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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Thermal Properties of Matter

BITSAT 2016 · Physics · Thermal Properties of Matter

Two spheres of different materials one with double the radius and...

Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is 25 minute and for smaller one is 16 minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

Options

  1. A4: 5
  2. B5: 4
  3. C25: 8
  4. D8: 25

Correct answer

D. 8: 25

Step-by-step solution

Radius of small sphere =r=\mathrm{r} Thickness of small sphere =t=\mathrm{t} Radius of bigger sphere =2r=2 \mathrm{r} Thickness of bigger sphere =t/4=\mathrm{t} / 4 Mass of ice melted =(=( volume of sphere )×) \times (density of ice) Let K1\mathrm{K}_{1} and K2\mathrm{K}_{2} be the thermal conductivities of larger and smaller sphere. For bigger sphere, K14π(2r)2×100t/4=43π(2r)3ρL25×60\frac{K_{1} 4 \pi(2 r)^{2} \times 100}{t / 4}=\frac{\frac{4}{3} \pi(2 r)^{3} \rho L}{25 \times 60} For smaller sphere, K2×4π\mathrmr2×100t=43π(2r)3ρL25×60\frac{\mathrm{K}_{2} \times 4 \pi \mathrmr^{2} \times 100}{\mathrm{t}}=\frac{\frac{4}{3} \pi(2 \mathrm{r})^{3} \rho \mathrm{L}}{25 \times 60}
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