Three rods of same dimensions have thermal co | vedclass

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

Question

(A) The amount of heat $Q$ transmitted through a rod in time $t$ is given by the formula:$Q = \frac{k A t (T_1 - T_2)}{d}$where $k$ is the thermal con

Vedclass · Accepted Answer

$\frac{200}{3}{ }^{\circ} C$

Vedclass · Answer

(A) The amount of heat $Q$ transmitted through a rod in time $t$ is given by the formula:$Q = \frac{k A t (T_1 - T_2)}{d}$where $k$ is the thermal conductivity,$A$ is the cross-sectional area,$d$ is the length of the rod,and $(T_1 - T_2)$ is the temperature difference.Since the rods have the same dimensions,$A$ and $d$ are constant for all rods.At the junction $P$,let the temperature be $T$. According to the principle of conservation of energy in the steady state,the heat flowing into the junction must equal the heat flowing out of the junction.Heat flowing from the $100^{\circ} C$ end to $P$ equals the sum of heat flowing from $P$ to the $50^{\circ} C$ end and from $P$ to the $0^{\circ} C$ end:$\frac{3K A (100 - T) t}{d} = \frac{2K A (T - 50) t}{d} + \frac{K A (T - 0) t}{d}$Canceling the common terms $\frac{K A t}{d}$ from both sides:$3(100 - T) = 2(T - 50) + (T - 0)$$300 - 3T = 2T - 100 + T$$300 - 3T = 3T - 100$$400 = 6T$$T = \frac{400}{6} = \frac{200}{3}{ }^{\circ} C$

Three rods of same dimensions have thermal conductivities $3K$,$2K$,and $K$. They are arranged as shown in the figure. In the steady state,the temperature of the junction $P$ is

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of $2x^3 - 2x - 1 = 0$,then $(\Sigma \alpha \beta)^2$ is equal to

$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal to

If $\tan \theta + \cot \theta = 2$,then $\sin \theta$ is equal to

Which one of the following is not a calcination reaction?

When a body of mass $1.0 ~kg$ is suspended from a certain light spring hanging vertically,its length increases by $5 ~cm$. By suspending a $2.0 ~kg$ block to the spring,if the block is pulled through $10 ~cm$ and released,the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 ~m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module