JEE PYQ: Current Electricity Question 31
Question 31 - 2020 (08 Jan 2020 Shift 1)
The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of $20,\Omega$, the null point on it is found to be at 1000 cm. The resistance of whole wire is:
(1) $80,\Omega$
(2) $120,\Omega$
(3) $60,\Omega$
(4) $100,\Omega$
Show Answer
Answer: (4)
Solution
Potential gradient $= \frac{I \times R}{L}$. $V_{AP} = \frac{60 \times 10^{-3} \times R}{1200} \times 1000 = 50R$ mV. At null point $V_{AP} = 5$ V. $50R \times 10^{-3} = 5$. $R = \frac{5}{50 \times 10^{-3}} = 100,\Omega$.
BETA
