Two rods (one semi-circular and other straight) of the same material and of same cross-sectional area are joined as shown in the diagram. The points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross-section of the straight rod in a given time is

Question

Two rods (one semi-circular and other straight) of the same material and of same cross-sectional area are joined as shown in the diagram. The points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross-section of the straight rod in a given time is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-koknwrqm16k7jxaw.jpg" /> (A) 2: π  (B) 1:2 (C) π: 2 (D) 3:2

Tardigrade Admin · Accepted Answer

(d Q/d t)=(K A Δ θ /l) For both rods K, A and Δ θ are same. ∴ (d Q/d t) propto (1/l) So, ((d Q / d t) textsemicircular/(d Q / d t) textstraight)=(l textstraight/l textsemicircular)=(2 r/π r)=(2/π )

$2 : π$

$1 : 2$

$π : 2$

$3 : 2$

$\frac{d Q}{d t} = \frac{K A Δ θ}{l}$
For both rods $K, A$ and $Δ θ$ are same.
$∴ \frac{d Q}{d t} \propto \frac{1}{l}$
So, $\frac{( d Q / d t ) _{semicircular}}{( d Q / d t ) _{straight}} = \frac{l _{straight}}{l _{semicircular}} = \frac{2 r}{π r} = \frac{2}{π}$

2:π

1:2

π:2

3:2

dtdQ​=lKAΔθ​ For both rods K,A and Δθ are same. ∴dtdQ​∝l1​ So, (dQ/dt)straight​(dQ/dt)semicircular​​=lsemicircular​lstraight​​=πr2r​=π2​

$2 : π$

$1 : 2$

$π : 2$

$3 : 2$

$\frac{d Q}{d t} = \frac{K A Δ θ}{l}$
For both rods $K, A$ and $Δ θ$ are same.
$∴ \frac{d Q}{d t} \propto \frac{1}{l}$
So, $\frac{( d Q / d t ) _{semicircular}}{( d Q / d t ) _{straight}} = \frac{l _{straight}}{l _{semicircular}} = \frac{2 r}{π r} = \frac{2}{π}$