Question

Two resistances are connected in the two gaps of a metre bridge. The balance point is $20cm$ from the zero end. When a resistance $15\Omega $ is connected in series with the smaller of two resistances, the null point shifts to $40cm$. The smaller of the two resistances has the value.
(A) $8\Omega $
(B) $9\Omega $
(C) $10\Omega $
(D) $12\Omega $

Answer 1

Hint: In order to solve this question, we will use the general balance point formula for a meter bridge and therefore form a relation between two resistance values and then solve for a smaller resistance value using given parameter values and conditions.

Formula Used:
If $l$ is the balance point of a metre bridge having two resistances ${R_1},{R_2}$ then Balance point equation is written as $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{l}{{100 - l}}$

Complete answer:
Let us assume that ${R_1},{R_2}$ be the two resistances used in metre bridge and their null point is given to us $l = 20cm$ then, using the formula $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{l}{{100 - l}}$ we get,
$
  \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{20}}{{100 - 20}} \\
  \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{4} \\
  {R_2} = 4{R_1} \to (i) \\
 $

Equation (i) shows that ${R_1}$ is the smaller resistance. Now, when a resistance of $15\Omega $ is connected in series with the smaller of two resistance the net resistance across old ${R_1}$ will become ${R_1}' = {R_1} + 15$ and keeping ${R_2} = 4{R_1}$ from equation (i) with new null point as $l = 40cm$.

so using formula $\dfrac{{{R_1}'}}{{{R_2}}} = \dfrac{l}{{100 - l}}$ and solving for ${R_1}$ we get,
$
  \dfrac{{{R_1} + 15}}{{4{R_1}}} = \dfrac{{40}}{{100 - 40}} \\
  \dfrac{{{R_1} + 15}}{{4{R_1}}} = \dfrac{2}{3} \\
  3{R_1} + 45 = 8{R_1} \\
  5{R_1} = 45 \\
   \Rightarrow {R_1} = 9\Omega \\
 $
So, the smaller resistance has a value of ${R_1} = 9\Omega $

Hence, the correct answer is option (B) $9\Omega $

Note: It should be remembered that when two resistances are added in series together then their net resistance is simply the sum of individual resistance if they were added in parallel combination then their net resistance is calculated using the formula $\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ and most commonly the length of the meter bridge is taken as $1m$ that’s why it’s called a meter bridge.