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JEE Main 202228 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
28 July
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2022 · 28 July · Evening Shift · Physics · Current Electricity

Given below are two statements : Statement I : A uniform wire of...

Given below are two statements :
Statement I : A uniform wire of resistance 80Ω80Ω is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5Ω5Ω.

Statement II : Two resistance 2R2R and 3R3R are connected in parallel in an electric circuit. The value of thermal energy developed in 3R3R and 2R2R will be in the ratio 3:23: 2.

In the light of the above statements, choose the most appropriate answer from the options given below

Given below are two statements : Statement I : A uniform wire of resistance 80Ω is cut into four equal parts. These

Options

  1. ABoth statement I and statement II are correct
  2. BBoth statement I and statement II are incorrect
  3. CStatement I is correct but statement II is incorrect
  4. DStatement I is incorrect but statement II is correct.

Correct answer

C. Statement I is correct but statement II is incorrect

Step-by-step solution

Statement 1−1 - R=80ΩR = 80Ω

When it is cut into 44 equal parts, the resistance of each part will be,

R1=R2=R3=R4=20Ω{R}_{1} = {R}_{2} = {R}_{3} = {R}_{4} = 20Ω

When these resistances are connected in parallel,

Req=204=5Ω{R}_{eq} = \dfrac{20}{4} = 5Ω

Hence, the statement I is correct.

Statement 2 -

Given below are two statements : Statement I : A uniform wire of resistance 80Ω is cut into four equal parts. These

When the resistances are connected in parallel, the potential difference between them is same. Therefore, the power consumed can be written as,

P=V2RP = \dfrac{{V}^{2}}{R}

⇒P1P2=(R2R1)=23⇒ \dfrac{{P}_{1}}{{P}_{2}} = \left(\dfrac{{R}_{2}}{{R}_{1}}\right) = \dfrac{2}{3}

Hence, statement II is incorrect.

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