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Properties of Matter question

2002 · Q163
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Properties of Matter question

2002 · Q163

NEETPhysicsProperties of MatterMCQ+4 / −1
Consider two rods of same length and different specific heats (S1, S2), conductivities (K1, K2) and area of cross-sections (A1, A2) and both having temperatures T1 and T2 at their ends. If rate of loss of heat due to conduction is equal, then
  1. A
    K1A1 = K2A2
  2. B
    K1A1S1=K2A2S2{{{K_1}{A_1}} \over {{S_1}}} = {{{K_2}{A_2}} \over {{S_2}}}S1​K1​A1​​=S2​K2​A2​​
  3. C
    K2A1 = K1A2
  4. D
    K2A1S2=K1A2S1{{{K_2}{A_1}} \over {{S_2}}} = {{{K_1}{A_2}} \over {{S_1}}}S2​K2​A1​​=S1​K1​A2​​
View written solutionFree

Correct answer: A

Rate of heat loss in rod 1 = Q1 =K1A1(T1−T2)l1 = {{{K_1}{A_1}\left( {{T_1} - {T_2}} \right)} \over {{l_1}}}=l1​K1​A1​(T1​−T2​)​

Rate of heat loss in rod 2 = Q2 =K2A2(T1−T2)l2 = {{{K_2}{A_2}\left( {{T_1} - {T_2}} \right)} \over {{l_2}}}=l2​K2​A2​(T1​−T2​)​

By problem, Q1 = Q2.

∴K1A1(T1−T2)l1=K2A2(T1−T2)l2 \therefore {{{K_1}{A_1}\left( {{T_1} - {T_2}} \right)} \over {{l_1}}} = {{{K_2}{A_2}\left( {{T_1} - {T_2}} \right)} \over {{l_2}}}∴l1​K1​A1​(T1​−T2​)​=l2​K2​A2​(T1​−T2​)​

∴\therefore∴ K1A1 = K2A2.   [∵l1=l2]\left[ \because{{l_1} = {l_2}} \right][∵l1​=l2​]

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