Aring consisting of two parts $ADB$ and $ACB$ of same conductivity $k$ carries an amount of heat $H$. The $ADB$ part is now replaced with another metal keeping the temperatures $T_1$ and $T_2$ constant. The heat carried increases to $2H$. What $ACB$ should be the conductivity of the new$ADB$ part? Given $\frac{{ACB}}{{ADB}}= 3$
A$\frac{7}{3} k$
B$2 k$
C$\frac{5}{2}k$
D$3 k$
Diffcult
Download our app for free and get started
A$\frac{7}{3} k$
a $H_{1}+H_{2}=\frac{K A\left(T_{1}-T_{2}\right)}{3 l}+\frac{K A\left(T_{1}-T_{2}\right)}{l}=\frac{4}{3 l} K A\left(T_{1}-T_{2}\right)$
Now, $H_{2}=2 H-H_{1}=\frac{7 K A\left(T_{1}-T_{2}\right)}{3 l}=\frac{K^{\prime} A\left(T_{1}-T_{2}\right)}{l}$
$\therefore \quad K^{\prime}=\frac{7}{3} K$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,
A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities $K_1$ and $K_2$ . The equivalent conductivity of the combination is
The wall with a cavity consists of two layers of brick separated by a layer of air.All three layers have the same thickness and the thermal conductivity of the brick is much greater than that of air. The left layer is at a higher temperature than the right layer and steady state condition exists. Which of the following graphs predicts correctly the variation of temperature $T$ with distance $d$ inside the cavity?
A body cools in $7$ minutes from $60^{\circ}\,C$ to $40^{\circ}\,C$. The temperature of the surrounding is $10^{\circ}\,C$. The temperature of the body after the next $7$ minutes will be
It takes $10$ minutes to cool a liquid from $61^oC$ to $59^oC$. If room temperature is $30^oC$ then time taken in cooling from $51^oC$ to $49^oC$ is ....... $\min$
A body takes $10$ minutes to cool down from $62^o C$ to $50^o C$. If the temperature of surrounding is $26^o C$ then in the next $10$ minutes temperature of the body will be ......... $^oC$
If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ and $L _{2}$ are the lengths and $A _{1}$ and $A _{2}$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}}=9, \frac{A_{1}}{A_{2}}=2, \frac{L_{1}}{L_{2}}=2$.
Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steel - copper junction in the steady state will be ........... $^{\circ} C$