In a meter-bridge,if the left and right gaps | vedclass

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(D) Key Idea: In a meter-bridge,if the balancing point is at the midpoint of the bridge wire,the resistances in the two gaps must be equal.Given:Left

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$6 \Omega$ in parallel

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(D) Key Idea: In a meter-bridge,if the balancing point is at the midpoint of the bridge wire,the resistances in the two gaps must be equal.Given:Left gap resistance $R_1 = 2 \Omega$Right gap resistance $R_2 = 3 \Omega$For a meter-bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{l_1}{L - l_1}$.If the balancing point is at the midpoint,$l_1 = L - l_1$,which implies $R_1 = R_2$.Since $R_1 = 2 \Omega$,we need the effective resistance in the right gap to be $2 \Omega$.Let a resistance $x$ be connected in parallel with the $3 \Omega$ resistor to make the equivalent resistance $2 \Omega$.$\frac{3 	imes x}{3 + x} = 2$$3x = 6 + 2x$$x = 6 \Omega$Thus,a $6 \Omega$ resistor must be connected in parallel with the $3 \Omega$ resistor. Hence,the correct option is $D$.

In a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?

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