In given figure, the potentiometer wire AB has a resistance of 5Ω and length 10m. The balancing length AM for the emf of 0.4V is:-

Question

In given figure, the potentiometer wire AB has a resistance of 5Ω and length 10m. The balancing length AM for the emf of 0.4V is:- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pjri47ihzwfevno6.jpg" /> (A) 0.4m (B) 4m (C) 0.8m (D) 8m

Tardigrade Admin · Accepted Answer

Emf of cell, E=Xl=(V/l)=(i R/L)× l ∴ E=(e/(R1 + R2 + r))× (R/L)× l 0.4=(e/(R1 + R2 + r))× (R/L)× l ∴ l=8m

Q. In given figure, the potentiometer wire $A B$ has a resistance of $5Ω$ and length $10 m .$ The balancing length $A M$ for the emf of $0.4 V$ is:-

$0.4 m$

$4 m$

$0.8 m$

$8 m$

Emf of cell, $E = Xl = \frac{V}{l} = \frac{i R}{L} \times l$
$∴ E = \frac{e}{( R _{1} + R _{2} + r )} \times \frac{R}{L} \times l$
$0.4 = \frac{e}{( R _{1} + R _{2} + r )} \times \frac{R}{L} \times l$
$∴ l = 8 m$

Q. In given figure, the potentiometer wire AB has a resistance of 5Ω and length 10m. The balancing length AM for the emf of 0.4V is:-

0.4m

4m

0.8m

8m