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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. Q.8

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Give me the best method to solve the surface areas and volumes of ncert class 10 , Find the easiest solution of the exercise 13.1 question number 8 . Also find the simplest method of this question . 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.

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  1. The diagram for the question is as follows:

    Ncert solutions class 10 chapter 13-10

    From the question we know the following:

    The diameter of the cylinder = diameter of conical cavity = 1.4 cm

    So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

    Also, the height of the cylinder = height of the conical cavity = 2.4 cm

    Ncert solutions class 10 chapter 13-11

    Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder

    = πrl+(2πrh+πr2)

    = πr(l+2h+r)

    = (22/7)× 0.7(2.5+4.8+0.7)

    = 2.2×8 = 17.6 cm2

    So, the total surface area of the remaining solid is 17.6 cm2

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