Find the change in the volume of 1-0 litre kerosene when it is subject to an extra pressure of 2-0- 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-

We know that v=√(K/ ρ) where K = bulk modulus of elasticity #160; ⇒ K=v^2ρ =(1330)^2× 800 N/m^2 We know K=( F/AΔ V/V) Δ V=PressureK=2× ...

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Standard XII

Physics

1st Equation of Motion

Find the chan...

Question

Find the change in the volume of $1.0$ litre kerosene when it is subject 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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Solution

We know that $v = \sqrt{K / ρ}$
where $K$ = bulk modulus of elasticity

$\Rightarrow K = v^{2} ρ = (1330)^{2} \times 800 N / m^{2}$

We know $K = (\frac{F / A}{Δ V / V})$

$Δ V = \frac{P r e s s u r e}{K} = \frac{2 \times 10^{5}}{1330 \times 1330 \times 800}$

So, $Δ V = 0.15 c m^{3}$

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Find the change in the volume of $1.0 litre$ kerosene when it is subjected to an extra pressure of $2.0 \times 10^{5} {Nm}^{- 2}$ from the following data. Density of kerosene = $800 {kgm}^{- 3}$ and speed of sound in kerosene $1300 {ms}^{- 1}$ .

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

We know that v=√K/ρwhere K= bulk modulus of elasticity ⇒K=v2ρ=(1330)2×800 N/m2We know K=(F/AΔV/V)ΔV=PressureK=2×1051330×1330×800So, ΔV=0.15 cm3

Find the change in the volume of $1.0$ litre kerosene when it is subject 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 $= 1330 m s^{- 1}$ .

We know that $v = \sqrt{K / ρ}$
where $K$ = bulk modulus of elasticity

$\Rightarrow K = v^{2} ρ = (1330)^{2} \times 800 N / m^{2}$

We know $K = (\frac{F / A}{Δ V / V})$

$Δ V = \frac{P r e s s u r e}{K} = \frac{2 \times 10^{5}}{1330 \times 1330 \times 800}$

So, $Δ V = 0.15 c m^{3}$