Question
The height of a right circular cylinder is 6 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface, then the radius of its base is :
Answer: Option D
$$\eqalign{
& 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr
& \Rightarrow 6r = 4h \cr
& \Rightarrow r = \frac{2}{3}h \cr
& \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr
& \Rightarrow r = 4\,m \cr} $$
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$$\eqalign{
& 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr
& \Rightarrow 6r = 4h \cr
& \Rightarrow r = \frac{2}{3}h \cr
& \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr
& \Rightarrow r = 4\,m \cr} $$
Was this answer helpful ?
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