One end of a long cylindrical bar is kept at $200 ° C$ and the other end is maintained at $40 ° C$ . In the steady state the graph of temperature of bar against distance from hot end is shown here. Assuming, no losses from sides, from the graph which of the following statements can be concluded true?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The quantity of heat entering the hot end per second is greater than that reaching the other end per second.

The cross-sectional area of the bar decreases as the distance from the hot end increases.

The quantity of heat entering the hot end per second is equal to that leaving the other end per second.

The cross-sectional area of the bar increases as the distance from the hot end increases.

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We know,

$\frac{dQ}{dt} = kA \frac{dT}{dx}$

Since, the heat current $\frac{dQ}{dt}$ has to remain constant throughout the length of bar (as there are no losses from sides), $A \frac{dT}{dx}$ is constant. From graph, magnitude of $\frac{dT}{dx}$ decreases with x and hence area A increases with x, i.e. from hot end to cold end.

Watch 3-min video & get full concept clarity