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 $60cm$ from $A$. Now an unknown resistance $R$ is connected in series to $P$ and the new position of the neutral point is at $80cm$ from $A$. The value of unknown resistance $R$ is :

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

Answer 1

Answer: (D)

$\frac{20}{3}\Omega$

$\Rightarrow \frac{4}{60}=\frac{R_P}{AN}=\frac{R_Q}{NB}$

$\Rightarrow R_Q=\frac{40\times 4}{60R_P+R}$

$\Rightarrow \frac{R_Q}{NB}=\frac{R_P+R}{80}$

$\Rightarrow \frac{40\times \frac{4}{60}}{20}=\frac{4+R}{80}$

$\Rightarrow \frac{16}{20\times 6}=\frac{4+R}{80}=\frac{32}{3}=4+R$

$\Rightarrow R=\frac{20}{3}\Omega$