Iift 2010 Question 27
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IIFT 2010
$$\dfrac{1}{2}$$*(Total volume - Volume of the hole) = Volume of the hole
Volume of the hole = $$\dfrac{1}{3}*\text{Total volume of the rocket}$$
$$\pi*R^2*(8/3)x$$ = $$\dfrac{1}{3}*(\pi*8^2*5x+\dfrac{1}{3}*\pi*8^2*3x)$$
$$R^2*(8/3)$$ = $$\dfrac{1}{3}*(8^2*5+\dfrac{1}{3}*8^2*3)$$
$$R^2 = 48$$
$$R$$ = $$4\sqrt{3}$$ cm. Hence, option A is the correct answer.