Rd Sharma 2022 for Class 9 Maths Chapter 21 - Surface Area And Volume Of A Sphere

Answer:

Here we need to find the surface area of spheres of different radii

We solve it using the formula,

Curved surface area of the sphere

(i) Radius = 10.5cm

Therefore, the surface area of the sphere of radius 10.5cm is

(ii) Radius = 5.6cm

Therefore, the surface area of the sphere of radius 5.6cm is

(iii) Radius = 14cm

Therefore, the surface area of the sphere of radius 14cm is

Answer:

Here we need to find the surface area of spheres of different diameter.

We solve it using the formula,

Curved surface area of the sphere =

(i) Diameter = 14 cm

Therefore, the surface area of the sphere of diameter 14 cm is.

(ii) Diameter = 21 cm

Therefore, the surface area of the sphere of diameter 21 cm is.

(iii) Diameter = 3.5 cm

Therefore, the surface area of the sphere of diameter 3.5 cm is.

Answer:

In the given problem,

Radius of the hemisphere = 10 cm

Now, the total surface area of a hemisphere =

Therefore, the total surface area of the hemisphere is

Also,

The total surface area of the solid hemisphere =

Therefore, the total surface area of the hemisphere is

Answer:

In the given problem,

Surface area of a sphere = 5544 cm²

Also, surface area of a sphere =

So according to the problem,

Now,

Therefore, diameter of the sphere = cm

= 42 cm

Therefore, the diameter of the sphere is

Answer:

In the given problem, we have a hemispherical bowl.

Here,

Diameter of the bowl = 10.5 cm

So, Radius of the bowl = cm

So, the curved surface area of the bowl

Now, the rate of tin plating per 100 cm² = Rs 4

The rate of tin plating per 1 cm² = Rs

So, the cost of tin plating the bowl

= Rs 6.93

Therefore, the cost of tin plating the hemispherical bowl from inside is

Answer:

In the given problem, the dome of the building is in a form of a hemisphere.

Here,

Radius of the hemisphere = 63 dm

= 6.3 m

So, to find the cost of painting, we first find the curved surface area of the hemisphere.

Curved surface area of the sphere

Now, the rate of painting per m² = Rs 2

So, the cost of painting the dome

Therefore, the cost of painting the dome is

Answer:

In the given problem, we have two objects; moon and earth.

Let us take the diameter of the earth = d km

Therefore, the radius of the earth = km

So, the surface area of the earth (S_e) =

Now, according to the problem, the diameter of moon is one-fourth of the diameter of earth.

So, the diameter of the moon = km

Therefore, the radius of the moon = km

So, the surface area of the moon (S_m) =

Ratio of surface areas =

Therefore, the ratio of the surface areas of moon to surface area of earth is

Answer:

In the given problem, a hemispherical dome of the building needs to be painted. So, we need to find the surface area of the dome.

Here, we are given the circumference of the hemispherical dome as 17.6 m and as we know that circumference of the hemisphere is given by. So, we get

So, now we find the surface area of the hemispherical dome.

So, the curved surface area of the dome is 49.28 m²

Since the rate of the painting is given in cm², we have to convert the surface area from m² to cm².

So, we get

Curved surface area =

= 492800 cm²

Now, the rate of painting per 100 cm² = Rs 5

The rate of painting per 1 cm² =

So, the cost of painting the dome =

Therefore, the cost of painting the hemispherical dome of the building is

Answer:

In the given problem, let us assume that the earth is a sphere and the radius of the earth is given equal to 6370 km.

The total surface area of earth can be calculated using the formula,

So, the total surface area of earth is equal to 510109600 km².

Now, according to the question three-fourth of the earth’s surface is covered with water. This means that one-fourth of the surface is covered with land.

Therefore, we get,

The area of land covered by water

Therefore the area of earth’s surface covered by land is equal to

Answer:

In the given problem, we are given a shape in which a cylinder is placed over a hemisphere. The radius of the hemisphere and the cylinder is equal and also the height of the cylinder is equal to the radius.

So, let draw a figure representing this shape and take the radius as r cm.

So, let us find the value of r first. From the figure, we can see that,

Also, height of the sphere = 3.5 cm (as radius and height of the cylinder are equal)

Now, the surface area of the shape is equal to the surface area of the hemisphere and surface area of cylinder together.

Surface area = surface area of hemisphere + surface area of cylinder

Therefore, the surface area of the given shape is

Answer:

In the given problem, we have a toy in the form of a cone surmounted on a hemisphere. So, let us first draw a diagram of the toy with the given dimensions.

So here,

Height of the cone = 15 cm

Diameter of the cone base and hemisphere = 16 cm

Therefore, radius of the cone base and hemisphere =

To find the lateral surface area of the cone, we first need to find the slant height of the cone, which is given by the following formula,

So, we get

Now, let us find the surface area of the toy first,

Total Surface area = lateral surface area of the cone + surface area of the hemisphere

So, the total surface area of the toy is 829.71 cm²

Now, the rate of painting the toy = Rs 7 per 100 cm²

= Rs per cm²

So, the cost of painting the given toy = total surface area x rate of painting

Therefore, the cost of painting the toy is

Answer:

In the given problem, we are given a storage tank which consists of a circular cylinder with a hemisphere adjoined on either side.

So here,

External diameter of the cylinder = 1.4 m

External radius of the cylinder = m

m

Height of the cylinder = 8 m

Now, we find the total surface area of the storage tank.

Total surface area = Surface area of cylinder + surface area of the hemisphere

Therefore, the external surface area of the storage tank is 38.28 m² .

Now, the rate of painting the storage tank on outside = Rs 10 per m²

So, the cost of painting the storage tank on outside =

Therefore, the cost of painting the storage tank on outside is

Answer:

In the given problem, we have 8 spheres mounted upon small cylindrical supports, as shown in the diagram.

Now, the spheres are to be painted silver. So, let us first find the surface area of the spheres first.

Diameter of the sphere = 21 cm

So, radius of the sphere = 10.5 cm

Now,

This is the surface area of one sphere. We have 8 such spheres. So,

Total surface area of 8 spheres =

cm²

At the junction of the sphere and the cylinder, there is some portion which would be covered. Hence, that would not be painted silver.

The covered area

So, for 8 spheres,

The covered area

Therefore, the total area to be painted

Now, the rate of silver paint per cm² = 25 paisa

=Rs 0.25

So, the cost of silver paint for 8 spheres =

= Rs 2757.86

Now, the cylinders are to be painted black. So, let us find the surface area of the cylinders.

Radius of the cylinder = 1.5 cm

Height of the cylinder = 7 cm

Now,

This is the surface area of one cylinder. We have 8 such cylinders. So,

Total surface area of 8 cylinder =

cm²

Now, the rate of black paint per cm² = 5 paise

= Rs 0.05

So, the cost of silver paint for 8 cylinders =

Therefore,

The total cost of the paint = cost of silver paint + cost of black paint

Therefore, the cost of painting would be

Answer:

In the given problem, we have to find the volume of the sphere of given radii.

(i) Radius of the sphere = 2 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 2 cm is

(ii) Radius of the sphere = 3.5 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 3.5 cm is

(iii) Radius of the sphere = 10.5 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 10.5 cm is

Answer:

In the given problem, we have to find the volume of the spheres of different diameter.

(i) Diameter of the sphere = 14 cm

So, volume of the sphere =

So, the volume of the sphere of diameter 14 cm is.

(ii) Diameter of the sphere = 3.5 dm

So, volume of the sphere =

So, the volume of the sphere of diameter 3.5 dm is.

(iii) Diameter of the sphere = 2.1 m

So, volume of the sphere =

So, the volume of the sphere of diameter is.

Answer:

In the given problem, we have a hemispherical tank of the following dimensions:

Inner radius of the tank (r) = 2.8 m

So for the capacity of the tank, we need to find the volume of the hemispherical tank.

Volume of the tank =

Now, as we know,

So,

Therefore, the capacity of the tank is.

Answer:

In the given problem, we have a hemispherical bowl of the following dimensions:

Inner radius of the bowl (r) = 5 cm

Thickness of the iron sheet = 0.25 cm

So, the outer radius of the tank (R) = inner radius + thickness of the sheet

Now, we need to find the volume of steel used to make the bowl. For that we can use the formula for calculating the volume of a hemispherical shell.

Volume of a hemispherical shell =

Therefore, the volume of the iron used to make the tank is

Answer:

In the given problem, we have a lead cube which is remolded into small spherical bullets.

Here, edge of the cube (s) = 22 cm

Diameter of the small spherical bullets (d) = 2 cm

Now, let us take the number of small bullets be x

So, the total volume of x spherical bullets is equal to the volume of the lead cube.

Therefore, we get,

Volume of the x bullets = volume of the cube

Therefore, small bullets can be made from the given lead cube.

Answer:

In the given problem, we have a spherical ball of lead which is remolded into 3 smaller balls.

Here, Diameter of the bigger ball (d) = 3 cm

Diameter of 1^st small ball (d₁) = cm

Diameter of 2^nd small ball (d₂) = cm

Let the diameter of the 3^rd ball = x cm.

So now,

Volume of the bigger ball = sum of the volumes of the smaller balls

Further, solving for x, we get

Further,

Therefore, the diameter of the third ball is.

Answer:

In the given problem, a sphere is immersed in a water filled cylinder and this leads to a rise in the water level by 5/3 cm. Here, we need to find the radius of the cylinder.

Given here,

Radius of the sphere (r_s) = 5 cm

Rise in the level of water in cylinder (h) = 5/3 cm

So, let us take the radius of the cylinder (r_c) = x cm

Now, according to the problem, the volume of the sphere will be equal to the increase in the volume of the cylinder.

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore, the radius of the cylinder is.

Answer:

In the given problem,

Let us take the radius of 1^st sphere (r₁) = x cm

So, the radius of 2^nd sphere (r₂) = 2x cm

So, volume of the 1^st sphere (V₁) =

Volume of the 2^nd sphere (V₂) =

Now, the ratio of the volumes of two balls =

Therefore, the ratio of the volumes is

Answer:

In the given problem, we have a cylinder of the following dimensions:

Let the diameter of the cylinder be d.

So, the height of the cylinder (h) =

Now, the volume of the cylinder (V₁) =

Also, volume of the cylinder is equal to the volume of a sphere of radius 4 cm.

So, volume of the sphere =

Now, it is given that the volume of the sphere is equal to the volume of the cylinder.

So,

So, the diameter of the base of cylinder is 8 cm

Therefore, radius of the base of cylinder is.

Answer:

In the given problem, we are given a cone and a hemisphere which have equal bases and have equal volumes. We need to show that the ratio of their heights.

So,

Let the radius of the cone and hemisphere be x cm.

Also, height of the hemisphere is equal to the radius of the hemisphere.

Now, let the height of the cone = h cm

So, the ratio of the height of hemisphere to the height of the cone =

Here, Volume of the hemisphere = volume of the cone

Therefore, the ratio of the heights is.

Answer:

In the given problem, we need to find the height of water in a given cylinder when a hemispherical bowl full of water is emptied into it.

Here,

Radius of the hemispherical bowl (r_s) = 3.5 cm

Radius of the cylinder (r_c) = 7 cm

So, according to the question, the volume of the water in the hemispherical bowl will be equal to the volume of the water in cylinder.

Let us take the height of the water level in cylinder as h cm

So,

Volume of hemispherical bowl = volume of the water in cylinder

Therefore, the height of the water in the cylinder is equal to .

Answer:

In the given problem, we need to find the height of water in a given cylinder when a hemispherical bowl full of water is emptied into it.

Here,

Radius of the hemispherical bowl (r_s) = 6 cm

Radius of the cylinder (r_c) = 4 cm

So, according to the question, the volume of the water in the hemispherical bowl will be equal to the volume of the water in cylinder.

Let us take the height of the water level in cylinder as h cm

So,

Volume of hemispherical bowl = volume of the water in cylinder

Therefore, the height of the water in the cylinder is equal to .

Answer:

In the given problem, the first step is to find the volume of the sphere.

Here,

The diameter of the sphere = 18 cm

So, the radius of the sphere =

Therefore, the volume of the sphere =

Now, the sphere is melted and drawn into a long wire of uniform cross section. So, the volume of both the sphere and the wire will be equal.

For the wire,

Length of the wire = 108 m

= 10800 cm

Let the radius of the wire = x cm

So the volume of the wire =

According to the question,

So, the radius of the wire is 0.3 cm.

Therefore, the diameter of the wire is

Answer:

In the given problem, we are given a sphere which is melted and drawn into a wire. So, here, the first step is to find the volume of the sphere.

Here,

The diameter of the sphere = 6 cm

So, the radius of the sphere =

Therefore, the volume of the sphere =

Now, the sphere is melted and drawn into a long wire of uniform cross section. So, the volume of both the sphere and the wire will be equal.

For the wire,

Diameter of the wire = 0.2 cm

Radius of the wire = 0.1 cm

Let the length of the wire = x cm

So, the volume of the wire =

Now, volume of sphere = volume of wire,

So, the length of the wire is

Answer:

In the given problem, we have a lead hemisphere which is remolded into a right circular cone.

Here,

Radius of the hemisphere (r_h) = 7 cm

Height of the cone (h) = 49 cm

So, let the base radius of the cone (r_c) = x cm

Now, the volume of the hemisphere will be equal to volume of the volume of the cone.

So, we get,

Volume of the hemisphere = volume of the cone

Further, solving for x

Further,

Therefore, base radius of the cone is .

Answer:

In the given problem, we have a hollow sphere of given dimensions;

Internal radius of the sphere (r) = 2 cm

External radius of the sphere (R) = 4 cm

Now, the given sphere is molded into a cone,

So, radius of the cone (r_c) = 4 cm

Now, the volume of hollow sphere is equal to the volume of the cone.

So, let the height of cone = h cm

Therefore, we get

Volume of hollow sphere = the volume of cone

Further, solving for h,

So, height of the cone is

Now, slant height (l) of a cone is given by the formula:

So, we get,

Therefore, slant height of the cone is

Answer:

In the given problem, we have a metallic sphere which is remolded into small cones.

Here, radius of the metallic sphere (r_s) = 10.5 cm

Radius of the small cone (r_c) = 3.5 cm

Height of the small cone (h) = 3 cm

Now, let us take the number of small cones be x

So, the total volume of x cones will be equal to the volume of the metallic sphere.

Therefore, we get,

Therefore, small cones can be made from the given bigger metallic sphere.

Answer:

In the given problem, we have a hemispherical tank of the following dimensions:

Inner radius of the tank (r) = 1 m

Thickness of the iron sheet = 1 cm

= 0.01 m

So, the outer radius of the tank (R) = inner radius + thickness of the sheet

Now, we need to find the volume of iron used to make the tank. For that we can use the formula for calculating the volume of a hemispherical shell.

Volume of a hemispherical shell =

Therefore, the volume of the iron used to make the tank is

Answer:

In the given problem, we have a capsule in the form of a sphere. We need to find out how much medicine is required to fill in the given capsule. So, for that we find the volume of the capsule.

Here,

Diameter of the capsule = 3.5 mm

Therefore, Radius of the capsule =

mm

Now, Volume of the sphere =

Therefore, the amount of medicine needed to fill the given capsule is

Answer:

In the given problem, we have two objects; moon and earth.

Let us take the diameter of the earth = d km

Therefore, the radius of the earth = km

So, the volume of the earth (V_e) =

Now, according to the problem, the diameter of moon is one-fourth of the diameter of earth.

So, the diameter of the moon = km

Therefore, the radius of the moon = km

So, the volume of the moon (V_m) =

So, fraction of the volumes =

Therefore, the fraction of the volume of moon to volume of earth is

Answer:

In the given problem, a sphere is immersed in a water filled cylinder and this leads to a rise in the water level by 9 cm. Here, we need to find the radius of the cylinder.

Given here,

Radius of the cylinder (r_c) = 16 cm

Rise in the level of water in cylinder (h) = 9 cm

So, let us take the radius of the sphere (r_s) = x cm

Now, according to the problem, the volume of the ball will be equal to the increase in the volume of the cylinder; as the volume of water replaced by the spherical ball will lead to rise in the level of water.

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore, the radius of the spherical ball is.

Answer:

In the given problem, a spherical iron ball is immersed in a water filled cylinder and this leads to a rise in the water level by 6.75 cm. Here, we need to find the radius of the ball.

Given here,

Radius of the cylinder (r_c) = 12 cm

Rise in the level of water in cylinder (h) = 6.75 cm

So, let us take the radius of the spherical ball (r_s) = x cm

Now, according to the problem, the volume of the spherical ball will be equal to the increase in the volume of the cylinder as the volume of water replaced by the ball is increases the level of water in the cylinder.

So,

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore the radius of the spherical ball is.

Answer:

In the given problem, let us take the number of spheres required to raise the level of oil by 2 cm be x.

Also, it is given that,

The radius of sphere (r_s) = 1.5 cm

The radius of the cylinder (r_c) = 6 cm

So, according to the question; the volume of the iron spheres will be equal to the volume of the oil increased.

We get,

Volume of x iron spheres = volume of the oil raised

Therefore, the number of iron spheres required to raise the level of oil by 2 cm is.

Answer:

In the given problem, let us take the change in the level of water be h cm.

Also, it is given that,

The diameter of the spherical balls = 2 cm

So, radius of spherical balls (r_s) = 1 cm

The diameter of the measuring cylinder = 10 cm

So, radius of the measuring cylinder (r_c) = 5 cm

So, according to the question; the volume of the 4 spherical balls will be equal to the volume of the water increased.

We get,

Volume of 4 spheres = volume of the water raised

Therefore, the water level increases by.

Answer:

In the given problem, we are given a cone, a hemisphere and a cylinder which stand on equal bases and have equal heights. We need to show that their volumes are in the ratio

1 : 2 : 3

So,

Let the radius of the cone, cylinder and hemisphere be x cm.

Now, the height of the hemisphere is equal to the radius of the hemisphere. So, the height of the cone and the cylinder will also be equal to the radius.

Therefore, the height of the cone, hemisphere and cylinder = x cm

Now, the next step is to find the volumes of each of these.

Volume of a cone (V₁) =

Volume of a hemisphere (V₂) =

Volume of a cylinder (V₃) =

So, now the ratio of their volumes = (V₁) : (V₂) : (V₃)

Hence proved, the ratio of the volumes of the given cone, hemisphere and the cylinder is .

Answer:

In the given problem, we have a spherical shell which is remolded into a solid cylinder. So here, we first find the volume of the spherical shell.

We are given that,

Radius of internal surface of the spherical shell = 3 cm

Radius of external surface of the spherical shell = 5 cm

So,

The volume of the spherical shell =

Where, R = external radius

r = internal radius

So,

Volume of the shell =

Next we find the volume of the cylinder.

Height of the cylinder =

Let us take the radius of the cylinder as r cm.

So,

Volume of the cylinder =

Now, according to the problem, the volume of shell will be equal to the volume of the solid cylinder. So, we get

Volume of spherical shell = Volume of cylinder

Since, the radius is 7 cm; the diameter of the cylinder will be 14 cm.

Therefore, the diameter of the cylinder is.

Answer:

In the given problem, a spherical form ball is immersed in a water filled cylinder and this leads to a rise in the water level by 6.75 cm. Here, we need to find the radius of the ball.

Given here,

Radius of the cylinder (r_c) = 12 cm

Rise in the level of water in cylinder (h) = 6.75 cm

So, let us take the radius of the spherical ball (r_s) = x cm

Now, according to the problem, the volume of the spherical ball will be equal to the increase in the volume of the cylinder as the volume of water replaced by the ball is increases the level of water in the cylinder.

So,

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore the radius of the spherical ball is.

Answer:

In the given problem, we are given a cone, a sphere and a cylinder which have equal diameter. Also, the height of cylinder and cone is equal to the diameter of the sphere. We need to find the ratio of their volumes.

So,

Let the diameter of the cone, cylinder and sphere be x cm.

Now, the height of the cone and cylinder is equal to the diameter of the hemisphere. Therefore, the height of the cone and cylinder = x cm

Now, the next step is to find the volumes of each of these.

Volume of a cone (V₁) =

Volume of a sphere (V₂) =

Volume of a cylinder (V₃) =

So, now the ratio of their volumes = (V₂) : (V₃) : (V₁)

Therefore, the ratio of the volumes of the given sphere, cylinder and cone is.

Answer:

In the given problem, we have to find the surface area of a sphere of a given radii.

Radius of the sphere (r) = 14 cm

So, surface area of the sphere =

Therefore, the surface area of the given sphere of radius 14 cm is.

Answer:

In the given problem, we have to find the total surface area of a hemisphere of a given radii.

Radius of the hemisphere (r) = 10 cm

So, total surface area of the hemisphere =

Therefore, the total surface area of the given hemisphere of radius 10 cm is.

Answer:

In the given problem, we have to find the radius of a sphere whose surface area is given.

Surface area of the sphere (S) = 154 cm²

Let the radius of the sphere be r cm

Now, we know that surface area of the sphere =

So,

Further, solving for r

Therefore, the radius of the given sphere is.

Answer:

In the given problem, the area available for the motorcyclist for riding will be equal to the surface area of the hollow sphere. So here, we have to find the surface area of a hollow sphere of a given diameter.

Diameter of the sphere (d) = 7 m

So, surface area of the sphere =

Therefore, the area available for the motorcyclist for riding is.

Answer:

In the given problem, we have to find the volume of a sphere whose surface area is given.

So, let us first find the radius of the given sphere.

Surface area of the sphere (S) = 154 cm²

Let the radius of the sphere be r cm

Now, we know that surface area of the sphere =

So,

Further, solving for r

Therefore, the radius of the given sphere is 3.5 cm.

Now, the volume of the sphere =

Therefore, the volume of the given sphere is.

Answer:

In the given problem, we have a lead cube which is remolded into small spherical bullets.

Here, edge of the cube (s) = 44 cm

Diameter of the small spherical bullets (d) = 4 cm

Now, let us take the number of small bullets be x

So, the total volume of x spherical bullets is equal to the volume of the lead cube.

Therefore, we get,

Volume of the x bullets = volume of the cube

Therefore, small bullets can be made from the given lead cube.

Answer:

In the given problem, we are given a cone and a sphere which have equal volumes. The dimensions of the two are;

Radius of the cone (r_c) = r

Radius of the sphere (r_s) = 2r

Now, let the height of the cone = h

Here, Volume of the sphere = volume of the cone

Further, solving for h

Therefore, the height of the cone is.

Answer:

In the given problem, we have a hollow sphere of given dimensions;

Internal diameter of the sphere (d) = 4 cm

External diameter of the sphere (D) = 8 cm

Now, the given sphere is molded into a cone,

Diameter of the base of cone (d_c) = 8 cm

Now, the volume of hollow sphere is equal to the volume of the cone.

So, let the height of cone = h cm

Therefore, we get

Volume of cone = the volume of hollow sphere

Further, solving for h,

So, height of the cone is

Answer:

In the given problem, we are given a sphere and a cone of the following dimensions:

Radius of the sphere (r_s) = 5 cm

So, surface area of the sphere =

Also, radius of the cone base (r_c) = 4 cm

So, curved surface area of the cone =

Now, it is given that the surface area of the sphere is 5 times the curved surface are of the cone. So, we get

Now, slant height (l) of a cone is given by the formula:

So, let us take the height of the cone as h,

We get,

Squaring both sides,

Further, solving for h

Therefore, height of the cone is.

Answer:

In the given problem, we are given a sphere inscribed in a cube. So, here we need to find the ratio between the volume of a cube and volume of sphere. This means that the diameter of the sphere will be equal to the side of the cube. Let us take the diameter as d.

Here,

Volume of a cube (V₁) =

Volume of a sphere (V₂) =

Now, the ratio of the volume of sphere to the volume of the cube =

So, the ratio of the volume of cube to the volume of the sphere is.

Answer:

We know

Volume of the sphere of radius R = $\frac{4}{3}\mathrm{\pi }{R}^3$

Now,

Radius of the sphere = 2r         (Given)

∴ Volume of the sphere

$=\frac{4}{3}\mathrm{\pi }{\left(2r\right)}^3$               

$=\frac{4}{3}\mathrm{\pi }\times 8r^3$

$=\frac{32}{3}\mathrm{\pi }{r}^3$

Thus, the volume of the sphere is $\frac{32}{3}\mathrm{\pi }{r}^3$ cu. units.

If the radius of a sphere is 2r, then its volume is $\underline{ \frac{32}{3}\mathrm{\pi }{r}^3 \mathrm{cu}. \mathrm{units} }$.

Answer:

We know

Surface area of the hemisphere having radius r = 3$\mathrm{\pi }$r²

It is given that the radius of a hemispherical balloon increases from 6 cm to 12 cm when air is pumped into it.

Let S₁ be the surface area of the hemispherical balloon when its radius is 6 cm and S₂ be the surface area of the hemispherical balloon when its radius is 12 cm.

∴ S₁ = Surface area of the hemispherical balloon when its radius is 6 cm = 3$\mathrm{\pi }$(6 cm)²

S₂ = Surface area of the hemispherical balloon when its radius is 12 cm = 3$\mathrm{\pi }$(12 cm)²

Now,

$\frac{S_1}{S_2}=\frac{3\mathrm{\pi }{\left(6 \mathrm{cm}\right)}^2}{3\mathrm{\pi }{\left(12 \mathrm{cm}\right)}^2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{6\times 6}{12\times 12}=\frac{1}{4}$

⇒ S₁ : S₂ = 1 : 4

Thus, the ratio of the surface areas of the hemispherical balloon in two cases is 1 : 4.

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into. The ratio of the surface areas of the balloon in two cases is ___1 : 4___.

Answer:

It is given that a cone, a hemisphere and a cylinder stand on equal bases and have the same height.

Let the radius of cone, hemisphere and cylinder be r units.

Radius of the cone = Radius of hemisphere = Radius of cylinder = r

Also,

Height of the cone = Height of the cylinder = Height of the hemisphere

We know that, the height of a hemisphere is same as its radius.

∴ Height of the hemisphere = r

⇒ Height of the cone = Height of the cylinder = Height of the hemisphere = r

Now,

Volume of the cone = $\frac{1}{3}\mathrm{\pi }$ × (Radius)² × Height = $\frac{1}{3}\mathrm{\pi }$ × r² × r = $\frac{1}{3}\mathrm{\pi }$r³

Volume of the hemisphere = $\frac{2}{3}\mathrm{\pi }$ × (Radius)³ = $\frac{2}{3}\mathrm{\pi }$r³

Volume of the cylinder = $\mathrm{\pi }$ × (Radius)² × Height = $\mathrm{\pi }$ × r² × r = $\mathrm{\pi }$r³

∴ Volume of the cone : Volume of the hemisphere : Volume of the cylinder

= $\frac{1}{3}\mathrm{\pi }$r³ : $\frac{2}{3}\mathrm{\pi }$r³ : $\mathrm{\pi }$r³

= $\frac{1}{3}$ : $\frac{2}{3}$ : 1

= 1 : 2 : 3

Thus, the ratio of their volumes is 1 : 2 : 3.

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is ____1 : 2 : 3____.

Answer:

The largest right circular cone that can be fitted in a cube of given edge is such that the diameter of the base of the cone is equal to the edge of the cube and the height of the cone is equal to the edge of the cube.

It is given that the edge of cube is 2r.

Let R be the radius and H be the height of the largest right circular cone that can be fitted in the given cube.

∴ Diameter of the base of the cone = Edge of the cube

⇒ 2R = 2r

⇒ R = r

Height of the cone = Edge of the cube

⇒ H = 2r

∴ Volume of the largest cone that can be fitted in the given cube

$=\frac{1}{3}\mathrm{\pi }{R}^{2}H$

$=\frac{1}{3}\mathrm{\pi }\times r^2\times 2r$

$=\frac{2}{3}\mathrm{\pi }{r}^3$

= Volume of the hemisphere of radius r

Thus, the volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r.

The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius ____r____.

Answer:

Let r and R be the radii of the two spheres.

Suppose V₁ be the volume and S₁ be the surface area of the sphere of radius r & V₂ be the volume and S₂ be the surface area of the sphere of radius R.

It is given that the ratio of the volumes of two spheres is 8 : 27.

$\therefore \frac{V_1}{V_2}=\frac{8}{27}$

$\Rightarrow \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3}=\frac{8}{27}$

$\Rightarrow {\left(\frac{r}{R}\right)}^3={\left(\frac{2}{3}\right)}^3$

$\Rightarrow \frac{r}{R}=\frac{2}{3}$          .....(1)

Now,

$\frac{S_1}{S_2}=\frac{4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{R}^2}$

$\Rightarrow \frac{S_1}{S_2}={\left(\frac{r}{R}\right)}^2$

$\Rightarrow \frac{S_1}{S_2}={\left(\frac{2}{3}\right)}^2$            [Using (1)]

$\Rightarrow \frac{S_1}{S_2}=\frac{4}{9}$

⇒ S₁ : S₂ = 4 : 9

Thus, the ratio of the surface areas of the two spheres is 4 : 9.

If the ratio of the volumes of two spheres is 8 : 27, then the ratio of their surface areas is ___4 : 9___.

Answer:

Let r be the radius of the hemisphere.

Volume of the hemisphere = 18$\mathrm{\pi }$ cm³         (Given)

$\Rightarrow \frac{2}{3}\mathrm{\pi }{r}^3=18\mathrm{\pi }$

$\Rightarrow r^3=\frac{18\times 3}{2}=27$

$\Rightarrow r^3={\left(3\right)}^3$

$\Rightarrow r=3 \mathrm{cm}$

∴ Total surface area of the hemisphere

= 3$\mathrm{\pi }$r²

= 3$\mathrm{\pi }$× (3 cm)²

= 27$\mathrm{\pi }$ cm²

Thus, the total surface area of the hemisphere is 27$\mathrm{\pi }$ cm².

If the volume of a hemisphere is 18π cm³, then its total surface area is ___27$\mathrm{\pi }$ cm²___.

Answer:

Radius of the sphere = r

∴ Volume of the sphere, V₁ = $\frac{4}{3}\mathrm{\pi }{r}^3$         .....(1)

If a sphere is placed inside a right circular cylinder so as to touch the top, base and the lateral surface of the cylinder, then the height of the cylinder is equal to the diameter of the sphere and the diameter of the base of cylinder is equal to the diameter of the sphere.

Let R be the radius and H be the height of the cylinder.

Height of the cylinder = Diameter of the sphere

⇒ H = 2r      .....(2)

Diameter of the base of cylinder = Diameter of the sphere

⇒ 2R = 2r

⇒ R = r        .....(3)

∴ Volume of the cylinder, V₂ = $\mathrm{\pi }$R²H = $\mathrm{\pi }$ × r² × 2r = 2$\mathrm{\pi }$r³           .....(4)             [Using (2) and (3)]

Now,

$\frac{V_2}{V_1}=\frac{2\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}=\frac{3}{2}$

⇒ V₂ : V₁ = 3 : 2

Thus, the ratio of volume of cylinder to volume of sphere is 3 : 2.

If a sphere is placed inside a right circular cylinder so as to touch the top, base and the lateral surface of the cylinder. If the radius of the sphere is r, the volume of the cylinder and the sphere are in the ratio ___3 : 2___.

Answer:

Let r be the radius of sphere.

Volume of the sphere = $\frac{4}{3}\mathrm{\pi }$r³ cu. units

Surface area of the sphere = 4$\mathrm{\pi }$r² sq. units

It is given that the volume and surface area of sphere are numerically same.

$\therefore \frac{4}{3}\mathrm{\pi }{r}^3=4\mathrm{\pi }{r}^2$

⇒ r = 3 units

Thus, the radius of sphere is 3 units.

If the volume and surface area of a sphere are numerically the same, then its radius is ___3 units___.

Answer:

Let the radius of right circular cylinder and sphere be r.

Suppose h be the height of the cylinder.

∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3$

Volume of the cylinder = $\mathrm{\pi }{r}^{2}h$

Now,

Volume of the cylinder = Volume of the sphere            (Given)

$\Rightarrow \mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }{r}^3$

$\Rightarrow h=\frac{4}{3}r$         .....(1)

Now,

Curved surface area of the cylinder, S₁ = 2$\mathrm{\pi }$rh

Curved surface area of the sphere, S₂ = 4$\mathrm{\pi }$r²

$\therefore \frac{S_1}{S_2}=\frac{2\mathrm{\pi }rh}{4\mathrm{\pi }{r}^2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{2\mathrm{\pi }r\times {\displaystyle \frac{4}{3}}r}{4\mathrm{\pi }{r}^2}$        [From (1)]

$\Rightarrow \frac{S_1}{S_2}=\frac{2}{3}$

⇒ S₁ : S₂ = 2 : 3

Thus, the ratio of the curved surface of cylinder to the curved surface area of sphere is 2 : 3.

A right circular cylinder and a sphere have the same volume and same radius. The ratio of the areas of their curved surfaces is ___2 : 3___.

Answer:

Let the radius of the original sphere be r.

∴ Surface area of the original sphere, S₁ = 4$\mathrm{\pi }$r²

If the radius of the sphere is doubled, then the radius of the new sphere is 2r.

∴ Surface area of the new sphere, S₂ = 4$\mathrm{\pi }$(2r)² = 16$\mathrm{\pi }$r²

Now,

Percentage increase in the surface area of sphere

$=\frac{\mathrm{Increase} \mathrm{in} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{sphere}}{\mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{original} \mathrm{sphere}}\times 100\%$

$=\frac{S_2-S_1}{S_1}\times 100\%$

$=\frac{16\mathrm{\pi }{r}^2-4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=\frac{12\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=300\%$

Thus, the percentage increase in the surface area of the sphere is 300%.

If the radius of a sphere is doubled, then the percentage increase in the surface area is ___300%___.

Answer:

Radius of the solid hemisphere, R = 8 cm

Radius of each sphere, r = 2 cm

It is given that, solid hemisphere of radius 8 cm is melted and recast into n spheres of radius 2 cm each.

∴ n × Volume of each sphere = Volume of the solid hemisphere

$\Rightarrow n\times \frac{4}{3}\mathrm{\pi }{r}^3=\frac{2}{3}\mathrm{\pi }{R}^3$

$\Rightarrow n=\frac{1}{2}\left(\frac{R^3}{r^3}\right)$

$\Rightarrow n=\frac{1}{2}{\left(\frac{R}{r}\right)}^3$

$\Rightarrow n=\frac{1}{2}\times {\left(\frac{8 \mathrm{cm}}{2 \mathrm{cm}}\right)}^3$                   (R = 8 cm and r = 2 cm)

$\Rightarrow n=\frac{1}{2}\times 64$

$\Rightarrow n=32$

Thus, the value of n is 32.

If a solid hemisphere of radius 8 cm is melted and recast into n spheres of radius 2 cm each, then n = ___32___.

Answer:

Radius of the solid sphere, R = 4 cm

Radius of each solid hemisphere, r = 2 cm

It is given that, the solid sphere of radius 4 cm is melted and recast into n solid hemispheres of radius 2 cm each.

∴ n × Volume of each solid hemisphere = Volume of the sphere

$\Rightarrow n\times \frac{2}{3}\mathrm{\pi }{r}^3=\frac{4}{3}\mathrm{\pi }{R}^3$

$\Rightarrow n=2\times \frac{R^3}{r^3}$

$\Rightarrow n=2\times {\left(\frac{R}{r}\right)}^3$

$\Rightarrow n=2\times {\left(\frac{4 \mathrm{cm}}{2 \mathrm{cm}}\right)}^3$           (R = 4 cm and r = 2 cm)

$\Rightarrow n=2\times 8$

$\Rightarrow n=16$

Thus, the value of n is 16.

If a solid sphere of radius 4 cm is melted and recast into n solid hemispheres of radius 2 cm each, then n = ___16___.

Answer:

Let the radius of each sphere be r.

Radius of the cylinder, R = 10 cm

Height of the cylinder, H = 5.4 cm

It is given that, 15 identical spheres are made by melting a solid cylinder of radius 10 cm and height 5.4 cm.

∴ 15 × Volume of each sphere = Volume of the cylinder

$\Rightarrow 15\times \frac{4}{3}\mathrm{\pi }{r}^3=\mathrm{\pi }{R}^{2}H$

$\Rightarrow r^3=\frac{{\left(10 \mathrm{cm}\right)}^2\times 5.4 \mathrm{cm}}{20}$

$\Rightarrow r^3=27 {\mathrm{cm}}^3$

$\Rightarrow r^3={\left(3 \mathrm{cm}\right)}^3$

$\Rightarrow r=3 \mathrm{cm}$

∴ Diameter of each sphere = 2r = 2 × 3 cm = 6 cm

Thus, the diameter of each sphere is 6 cm.

Fifteen identical spheres are made by melting a solid cylinder of radius 10 cm and height 5.4 cm, the diameter of each sphere is ___6 cm___.