QJEEM16Ph7a12: JEE Main 2016 Physics - Class 12 JEEM16Ph7a12 Solution: A galvanometer has a 50 division scale. Battery has no internal …

Let full scale deflection of current = I

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

$\therefore$     ${{40} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$     ${4 \over 5}{\rm I}$ = ${2 \over {G + 2400}}$           . . .(1)

Incase 2, when R = 4900 $\Omega$ and deflection of 20 divisions present

$\therefore$     ${{20} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$     ${2 \over 5}{\rm I}$ = ${2 \over {G + 4900}}$           . . .(2)

From (1) and (2) we get,

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

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

$\Rightarrow$    G $=$ 100 $\Omega$

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

${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$

$\Rightarrow$     ${\rm I} = 1$ mA

Current sensitivity $=$ ${{\rm I} \over {number\,\,of\,\,divisions}}$

$=$ ${1 \over {50}}$

$=$ 0.02 mA / division

$=$ 20 $\mu$ A / division

Resistance required for deflection of 10 divisions

${{10} \over {50}}{\rm I} = {V \over {G + R}}$

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

$\Rightarrow$    R $=$ 9900 $\Omega$