Two rods $A$ and $B$ of same cross-sectional are $A$ and length $l$ connected in series between a source $(T_1 = 100^o C)$ and a sink $(T_2 = 0^o C)$ as shown in figure. The rod is laterally insulated If $T_A$ and $T_B$ are the temperature drops across the rod $A$ and $B$, then
$\frac{{{T_A}}}{{{T_B}}} = \frac{3}{1}$
$\frac{{{T_A}}}{{{T_B}}} = \frac{1}{3}$
$\frac{{{T_A}}}{{{T_B}}} = \frac{3}{4}$
$\frac{{{T_A}}}{{{T_B}}} = \frac{4}{3}$
Assuming newton's law of cooling to be valid, body at temperature $50^o\ C$ in surrounding of temperature $20^o\ C$ , achieve steady state with help of $100\ W$ heater. If same body has temperature $35^o\ C$ in same surrounding, then power of heater required to maintain steady state ........ $W$
Two rods $A$ and $B$ of same cross-sectional are $A$ and length $l$ connected in series between a source $(T_1 = 100^o C)$ and a sink $(T_2 = 0^o C)$ as shown in figure. The rod is laterally insulated The ratio of the thermal resistance of the rod is
Five wires each of cross-sectional area $A$ and length $l$ are combined as shown. The thermal conductivity of copper and steel are $k_1$ and $k_2$ respectively. The equivalent thermal resistance between $A$ and $C$ is
The rate of emission of radiation of a black body at $273^o C$ is $E$, then the rate of emission of radiation of this body at $0^o C$ will be
If the temperature of the sun were to increase from $T$ to $2T$ and its radius from $R$ to $2R$, then the ratio of the radiant energy received on earth to what it was previously will be