A rod of 40 cm in length and temperature difference of 80- C between its two ends- Another rod B of length 60 cm and of temperature difference 90- C- have the same area of cross-section- If the rate of flow of heat is the same- then the ratio of their thermal conductivities will be

DQ/dt=KA(θ _1 θ _2)/d ⇒K_1Δθ _1/l_1=K_2Δθ _2/l_2 ( ∵dQ/dt and A are same ) ⇒K_1× 80/40=K_2× 90/60⇒K_1/K_2=3/4 ...

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Standard XI

Physics

Conduction Law

A rod of 40 c...

Question

A rod of 40 cm in length and temperature difference of $80^{\circ}$ C between its two ends. Another rod B of length 60 cm and of temperature difference $90^{\circ}$ C, have the same area of cross-section. If the rate of flow of heat is the same, then the ratio of their thermal conductivities will be

3 : 4

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4 : 3

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1 : 2

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2 : 1

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Solution

The correct option is A 3 : 4
$\frac{d Q}{d t} = \frac{K A (θ_{1} - θ_{2})}{d}$
$\Rightarrow \frac{K_{1} Δ θ_{1}}{l_{1}} = \frac{K_{2} Δ θ_{2}}{l_{2}}$ $(∵ \frac{d Q}{d t} a n d A a r e s a m e)$
$\Rightarrow \frac{K_{1} \times 80}{40} = \frac{K_{2} \times 90}{60} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{3}{4}$

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A rod of 40 cm in length and temperature difference of 80∘ C between its two ends. Another rod B of length 60 cm and of temperature difference 90∘ C, have the same area of cross-section. If the rate of flow of heat is the same, then the ratio of their thermal conductivities will be

The correct option is A 3 : 4 dQdt=KA(θ1−θ2)d ⇒K1Δθ1l1=K2Δθ2l2 (∵dQdt and A are same) ⇒K1×8040=K2×9060⇒K1K2=34