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(B) In a tangent galvanometer,the current $I$ is proportional to the tangent of the deflection angle $	heta$,given by $I = K 	an 	heta$,where $K$ i

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Decrease by $15^{\circ}$

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(B) In a tangent galvanometer,the current $I$ is proportional to the tangent of the deflection angle $	heta$,given by $I = K 	an 	heta$,where $K$ is the reduction factor.Given $I_1 \propto 	an 	heta_1$ and $I_2 \propto 	an 	heta_2$.We have $\frac{I_1}{I_2} = \frac{	an 	heta_1}{	an 	heta_2}$.Given $	heta_1 = 45^{\circ}$ and $I_2 = \frac{I_1}{\sqrt{3}}$,so $\frac{I_1}{I_1/\sqrt{3}} = \frac{	an 45^{\circ}}{	an 	heta_2}$.This simplifies to $\sqrt{3} = \frac{1}{	an 	heta_2}$,which means $	an 	heta_2 = \frac{1}{\sqrt{3}}$.Therefore,$	heta_2 = 30^{\circ}$.The change in deflection is $	heta_1 - 	heta_2 = 45^{\circ} - 30^{\circ} = 15^{\circ}$.Thus,the deflection decreases by $15^{\circ}$.

$A$ certain amount of current when flowing in a properly set tangent galvanometer,produces a deflection of $45^{\circ}$. If the current is reduced by a factor of $\sqrt{3}$,the deflection would:

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