Question

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance $P = 4 \, \Omega$ and the neutral point $N$ is at $60\, cm$ from $A$. Now an unknown resistance $R$ is connected in series to $P$ and the new position of the neutral point is at $80 \,cm$ from $A$. The value of unknown resistance $R$ is :

• A. $\frac{33}{5} \Omega$
• B. $6 \Omega$
• C. $7 \Omega$
• D. $\frac{20}{3} \Omega$

Answer 1

Answer: (D)

Initially $\frac{4}{60}=\frac{Q}{40} Q=\frac{16}{6}=\frac{8}{3}\,\Omega$
Finally $\frac{4+R}{80}=\frac{Q}{20}=\frac{8}{60}$
$4+R=\frac{64}{6}$
$R=\frac{64}{6}-4=\frac{64-24}{6}=\frac{40}{6}$
$=\frac{20}{3}\,\Omega$