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From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. $(Use π = \frac{22}{7})$

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14 {cm}^{2}

16 {cm}^{2}

18 {cm}^{2}

20 {cm}^{2}

answer is C.

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Detailed Solution

Given,
The cylinder height h= 2.4 cm
And diameter of cylinder =1.4 cm
We know that,

Radius = \frac{Diameter}{2}

Radius = \frac{1.4}{2}

Radius = 0.7

cm
Now, conical part slant height

l = \sqrt{r^{2} + h^{2}}

l = \sqrt{{0.7}^{2} + {2.4}^{2}}

l = \sqrt{0.49 + 5.76}

l = \sqrt{6.25}

l = 2.5

Therefore,
The remaining solid of surface area,

= 2 πrh + πrl + π r^{2}

= 2 \times \frac{22}{7} \times 0.7 \times 2.4 + \frac{22}{7} \times 0.7 \times 2.5 + \frac{22}{7} \times 0.7 \times 0.7

= 4.4 \times 2.4 + 2.2 \times 2 . + 2.2 \times 0.7

= 10.56 + 5.50 + 1.54

= 17.60 {cm}^{2}

So, the total surface area of the remaining solid to the nearest =

18 {cm}^{2}

.
Thus, the option 3 is correct.

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