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Home » 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 and height and the cone has the slant height . Calculate the total surface area and the volume of the rocket.
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 the slant height 8m. Calculate the total surface area and the volume of the rocket.
CBSE | Class 10 | Exercise 16.2 | Maths | RD Sharma | Surface Areas And Volumes
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 the slant height 8m. Calculate the total surface area and the volume of the rocket.

According to the question it is given that,

Radius of the cylindrical portion of the rocket (R) =2.5m

Height of the cylindrical portion of the rocket (H) =21m

Slant Height of the Conical surface of the rocket (L) =8m

Curved Surface Area of the Cone \left( {{S}_{1}} \right)=\pi rL=\pi \left( 2.5 \right)\left( 8 \right)=20\pi

Then,

Curved Surface Area of the Cone \left( {{S}_{2}} \right)=2\pi RH+\pi {{R}^{2}}

{{S}_{2}}=\left( 2\pi \times 2.5\times 21 \right)+\pi {{\left( 2.5 \right)}^{2}}

{{S}_{2}}=\left( \pi \times 105 \right)+\left( \pi \times 6.25 \right)

Thus, the total curved surface area S is

S={{S}_{1}}+{{S}_{2}}

S=\left( \pi 20 \right)+\left( \pi 105 \right)+\left( \pi 6.25 \right)

S=\left( 22/7 \right)\left( 20+105+6.25 \right)=22/7\times 131.25

S=412.5{{m}^{2}}

Now, the total Surface Area of the Conical Surface =412.5{{m}^{2}}

Then, calculating the volume of the rocket

Volume of the conical part of the rocket \left( {{V}_{1}} \right)=1/3\times 22/7\times {{R}^{2}}\times h

{{V}_{1}}=1/3\times 22/7\times {{\left( 2.5 \right)}^{2}}\times h

Assume, h be the height of the conical portion in the rocket.

We all know that,

{{L}^{2}}={{R}^{2}}+{{h}^{2}}

{{h}^{2}}={{L}^{2}}-{{R}^{2}}={{8}^{2}}-{{2.5}^{2}}

h=7.6m

Using the value of h, we will get

Volume of the conical part \left( {{V}_{1}} \right)=1/3\times 22/7\times {{2.5}^{2}}\times 7.6{{m}^{2}}=49.67{{m}^{2}}

Then,

Volume of the Cylindrical Portion \left( {{V}_{2}} \right)=\pi {{R}^{2}}h

{{V}_{2}}=22/7\times {{2.5}^{2}}\times 21=412.5{{m}^{2}}

Thus, the total volume of the rocket ={{V}_{1}}+{{V}_{2}}

V=412.5+49.67=462.17{{m}^{2}}

Hence, the total volume of the Rocket is 462.17{{m}^{2}}

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