Work Power And Energy Ques 2
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Answer:
Correct Answer: 2.$u=\sqrt{g L\left(2+\frac{3 \sqrt{3}}{2}\right)}$
Solution:
Formula:
Now, we have the following equations
(1) $T_Q=0$. Therefore, $m g \sin \theta=\frac{m v^2}{L}$
(2) $v^2=u^2-2 g h=u^2-2 g L(1+\sin \theta)$
(3) $Q D=\frac{1}{2}$ (Radius)
$ \Rightarrow\left(L \cos \theta-\frac{L}{8}\right)=\frac{v^2 \sin 2\left(90^{\circ}-\theta\right)}{2 g}=\frac{v^2 \sin 2 \theta}{2 g} $
Eq. (iii) can be written as $\left(\cos \theta-\frac{1}{8}\right)=\left(\frac{v^2}{g L}\right) \sin \theta \cos \theta$
Substituting value of $\left(\frac{v^2}{g L}\right)=\sin \theta$ from Eq. (i), we get
$ \left(\cos \theta-\frac{1}{8}\right)=\sin ^2 \theta \cdot \cos \theta=\left(1-\cos ^2 \theta\right) \cos \theta $
or $\quad \cos \theta-1 / 8=\cos \theta-\cos ^3 \theta$
$\therefore \cos ^3 \theta=1 / 8$ or $\cos \theta=1 / 2$ or $\theta=60^{\circ} + 360^{\circ}n$
From Eq. (i), $v^2=g L \sin \theta=g L \sin 60^{\circ}$
or
$ v^2=\frac{1}{2} g L $
$\therefore$ Substituting this value of $v^2$ in Eq. (ii)
$ u^2=v^2+2 g L(1-\sin \theta) $
BETA
