To solve this problem, we need to determine the resistance of an ammeter given that 5% of the main current passes through the galvanometer whose resistance is \( G \). Let's work through this step-by-step.
Thus, the correct answer is: None of these.
Given: - 5% of the main current passes through the galvanometer. - The resistance of the galvanometer is \( G \).
The shunt resistance \( S \) is connected in parallel with the galvanometer such that 95% of the main current passes through the shunt. The current division formula for parallel resistances gives:
\[ \frac{I_g}{I} = \frac{S}{S + G} \]
where \( I_g \) is the current through the galvanometer and \( I \) is the total current. Given that:
\[ \frac{I_g}{I} = 0.05 \]
Substituting this value:
\[ 0.05 = \frac{S}{S + G} \]
Rearranging:
\[ 0.05(S + G) = S \] \[ 0.05G = 0.95S \] \[ S = \frac{G}{19} \]
The resistance of the ammeter \( R_a \) is the equivalent resistance of the galvanometer and the shunt connected in parallel:
\[ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{S} \]
Substituting the value of \( S \):
\[ \frac{1}{R_a} = \frac{1}{G} + \frac{19}{G} = \frac{20}{G} \] \[ R_a = \frac{G}{20} \]
Since the resistance values provided in the options differ from this result, it is possible that additional context or conditions may influence the choice of answer.
The problem seems to indicate that the correct answer is marked as a bonus question, suggesting that there may be additional considerations or assumptions needed for a precise determination.


A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.