हिंदी

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 2.5m and Height 21m and the Cone Has a Slant Height 8m. Calculate Total Surface Area and Volume of the Rocket? - Mathematics

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प्रश्न

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 2.5m and height 21m and the cone has a slant height 8m. Calculate total surface area and volume of the rocket?

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उत्तर

Given radius of cylinder (a) = 2.5m

Height of cylinder (h) = 21m

Slant height of cylinder (l) = 8m

Curved surface area of cone(S1) = πrl

S1 = π(2.5)(8)cm2            ...........(1)

Curbed surface area of a cone`= 2pirh+pir^2`

`S_2=2pi(2.5)(21)+pi(2.5)^2cm^2`         ...........(2)

∴Total curved surface area = (1) + (2)

S = S1 + S2

S = π(2.5)(8) + 2π(2.5)(21) + π(2.5)2

S = 62.831 + 329.86 + 19.63

S = 412.3m2

∴Total curved surface area = 412.3m2

Volume of a cone `=1/3pir^2h`

`V_1=1/3xxpi(2.5)^2h cm^3`      .........(3)

Let ‘h’ be height of cone

`l^2=r^2+h^2`

⇒ `l^2-r^2=h^2`

⇒ `h=sqrt(l^2-r^2)`

⇒ `h=sqrt(8^2-25^2)`

⇒ h =23.685m

Subtracting ‘h’ value in(3)

Volume of a cone `(V_1)=1/3xxpi(2.5)^2(23.685)  cm^2`      ........(4)

Volume of a cylinder `(V_2)=pir^2h`

`=pi(2.5)^2 21m^3`           ...........(5)

Total volume = (4) + (5)

V = V1 + V2

⇒ `V=1/3xxpi(2.5)^2(23.685)+pi(2.5)^2=1`

⇒ V = 461.84m2

Total volume (V) = 461.84m2

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Surface Areas and Volumes - Exercise 14.2 [पृष्ठ ६०]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 14 Surface Areas and Volumes
Exercise 14.2 | Q 2 | पृष्ठ ६०
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