Question

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is

(a) 3:4 (b) 3:2 (c) 5:1 (d) 5:4

Answer 1

Correct option (b) \(3: 2\)
Explanation:
Suppose two cells have emfs \(\varepsilon_{1}\) and \(\varepsilon_{2}\) (also \(\varepsilon 1>\varepsilon 2\) ).
Potential difference per unit length of the potentiometer wire= \(k\) (say)
When \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are in series and support each other then
$$
\varepsilon_{1}+\varepsilon_{2}=50 \times k
$$
When \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are in opposite direction
$$
\varepsilon_{1}-\varepsilon_{2}=10 \times k
$$
On adding eqn. (i) and eqn. (ii)
\(2 \varepsilon_{1}=60 k \Rightarrow \varepsilon_{1}=30 k\) and
$$
\begin{aligned}
&\varepsilon_{2}=50 k-30 k=20 k \\
&\therefore \quad \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{30 k}{20 k}=\frac{3}{2}
\end{aligned}
$$

Answer 2

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Answer 3

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