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NEET PHYSICSThermal Properties of MatterMedium

Question

The quantities of heat required to raise the temperature of two solid copper spheres of radii r1r_1 and r2r_2 (r1=1.5r2r_1=1.5 r_2) through 1 K1\text{ K} are in the ratio:

A

94\frac{9}{4}

B

32\frac{3}{2}

C

53\frac{5}{3}

D

278\frac{27}{8}

Step-by-Step Solution

The quantity of heat required to raise the temperature of a body is given by ΔQ=mcΔT\Delta Q = mc\Delta T. The mass of a solid sphere is m=Volume×Density=43πr3ρm = \text{Volume} \times \text{Density} = \frac{4}{3}\pi r^3 \rho. Therefore, ΔQ=43πr3ρcΔT\Delta Q = \frac{4}{3}\pi r^3 \rho c \Delta T. For both copper spheres, the density ρ\rho, specific heat capacity cc, and the change in temperature ΔT\Delta T (1 K1\text{ K}) are the same. Thus, the heat required is directly proportional to the cube of the radius: ΔQr3\Delta Q \propto r^3. The ratio of the quantities of heat required is: ΔQ1ΔQ2=(r1r2)3\frac{\Delta Q_1}{\Delta Q_2} = \left(\frac{r_1}{r_2}\right)^3 Given r1=1.5r2=32r2r_1 = 1.5 r_2 = \frac{3}{2} r_2, we have r1r2=32\frac{r_1}{r_2} = \frac{3}{2}. Substituting this value: ΔQ1ΔQ2=(32)3=278\frac{\Delta Q_1}{\Delta Q_2} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}

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 Matterquantitiesrequiredtemperaturecopperspheres

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