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Question

$1.$ $\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{K}_{\mathrm{i}}}=\frac{\frac{1}{2} \mathrm{m}(\mathrm{u} \cos 	heta)^{2}}{\frac{1}{2} \mathrm{mu}^{2}}=\

Vedclass · Accepted Answer

$A-4,\,\,B-1,\,\,C-3,\,\,D-2$

Vedclass · Answer

$1.$ $\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{K}_{\mathrm{i}}}=\frac{\frac{1}{2} \mathrm{m}(\mathrm{u} \cos 	heta)^{2}}{\frac{1}{2} \mathrm{mu}^{2}}=\cos ^{2} 	heta=\frac{1}{4} \Rightarrow 	heta=60^{\circ}$

$2.$ $\frac{\mathrm{gT}^{2}}{\mathrm{R}}=\frac{\mathrm{g}\left(\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}ight)^{2}}{\frac{\mathrm{u}^{2}}{\mathrm{g}} 2 \sin 	heta \cos 	heta}=2 	an 	heta=8$

$\Rightarrow 	an 	heta=4 \Rightarrow 	heta=	an ^{-1}(4)$

$\frac{\mathrm{R}}{\mathrm{H}}=\frac{24^{2} \sin 	heta \cos 	heta}{\mathrm{g}} 	imes \frac{2 \mathrm{g}}{4^{2} \sin ^{2} 	heta}=4 \cot 	heta$

$ycot$ $	heta=y \sqrt{3}$

$\cot 	heta=\sqrt{3}$

$	heta=30^{\circ}$

$ycot$ $	heta=y$

$\cot 	heta=1$

$	heta=45^{\circ}$

Column $-I$ Angle of projection	Column $-II$
$A.$ $\theta \, = \,{45^o}$	$1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$
$B.$ $\theta \, = \,{60^o}$	$2.$ $\frac{{g{T^2}}}{R} = 8$
$C.$ $\theta \, = \,{30^o}$	$3.$ $\frac{R}{H} = 4\sqrt 3 $
$D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$	$4.$ $\frac{R}{H} = 4$

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