$ \Rightarrow {\rm{N}}i{\rm{AB}} = {\rm{C}}\frac{\pi }{2} \Rightarrow {\rm{C}} = \frac{{2{\rm{N}}i{\rm{AB}}}}{\pi }$ ........$(1)$
When $Q$ is passed,
$\int \tau dt = \int {Ni} ABdt$
$\mathrm{I} \omega=\mathrm{N} \mathrm{A} \mathrm{B} \mathrm{Q}$ ........$(2)$
Also, maximum deflection happens when entire $\frac{1}{2} \mathrm{I} \omega^{2}$ converts to $\frac{1}{2} \mathrm{C} \theta_{\mathrm{max}}^{2}$
So ${\theta _{\max }} = \sqrt {\frac{{{\rm{I}}{\omega ^2}}}{{\rm{C}}}} = \frac{1}{{\sqrt {{\rm{IC}}} }}{\rm{I}}\omega = \frac{{{\rm{NABQ}}}}{{\sqrt {{\rm{IC}}} }}$ (from $(1)$)
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Statements $I$ : Astronomical unit $(Au)$. Parsec $(Pc)$ and Light year $(ly)$ are units for measuring astronomical distances.
Statements $II:$ $Au < Parsec ( Pc ) < ly$
In the light of the above statements. choose the most appropriate answer from the options given below:
$A.$ The force acting on a planet is inversely proportional to square of distance from sun.
$B.$ Force acting on planet is inversely proportional to product of the masses of the planet and the sun
$C.$ The centripetal force acting on the planet is directed away from the sun.
$D.$ The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.
Choose the correct answer from the options given below :
| Column - l | Column - II |
| (A) Density of nucleus | (l) 85 |
| (B) Time period of infinite length pendulum | (II) $10^{17}$ |
| (C) Speed of satellite very close of earth | (III) 1 |
| (D) Length of second pendulum in earth | (IV) 8 |
