In this question we have to find the the total surface area
and the volume of the rocket given that a rocket is in the form
of a circular cylinder closed at the lower end with a
cone of the same radius attached to the top. The cylinder
is of radius 2.5 m andheight 21 m and the cone has the slant height 8 m.
This question is from r d sharma class 10 maths.
I found this question while doing maths class 10.
I need help in getting the solution.
Given,
Radius of the cylindrical portion of the rocket (R) = 2.5 m
Height of the cylindrical portion of the rocket (H) = 21 m
Slant Height of the Conical surface of the rocket (L) = 8 m
Curved Surface Area of the Cone (S1) = πRL = π(2.5)(8)= 20π
And,
Curved Surface Area of the Cone (S2) = 2πRH + πR2
S2 = (2π × 2.5 × 21) + π (2.5)2
S2 = (π × 105) + (π × 6.25)
Thus, the total curved surface area S is
S = S1 + S2
S = (Ï€20) + (Ï€105) + (Ï€6.25)
S = (22/7)(20 + 105 + 6.25) = 22/7 x 131.25
S = 412.5 m2
Therefore, the total Surface Area of the Conical Surface = 412.5 m2
Now, calculating the volume of the rocket
Volume of the conical part of the rocket (V1) = 1/3 × 22/7 × R2 × h
V1 = 1/3 × 22/7 × (2.5)2 × h
Let, h be the height of the conical portion in the rocket.
We know that,
L2 = R2 + h2
h2 = L2 – R2 = 82 – 2.52
h = 7.6 m
Using the value of h, we will get
Volume of the conical part (V1) = 1/3 × 22/7 × 2.52 × 7.6 m2 = 49.67 m2
Next,
Volume of the Cylindrical Portion (V2) = πR2h
V2 = 22/7 × 2.52 × 21 = 412.5 m2
Thus, the total volume of the rocket = V1 + V2
V = 412.5 + 49.67 = 462.17 m2
Hence, the total volume of the Rocket is 462.17 m2