Question

In the experiment of calibration of voltmeter, a 1.1 volt standard cell gets balanced at 440 cm length of the wire. The balancing length corresponding to a potential difference between the ends of a resistance comes out to be 190cm. A voltmeter shows 0.5 volt for this potential difference. The error in the reading of voltmeter will be:
A) 0.025volt
B) 25volt
C) 2.5volt
D) 0.25volt

Answer 1

Hint: We have the relationship between electromotive force(reading of the voltmeter) and potential difference:
$E=\dfrac{V}{d}$, where E is the emf of the cell, V is the potential difference of the cell, d is the length of balance for $E$.
Also formula for potential gradient(relation between potential difference and length across which potential difference is measured in voltmeter) is:
Let’s denote potential gradient (it is the potential per unit length) by g:
$g = \dfrac{V}{d}$ (used for determining potential at per unit length)
Using the above relations we will solve the problem.

Complete step by step answer:
Now, we will define voltmeter and potential difference.
Voltmeter: a voltmeter is an instrument used for measuring electric potential difference between two points in an electric circuit. Voltmeters are of two types analog voltmeter and digital voltmeter. Analog voltmeter uses a pointer which on deflection detects a voltage while digital voltmeter gives numerical digits on the display panel of the voltmeter.
Potential difference: potential difference is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points.
Let’s do the calculation part now:
Potential per unit length is given by g (potential gradient) :
$ \Rightarrow g = \dfrac{V}{d}$
After substituting the given values
$
   \Rightarrow g = \dfrac{{1.1}}{{440}} \\
   \Rightarrow g = 0.0025 \\
 $
When the length is 190cm
Voltmeter reading will be
$
   \Rightarrow g \times l \\
   \Rightarrow 0.0025 \times 190 \\
   \Rightarrow. 475V \\
 $
Error in the reading of the voltmeter is :
$E = {E_g} - {E_c}$ ( $E_g$ is the given value and $E_c$ is calculated value)
$
   \Rightarrow E = 0.5 - 0.475 \\
   \Rightarrow E = 0.025V \\
 $

Therefore, the error comes out to be 0.025V, option A is correct.

Note:
Potential gradient is defined as the rate of change of potential with respect to distance. It is represented as $\dfrac{{\delta V}}{{\delta d}}$, $V$ is the potential and d is the distance. Potential gradient is a vector quantity which has both the magnitude and direction.