myhelper: RD Sharma 2020 solution class 9 chapter 20 Surface Area and Volume of a Right Circular Cone FBQS

RD Sharma 2020 solution class 9 chapter 20 Surface Area and Volume of a Right Circular Cone FBQS

FBQS

Page-20.25

Question 1:

The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height is __________.

Answer 1:

Let r be the radius of base and h be the height of both right circular cylinder and right circular cone.

∴ Volume of the cylinder, V₁ = $π r^{2} h$

Volume of the cone, V₂ =

\frac{1}{3} π r^{2} h

Now,

\frac{V_{1}}{V_{2}} = \frac{π r^{2} h}{\frac{1}{3} π r^{2} h} = 3

⇒ V₁ : V₂ = 3 : 1

Thus, the ratio of the volume of a right circular cylinder and a right circular cone of the same base and height is 3 : 1.

The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height is ___3 : 1___.

Question 2:

From a solid right circular cylinder of height h and base radius r, a conical cavity of the same height and base is scooped out. Then the ratio of the volume of the cone and the remaining solid is __________.

Answer 2:

It is given that, the radius and height of the solid right circular cylinder is r and h, respectively.

Now,

Radius of the conical cavity scooped out = Radius of the solid right circular cylinder = r

Height of the conical cavity scooped out = Height of the solid right circular cylinder = h

∴ Volume of the cone, V₁ =

\frac{1}{3} π r^{2} h

Also,

Volume of the remaining solid, V₂

= Volume of the cylinder − Volume of the conical cavity scooped out

= π r^{2} h - \frac{1}{3} π r^{2} h

= \frac{2}{3} π r^{2} h

∴ \frac{V_{1}}{V_{2}} = \frac{\frac{1}{3} π r^{2} h}{\frac{2}{3} π r^{2} h} = \frac{1}{2}

⇒ Volume of the cone, V₁ : Volume of the remaining solid, V₂ = 1 : 2

Thus, the the ratio of the volume of the cone and the remaining solid is 1 : 2.

From a solid right circular cylinder of height h and base radius r, a conical cavity of the same height and base is scooped out. Then the ratio of the volume of the cone and the remaining solid is ___1 : 2___.

Question 3:

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4. If their height are in the ratio 2 : 3, then their volumes are in the ratio ________.

Answer 3:

Let r be the radius and h be the height of the cylinder & R be the radius and H be the height of the cone.

It is given that,

\frac{r}{R} = \frac{3}{4}

and

\frac{h}{H} = \frac{2}{3}

.....(1)

Suppose V₁ be the volume of the cylinder and V₂ be the volume of the cone.

∴ \frac{V_{1}}{V_{2}}

= \frac{π r^{2} h}{\frac{1}{3} π R^{2} H}

= 3 \times {(\frac{r}{R})}^{2} \times \frac{h}{H}

= 3 \times {(\frac{3}{4})}^{2} \times \frac{2}{3}

[Using (1)]

= \frac{9}{8}

∴ Volume of the cylinder, V₁ : Volume of the cone, V₂ = 9 : 8

Thus, the ratio of volume of cylinder to the volume of cone is 9 : 8.

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4. If their height are in the ratio 2 : 3, then their volumes are in the ratio __9 : 8__.

Question 4:

A conical tent is 12 m high and the radius of its base is 9 m. The cost of canvas required to make the tent at the rate of ₹14 per m², is __________.

Answer 4:

Radius of the conical tent, r = 9 m

Height of the conical tent, h = 12 m

Let l be the slant height of the conical tent.

∴ l = \sqrt{r^{2} + h^{2}} = \sqrt{{(9)}^{2} + {(12)}^{2}} = \sqrt{225} = 15 m

Now,

Area of the canvas required to make the tent = Curved surface area of the conical tent

= π r l = \frac{22}{7} \times 9 \times 15 = \frac{2970}{7}

m²

Rate of the canvas = ₹14 per m²

∴ Cost of the canvas required to make the tent

= Area of the canvas required to make the tent × Rate of the canvas

= ₹ \frac{2970}{7} \times 14

= ₹5,940

Thus, the cost of canvas required to make the tent is ₹5,940.

A conical tent is 12 m high and the radius of its base is 9 m. The cost of canvas required to make the tent at the rate of ₹14 per m², is __₹5,940__.

Question 5:

A metal cuboid of dimensions 49 m, 22 m and 14 m is melted and cast into 7 identical cylinders of radius 7 m. These cylinders are again melted and cast into cubes such that the side of each cube is equal to half of the height of each cylinder. The numbers of cubes thus formed is __________.

Answer 5:

Let h be the height of each cylinder.

Radius of each cylinder, r = 7 m

It is given that, the metal cuboid of given dimensions is melted and re-cast into 7 identical cylinders.

∴ 7 × Volume of each cylinder = Volume of the cuboid

\Rightarrow 7 \times π r^{2} h = Length \times Breadth \times Height

\Rightarrow 7 \times \frac{22}{7} \times {(7 m)}^{2} \times h = 49 m \times 22 m \times 14 m

\Rightarrow h = \frac{49 \times 22 \times 14}{22 \times 49} = 14 m

So, the height of each cylinder is 14 m.

It is also given that these cylinders are again melted and re-cast into cubes such that the side of each cube is half of the height of each cylinder.

Let the side of each cube be a.

∴ a = \frac{h}{2} = \frac{14}{2} = 7 m

Suppose the number of cubes formed be n.

∴ n × Volume of each cube = Volume of 7 identical cylinder = Volume of the cuboid

\Rightarrow n = \frac{Volume of the cuboid}{Volume of each cube}

\Rightarrow n = \frac{49 m \times 22 m \times 14 m}{{(7 m)}^{3}}

[Volume of the cube = (Side)³]

\Rightarrow n = 44

Thus, the number of cubes formed is 44.

A metal cuboid of dimensions 49 m, 22 m and 14 m is melted and cast into 7 identical cylinders of radius 7 m. These cylinders are again melted and cast into cubes such that the side of each cube is equal to half of the height of each cylinder. The numbers of cubes thus formed is ___44___.

Question 6:

A solid cylinder and a solid cone have equal bases and equal heights. If the radius and height be in the ratio 4 : 3, the ratio of the total surface area of the cylinder to that of the cone is __________.

Answer 6:

Let r be the radius of the solid cylinder and solid cone & h be the height of the solid cylinder and solid cone.

It is given that, r : h = 4 : 3

Suppose r = 4x units and h = 3x units, where x is constant

Let l be the slant height of the cone.

∴ l = \sqrt{r^{2} + h^{2}} = \sqrt{{(4 x)}^{2} + {(3 x)}^{2}} = \sqrt{25 x^{2}} = 5 x

units

Now,

Total surface area of the cylinder = 2 $π$ r(r + h) = 2

π

× 4x × (4x + 3x) = 56

π

x² sq. units

Total surface area of the cone = $π$ r(r + l) =

π

× 4x × (4x + 5x) = 36

π

x² sq. units

∴ \frac{Total surface area of the cylinder}{Total surface area of the cone} = \frac{56 π x^{2}}{36 π x^{2}} = \frac{14}{9}

Thus, the ratio of total surface area of the cylinder to the total surface area of the cone is 14 : 9.

A solid cylinder and a solid cone have equal bases and equal heights. If the radius and height be in the ratio 4 : 3, the ratio of the total surface area of the cylinder to that of the cone is ___14 : 9___.

Question 7:

If h, S and V be the height curved surface area and volume of a cone respectively, then 3πVh³ – S²h²+ 9V² is equal to ________.

Answer 7:

Height of the cone = h

Let r be the radius and l be the slant height of the cone.

∴ l² = r² + h² .....(1)

Volume of the cone = V =

\frac{1}{3} π r^{2} h

Curved surface area of the cone = S =

π

∴ 3 π V h^{3} - S^{2} h^{2} + 9 V^{2}

= 3 π \times \frac{1}{3} π r^{2} h \times h^{3} - π^{2} r^{2} l^{2} \times h^{2} + 9 \times \frac{1}{9} π^{2} r^{4} h^{2}

= π^{2} r^{2} h^{4} - π^{2} r^{2} (r^{2} + h^{2}) \times h^{2} + π^{2} r^{4} h^{2}

[Using (1)]

= π^{2} r^{2} h^{4} - π^{2} r^{2} h^{4} - π^{2} r^{4} h^{2} + π^{2} r^{4} h^{2}

= 0

Thus, the value of

3 π V h^{3} - S^{2} h^{2} + 9 V^{2}

is 0.

If h, S and V be the height, curved surface area and volume of a cone respectively, then 3πVh³ – S²h²+ 9V² is equal to ___0___.

Question 8:

If the heights of two cones are in the ratio 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is _________.

Answer 8:

Let r be the radius and h be the height of the first cone & R be the radius and H be the height of the second cone.

It is given that,

\frac{h}{H} = \frac{1}{4}

and

\frac{r}{R} = \frac{4}{1}

.....(1)

Suppose V₁ be the volume of the first cone and V₂ be the volume of the second cone.

∴ \frac{V_{1}}{V_{2}}

= \frac{\frac{1}{3} π r^{2} h}{\frac{1}{3} π R^{2} H}

= {(\frac{r}{R})}^{2} \times \frac{h}{H}

[Using (1)]

= {(\frac{4}{1})}^{2} \times \frac{1}{4}

= 4

Thus, the ratio of the volume of first cone to the volume of second cone is 4 : 1.

If the heights of two cones are in the ratio 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is ___4 : 1___.

Question 9:

The diameters of two cones are equal and their slant heights are in the ratio 5 : 4. If the curved surface area of the smaller cone is 200 cm², then the curved surface area of the bigger cone is _________.

Answer 9:

Let l₁ be the slant height of first cone and l₂ be the slant height of the second cone.

Given: l₁ : l₂ = 5 : 4

∴ l₁ = 5x and l₂ = 4x, where x is constant

It is given that the diameters of the two cones are equal. Therefore, the radii of the two cones are equal.

Let the radius of each cone be r cm.

Since the radii of two cones are equal, so the cone with bigger slant height would be bigger than the other.

Now,

Curved surface area of the bigger cone = $π$ rl₁

Curved surface area of the smaller cone = $π$ rl₂

∴ \frac{Curved surface area of the bigger cone}{Curved surface area of the smaller cone} = \frac{π r l_{1}}{π r l_{2}}

\Rightarrow \frac{Curved surface area of the bigger cone}{200 {cm}^{2}} = \frac{l_{1}}{l_{2}} = \frac{5 x}{4 x}

⇒ Curved surface area of the bigger cone =

\frac{5}{4} \times 200 {cm}^{2}

= 250 cm²

Thus, the curved surface area of the bigger cone is 250 cm².

The diameters of two cones are equal and their slant heights are in the ratio 5 : 4. If the curved surface area of the smaller cone is 200 cm², then the curved surface area of the bigger cone is ____250 cm²____.

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UnknownFebruary 27, 2021 at 11:02 AM
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RD Sharma 2020 solution class 9 chapter 20 Surface Area and Volume of a Right Circular Cone FBQS

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