Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of 2.0 x 10 $^{5}$ N m $^{- 2}$ from the following data. Density of kerosene = 800 kg m $^{- 3}$ and speed of sound in kerosene = 1330 m s $^{- 1}$ .

0.14 cm³

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2 cm³

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14 cm³

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cannot be determined

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Solution

The correct option is A
0.14 cm³

$v = \sqrt{\frac{B}{ρ}} \Rightarrow 1330 = s q r t \frac{B}{800} \Rightarrow (1330)^{2} \times 800 = B$

$B = 1.41 \times 10^{9} N m^{- 2}$

Now

$B = \frac{Δ P}{\frac{(Δ v}{v}}$ (By definition)

$\frac{1.41 \times 10^{9} \times D e l t a v}{10^{- 3} m^{3}} = - (2 \times 10^{5})$

$Δ v = - \frac{2 \times 10^{2}}{1.41 \times 10^{9}}$

$D e l t a = - 1.41 \times 10^{- 7} m^{3}$

$Δ = - 0.14 c m^{3}$

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Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of $2.0 \times 10^{5} N m^{- 2}$ from the following data. Density of kerosene $= 800 k g m^{- 3}$ and speed of sound in kerosene = $1330 m s^{- 1}$ .

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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 x 105 N m−2 from the following data. Density of kerosene = 800 kg m−3 and speed of sound in kerosene = 1330 m s−1.

The correct option is A 0.14 cm3 v=√Bρ⇒ 1330=sqrtB800⇒ (1330)2 × 800=B B=1.41×109Nm−2 Now B=ΔP(Δvv (By definition) 1.41×109×Deltav10−3m3=−(2×105) Δv=−2×1021.41×109 Delta=−1.41×10−7m3 Δ=−0.14 cm3

Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of 2.0 x 10 $^{5}$ N m $^{- 2}$ from the following data. Density of kerosene = 800 kg m $^{- 3}$ and speed of sound in kerosene = 1330 m s $^{- 1}$ .