Question

Find the change in the volume of 1.0-liter kerosene when it is subjected to an extra pressure of 2.0 x 10⁵ N/m² from the following data. Density of kerosene = 800 kg/m³ and speed of sound in kerosene = 1330 m/s.

Answer 1

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{0.15cm^{-3}}$

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EXPLAINATION ➣

$Speed \:of \:sound, v=\sqrt{\dfrac{K}{\rho}}</p><p>\\\Rightarrow\:K=v^2\rho=1330^2\times 800Nm^2\\\\As\: we\: know, K=\dfrac{P}{\Delta V/V}\\\\putting \:the\: values\:we\:get\\\Rightarrow\Delta V=0.15cm^{-3}$

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Answer 2

Answer ⇒ Change in volume is 0.14 cm³.

Explanation ⇒ We know that,

the speed of sound is given by the relation,

V =√(B/ρ)

where B = bulk modulus and ρ = density of the liquid.

On squaring both sides of the equation we get,

B =ρV²

∴ B = 800 × (1330)²

∴ B = 1.42 × 10⁹ N/m²

Now,volume of Kerosene, V = 1 liter

= 10⁻³ m³

Let the change in volume due to the extra pressure be ΔV

Thus,  B = p/(ΔV/V),  

ΔV/V = p/B

ΔV = pv/B

ΔV = 1.4 × 10⁻⁷ m³

ΔV = 0.14 cm³

Hence, change in volume is 0.14 cm³.

Hope it helps.