$\left(\text { also } \varepsilon_{1}>\varepsilon_{2}\right)$
Potential difference per unit length of the potentiometer wire $=k$ (say)
When $\varepsilon_{1}$ and $\varepsilon_{2}$ are in series and supporteach other then
$\varepsilon_{1}+\varepsilon_{2}=50\, \times k$ .....$(i)$
When $\varepsilon_{1}$ and $\varepsilon_{2}$ are in opposite direction
$\varepsilon_{1}-\varepsilon_{2}=10 \times k$ ....$(ii)$
On adding eqn. $(i)$ and eqn. $(ii)$
$2 \varepsilon_{1}=60\, k \Rightarrow \varepsilon_{1}=30 \,k$ and $\varepsilon_{2}=50\, k-30 \,k=20\, k$
$\therefore \quad \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{30 \,k}{20 \,k}=\frac{3}{2}$
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