Question

Potentiometer wire of length $1\, m$ is connected in series with $490 \,\Omega$ resistance and $2 \,V$ battery. If $0.2 \,mV / cm$ is the potential gradient, then resistance of the potentiometer wire is

• A. $4.9 \Omega$
• B. $7.9\, \Omega$
• C. $5.9\, \Omega$
• D. $6.9\, \Omega$

Answer 1

Answer: (A)

Pot. gradient $=0.2 \,mV / cm$
$=\frac{0.2 \times 10^{-3}}{10^{-2}}=2 \times 10^{-2} \,V / m$
Emf of cell $=2 \times 10^{-2} \times 1\, m $
$=2 \times 10^{-2} V =0.02\, V$
As per the condition of potentiometer
$0.02(R+490)=2(R)$
or $1.98 R=9.8$
$\Rightarrow R =\frac{9.8}{1.98}=4.9\, \Omega$