Question

The resistances in the two arms of the meter bridge are $5\, \Omega$ and $R\, \Omega$ respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6l_1$. The resistance $R$ is

• A. $10\, \Omega$
• B. $15\, \Omega$
• C. $20\, \Omega$
• D. $25\, \Omega$

Answer 1

Answer: (B)

In the first case,

At balance point
$\frac{5}{R}=\frac{l_1}{100-l_1}$...(i)
In the second case,

At balance point
$\frac{5}{(R/2)}=\frac{1.6l_1}{100-1.6l_1}$ ..(ii)
Divide eqn. (i) by eqn. (ii), we get
$\frac{1}{2}=\frac{100-1.6l_1}{1.6(100-l_1)}$
$160-1.6l_1=200-3.2l_1$
$ 1.6l_1 = 40$
or $l_1=\frac{40}{1.6}=25\, cm $
Substituting this value in eqn. (i), we get
$\frac{5}{R}=\frac{25}{75}$
$R=\frac{375}{25}\, \Omega = 15\, \Omega$