JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Heat & Thermodynamics

WhatsApp Email

Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity

κ

and the other

2 κ

. The temperature difference between the ends along the

x

-axis is the same in both the configurations. It takes

9 s

to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is

(A) $2.0 s$ ✓ Correct
(B) $3.0 s$
(C) $4.5 s$
(D) $6.0 s$

Answer: (A)

Solution

In configuration I (series arrangement), the total thermal resistance is:

R_{e q} = \frac{ℓ}{κ A} + \frac{ℓ}{2 κ A} = \frac{3 ℓ}{2 κ A}

In configuration II (parallel arrangement), the equivalent thermal resistance is:

R_{e q}^{'} = \frac{1}{\frac{κ A}{ℓ} + \frac{2 κ A}{ℓ}} = \frac{ℓ}{3 κ A}

Since

Q = \frac{( T _{1} - T _{2} )}{R} t

, we have

t \propto R

.

Therefore:

\frac{t _{2}}{t _{1}} = \frac{R _{2}}{R _{1}} = \frac{ℓ / ( 3 κ A )}{3 ℓ / ( 2 κ A )} = \frac{2}{9}

t_{2} = \frac{2}{9} \times 9 = 2 s

Hence the answer is (A).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option

Practise more: all JEE Advanced PYQs on Heat & Thermodynamics →

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Heat & Thermodynamics

Solution

More JEE Advanced questions on Heat & Thermodynamics