Question

Two unknown resistances $X$ and $Y$ are placed on the left and right gaps of a meter bridge. The null point in the galvanometer is obtained at a distance of $80\text{cm}$ from the left. A resistance of $100\Omega$ is now connected in parallel across $X$. The null point is then found by shifting the sliding contact left by $20\text{cm}$. Calculate $X$ and $Y$. [Neglect the end corrections]

• A.
  $X=\frac{500}{30}\Omega$ , $Y=\frac{125}{3}\Omega$
• B.
  $X=\frac{200}{3}\Omega$ , $Y=\frac{25}{3}\Omega$
• C.
  $$X=\frac{200}{50}\Omega ,Y=\frac{200}{50}\Omega$$
• D.
  $X=\frac{100}{3}\Omega$ , $Y=\frac{25}{3}\Omega$

Answer 1

The correct option is A

$X=\frac{500}{30}\Omega$ , $Y=\frac{125}{3}\Omega$
From the first null point:


$\frac{X}{Y}=\frac{80}{20}.....\left(1\right)$

Now, when $100\Omega$ is connected across $X$, then


$\frac{\left(\frac{100X}{100+X}\right)}{Y}=\frac{60}{40}.....\left(2\right)$

From eqs.$\left(1\right)$ and $\left(2\right)$, we get

$\frac{\left(\frac{100\times 4Y}{100+4Y}\right)}{Y}=\frac{60}{40}$

$\Rightarrow 3Y(100+4Y)=800Y$

$\Rightarrow 12Y^2-500Y=0$

$\Rightarrow Y(12Y-500)=0$

$\therefore Y=\frac{500}{12}=\frac{125}{3}\Omega$

Using equation $\left(1\right)$,

$X=\frac{500}{3}\Omega$

Hence, option (b) is the correct answer.