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A wire has a resistance of $6 \Omega$. It is cut into two parts and both half values are connected in parallel. The new resistance is
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The correct answer is:
$1 . \mathrm{\Omega}$
Given, resistance of the wire, $R=6 \Omega$
When it is cut in two equal parts, then resistance of the cach part of wire is given by
$$
r=\frac{R}{2}=3 \Omega
$$
$(\because R \propto \bar{l})$
Now, these new resistances are connected in parallel wmbination. The equivalent resistance is given bÿ
$$
\begin{aligned}
\frac{1}{R_{\mathrm{eq}}} &=\frac{1}{r}+\frac{1}{r} \\
\Rightarrow \quad \frac{1}{R_{\mathrm{eq}}} &=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \\
\Rightarrow \quad R_{\mathrm{eq}} &=1.5 \Omega
\end{aligned}
$$
When it is cut in two equal parts, then resistance of the cach part of wire is given by
$$
r=\frac{R}{2}=3 \Omega
$$
$(\because R \propto \bar{l})$
Now, these new resistances are connected in parallel wmbination. The equivalent resistance is given bÿ
$$
\begin{aligned}
\frac{1}{R_{\mathrm{eq}}} &=\frac{1}{r}+\frac{1}{r} \\
\Rightarrow \quad \frac{1}{R_{\mathrm{eq}}} &=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \\
\Rightarrow \quad R_{\mathrm{eq}} &=1.5 \Omega
\end{aligned}
$$
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