JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 22)
Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length $${{3L} \over 2},$$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to :
_9th_January_Morning_Slot_en_22_1.png)
45oC
75oC
60oC
35oC
Explanation
We know,
Resistance, R = $${{\rho L} \over A}$$
$$ \therefore $$ R $$ \propto $$ L
So, Resistance is directly proportional to lengt5h of the rod.
_9th_January_Morning_Slot_en_22_2.png)
Equivalent resistance between A and B is,
Req = $${R \over 2} + {{R \times {{3R} \over 2}} \over {R + {{3R} \over 2}}} + {R \over 2}$$
= R + $${{3{R^2}} \over 2} \times {2 \over {5R}}$$
= R + $${{3R} \over 5}$$
= $${{8R} \over 5}$$
Thermal current between point A and B is,
I = $${{\Delta {T_{AB}}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$$
= $${{120 - 0} \over {{{8R} \over 5}}}$$
= $${{120 \times 5} \over {8R}}$$
Resistance between P and Q = $${{3R} \over 5}$$
$$ \therefore $$ $$\Delta $$TPQ = $${\rm I} \times {{3R} \over 5}$$
= $${{120 \times 5} \over {8R}} \times {{3R} \over 5}$$
= 45oC
Resistance, R = $${{\rho L} \over A}$$
$$ \therefore $$ R $$ \propto $$ L
So, Resistance is directly proportional to lengt5h of the rod.
Equivalent resistance between A and B is,
Req = $${R \over 2} + {{R \times {{3R} \over 2}} \over {R + {{3R} \over 2}}} + {R \over 2}$$
= R + $${{3{R^2}} \over 2} \times {2 \over {5R}}$$
= R + $${{3R} \over 5}$$
= $${{8R} \over 5}$$
Thermal current between point A and B is,
I = $${{\Delta {T_{AB}}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$$
= $${{120 - 0} \over {{{8R} \over 5}}}$$
= $${{120 \times 5} \over {8R}}$$
Resistance between P and Q = $${{3R} \over 5}$$
$$ \therefore $$ $$\Delta $$TPQ = $${\rm I} \times {{3R} \over 5}$$
= $${{120 \times 5} \over {8R}} \times {{3R} \over 5}$$
= 45oC
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