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Q.

From a solid cylinder whose height is 2.4cm and diameter 1.4cm, a conical cavity of the same height and the same diameter is hollowed out. The total surface area of the remaining solid to the nearest  c m 2  is ____.


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Detailed Solution

 IMG_256
According to the given information, a height of 2.4 cm of a solid cylinder with a diameter of 1.4 cm, where the same conical cavity is hollowed out at the same height and diameter
So the dimensions we know are
Height of full cylinder (v) = 2.4 cm
Diameter of solid cylinder (d) = 1.4 cm
Therefore, the radius of the solid cylinder (r) = 0.7 cm
As we know, the formula of the total surface of the remaining solid is equal to the curved surface of the cylinder + the area of ​​​​the base of the cylinder + the curved surface of the cone
Now we know that the curved surface of the cylinder = 2πrh  
Area of the base of the cylinder  π r 2  and Curved surface area of the cone  πrl  .
Where  l  slant height which is given as l= r 2 + h 2  
Therefore, Total surface area of the remaining solid =
2πrh+π r 2 +πr r 2 + h 2  
Now substituting the values in the above formula, we get
2× 22 7 ×0.7×2.4+ 22 7 × (0.7) 2 × (2.4) 2 + (0.7) 2 10.56+1.54+5.5c m 2 17.6c m 2  
So, Total surface area of the remaining solid is equal to 17.6c m 2  .
 
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