In the given potentio-meter circuit, the resistance of uniform cross = section potentiometer wire A B of length 1 m is 10 Ω. When the variable resistance R is 10 Ω, the balance point is obtained for length l as shown. If the variable resistance is doubled, the new balance length is k l. Find k.

Question

In the given potentio-meter circuit, the resistance of uniform cross = section potentiometer wire A B of length 1 m is 10 Ω. When the variable resistance R is 10 Ω, the balance point is obtained for length l as shown. If the variable resistance is doubled, the new balance length is k l. Find k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

VA J=(E/R+R) × Rl=e and (E/2 R+R) ⋅ Rl'=e ⇒ (Rl'/Rl)=(l prime/l)=(3/2) (since .R propto l)

Q. In the given potentio-meter circuit, the resistance of uniform cross = section potentiometer wire AB of length 1m is 10Ω. When the variable resistance R is 10Ω, the balance point is obtained for length l as shown. If the variable resistance is doubled, the new balance length is kl. Find k.

VAJ​=R+RE​×Rl​=e and 2R+RE​⋅Rl′​=e ⇒Rl​Rl′​​=ll′​=23​ (since R∝l)

$V_{A J} = \frac{E}{R + R} \times R_{l} = e$
and $\frac{E}{2 R + R} \cdot R_{l^{'}} = e$
$\Rightarrow \frac{R _{l^{'}}}{R _{l}} = \frac{l ^{'}}{l} = \frac{3}{2}$
(since $R \propto l)$