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A hollow body as shown in figure consist of right circular portion attacted to a hemisphere portion of Rdius R. Determine the height H of cone if the centre of mass of the composite body considers with the centre O of the circular base (Take $R = \sqrt{x}H$ )

Option: 1
$\frac{1-\sqrt{37}}{18}$

Option: 2

$\frac{1+\sqrt{37}}{18}$

Option: 3

$\frac{1+3\sqrt{2}}{5}$

Option: 4

$\frac{3\sqrt{2}-1}{5}$

Answers (1)

best_answer

$m_1 = \sigma \pi R(\sqrt{H^2+R^2})$

$m_2 = \sigma (2\pi R^2)$

$y_{cm}= \frac{m_1y_1+m_2y_2}{m_1+m_2}$

$0= m_1\left ( \frac{+H}{3} \right )+m_2\left ( \frac{-R}{2} \right )$

$m_1\left ( \frac{H}{3} \right )=m_2\left ( \frac{R}{2} \right )$

$\sigma \pi R(\sqrt{H^2+R^2})\left (\frac{H}{3} \right )= \sigma (2\pi R^2)\left (\frac{R}{2} \right )$

$\left ( \sqrt{H^2+R^2} \right )H=\frac{3R^3}{R}=3R^2$

$(H^2+R^2)H^2=9R^4$

$(1+(\sqrt{x})^2)=9(\sqrt{x})^4$

$1+x=9x^2; 9x^2-x-1=0$

$x=\frac{1\pm \sqrt{1-4\times 9\times -1}}{2\times 1}=\frac{1+\sqrt{37}}{18}$

Posted by

rishi.raj

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