Question

A standard cell of emf $E_0=1.11\text{V}$ is balanced against $72\text{cm}$ length of a potentiometer. The same potentiometer is used to measure the potential difference across the standard resistance $R=120\Omega$. When the ammeter shows a current of $7.8\text{mA}$, a balanced length of $60\text{cm}$ is obtained on the potentiometer. Estimate the percentage error in the measurement of ammeter:

• A. $1.3$
• B. $2.75$
• C. $2.5$
• D. $0.5$

Answer 1

The correct option is A $1.3$

The balancing length obtained in two cases are

$l_0=72\text{cm}\&l=60\text{cm}$

The emf of standard cell is

$E_0=1.11\text{V},R=120\Omega$

Let $K$ be the potential gradient in the wire of potentiometer.

By using formula,

$E_0=K{l}_0.....\left(1\right)$

Also, the potential difference across resistance is,

$V=IR=Kl.......\left(2\right)$

From eq. $\left(1\right)\&\left(2\right)$,

$I=\frac{E_0}{R}\left(\frac{l}{l_0}\right)$

$I=\frac{1.11}{120}\times \frac{60}{72}=0.0077\text{A}$

$\therefore I=7.7\text{mA}$

The reading obtained by ammeter is

$I^{\prime }=7.8\text{mA}$

So, error in measurement

$\Delta I=7.8-7.7=0.1\text{mA}$

$\%$ error $=\frac{\Delta I}{I}\times 100$

$\therefore \%\text{error}=\frac{0.1}{7.7}\times 100\approx 1.3\%$

Hence, option $\left(a\right)$ is correct.

Why this question ? Tip: we can equate the potential difference of potentiometer wire and the potential difference across theresistance in the balanced state.