Question

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10cm$. The resistance of their series combination is $1k\Omega $. How much was the resistance on the left slot before interchanging the resistances?
(A) $550\Omega $
(B) $910\Omega $
(C) $990\Omega $
(D) $505\Omega $

Answer 1

Hint:You are given a meter bridge which consists of two resistors in two slots of the meter bridge. Let the left slot resistor have a resistance of ${R_1}$ and the right slot resistor have a resistance of ${R_2}$. Now, there are two unknowns which need to be sorted out and hence, you obviously need two equations in order to obtain the value of the two unknown resistances. Go through the question one more time and try to use it in order to obtain the equations.

Complete step by step answer:
Now, the first equation you need is given in the question, that is, the series combination of the two resistances is equal to $1k\Omega $. So, we have $P + Q = 1000$$...\left( 1 \right)$ as our first equation.Now, the meter bridge works on the principle of Wheatstone bridge. A Wheatstone bridge is used to find the resistance of an unknown resistance if we take a resistance whose value is known to us. Mathematically, we have, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_3}}}{{{R_4}}}$, where ${R_1} = P$,${R_2} = \rho \dfrac{{{l_1}}}{A}$, where ${l_1}$ is the length of wire which is at the left side of the jockey, ${R_3} = Q$ and ${R_4} = \rho \dfrac{{{l_2}}}{A}$, where ${l_2} = 100 - {l_1}$.Here, $\rho $ is the resistivity of the wire and $A$ is the cross section of the wire.

So, let us substitute the values for the first case when the resistances are not interchanged.
$\dfrac{P}{{{l_1}}} = \dfrac{Q}{{100 - {l_1}}}$
Now, when the resistances are interchanged, the balance point is shifted to left, meaning the ${l_1} \to {l_1} - 10\& 100 - {l_1} \to 110 - {l_1}$
\[\dfrac{Q}{{{l_1} - 10}} = \dfrac{P}{{110 - {l_1}}}\]
Let us try to get relation between $P\& Q$
$\dfrac{P}{Q} = \dfrac{{{l_1}}}{{100 - {l_1}}}$ and $\dfrac{P}{Q} = \dfrac{{110 - {l_1}}}{{{l_1} - 10}}$
$
\dfrac{{{l_1}}}{{100 - {l_1}}} = \dfrac{{110 - {l_1}}}{{{l_1} - 10}} \\
\Rightarrow{l_1}^2 - 10{l_1} = 11000 - 100{l_1} - 110{l_1} + {l_1}^2 \\
\Rightarrow {l_1} = 55cm \\ $
$
\Rightarrow\dfrac{P}{Q} = \dfrac{{{l_1}}}{{100 - {l_1}}} \\
\Rightarrow P = Q\left( {\dfrac{{{l_1}}}{{100 - {l_1}}}} \right) \\
\Rightarrow P = Q\left( {\dfrac{{55}}{{100 - 55}}} \right) \\
\Rightarrow P = \dfrac{{55}}{{45}}Q \\
\Rightarrow P = \dfrac{{11Q}}{9} \\
\Rightarrow Q = \dfrac{{9P}}{{11}}...\left( 2 \right) \\ $
This is our second equation. Solve these two equations, we will get,
$
P + Q = 1000 \\
\Rightarrow P + \dfrac{{9P}}{{11}} = 1000 \\
\Rightarrow\dfrac{{20P}}{{11}} = 1000 \\
\therefore P = 550\Omega \\ $
Therefore, the resistance on the left slot before interchanging the resistances is $550\Omega $.

Hence, option A is correct.

Note: Here, we have used the property of meter bridge that it works on the principle of Wheatstone bridge. It lets you find the value of one resistance if you take a resistance whose value is known to you. Also, remember that while finding the resistance of the wire, it might be possible that the resistivity and the cross section will keep on changing. But in this case, you need not worry about it. The jockey is kept moving on the wire until there is zero deflection on the galvanometer, this is known as the null point.