JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 12)
The galvanometer deflection, when key K1 is closed but K2 is open, equals $$\theta $$0 (see figure). On closing K2 also and adjusting R2 to 5$$\Omega $$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :
_12th_January_Morning_Slot_en_12_1.png)
5 $$\Omega $$
25 $$\Omega $$
12 $$\Omega $$
22 $$\Omega $$
Explanation
case I :
ig = $${E \over {220 + {R_g}}}$$ = C$$\theta $$0 . . .(i)
Case II :
ig = $$\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$$ $$ \times $$ $${5 \over {\left( {{R_g} + 5} \right)}}$$ = $${{C{\theta _0}} \over 5}$$ . . .(ii)
$$ \Rightarrow $$ $${{5E} \over {225{R_g} + 1100}}$$ = $${{C{\theta _0}} \over 5}$$ . . .(ii)
$${E \over {220 + {R_g}}} = C\theta $$ . . .(i)
$$ \Rightarrow $$ $${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$$
$$ \Rightarrow $$ 5500 + 25Rg = 225Rg + 1100
200Rg = 4400
Rg = 22$$\Omega $$
ig = $${E \over {220 + {R_g}}}$$ = C$$\theta $$0 . . .(i)
Case II :
ig = $$\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$$ $$ \times $$ $${5 \over {\left( {{R_g} + 5} \right)}}$$ = $${{C{\theta _0}} \over 5}$$ . . .(ii)
$$ \Rightarrow $$ $${{5E} \over {225{R_g} + 1100}}$$ = $${{C{\theta _0}} \over 5}$$ . . .(ii)
$${E \over {220 + {R_g}}} = C\theta $$ . . .(i)
$$ \Rightarrow $$ $${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$$
$$ \Rightarrow $$ 5500 + 25Rg = 225Rg + 1100
200Rg = 4400
Rg = 22$$\Omega $$
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