In the experiment to determine the galvanomet | vedclass

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Question

(A) The current $I$ in the galvanometer is given by $I = K 	heta$,where $K$ is the figure of merit.From the circuit,the current is $I = \frac{E}{G+R}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The current $I$ in the galvanometer is given by $I = K 	heta$,where $K$ is the figure of merit.From the circuit,the current is $I = \frac{E}{G+R}$,where $E = 2 	ext{ V}$ is the emf of the source,$G$ is the galvanometer resistance,and $R$ is the external resistance.Equating the two expressions: $K 	heta = \frac{E}{G+R} \Rightarrow \frac{1}{	heta} = \frac{G+R}{E} = \frac{1}{E} R + \frac{G}{E}$.Comparing this with the equation of a straight line $y = mx + c$,we have the slope $m = \frac{1}{E}$.From the graph,the slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{6 - 2} = \frac{1}{4} 	ext{ } \Omega^{-1}$.Since $m = \frac{K}{E}$,we have $\frac{K}{2} = \frac{1}{4} \Rightarrow K = 0.5 	ext{ A/division}$.Expressing $K$ in terms of $10^{-1}$,we get $K = 5 	imes 10^{-1} 	ext{ A/division}$.Thus,the correct option is $A$.

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