- Home
- Standard 12
- Physics
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 $1.6\,l_1$. The resistance $R$ is .................. $\Omega$
$10$
$15$
$20$
$25$
Solution
At balance point
$\frac{5}{R}=\frac{l_{1}}{100-l_{1}}$ $\ldots(i)$
In the second case,
Substituting this value in eqn. $(i)$, we get
${\frac{5}{R}=\frac{25}{75}} $
${R=\frac{375}{25}\, \Omega=15\, \Omega}$
At balance point
$\frac{5}{(R / 2)}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ ….$(ii)$
Divide eqn. $(i)$ by eqn. $(ii),$ we get
${\frac{1}{2}=\frac{100-1.6 l_{1}}{1.6\left(100-l_{1}\right)}}$
${160-1.6 l_{1}=200-3.2 l_{1}} $
${1.6 l_{1}=40 \quad \text { or } \quad l_{1}=\frac{40}{1.6}=25\, \mathrm{cm}}$
