Two conducting rods A and B of same length and cross-sectional area are connected i In series ii In parallel as shown- In both combination a temperature difference of 100- C- is maintained- If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is

Heat current H=Δθ/R⇒H_p/H_s=R_s/R_p In first case : R_s=R_1+R_2=l/(3K)A+l/KA =4/3l/KA In second case : R_p=R_1R_2/R_1+R_2=1/(3K)A×l/KA/ ( l/(3K)A+l/KA )=l/4KA ...

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Byju's Answer

Standard XII

Physics

Thermal Resistance

Two conductin...

Question

Two conducting rods A and B of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of $100^{\circ}$ C is maintained. If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is

$\frac{16}{3}$

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$\frac{3}{16}$

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$\frac{1}{1}$

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$\frac{1}{3}$

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Solution

The correct option is A
$\frac{16}{3}$

Heat current $H = \frac{Δ θ}{R} \Rightarrow \frac{H_{p}}{H_{s}} = \frac{R_{s}}{R_{p}}$
In first case : $R_{s} = R_{1} + R_{2} = \frac{l}{(3 K) A} + \frac{l}{K A} = \frac{4}{3} \frac{l}{K A}$
In second case : $R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{\frac{1}{(3 K) A} \times \frac{l}{K A}}{(\frac{l}{(3 K) A} + \frac{l}{K A})} = \frac{l}{4 K A}$
$∴ \frac{H_{p}}{H_{s}} = \frac{\frac{4 l}{3 K A}}{\frac{l}{4 K A}} = \frac{16}{3}$

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The correct option is A 163 Heat current H=ΔθR⇒HpHs=RsRp In first case : Rs=R1+R2=l(3K)A+lKA=43lKA In second case : Rp=R1R2R1+R2=1(3K)A×lKA(l(3K)A+lKA)=l4KA ∴HpHs=4l3KAl4KA=163