Current Electricity

JEE Main level, single correct, medium

306 of 380

Question 306 of 380

In a metre bridge experiment null point is obtained at 40 cm40 \mathrm{~cm} from one end of the wire when resistance X\mathrm{X} is balanced against another resistance Y\mathrm{Y}. If X<Y\mathrm{X} \lt \mathrm{Y}, then the new position of the null point from the same end, if one decides to balance a resistance of 3X3 \mathrm{X} against Y\mathrm{Y}, will be close to :

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Answer: (C) 67 cm67 \mathrm{~cm}

From question, xy=40100−40=23\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}
⇒x=23y\Rightarrow \mathrm{x}=\frac{2}{3} \mathrm{y}
Again, 3xy=Z100−Z\frac{3 x}{y}=\frac{Z}{100-Z}
or 3×2y3y=Z100−Z\frac{3 \times \frac{2 y}{3}}{y}=\frac{Z}{100-Z}
Solving we get Z=67 cm\mathrm{Z}=67 \mathrm{~cm}
Therefore new position of null point
≅67 cm\cong 67 \mathrm{~cm}

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