On interchanging the resistances,the balance | vedclass

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(B) Let the resistances be $R_1$ and $R_2$. Given $R_1 + R_2 = 1000 \, \Omega$.In the first case,the balance point is at length $l$ from the left end:

Vedclass · Accepted Answer

$550$

Vedclass · Answer

(B) Let the resistances be $R_1$ and $R_2$. Given $R_1 + R_2 = 1000 \, \Omega$.In the first case,the balance point is at length $l$ from the left end:$\frac{R_1}{R_2} = \frac{l}{100-l} \implies \frac{R_1}{1000-R_1} = \frac{l}{100-l} \quad ... (i)$After interchanging the resistances,the balance point shifts to the left by $10 \, cm$,so the new balance length is $(l-10) \, cm$:$\frac{R_2}{R_1} = \frac{l-10}{100-(l-10)} = \frac{l-10}{110-l} \quad ... (ii)$From $(i)$,$\frac{R_2}{R_1} = \frac{100-l}{l}$. Substituting this into $(ii)$:$\frac{100-l}{l} = \frac{l-10}{110-l}$$(100-l)(110-l) = l(l-10)$$11000 - 100l - 110l + l^2 = l^2 - 10l$$11000 = 200l \implies l = 55 \, cm$.Substituting $l = 55$ into $(i)$:$\frac{R_1}{1000-R_1} = \frac{55}{100-55} = \frac{55}{45} = \frac{11}{9}$$9R_1 = 11000 - 11R_1$$20R_1 = 11000 \implies R_1 = 550 \, \Omega$.

On interchanging the resistances,the balance point of a meter bridge shifts to the left by $10 \, cm$. The resistance of their series combination is $1 \, k\Omega$. How much was the resistance on the left slot before interchanging the resistances? ................... $\Omega$

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