A wall is made up of two layers A and B. The | vedclass

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

Question

(B) Let the thickness of each layer be $x$ and the thermal conductivity of layer $B$ be $K$. Then the thermal conductivity of layer $A$ is $2K$.In ste

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let the thickness of each layer be $x$ and the thermal conductivity of layer $B$ be $K$. Then the thermal conductivity of layer $A$ is $2K$.In steady state,the rate of heat flow $H$ through both layers is the same.$H = \frac{Q}{t} = \frac{K_A A \Delta T_A}{x} = \frac{K_B A \Delta T_B}{x}$Since the cross-sectional area $A$ and thickness $x$ are the same for both layers,we have:$K_A \Delta T_A = K_B \Delta T_B$$(2K) \Delta T_A = (K) \Delta T_B$$\Delta T_B = 2 \Delta T_A$The total temperature difference is given as $\Delta T_A + \Delta T_B = 36\,^{\circ}C$.Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:$\Delta T_A + 2 \Delta T_A = 36\,^{\circ}C$$3 \Delta T_A = 36\,^{\circ}C$$\Delta T_A = 12\,^{\circ}C$Therefore,the temperature difference across layer $A$ is $12\,^{\circ}C$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

Similar Questions

The smallest positive root of the equation $\tan x - x = 0$ lies in the interval:

$6 \Omega$ and $12 \Omega$ resistors are connected in parallel. This combination is connected in series with a $10 \text{ V}$ battery and a $6 \Omega$ resistor. What is the potential difference between the terminals of the $12 \Omega$ resistor (in $\text{ V}$)?

The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5 \, A$ and is decreasing at the rate of $10^3 \, A/s$,then $V_A - V_B$ is....... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module