Question

In a potentiometer experiment, the balancing length of potentiometer of a cell of e.m.f 1.5V in the secondary is 440 cm. A resistance 5$\Omega$ is connected between the terminals of cell, the balancing length is 400 cm.Then
a) internal resistance of the cell is 0.5$\Omega$
b) terminal voltage of the cell is 15/11V
c) Potential gradient of the potentiometer wire is $\frac{1.5}{440}V/cm$
d) potential difference across the potoentiometer wire of length 10m is nearly 3.4V

• A. a,b are correct
• B. a,b and c are correct
• C. a,b and d are correct
• D. a,b,c and d are correct

Answer 1

The correct option is D a,b,c and d are correct

$r=R(\frac{l_1}{l_2}-1)$

$=5(\frac{440}{400}-1)=0.5\Omega$

$\Rightarrow 1.5=I\left(5.5\right)$

$I=\frac{1.5}{5.5}$

$\Rightarrow TerminalVoltage=E-IR$

$=1.5-\frac{1.5}{5.5}\times 0.5$

$=1.5\left(\frac{5.5-0.5}{5.5}\right)=\frac{15}{11}V$

P.G $=\frac{emfbalanced}{lengthrequired}$

$=\frac{1.5}{440}\frac{V}{cm}$

$\Rightarrow \frac{\Delta V}{l}=\frac{1.5}{440}$

$\Delta V=\frac{1.5}{440}\times 1000$

$=3.4V$

$\therefore a,b,c,d$

are true