Question

In a meter bridge $R_{1}and{R}_2$ are balanced at point J which is 40 cm from A , when $10\Omega$ resistor connected in series with $R_1$ then balance shift point 10 cm towards right . How much resistor connected in parallel with $(R_1+10)$ so that balance point regain it previous position

• A. $60\Omega$
• B. $80\Omega$
• C. $120\Omega$
• D. $10\Omega$

Answer 1

The correct option is A $60\Omega$

Given: In a meter bridge R1andR2R1andR2 are balanced at point J which is 40 cm from A , when 10Ω10Ω resistor connected in series with R1R1 then balance shift point 10 cm towards right . How much resistor connected in parallel with (R1+10)(R1+10) so that balance point regain it previous position.

Solution:

Case 1 (when balance point J is at 40cm from A)

We have a relation for meter bridge that is $\frac{R_1}{R_2}=\frac{l}{100-l}$

we have $l=40cm$

then $\frac{R_1}{R_2}=\frac{40}{100-40}\Rightarrow$ $\frac{R_1}{R_2}=\frac{2}{3}$ (eq-1)

Case 2 (When 10$\Omega$ resistance is added to $R_1$ in series)

Then we can write $\frac{R_1+10}{R_2}=\frac{l}{100-l}$

We have a new balance point which is 10 cm right of 0ld balance point that is 50cm

$\frac{R_1+10}{R_2}=\frac{50}{100-50}$

$R_1+10=R_2$ (eq-2)

Solving equation 1 and 2 we get

$R_1=20\Omega$ and $R_2=30\Omega$

Case 3 (when an unknown resistance x is added to $R_1+10$ ohm resistor in parellal)

We have now our balance point at first position that is at 40cm

we have equivalent resistance now $\frac{30x}{30+x}$

new relation will be from eq-1

$\frac{\frac{30x}{30+x}}{R_2}=\frac{2}{3}$ $\Rightarrow$ $\frac{30x}{30+x}=\frac{2}{3}\times 30$

by solving we get $x=60\Omega$

So the correct option is A