JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 11)
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 $$\times$$ 10$$^{-3}$$ kg, density is 8.92 $$\times$$ 10$$^{3}$$ kg/m$$^3$$ and resistivity is 1.7 $$\times$$ 10$$^{-8}~\Omega$$-$$\mathrm{m}$$. The length of wire is :
$$l=100$$ m
$$l=6.8$$ m
$$l=5$$ m
$$l=10$$ m
Explanation
$m=8.92 \times 10^{-3} \mathrm{~kg}$
Density $=8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
Volume $=\frac{8.92 \times 10^{-3}}{8.92 \times 10^{3}}=\left(10^{-6}\right) \mathrm{m}^{3}$
Resistance $=\frac{3.4}{2}=1.7 \Omega=\left(\frac{\rho l}{A}\right)$
$1.7=\frac{\rho l^{2}}{(A l)}$
$\Rightarrow 1.7=\frac{1.7 \times 10^{-8} \times l^{2}}{10^{-6}}$
$l^2=100$
$l=10 \mathrm{~m}$
Density $=8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
Volume $=\frac{8.92 \times 10^{-3}}{8.92 \times 10^{3}}=\left(10^{-6}\right) \mathrm{m}^{3}$
Resistance $=\frac{3.4}{2}=1.7 \Omega=\left(\frac{\rho l}{A}\right)$
$1.7=\frac{\rho l^{2}}{(A l)}$
$\Rightarrow 1.7=\frac{1.7 \times 10^{-8} \times l^{2}}{10^{-6}}$
$l^2=100$
$l=10 \mathrm{~m}$
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