Q. A galvanometer, having a resistance of 50$\ \Omega,\ $gives a full scale deflection for a current of 0.05A. The length in meter of a resistance wire of area of cross-section $2.97 \times 10^{- 2}cm^{2}$ that can be used to convert the galvanometer into a ammeter which can read a maximum of 5A current is (Specific resistance of the wire $= 5 \times 10^{- 7}\ \Omega m$)
A) 9
B) 6
C) 3
D) 1.5
Explanation
(c) Resistance of galvanometer
$G = 50\Omega$
Full scale current $i_{g} = 0.05$
$A = 2.97 \times 10^{- 2}cm^{2}$
$\ \ \ \ = 2.97 \times 10^{- 2} \times 10^{- 4}m^{2}$
$\ \ \ \ = 2.97 \times 10^{- 6}m^{2}$
$i = 5A$
$\rho = 5 \times 10^{- 7}\Omega m$
Required resistance to convert the galvanometer into ammeter.
$R = \frac{i_{g}G}{i - i_{g}} = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95}$
$\rho\frac{l}{A} = \frac{50}{99}$
$l = \frac{50}{99} \times \frac{A}{\rho} = \frac{50}{99} \times \frac{2.97 \times 10^{- 6}}{5 \times 10^{- 7}} = \frac{50}{99} \times \frac{29.7}{5} = 10 \times 0.3 = 3m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.