A galvanometer has a resistance of $50\ \Omega$ and it allows maximum current of $5 \mathrm{~mA}$. It can be converted into voltmeter to measure upto $100 \mathrm{~V}$ by connecting in series a resistor of resistance
JEE MAIN 2024, Diffcult
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R= $\frac{V}{I_g}-R_g=\frac{100}{5 \times 10^{-3}}-50$
$=20000-50$
$=19950 \Omega$
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