# Find the Change in the Volume of 1.0 Litre Kerosene When It is Subjected to an Extra Pressure of 2.0 × 105 N M−2 from the Following Data. - Physics

Sum

Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of 2.0 × 105 N m−2 from the following data. Density of kerosene = 800 kg m−3and speed of sound in kerosene = 1330 ms−1.

#### Solution

Given:
Volume of kerosene V = 1 litre =$1 \times {10}^{- 3} m^3$

Pressure applied P = 2.0 × 10^5 "Nm"^(- 2)

Density of kerosene ρ  = 800 kgm−3
Speed of sound in kerosene v  = 1330 ms−1
Change in volume of kerosene $∆ V$= ?
The velocity in terms of the bulk modulus $\left( K \right)$ and density $\left( \rho \right)$is given by :

$v = \sqrt{\left( \frac{K}{p} \right)}$,

$\text { where }$

$K = v^2 \rho .$

$\Rightarrow K = \left( 1330 \right)^2 \times 800 N/ m^2$

$\text { As we know, }$

$K = \frac{\left( \frac{F}{A} \right)}{\left( \frac{∆ V}{V} \right)} .$

$\therefore ∆ V = \frac{\text { Pressure }\times V}{K} \left( \because P = \frac{F}{A} \right)$

$\text { On substituting the respective values, we get: }$ $∆ V = \frac{2 \times {10}^5 \times 1 \times {10}^{- 3}}{1330 \times 1330 \times 800} = 0 . 14 {cm}^3$

Therefore, the change in the volume of kerosene ∆V = 0.14 cm3.

Concept: Speed of Wave Motion
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 15 | Page 353