JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
Given below are two statements :
Statement I : A uniform wire of resistance 80 Ω is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 Ω.
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: 2.
In the light of the above statements, choose the most appropriate answer from the option given below
- (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
Answer and solution
Answer (c) Statement I is correct but statement II is incorrect
Solution
Physics sir · 4 steps
Cut first, then join in parallel. And in parallel the voltage is shared, so less resistance means more heat.
- Cut the wire. Resistance is proportional to length, so each quarter of the 80 Ω wire has 80/4 = 20 Ω.
- Join in parallel. Four equal 20 Ω resistors in parallel give 20/4 = 5 Ω. Statement I is correct.
- Heat in parallel. Resistors in parallel have the same voltage V across them. So the right formula is H = (V² t)/R, and the heat is inversely proportional to R.
H_(3R)/H_(2R) = (V² t/3R)/(V² t/2R) = 2R/3R = 2/3
- So. The heat in 3R and 2R is in the ratio 2:3, not 3:2, so Statement II is incorrect. Statement I is correct and Statement II is incorrect: option (c).
The idea to keep. Pick the heating formula by what is common. In series the current is the same, so use I²Rt and heat grows with R. In parallel the voltage is the same, so use V²t/R and heat falls with R.
Answer from the NTA final answer key. Solution by PrepWiser.