Maths-

General

Easy

Question

# A hemisphere has 3 cm radius , find its volume?

Hint:

### Volume of hemisphere is .

## The correct answer is: 56.54

### Explanation:

- We have given radius of hemisphere 3cm.
- We have to find the volume of the hemisphere.

Step 1 of 1:

We have given the radius of the hemisphere 3cm

So, The Volume will be

### Related Questions to study

Maths-

### Bloom’s category: Understanding

Level: Medium

Identify and name three collinear points.

ANS :- Point F, point O ,point G are three collinear points.

The points that lie on the same straight line are Collinear points.

From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.

We get Point F, point O ,point G are three collinear points .

∴ Point F, point O ,point G are three collinear points.

The points that lie on the same straight line are Collinear points.

From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.

We get Point F, point O ,point G are three collinear points .

∴ Point F, point O ,point G are three collinear points.

### Bloom’s category: Understanding

Level: Medium

Identify and name three collinear points.

Maths-General

ANS :- Point F, point O ,point G are three collinear points.

The points that lie on the same straight line are Collinear points.

From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.

We get Point F, point O ,point G are three collinear points .

∴ Point F, point O ,point G are three collinear points.

The points that lie on the same straight line are Collinear points.

From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.

We get Point F, point O ,point G are three collinear points .

∴ Point F, point O ,point G are three collinear points.

Maths-

### A dome of a building is in the form of a hemisphere. The total cost of white washing it from inside, was '1330.56. The rate at which it was white washed is '3 per square metre. Find the volume of the dome approximately

Explanation:

And the rate of white washing is 3 per square meter

So, The surface area of the dome is

= 443.52m

We know that white washing is only done on curved surface area

So,

Now the volume of dome will be

- We have given total cost of white washing the dome is 1330.56
- We have to find the volume of dome.

And the rate of white washing is 3 per square meter

So, The surface area of the dome is

= 443.52m

^{2}We know that white washing is only done on curved surface area

So,

Now the volume of dome will be

### A dome of a building is in the form of a hemisphere. The total cost of white washing it from inside, was '1330.56. The rate at which it was white washed is '3 per square metre. Find the volume of the dome approximately

Maths-General

Explanation:

And the rate of white washing is 3 per square meter

So, The surface area of the dome is

= 443.52m

We know that white washing is only done on curved surface area

So,

Now the volume of dome will be

- We have given total cost of white washing the dome is 1330.56
- We have to find the volume of dome.

And the rate of white washing is 3 per square meter

So, The surface area of the dome is

= 443.52m

^{2}We know that white washing is only done on curved surface area

So,

Now the volume of dome will be

Maths-

### Bloom’s category: Understanding

Level: Medium

Through any three points not on the same line,

ANS :- Option B

If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.

i.e three points lie on same plane

∴ Option B

If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.

i.e three points lie on same plane

∴ Option B

### Bloom’s category: Understanding

Level: Medium

Through any three points not on the same line,

Maths-General

ANS :- Option B

If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.

i.e three points lie on same plane

∴ Option B

If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.

i.e three points lie on same plane

∴ Option B

Maths-

### A hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the volume of steel used.

Explanation:

Inner radius is equal to 5cm

So, Outer radius is equal to 5 + 0.25 = 5.25cm

Outer Volume of the hemisphere will be

- We have given a hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm.
- We have to find the volume of the steel

Inner radius is equal to 5cm

So, Outer radius is equal to 5 + 0.25 = 5.25cm

Outer Volume of the hemisphere will be

Inner volume will be

Therefore, the volume of steel used will be

### A hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the volume of steel used.

Maths-General

Explanation:

Inner radius is equal to 5cm

So, Outer radius is equal to 5 + 0.25 = 5.25cm

Outer Volume of the hemisphere will be

- We have given a hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm.
- We have to find the volume of the steel

Inner radius is equal to 5cm

So, Outer radius is equal to 5 + 0.25 = 5.25cm

Outer Volume of the hemisphere will be

Inner volume will be

Therefore, the volume of steel used will be

Maths-

### Bloom’s category: Understanding

Level: Medium

A plane has ____ dimensions.

ANS :- Option B

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)

∴ Option B

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)

∴ Option B

### Bloom’s category: Understanding

Level: Medium

A plane has ____ dimensions.

Maths-General

ANS :- Option B

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)

∴ Option B

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)

∴ Option B

Maths-

### If the Diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn. Then , what would be the length of the wire ?

Explanation:

We have diameter of metallic sphere 6cm

Radius of metallic sphere 3cm

Also we have diameter of cross-section of cylinder wire 0.2cm

So, Radius of cross-section of cylinder wire 0.1cm

Let the length of the wire be hcm

Since the metallic sphere is converted into cylindrical shaped wire of length hcm

Volume of metal used in wire is equal to volume of sphere

So,

0.01h = 36

h = 3600 cm

Therefore, The length of the wire is 3600cm.

- We have given diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn.
- We have to find the length of the wire.

We have diameter of metallic sphere 6cm

Radius of metallic sphere 3cm

Also we have diameter of cross-section of cylinder wire 0.2cm

So, Radius of cross-section of cylinder wire 0.1cm

Let the length of the wire be hcm

Since the metallic sphere is converted into cylindrical shaped wire of length hcm

Volume of metal used in wire is equal to volume of sphere

So,

0.01h = 36

h = 3600 cm

Therefore, The length of the wire is 3600cm.

### If the Diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn. Then , what would be the length of the wire ?

Maths-General

Explanation:

We have diameter of metallic sphere 6cm

Radius of metallic sphere 3cm

Also we have diameter of cross-section of cylinder wire 0.2cm

So, Radius of cross-section of cylinder wire 0.1cm

Let the length of the wire be hcm

Since the metallic sphere is converted into cylindrical shaped wire of length hcm

Volume of metal used in wire is equal to volume of sphere

So,

0.01h = 36

h = 3600 cm

Therefore, The length of the wire is 3600cm.

- We have given diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn.
- We have to find the length of the wire.

We have diameter of metallic sphere 6cm

Radius of metallic sphere 3cm

Also we have diameter of cross-section of cylinder wire 0.2cm

So, Radius of cross-section of cylinder wire 0.1cm

Let the length of the wire be hcm

Since the metallic sphere is converted into cylindrical shaped wire of length hcm

Volume of metal used in wire is equal to volume of sphere

So,

0.01h = 36

h = 3600 cm

Therefore, The length of the wire is 3600cm.

Maths-

### Bloom’s category: Understanding

Level: Medium

A point has ____ dimension.

ANS :- Option C

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)

∴ Option C

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)

∴ Option C

### Bloom’s category: Understanding

Level: Medium

A point has ____ dimension.

Maths-General

ANS :- Option C

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)

∴ Option C

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)

∴ Option C

Maths-

### A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule ?

Explanation:

Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.

The volume of sphere =

Diameter of the spherical capsule, d = 3.5mm

Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm

Now, the medicine is needed to fill this capsule = The volume of the spherical capsule

The medicine needed to fill this capsule =

= 4/3 22/7 1.75mm 1.75mm 1.75mm

22 .46mm

- We have given a capsule with diameter 3.5mm
- We have to find the how much medicine require to fill the capsule.

Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.

The volume of sphere =

Diameter of the spherical capsule, d = 3.5mm

Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm

Now, the medicine is needed to fill this capsule = The volume of the spherical capsule

The medicine needed to fill this capsule =

= 4/3 22/7 1.75mm 1.75mm 1.75mm

22 .46mm

^{3}(approx)^{}### A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule ?

Maths-General

Explanation:

Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.

The volume of sphere =

Diameter of the spherical capsule, d = 3.5mm

Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm

Now, the medicine is needed to fill this capsule = The volume of the spherical capsule

The medicine needed to fill this capsule =

= 4/3 22/7 1.75mm 1.75mm 1.75mm

22 .46mm

- We have given a capsule with diameter 3.5mm
- We have to find the how much medicine require to fill the capsule.

Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.

The volume of sphere =

Diameter of the spherical capsule, d = 3.5mm

Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm

Now, the medicine is needed to fill this capsule = The volume of the spherical capsule

The medicine needed to fill this capsule =

= 4/3 22/7 1.75mm 1.75mm 1.75mm

22 .46mm

^{3}(approx)^{}Maths-

### Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

Solution :-

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

### Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

Maths-General

Solution :-

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Maths-

### Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

Explanation:

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

- We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
- We have to find the Radius of new sphere and the ratio of S and S’

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

### Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

Maths-General

Explanation:

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

- We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
- We have to find the Radius of new sphere and the ratio of S and S’

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

Maths-

### Bloom’s category: Application

Level: Medium

Identify points, lines and planes in the given image of a floor.

HINT :- Identify points, lines and planes in the given image using the definitions of them .

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

### Bloom’s category: Application

Level: Medium

Identify points, lines and planes in the given image of a floor.

Maths-General

HINT :- Identify points, lines and planes in the given image using the definitions of them .

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

Maths-

### A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

Explanation:

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

Step 2 of 2:

Now inside surface area is 246.48m

r

r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

- We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
- We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

^{2}Step 2 of 2:

Now inside surface area is 246.48m

^{2}r

^{2}= 39.22r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

^{2}and the inside volume is 82.07m^{3}### A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

Maths-General

Explanation:

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

Step 2 of 2:

Now inside surface area is 246.48m

r

r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

- We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
- We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

^{2}Step 2 of 2:

Now inside surface area is 246.48m

^{2}r

^{2}= 39.22r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

^{2}and the inside volume is 82.07m^{3}Maths-

### Find the volume of a sphere whose surface area is 154 sq.cm

Explanation:

We have given surface area 154cm

So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

- We have given surface area .
- We have to find the volume of the sphere.

We have given surface area 154cm

^{2}.So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

^{3}.### Find the volume of a sphere whose surface area is 154 sq.cm

Maths-General

Explanation:

We have given surface area 154cm

So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

- We have given surface area .
- We have to find the volume of the sphere.

We have given surface area 154cm

^{2}.So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

^{3}.Maths-

### A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

Explanation:

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

- We have given a hemisphere with thickness 1cm.
- We have to find the volume of iron used to make the tank.

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

^{3 }of iron used to make the tank### A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

Maths-General

Explanation:

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

- We have given a hemisphere with thickness 1cm.
- We have to find the volume of iron used to make the tank.

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

^{3 }of iron used to make the tankMaths-

### How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Explanation:

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

- We have given a hemispherical bowl of diameter 10.5cm
- We have to find how many litre of milk this bowl can hold.

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

### How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Maths-General

Explanation:

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

- We have given a hemispherical bowl of diameter 10.5cm
- We have to find how many litre of milk this bowl can hold.

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,