Question

In a meter bridge, as shown in the figure, it is given that resistance $Y=12.5\Omega$ and that the balance is obtained at a distance $39.5cm$ from end $A$ (by jockey J). After interchanging the resistances $X$ and $Y$, a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$?

• A. $19.15\Omega$ and $39.5cm$
• B. $8.16\Omega$ and $60.5cm$
• C. $19.15\Omega$ and $60.5cm$
• D. $8.16\Omega$ and $39.5cm$

Answer 1

Answer: (B)

$8.16\Omega$ and $60.5cm$

$\frac{X}{Y}$ = $\frac{39.5}{60.5}$

$X=\frac{39.5}{60.5}\times 12.5$

$=8.16\Omega$

$\frac{12.5}{8.16}=\frac{l_2}{100-l_2}$

$1.53(100+l_2)=l_2$

$153-1.53l_2=l_2$

$\frac{153}{2.53}=l_2$

$=60.5cm$