Given:
- \( i_2 = 200 \, \mu A \),
- \( \theta_2 = 60^\circ = \frac{\pi}{3} \, \text{radians} \).
The deflection \( \theta \) is proportional to the current \( i \). Therefore:
\(\frac{i_1}{i_2} = \frac{\theta_1}{\theta_2}.\)
For \( \theta_1 = \frac{\pi}{10} \, \text{radians} \):
\(\frac{i_1}{200} = \frac{\frac{\pi}{10}}{\frac{\pi}{3}}.\)
Simplify:
\(\frac{i_1}{200} = \frac{3}{10} \implies i_1 = 200 \times \frac{3}{10} = 60 \, \mu A.\)
The Correct answer is: 60 $\mu A$
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32