Question

A cell in secondary circuit gives null deflection for \(2.5\,m\) length of wire for a potentialmeter having \(10\,m\) length of wire. If the length of the potentiometer wire is increased by \(1\,m\) without changing the cell in the primary, the position of the null point now is

• A. \(3.5\,m\)
• B. \(3\,m\)
• C. \(2.75\,m\)
• D. \(2\,m\)

Hint:

If total length of potentiometer wire increases while supply stays same, potential gradient decreases and balancing length increases proportionally.

Answer 1

The Correct Option is C

Step 1: Use potentiometer principle.
At balance point:
\[ E = k\ell \]
where \(k\) is potential gradient.
Step 2: Express potential gradient.
\[ k = \frac{V}{L} \]
Since cell and primary circuit unchanged, total potential \(V\) across wire remains constant.
Step 3: Initial condition.
\[ L_1 = 10m,\quad \ell_1 = 2.5m \]
\[ E = \frac{V}{10}\cdot 2.5 \]
Step 4: New length.
\[ L_2 = 11m \]
New balance length \(\ell_2\):
\[ E = \frac{V}{11}\ell_2 \]
Step 5: Equate and solve.
\[ \frac{V}{10}\cdot 2.5 = \frac{V}{11}\ell_2 \Rightarrow \ell_2 = \frac{11}{10}\cdot 2.5 = 2.75m \]
Final Answer:
\[ \boxed{2.75\,m} \]