In the fig. shown for given values of R_1 and R_2 the balance point for jockey is at 40, cm from A. When R

Q: In the fig. shown for given values of $R_1$ and $R_2$ the balance point for jockey is at $40\, cm$ from $A$. When $R_2$ is shunted by a resistance of $10\,\Omega $, balance shifts to $50\, cm. R_1$ and $R_2$ are $(AB = 1\,m)$

a Let resistance per unit length of wire is $\rho$ then from given information $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\rho \times 40}{\rho \times 60}$ $\frac{\frac{\mathrm{R}}{10 \mathrm{R}_{2}}}{\frac{10 \mathrm{R}_{2}}{10+\mathrm{R}_{2}}}=\frac{50 \times \rho}{50 \times \rho}$ Solving above equation $\mathrm{R}_{1}=\frac{10}{3}\, \Omega \quad \mathrm{R}_{2}=5\, \Omega$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

In the fig. shown for given values of $R_1$ and $R_2$ the balance point for jockey is at $40\, cm$ from $A$. When $R_2$ is shunted by a resistance of $10\,\Omega $, balance shifts to $50\, cm. R_1$ and $R_2$ are $(AB = 1\,m)$

A$\frac{{10}}{3}\,\Omega ,5\,\Omega $
B$20\,\Omega ,30\,\Omega $
C$10\,\Omega ,15\,\Omega $
D$5\,\Omega ,\frac{{15}}{2}\,\Omega $

Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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