In the figure shown for gives values of R_1 and R_2 the balance point for Jockey is at 40,cm from A. When R

Q: In the figure shown for gives 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 For balancing condition we can write, $R_{1}=R_{2} \frac{l}{100-l}$ in first case, $R_{1}=R_{2} \frac{40}{100-40}=R_{2} \frac{4}{6}=R_{2} \frac{2}{3}$$...(1)$ in second case, $R_{1}=\left(\frac{10 R_{2}}{10+R_{2}}\right) \frac{l}{100-l}=\left(\frac{10 R_{2}}{10+R_{2}}\right)\left(\frac{50}{100-50}\right)$ $R_{1}=\frac{10 R_{2}}{10+R_{2}}$ from$(1)$ $R_{2} \frac{2}{3}=\frac{10 R_{2}}{10+R_{2}}$ $-10 R_{2}+2 R_{2}^{2}=0$ $2 R_{2}=10$ $R_{2}=5 \Omega$ hence, $R_{1}=R_{2} \frac{2}{3}=(5) \frac{2}{3}=\frac{10}{3} \Omega$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

In the figure shown for gives 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$5\, \Omega , \frac{{15}}{2}\, \Omega$

Advanced

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

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