Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity K1 , K2 and K3 , respecrtively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100oC and the other at 0oC (see figure). If the joints of the rod are at 70oC and 20oC in steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1 , K2 and K3 is :

K1 : K3 = 2 : 3, K2 : K3 = 2 : 5

K1 : K2 = 5 : 2, K1 : K3 = 3 : 5

Heat and Thermodynamics PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q38 with Solution

View written solutionFree

Correct answer: A

Use steady-state heat conduction

In steady state, the same heat current flows through all three rods because:

they are connected in series,
cross-sectional areas are identical,
there is no heat loss from the surface.

So,

H = \frac{K A \Delta T}{L}

is the same for each rod.

Since all rods have the same $A$ and $L$ , we get

K_1 \Delta T_1 = K_2 \Delta T_2 = K_3 \Delta T_3

Find temperature drops across each rod

From the figure/data:

left end = $100^\circ C$
first joint = $70^\circ C$
second joint = $20^\circ C$
right end = $0^\circ C$

Therefore,

\Delta T_1 = 100 - 70 = 30^\circ C

\Delta T_2 = 70 - 20 = 50^\circ C

\Delta T_3 = 20 - 0 = 20^\circ C

Apply equality of heat current

Thus,

K_1(30) = K_2(50) = K_3(20)

Let the common value be $C$ . Then,

K_1 = \frac{C}{30}, \quad K_2 = \frac{C}{50}, \quad K_3 = \frac{C}{20}

Hence,

K_1 : K_2 : K_3 = \frac{1}{30} : \frac{1}{50} : \frac{1}{20}

Taking LCM or simplifying,

K_1 : K_2 : K_3 = 10 : 6 : 15

Check the required relations

Now,

K_1 : K_3 = 10 : 15 = 2 : 3

and

K_2 : K_3 = 6 : 15 = 2 : 5

This matches Option A.

Also,

K_3 > K_1 > K_2

so options B, C, and D are incorrect.

Final answer

The correct relationship is:

K_1 : K_3 = 2:3, \qquad K_2 : K_3 = 2:5

So the correct option is A.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: A

Use steady-state heat conduction

In steady state, the same heat current flows through all three rods because:

they are connected in series,
cross-sectional areas are identical,
there is no heat loss from the surface.

So,

H = \frac{K A \Delta T}{L}

is the same for each rod.

Since all rods have the same $A$ and $L$ , we get

K_1 \Delta T_1 = K_2 \Delta T_2 = K_3 \Delta T_3

Find temperature drops across each rod

From the figure/data:

left end = $100^\circ C$
first joint = $70^\circ C$
second joint = $20^\circ C$
right end = $0^\circ C$

Therefore,

\Delta T_1 = 100 - 70 = 30^\circ C

\Delta T_2 = 70 - 20 = 50^\circ C

\Delta T_3 = 20 - 0 = 20^\circ C

Apply equality of heat current

Thus,

K_1(30) = K_2(50) = K_3(20)

Let the common value be $C$ . Then,

K_1 = \frac{C}{30}, \quad K_2 = \frac{C}{50}, \quad K_3 = \frac{C}{20}

Hence,

K_1 : K_2 : K_3 = \frac{1}{30} : \frac{1}{50} : \frac{1}{20}

Taking LCM or simplifying,

K_1 : K_2 : K_3 = 10 : 6 : 15

Check the required relations

Now,

K_1 : K_3 = 10 : 15 = 2 : 3

and

K_2 : K_3 = 6 : 15 = 2 : 5

This matches Option A.

Also,

K_3 > K_1 > K_2

so options B, C, and D are incorrect.

Final answer

The correct relationship is:

K_1 : K_3 = 2:3, \qquad K_2 : K_3 = 2:5

So the correct option is A.

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics