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Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at `0^@C` and `90^@C`, respectively. The temperature of junction of the three rods will be (a) `45^@C` (b) `60^@C` (c) `30^@C` (d) ` 20^@C`. A. `45^(@)C`B. `60^(@)C`C. `30^(@)C`D. `20^(@)C` |
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Answer» `(b)` Let `theta^(@)C` be the temperature at `B`. Let `Q` is the heat flowing per second from `A` to `B` on account of temperature difference by conductivity. `:. Q=(KA(90-theta))/(l)`…….`(i)` Where `K=` Thermal conducitivity of the rod `A=` Area of cross-section of the rod `l=` Length of the rod By symmetry, the same will be the case for heat flow from `C` to `B` `:.` The heat flowing per second from `B` to `D` will be `2Q=(KA(theta-0))/(l)`......`(ii)` Diveding equation `(ii)` by equation `(i)` `2=(theta)/(92-theta)` `implies theta=60^(@)` |
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