Current Electricity Question 23
Question 23 - 2024 (31 Jan Shift 2)
The resistance per centimeter of a meter bridge wire is $r$, with $X \Omega$ resistance in left gap. Balancing length from left end is at $40 cm$ with $25 \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2 r$ resistance per centimeter. The new balancing length for same settings will be at
(1) $20 cm$
(2) $10 cm$
(3) $80 cm$
(4) $40 cm$
Show Answer
Answer: (4)
Solution:
$\frac{25}{r \ell _1}=\frac{X}{r \ell _2}$
$\frac{25}{2 r \ell _1^{\prime}}=\frac{X}{2 r \ell _2^{\prime}}$
From (i) and (ii)
$\ell _2^{\prime}=\ell _2=40 cm$
