Question

In a potentiometer circuit, two wires of same material of resistivity , $\rho$ one of radius of cross-section a and other of radius of cross-section 2a are joined in series. They are of length l and 2l respectively. This combination act as potentiometer wire of length 3l. The emf of the cell in primary circuit is E and internal resistance is $\frac{\rho l}{2\pi a^2}$. This cell is connected to the potentiometer wire by a conducting wire of negligible resistance with positive terminal of the cell connected to one end (call it A) of longer wire.

The balanced length measured from point A obtained in measurement of emf of cell of emf $\frac{E}{2}$ is

• A. $\frac{3l}{2}$
• B. $\frac{5l}{2}$
• C. $\frac{2l}{3}$
• D. $\frac{5l}{7}$

Answer 1

The correct option is B $\frac{5l}{2}$

If $V_1,V_2$ and $V_3$ are the pds across r, AC and AD, then $V_1:V_2:V_3\frac{\rho l}{2\pi a^2}:\frac{\rho l}{2\pi a^2}:\frac{\rho l}{\pi a^2}=1:1:2$ as the current is the same.
Hence $V_2=\frac{E}{4}and{V}_3=\frac{E}{2}$. Since we require a pd of $\frac{E}{2}$ from A we need the full length of AC and half length of CD. Hence the balancing length will be $2l+\frac{l}{2}=\frac{5l}{2}$