1 \, \text{litre}$ and $2 \, \text{litre}$ directly in the ratio is incorrect because bulk modulus depends on fractional compression $\left(\frac{\Delta V}{V}\right)$ , not on the initial volume alone. Use percentage compression for each liquid.
Treating $0.01\%$ as $0.01$ is incorrect because percentage must be interpreted properly. The given solution writes it as division by $0.01$ followed by multiplication by $100$ . Do not ignore the percent sign.
Dropping the ratio comparison step is a common mistake. After finding that
$\frac{B_{\text{water}}}{B_{\text{liquid}}} = 3$
you must compare it with
$\frac{3}{x}$
to obtain $x = 1$ .

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