The coefficient of volume expansion of glycerine is 49 × 10^-50 C^-1 . What is the fractional change in density for a 30^° C rise in temperature?

Correct answer: (B) 0.0145. Here, γ=49 × 19^-5^° C^-1 Δ T =30^° C V^′ =V+Δ V=V(l+γ Δ T) ∴ V^′=V (1+49 × 10^-5 × 30 )=1.0147 ~V ∵ ρ=m/V and ρ^′=fracmV^′=m/0.0147=0.9855 ρ Hence, fractional change in density =fracρ-ρ^′ρ=(ρ-0.9855 ρ)/(ρ)=0.0145

Q131 / 180 · JIPMER 2018 · Physics

The coefficient of volume expansion of glycerine is $49 \times 10^{-50} \mathrm{C}^{-1}$. What is the fractional change in density for a $30^{\circ} \mathrm{C}$ rise in temperature?

A
$0.0155$
Correct answer
B
$0.0145$
C
$0.0255$
D
$0.0355$

Solution

Here, $\gamma=49 \times 19^{-5}{ }^{\circ} \mathrm{C}^{-1}$ $\begin{aligned} \Delta \mathrm{T} & =30^{\circ} \mathrm{C} \\ \mathrm{V}^{\prime} & =\mathrm{V}+\Delta \mathrm{V}=\mathrm{V}(\mathrm{l}+\gamma \Delta \mathrm{T})\end{aligned}$ $\therefore \quad \mathrm{V}^{\prime}=\mathrm{V}\left(1+49 \times 10^{-5} \times 30\right)=1.0147 \mathrm{~V}$ $\because \quad \rho=\frac{\mathrm{m}}{\mathrm{V}}$ and $\rho^{\prime}=\frac{\mathrm{m}}{\mathrm{V}^{\prime}}=\frac{\mathrm{m}}{0.0147}=0.9855 \rho$ Hence, fractional change in density $=\frac{\rho-\rho^{\prime}}{\rho}=\frac{\rho-0.9855 \rho}{\rho}=0.0145$

Hence, the correct answer is option: 'B'.

The coefficient of volume expansion of glycerine is $49 \times 10^{-50} \mathrm{C}^{-1}$. What is the fractional change in density for a $30^{\circ} \mathrm{C}$ rise in temperature?

Solution

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