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In the arrangement shown in figure, when the switch $S_2$ is open, the galvanometer, shows no deflection for $l=50 \mathrm{~cm}$ when the switch $S_2$ is closed, the galvanometer shows no deflection for $l=0.416 \mathrm{~m}$. The internal resistance $(r)$ of $6 \mathrm{~V}$ cell is
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Verified Answer
The correct answer is:
$2 \Omega$
$2 \Omega$
When $S_2$ open.
Assume resistance of $X Y=R$.
Resistance of wire per unit length, $x=\frac{R}{L}=R \Omega \mathrm{m}^{-1}$
$\because \quad I=E_0 / R$
Now, the potential drop across $50 \mathrm{~cm}$ length is $6 \mathrm{~V}$, So
$$
\begin{array}{cc}
& \frac{E_0}{R} \times R \times \frac{50}{100}=6 \\
\Rightarrow & E_0=12 \mathrm{~V}
\end{array}
$$
When $S_2$ closed, potential drop across $0.416 \mathrm{~cm}$ length,
$$
V_1=\frac{E_0}{R} \times R \times 0.416=12 \times 0.416 \approx 5 \mathrm{~V}
$$
Hence, $\quad E-I r=5 \mathrm{~V}$
$$
\begin{aligned}
\Rightarrow & & 6-I r & =5 \\
& \because & I & =\frac{5}{10} \\
\therefore & & 6-5 & =\frac{5}{10} r \\
\Rightarrow & & r & =2 \Omega
\end{aligned}
$$
Assume resistance of $X Y=R$.
Resistance of wire per unit length, $x=\frac{R}{L}=R \Omega \mathrm{m}^{-1}$
$\because \quad I=E_0 / R$
Now, the potential drop across $50 \mathrm{~cm}$ length is $6 \mathrm{~V}$, So
$$
\begin{array}{cc}
& \frac{E_0}{R} \times R \times \frac{50}{100}=6 \\
\Rightarrow & E_0=12 \mathrm{~V}
\end{array}
$$
When $S_2$ closed, potential drop across $0.416 \mathrm{~cm}$ length,
$$
V_1=\frac{E_0}{R} \times R \times 0.416=12 \times 0.416 \approx 5 \mathrm{~V}
$$
Hence, $\quad E-I r=5 \mathrm{~V}$
$$
\begin{aligned}
\Rightarrow & & 6-I r & =5 \\
& \because & I & =\frac{5}{10} \\
\therefore & & 6-5 & =\frac{5}{10} r \\
\Rightarrow & & r & =2 \Omega
\end{aligned}
$$
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