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QJEEM19Ph8178: JEE Main 2019 Physics - Class 11 JEEM19Ph8178 Solution: Temperature difference of 120 o C is maintained between ends of …

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 3L2,{{3L} \over 2}, is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Heat and Thermodynamics Question 342 English

Multiple Choice Options

  • A 45oC
  • B 75oC
  • C 60oC
  • D 35oC

Step-by-Step Verified Solution

We know,

Resistance, R = ρLA{{\rho L} \over A}

∴\therefore    R ∝\propto L

So, Resistance is directly proportional to lengt5h of the rod.

JEE Main 2019 (Online) 9th January Morning Slot Physics - Heat and Thermodynamics Question 342 English Explanation

Equivalent resistance between A and B is,

Req = R2+R×3R2R+3R2+R2{R \over 2} + {{R \times {{3R} \over 2}} \over {R + {{3R} \over 2}}} + {R \over 2}

= R + 3R22×25R{{3{R^2}} \over 2} \times {2 \over {5R}}

= R + 3R5{{3R} \over 5}

= 8R5{{8R} \over 5}

Thermal current between point A and B is,

      I = ΔTABReq{{\Delta {T_{AB}}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}

= 120−08R5{{120 - 0} \over {{{8R} \over 5}}}

= 120×58R{{120 \times 5} \over {8R}}

Resistance between P and Q = 3R5{{3R} \over 5}

∴\therefore     Δ\Delta TPQ = I×3R5{\rm I} \times {{3R} \over 5}

= 120×58R×3R5{{120 \times 5} \over {8R}} \times {{3R} \over 5}

= 45oC

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