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BITSAT 2016 Paper

Section: Physics

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Question : 7 of 150

Marks: +1, -0

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

$4:5$
$5:4$
$25:8$
$8:25$

Solution:

Radius of small sphere

=r

Thickness of small sphere

=t

Radius of bigger sphere

=2r

Thickness of bigger sphere

=\frac{t}{4}

Mass of ice melted = (volume of sphere) × (density ofice)

Let

K_1

and

K_2

be the thermal conductivities of largerand smaller sphere.

For bigger sphere,

\;\;\frac{ K_1 4\pi (2r)^2 \times 100 }{ \frac{t}{4} } = \;\;\frac{ \;\;\frac{4}{3} \pi (2r)^3 \rho L }{ 25 \times 60 }

For smaller sphere,

\;\;\frac{ K_2 \times 4\pi r^2 \times 100 }{ t } = \;\;\frac{ \;\;\frac{4}{3} \pi r^3 \rho L }{ 16 \times 60 }

\therefore \frac{K_1}{K_2} = \frac{8}{25}

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