Verified Solution • 2025–26 Session moderate

QJEEM22Ph6112: JEE Main 2022 Physics - Class 12 JEEM22Ph6112 Solution: Given below are two statements : Statement I : A uniform wire…

Given below are two statements :

Statement I : A uniform wire of resistance 80 Ω80 \,\Omega is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 Ω5 \,\Omega .

Statement II: Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3:23: 2 .

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

Multiple Choice Options

  • A Both statement I and statement II are correct
  • B Both statement I and statement II are incorrect
  • C Statement I is correct but statement II is incorrect
  • D Statement I is incorrect but statement II is correct

Step-by-Step Verified Solution

Statement I : R1 part=804=20 Ω{R_{1\,part}} = {{80} \over 4} = 20\,\Omega

⇒Reff=204=5 Ω\Rightarrow {R_{eff}} = {{20} \over 4} = 5\,\Omega

Statement II : Ratio =(ΔV)23R(ΔV)22R= {{{{{{(\Delta V)}^2}} \over {3R}}} \over {{{{{(\Delta V)}^2}} \over {2R}}}}

=23= {2 \over 3}

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