Two rectangular blocks,having identical dimen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the length of each block be $L$,cross-sectional area be $A$,and thermal resistance of the block with conductivity $k$ be $R = \frac{L}{kA}$. T

Vedclass · Accepted Answer

$2.0$

Vedclass · Answer

(A) Let the length of each block be $L$,cross-sectional area be $A$,and thermal resistance of the block with conductivity $k$ be $R = \frac{L}{kA}$. Then the resistance of the block with conductivity $2k$ is $R' = \frac{L}{2kA} = \frac{R}{2}$.In configuration $I$ (series): The blocks are in series. The equivalent thermal resistance is $R_{eq,I} = R + \frac{R}{2} = \frac{3R}{2}$.The heat current $H_I = \frac{\Delta T}{R_{eq,I}} = \frac{2\Delta T}{3R}$.In configuration $II$ (parallel): The blocks are in parallel. The equivalent thermal resistance is given by $\frac{1}{R_{eq,II}} = \frac{1}{R} + \frac{1}{R/2} = \frac{1}{R} + \frac{2}{R} = \frac{3}{R}$. So,$R_{eq,II} = \frac{R}{3}$.The heat current $H_{II} = \frac{\Delta T}{R_{eq,II}} = \frac{3\Delta T}{R}$.Since the amount of heat $Q$ transported is the same,$Q = H_I t_I = H_{II} t_II$.$t_{II} = t_I 	imes \frac{H_I}{H_{II}} = 9 	imes \frac{2\Delta T / 3R}{3\Delta T / R} = 9 	imes \frac{2}{9} = 2.0 \ s$.

Similar Questions

Find the effective thermal resistance between $A$ and $B$ of a cube made up of $12$ rods of the same dimensions,with the given thermal conductivities as shown in the figure. [ $l =$ length of rod,$a =$ cross-sectional area of rod ]

Two rods whose lengths are $l_1$ and $l_2$ with thermal conductivity coefficients $k_1$ and $k_2$ are placed end to end. The thermal conductivity coefficient of a uniform rod of length $l_1+l_2$ whose thermal resistance is the same as that of the system of these two rods is

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at $100^{\circ} C$ and $10^{\circ} C$ respectively. The temperature of junction $C$ will be $.... ^{\circ} C$.

$A$ rod $CD$ of thermal resistance $10.0 \; K/W$ is joined at the middle of an identical rod $AB$ as shown in the figure. The ends $A$,$B$,and $D$ are maintained at $200^{\circ}C$,$100^{\circ}C$,and $125^{\circ}C$ respectively. The heat current in $CD$ is $P$ watt. The value of $P$ is ... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module