JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 20)
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 :
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Resistance of galvanometer is 200 $$\Omega $$
Full scale deflection current is 2 mA.
Current sensitivity of galvanometer is 20 $$\mu $$A/division.
Resistance required on R.B. for a deflection of 10 divisions is 9800 $$\Omega $$.
Explanation
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 $$\Omega $$
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 $$\Omega $$
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