Question

Two equal resistances are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $10\%$ the balancing point shifts by

• A. $10\%$ to right
• B. $10\%$ to left
• C. $9.6\%$ to right
• D. $4.8\%$ to right

Answer 1

The correct option is D $4.8\%$ to right

$l_1=50cm{R}_1=R_2\frac{R_1}{R_2}=\frac{l_2}{100-l_2};\frac{110}{100}=\frac{l_2}{100-l_2}1100-11l_2=10l_{2}1100=21l_2\frac{l_2}{l_1}=\frac{1100}{21\times 50}=\frac{110}{105}(\frac{l_2}{l_1}-1)100\%=\frac{110-105}{105}100\%=\frac{5}{105}100\%=\frac{100}{21}\%=4.8\%right$