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### A particle moving in a circle of radius R ....

A particle moving in a circle of radius R with a uniform speed takes time T to complete one revolution. If this particle were projected with the same speed at an angle '$\theta$' to the horizontal, the maximum height attained by it equals 4R. The angle of projection '$\theta$' is then given by:

(1) $\theta = {\cos ^{ - 1}}\sqrt {\frac{{g{T^2}}}{{{\pi ^2}R}}}$
(2) $\theta = {\cos ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}}$
(3) $\theta = {\sin ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}}$
(4) $\theta = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}}$

Solution

For circular motion we have, $v=\frac {2\pi R}{T}$

For projectile motion we have, max. height = $H=\frac {v^2 sin^2 \theta }{2g}=4R$

$\Rightarrow {\left( {\frac{{2\pi R}}{T}} \right)^2}\frac{{{{\sin }^2}\theta }}{{2g}} = 4R$

$\therefore \sin \theta = \sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}}$ Or $\theta = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}}$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$