A galvanometer has a 50 division scale. Battery has no internal r… - Vidyadip

MCQ

A galvanometer has a $50$ $division$ scale. Battery has no internal resistance. It is found that there is deflection of $40$ $divisions$ when $R = 2400\,\Omega $. Deflection becomes $20$ $divisions$ when resistance taken from resistance box is $4900\,\Omega $ . Then we can conclude

A
Full scale deflection current is $2\, mA$.
B
Resistance of galvanometer is $200\,\Omega \,$
C
Resistance required on $R.B.$ for a deflection of $10$ $divisions$ is $9800\,\Omega \,$
D
Current sensitivity of galvanometer is $20\,\mu A$ $/ division$

✓

Answer

Let full scale deflection of current $=1$

In case $1,$ when $\mathrm{R}=2400 \Omega$ and deflection of 40 divisions present.

$\therefore \quad \frac{40}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \quad \frac{4}{5} \mathrm{I}=\frac{2}{G+2400}\dots (1)$

In case $2,$ when $\mathrm{R}=4900 \Omega$ and deflection of 20 divisions present

$\therefore \quad \frac{20}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \quad \frac{2}{5} \mathrm{I}=\frac{2}{G+4900}\dots (2)$

From $(1)$ and $(2)$ we get,

$\frac{4}{2}=\frac{G+4900}{G+2400}$

$\Rightarrow 2 G+4800=G+4900$

$\Rightarrow \mathrm{G}=100 \Omega$

Putting value of G in equation (1), we get.

$\frac{4}{5} \mathrm{I}=\frac{2}{100+2400}$

$\Rightarrow \mathrm{I}=1 \mathrm{mA}$

Current sensitivity $=\frac{\mathrm{I}}{\text { number of divisions }}$

$=\frac{1}{50}$

$=0.02 \mathrm{mA} /$ division

$=20 \mu \mathrm{A} /$ division

Resistance required for deflection of 10 divisions

$\frac{10}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \frac{1}{5} \times 1 \times 10^{-3}=\frac{2}{100+R}$

$\Rightarrow \mathrm{R}=9900 \Omega$

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