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A potentiometer wire has length of $5 \mathrm{~m}$ and resistance of $16 \Omega$. The driving cell has an e.m.f. of $5 \mathrm{~V}$ and an internal resistance of $4 \Omega$. When the two cells of e.m.f.s $1.3 \mathrm{~V}$ and $1.1 \mathrm{~V}$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively
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Verified Answer
The correct answer is:
$3 \mathrm{~m}, 0.25 \mathrm{~m}$
$$
\begin{aligned}
& K=\frac{E R}{(R+r) L} \\
& \mathrm{E}=5 \mathrm{~V}, \mathrm{r}=4 \Omega, \mathrm{L}=5 \mathrm{~m}, \mathrm{R}=16 \Omega \\
& \therefore \quad \mathrm{K}=\frac{5 \times 16}{(16+4) \times 5} \\
& \therefore \quad \mathrm{K}=0.8 \mathrm{~V} / \mathrm{m} \\
&
\end{aligned}
$$
When ' $E_1$ ' and ' $E_2$ ' are connected so as to assist each other
$$
\begin{array}{ll}
& \mathrm{E}_1+\mathrm{E}_2=\mathrm{K} l_1 \\
& 1.3+1.1=0.8 \times l_1 \\
\therefore \quad & l_1=3 \mathrm{~m}
\end{array}
$$
When ' $E_1$ ' and ' $E_2$ ' are connected so as to oppose each other,
$$
\begin{aligned}
& \mathrm{E}_1-\mathrm{E}_2=\mathrm{K}_2 \\
& 1.3-1.1=0.8 \times l_2 \\
\therefore \quad & l_2=0.25 \mathrm{~m}
\end{aligned}
$$
\begin{aligned}
& K=\frac{E R}{(R+r) L} \\
& \mathrm{E}=5 \mathrm{~V}, \mathrm{r}=4 \Omega, \mathrm{L}=5 \mathrm{~m}, \mathrm{R}=16 \Omega \\
& \therefore \quad \mathrm{K}=\frac{5 \times 16}{(16+4) \times 5} \\
& \therefore \quad \mathrm{K}=0.8 \mathrm{~V} / \mathrm{m} \\
&
\end{aligned}
$$
When ' $E_1$ ' and ' $E_2$ ' are connected so as to assist each other
$$
\begin{array}{ll}
& \mathrm{E}_1+\mathrm{E}_2=\mathrm{K} l_1 \\
& 1.3+1.1=0.8 \times l_1 \\
\therefore \quad & l_1=3 \mathrm{~m}
\end{array}
$$
When ' $E_1$ ' and ' $E_2$ ' are connected so as to oppose each other,
$$
\begin{aligned}
& \mathrm{E}_1-\mathrm{E}_2=\mathrm{K}_2 \\
& 1.3-1.1=0.8 \times l_2 \\
\therefore \quad & l_2=0.25 \mathrm{~m}
\end{aligned}
$$
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