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Step-by-Step Solution
Step 1: Identify the given data
• Resistance of galvanometer, $G = 50\,\Omega$
• Full-scale deflection current (galvanometer), $I_G = 0.05\,\text{A}$
• Desired maximum current for the ammeter, $I = 5\,\text{A}$
• Cross-sectional area of the wire, $A = 3 \times 10^{-2}\,\text{cm}^2 = 3 \times 10^{-6}\,\text{m}^2$
• Resistivity of the wire, $\rho = 5 \times 10^{-7}\,\Omega\,\text{m}$
Step 2: Calculate the required shunt resistance
To convert the galvanometer into an ammeter, a shunt (low resistance) is connected in parallel so that only $I_G$ flows through the galvanometer and the remainder goes through the shunt.
The required shunt resistance $S$ is given by:
$$S = \frac{I_G \times G}{I - I_G}.$$
Substituting the values:
$$S = \frac{(0.05)\,(50)}{5 - 0.05}
= \frac{2.5}{4.95}
= \frac{50}{99}
\approx 0.505\,\Omega.$$
Step 3: Determine the length of the wire
The resistance $S$ of a wire of length $l$, cross-sectional area $A$, and resistivity $\rho$ is:
$$S = \frac{\rho \, l}{A}.$$
Rearranging for $l$:
$$l = \frac{S \times A}{\rho}.$$
Substituting the known values $S = \frac{50}{99}\,\Omega$, $A = 3 \times 10^{-6}\,\text{m}^2$, and $\rho = 5 \times 10^{-7}\,\Omega\,\text{m}$:
$$l = \frac{\left(\frac{50}{99}\right) \times (3 \times 10^{-6})}{5 \times 10^{-7}}
= \left(\frac{50}{99}\right) \times \frac{3 \times 10^{-6}}{5 \times 10^{-7}}
\approx 3\,\text{m}.$$
Step 4: Final answer
The length of the wire required is $3\,\text{m}.$
