(B) In a series connection,the rate of heat flow $(H)$ through both rods is the same.The rate of heat flow is given by $H = KA \frac{\Delta T}{l}$,whe

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$\frac{G_A}{G_B} = \frac{1}{3}$

(B) In a series connection,the rate of heat flow $(H)$ through both rods is the same.The rate of heat flow is given by $H = KA \frac{\Delta T}{l}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,and $\Delta T/l$ is the temperature gradient $(G)$.Since $H$ is constant for both rods,we have $H = K_A A G_A = K_B A G_B$.Given $K_A = 3K$ and $K_B = K$,we get $(3K) A G_A = (K) A G_B$.Simplifying this,we get $3 G_A = G_B$.Therefore,the ratio of the temperature gradients is $\frac{G_A}{G_B} = \frac{1}{3}$.

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

Medium

Two rods $A$ and $B$ of the same cross-sectional area $A$ and length $l$ are connected in series between a source $(T_1 = 100^{\circ}C)$ and a sink $(T_2 = 0^{\circ}C)$ as shown in the figure. The rods are laterally insulated. If $G_A$ and $G_B$ are the temperature gradients across rod $A$ (thermal conductivity $3K$) and rod $B$ (thermal conductivity $K$),then:

A
$\frac{G_A}{G_B} = \frac{3}{1}$
B
$\frac{G_A}{G_B} = \frac{1}{3}$
C
$\frac{G_A}{G_B} = \frac{3}{4}$
D
$\frac{G_A}{G_B} = \frac{4}{3}$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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$A$ slab of stone of area $3600 \, cm^2$ and thickness $10 \, cm$ is exposed on the lower surface to steam at $100^{\circ} C$. $A$ block of ice at $0^{\circ} C$ rests on the upper surface of the slab. In one hour, $4.8 \, kg$ of ice is melted. The thermal conductivity of the stone in $J \, s^{-1} \, m^{-1} \, K^{-1}$ is (Latent heat of ice $= 3.36 \times 10^5 \, J/kg$)

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Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same,the rate of flow of heat becomes $H_2$. Then:

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Find the equivalent thermal conductivity of the system.

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Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same,the rate of flow of heat becomes $H_2$. Then:

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