
Using the equation:
\[ i = K\theta, \]
At half deflection:
\[ \frac{2}{G + R} = K\theta. \]
Rearranging:
\[ \frac{1}{\theta} = \frac{(G + R)K}{2} = RK + \frac{GK}{2}. \]
From the slope of the graph:
\[ \text{Slope} = K = 0.5 = 5 \times 10^{-1} \, \text{A/division}. \]

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
Let R = {(1, 2), (2, 3), (3, 3)}} be a relation defined on the set \( \{1, 2, 3, 4\} \). Then the minimum number of elements needed to be added in \( R \) so that \( R \) becomes an equivalence relation, is:
The Maximum number of RBr producing 2-methylbutane by above sequence of reaction