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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2021 · 1 September · Evening Shift · Physics · Kinetic Theory of Gases

A uniform heating wire of resistance 36Ω is connected across a...

A uniform heating wire of resistance 36Ω36Ω is connected across a potential difference of 240V.240V . The wire is then cut into half and a potential difference of 240V240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1:x,1: x, where xx isA uniform heating wire of resistance 36Ω is connected across a potential difference of 240V . The wire is then cut intoA uniform heating wire of resistance 36Ω is connected across a potential difference of 240V . The wire is then cut into

Correct answer

4

Step-by-step solution

A uniform heating wire of resistance 36Ω is connected across a potential difference of 240V . The wire is then cut into

P=V2/RP = {V}^{2} / R

A uniform heating wire of resistance 36Ω is connected across a potential difference of 240V . The wire is then cut into

P1=V2R2⇒P2=V2R2⇒P1+P2=4V2R{P}_{1} = \dfrac{{V}^{2}}{\dfrac{R}{2}} ⇒ {P}_{2} = \dfrac{{V}^{2}}{\dfrac{R}{2}} ⇒ {P}_{1} + {P}_{2} = \dfrac{4{V}^{2}}{R}

Now
P1+P2P=4V2RV2R=4\dfrac{{P}_{1} + {P}_{2}}{P} = \dfrac{\dfrac{4{V}^{2}}{R}}{\dfrac{{V}^{2}}{R}} = 4

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