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A galvanometer, having a resistance of $50 \Omega$, gives a full scale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \mathrm{~cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is :
(Specific resistance of the wire$\left.=5 \times 10^{-7} \Omega-\mathrm{m}\right)$
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(Specific resistance of the wire$\left.=5 \times 10^{-7} \Omega-\mathrm{m}\right)$
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The correct answer is:
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Resistance of galvanometer,
$G=50 \Omega$
Full scale current, $i_g=0.05$
$A=2.97 \times 10^{-2} \mathrm{~cm}^2$
$=2.97 \times 10^{-2} \times 10^{-4} \mathrm{~m}^2$
$=2.97 \times 10^{-6} \mathrm{~m}^2$
$i=5 \mathrm{~A}$
$\rho=5 \times 10^{-7} \Omega-\mathrm{m}$
Required resistance to convert the galvanometer into ammeter,
$R=\frac{i_g G}{i-i_g}=\frac{0.05 \times 50}{5-0.05}=\frac{25}{4.95}$
$\rho \frac{l}{A}=\frac{50}{99}$
$l=\frac{50}{99} \times \frac{A}{\rho}$
$=\frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}}$
$=\frac{50}{99} \times \frac{29.7}{5}$
$=10 \times 0.3$
$=3 \mathrm{~m}$
$G=50 \Omega$
Full scale current, $i_g=0.05$
$A=2.97 \times 10^{-2} \mathrm{~cm}^2$
$=2.97 \times 10^{-2} \times 10^{-4} \mathrm{~m}^2$
$=2.97 \times 10^{-6} \mathrm{~m}^2$
$i=5 \mathrm{~A}$
$\rho=5 \times 10^{-7} \Omega-\mathrm{m}$
Required resistance to convert the galvanometer into ammeter,
$R=\frac{i_g G}{i-i_g}=\frac{0.05 \times 50}{5-0.05}=\frac{25}{4.95}$
$\rho \frac{l}{A}=\frac{50}{99}$
$l=\frac{50}{99} \times \frac{A}{\rho}$
$=\frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}}$
$=\frac{50}{99} \times \frac{29.7}{5}$
$=10 \times 0.3$
$=3 \mathrm{~m}$
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