Verified Solution • 2025–26 Session moderate

QAIPM-2002-Ph-2831: NEET 2002 Physics - Class 11 AIPM-2002-Ph-2831 Solution: Consider two rods of same length and different specific heats…

Consider two rods of same length and different specific heats (

S_1, S_2

), conductivities (

K_1, K_2

) and area of cross-sections (

A_1, A_2

) and both having temperatures

T_1

and

T_2

at their ends. If rate of loss of heat due to conduction is equal, then

Multiple Choice Options

A $K_1A_1=K_2A_2$
B $\frac{K_1A_1}{S_1}=\frac{K_2A_2}{S_2}$
C $K_2A_1=K_1A_2$
D $\frac{K_2A_1}{S_2}=\frac{K_1A_2}{S_1}$

Step-by-Step Verified Solution

Rate of heat loss in rod 1 = Q 1 = K 1 A 1 ( T 1 _ T 2 ) l 1 Rate of heat loss in rod 2 = Q 2 = K 2 A 2 ( T 1 _ T 2 ) l 2 By problem, Q1 = Q2. _ K 1 A 1 ( T 1 _ T 2 ) l 1 = K 2 A 2 ( T 1 _ T 2 ) l 2 _ K 1 A 1 = K 2 A 2 . [ _ l 1 = l 2 ]

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