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 aligned Δ T & =30^ C V^ & =V+Δ V=V(l+γ Δ T) aligned V^ =V(1+49 × 10^ -5 × 30)=1.0147 ~V = m V and ^ = m V^ = m 0.0147 =0.9855 Hence, fractional change in density = - ^ =( -0.9855 )/( )=0.0145

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Thermal Properties of Matter

JIPMER 2018 · Physics · Thermal Properties of Matter

The coefficient of volume expansion of glycerine is 49 × 10^ -50 C^...

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?

Options

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

Correct answer

B. $0.0145$

Step-by-step 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

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