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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.

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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
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