A) \[5050\,\,\Omega \]
B) \[5550\,\,\Omega \]
C) \[6050\,\,\Omega \]
D) \[4450\,\,\Omega \]
Correct Answer: D
Solution :
Current through the galvanometer \[I=\frac{3}{(50+2950)}={{10}^{-3}}A\] Current for 30 divisions \[={{10}^{-3}}A\] Current for 20 divisions\[=\frac{{{10}^{-3}}}{30}\times 20\] \[=\frac{2}{3}\times {{10}^{-3}}A\] For the same deflection to obtain for 20 divisions, let resistance added be R \[\therefore \]\[\frac{2}{3}\times {{10}^{-3}}=\frac{3}{(50+1R)}\]or\[R=4450\Omega \]You need to login to perform this action.
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