Question 76: Current Electricity

Question

In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \mathrm{~cm}$ from the left. Now when an unknown resistance $x \Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is
$\_\_\_\_$ $\Omega$.

Accepted Answer

A meter bridge works on the principle of a balanced Wheatstone bridge. The ratio of the resistances in the gaps is equal to the ratio of the lengths of the wire segments.
Case 1 :
Null point from left is l cm . So, the remaining length from right (100 - l) cm
At balanced condition
$$ \frac{2}{3}=\frac{l}{100-l} \Rightarrow 2(100-l)=3l $$
$\Rightarrow 200-2l=3l$
$\Rightarrow 5 \mathrm{l}=200 \Rightarrow \mathrm{l}=40 \mathrm{~cm}$
Case 2 :
When an unknown resistance x is connected in parallel with the $3 \Omega$ resistor in the right gap. This changes the equivalent resistance of the right gap.
New right gap resistance $S^{\prime}=\frac{3 x}{3+x}$
New null point $\mathrm{l}^{\prime}=\mathrm{l}+10=40+10=50 \mathrm{~cm}$
So, the remaining length from right $=100-50=50 \mathrm{~cm}$
At balanced condition for the second case:
$$ \frac{\mathrm{R}}{\mathrm{~S}^{\prime}}=\frac{\mathrm{l}^{\prime}}{100-\mathrm{l}^{\prime}} $$
$$\Rightarrow $$ $\frac{2}{\left(\frac{3 x}{3+x}\right)}=\frac{50}{50} \Rightarrow \frac{6+2 x}{3 x}=1$
$$ \Rightarrow 3 x=6+2 x \Rightarrow x=6 \Omega $$
The value of unknown resistance x is $6 \Omega$.
Hence, the correct answer is 6.

Current Electricity

Integer / Numeric Answer Type