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Home » 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 .
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 cm^2.
CBSE | Class 10 | Exercise 13.1 | Maths | NCERT | Surface Areas And Volumes
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 cm^2.

The outline for the inquiry is as per the following:

Ncert solutions class 10 chapter 13-10

From the inquiry we know the accompanying:

The width of the chamber = measurement of cone shaped hole = 1.4 cm

Along these lines, the range of the chamber = sweep of the conelike hole = 1.4/2 = 0.7

Additionally, the tallness of the chamber = stature of the funnel shaped cavity = 2.4 cm

Ncert solutions class 10 chapter 13-11

Presently, the TSA of staying strong = surface space of funnel shaped pit + TSA of the chamber

    \[=\text{ }\pi rl+\left( 2\pi rh+\pi r^2 \right)\]

    \[=\text{ }\pi r\left( l+2h+r \right)\]

    \[=\text{ }\left( 22/7 \right)\times \text{ }0.7\left( 2.5+4.8+0.7 \right)\]

    \[=\text{ }2.2\times 8\text{ }=\text{ }17.6\text{ }cm^2\]

Along these lines, the absolute surface space of the leftover strong is 17.6 cm^2.

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