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NEET PHYSICSThermal Properties of MatterMedium

Question

The value of coefficient of volume expansion of glycerin is 5×104 K15 \times 10^{-4}\text{ K}^{-1}. The fractional change in the density of glycerin for a rise of 40C40^{\circ}\text{C} in its temperature is -

A

0.0150.015

B

0.0200.020

C

0.0250.025

D

0.0100.010

Step-by-Step Solution

Let the initial density be ρ\rho and volume be VV. With a rise in temperature ΔT\Delta T, the new volume V=V(1+γΔT)V' = V(1 + \gamma \Delta T). The new density ρ=mV=mV(1+γΔT)=ρ(1+γΔT)1ρ(1γΔT)\rho' = \frac{m}{V'} = \frac{m}{V(1 + \gamma \Delta T)} = \rho (1 + \gamma \Delta T)^{-1} \approx \rho (1 - \gamma \Delta T) (since γΔT1\gamma \Delta T \ll 1). The change in density is Δρ=ρρ=ρρ(1γΔT)=ργΔT\Delta \rho = \rho - \rho' = \rho - \rho (1 - \gamma \Delta T) = \rho \gamma \Delta T. The fractional change in density is Δρρ=γΔT\frac{\Delta \rho}{\rho} = \gamma \Delta T. Given γ=5×104 K1\gamma = 5 \times 10^{-4}\text{ K}^{-1} and ΔT=40C=40 K\Delta T = 40^{\circ}\text{C} = 40\text{ K}. Fractional change =5×104×40=200×104=0.020= 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Mattercoefficientvolumeexpansionglycerinfractional

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