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QJEEAdv21Ph2176: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph2176 Solution: In order to measure the internal resistance r 1 of a cell of…

In order to measure the internal resistance r1 of a cell of emf E, a meter bridge of wire resistance R0 = 50 Ω\Omega , a resistance R0/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r1 = __________ Ω\Omega .

JEE Advanced 2021 Paper 2 Online Physics - Current Electricity Question 19 English

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JEE Advanced 2021 Paper 2 Online Physics - Current Electricity Question 19 English Explanation
Resistance of potential wire, R0 = 50 Ω\Omega

Resistance of 100 m wire = 50 Ω\Omega

So, resistance of 72 cm wire =50100×72=36Ω= {{50} \over {100}} \times 72 = 36\Omega

Current, I=ε214Ω+25Ω=ε2r1+36ΩI = {{{\varepsilon \over 2}} \over {14\Omega + 25\Omega }} = {{{\varepsilon \over 2}} \over {{r_1} + 36\Omega }}

⇒r1=39−36=3Ω\Rightarrow {r_1} = 39 - 36 = 3\Omega

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