A galvanometer of resistance 50,Omega is conn | vedclass

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(B) The initial total resistance in the circuit is $R_{total,1} = R_G + R_1 = 50\,\Omega + 2950\,\Omega = 3000\,\Omega$.Given the battery voltage $\va

Vedclass · Accepted Answer

$4450$

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(B) The initial total resistance in the circuit is $R_{total,1} = R_G + R_1 = 50\,\Omega + 2950\,\Omega = 3000\,\Omega$.Given the battery voltage $\varepsilon = 3\,V$,the initial current $I_1$ is $I_1 = \frac{\varepsilon}{R_{total,1}} = \frac{3\,V}{3000\,\Omega} = 1 	imes 10^{-3}\,A = 1\,mA$.Since $30$ divisions correspond to $1\,mA$,the current per division is $\frac{1\,mA}{30}$.To get a deflection of $20$ divisions,the new current $I_2$ must be $I_2 = 20 	imes \frac{1\,mA}{30} = \frac{2}{3}\,mA$.Let the new total resistance be $R_{total,2}$. Using Ohm's law,$I_2 = \frac{\varepsilon}{R_{total,2}}$,so $R_{total,2} = \frac{\varepsilon}{I_2} = \frac{3\,V}{(2/3) 	imes 10^{-3}\,A} = 4500\,\Omega$.The new total resistance is $R_{total,2} = R_G + R_{new}$,where $R_{new}$ is the new series resistance.$4500\,\Omega = 50\,\Omega + R_{new} \Rightarrow R_{new} = 4450\,\Omega$.

List-$I$	List-$II$
$(1)$ Heavy water	$(a)$ Bicarbonates of $Ca$ and $Mg$ in water
$(2)$ Temporary hard water	$(b)$ No foreign ions in water
$(3)$ Soft water	$(c)$ $D_2O$
$(4)$ Permanent hard water	$(d)$ Sulphates and chlorides of $Ca$ and $Mg$ in water

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