Question

In an experiment to determine the emf $E_2$ of an unknown cell, its emf is compared with a standard cell of known emf $E_1=1.12\text{V}$. The balance point is obtained at $56\text{cm}$ with the standard cell and $80\text{cm}$ with the unknown cell. Determine the emf of the unknown cell.

• A. $1.6\text{V}$
• B. $1.2\text{V}$
• C. $1.0\text{V}$
• D. $0.8\text{V}$

Answer 1

The correct option is A $1.6\text{V}$

Given:
$E_1=1.12\text{V};l_1=56\text{cm};l_2=80\text{cm}$

Let $K$ be the potential gradient across the wire of primary circuit.

Using equation,

$E_1=K{l}_1$ and $E_2=K{l}_2$

$\therefore \frac{E_1}{E_2}=\frac{l_1}{l_2}\Rightarrow E_2=E_1\left(\frac{l_2}{l_1}\right)$

taking the value of given data, we get

$E_2=1.12\left(\frac{80}{56}\right)=1.6\text{V}$

Hence, option (a) is the correct answer.

Why this question ? Tip: For comparison of cells, we can use directly, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$