RS Aggarwal Solutions: Volume and Surface Area of Solids- 5 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Pencil is a combination of Cylinder + Cone

Shuttle is a combination of Frustum of a cone + Hemisphere

Funnel is a combination of Frustum of a cone + Cylinder

Sarahi is a combination of Sphere + Cylinder

Glass is in the shape of a Frustum of a cone.

Gilli in the gili-Danda is a combination of Frustum of a frustum of a cone + Cylinder + cone

Plumbline (sahul) is a combination of a Cone + hemisphere.

When a cone is cut by a plane parallel to its base and the upper part is removed. The part that is left over is called frustum of a cone.

When a object of certain volume is melted and converted to some other shape, the volume of the new object formed will be the same as the volume of the old object.

In a right circular cone, the cross section made by a plane parallel to the base is a circle.

Given: Dimension of cuboid (49 cm × 33 cm × 24 cm)
Volume of cuboid is : length × breadth × height
Volume of Solid Sphere is : 4/3 × π × r³ (here r is radius of the sphere)
Let v₁ be the volume of given cuboid.
∴ v₁ = 49 cm × 33 cm × 24 cm = 38808 cm³
Let v₂ be the volume of Solid Sphere.
We know that when a object is moulded from one shape to other its volume does not change.
∴ v₁ = v₂
That is,
38808 =  × π × r³
r³ =  × 38808
r³ = 9261 r = ∛9261 = 21
∴ r = 21cm
That is, radius of the Solid sphere = 21cm

Given: edge of the cube = 4.2 cm
A right circular cone is a Cone whose height is perpendicular to the diameter (radius) of the base circle.
In a cube, a largest right circular Cone is formed when its base lies on one of the faces of the Cube and its tip lies on the opposite face.

∴ Diameter of largest right circular Cone in Cube = edge length of cube.
∴ Diameter = 4.2 cm
⇒ Radius = = 2.1 cm
∴ Radius of the largest right circular Cone in Cube is 2.1 cm.

Given: Radius of the solid sphere = 9 cm
Radius of the cylinder = 9 cm
Volume of Solid Sphere is: 4/3 × π × r3 (here r is radius of the sphere)
Volume of Solid Cylinder is: π × r² × h (here r is radius and h is height of the cylinder)
Let v₁ be the volume of given Solid Sphere.
∴ v₁ = 4/3 × π × 9³ cm³
Let v₂ be the volume of Solid Cylinder.
∴ v₂ = π × 9² × h
We know that when a object is moulded from one shape to other its volume does not change.
∴ v₁ = v₂
That is, 4/3× π × 9³ = π × 9² × h

∴ h = 12 cm
That is Height of the Cylinder is 12 cm.

Given: Dimensions of rectangular sheet: 40cm × 22cm
Height of the Hollow Cylinder : 40cm
Area of the Rectangle is = length × breadth
Curved surface Area of the Cylinder = 2πrh (where r and h are radius and height of cylinder respectively)
Let a₁ be the area of Rectangle
∴ a₁ = 40 × 22 cm²
Let a₂ be the Curved surface area of Cylinder
∴ a₂ = 2 × π × r × 40 cm²
We know that when area of a surface doesn’t change even if its shape is changed.
∴ a₁ = a₂
⇒ 40 × 22 = 2 × π × r × 40

∴ Radius of the Cylinder is 3.5cm

Given: Diameter of the Solid Sphere is: 6 cm
Height of the Cylinder is: 45cm
Diameter of the Cylinder is: 4cm
Volume of Solid Cylinder is: π × r² × h (here r is radius and h is height)
Volume of the Solid Sphere is: 4/3 × π × r³ (here r is the radius of the Sphere)
Let v₁ be the volume of given Cylinder
∴ v₁ = π × 2² × 45 cm³ (4cm is diameter, ∴ 2cm is the radius of cylinder)
Let v₂ be the volume of Solid Sphere.
V₂ = 4/3 × π × 3³ cm³ (6cm is the diameter, ∴ 3cm is the radius of the Sphere)
We know that when a object is moulded from one shape to other its volume does not change.
Let n be the number of Solid Sphere of diameter 6cm required.
∴ v₁=n × v₂ (volume of n Spheres = volume of Cylinder)
That is,
π × 2² × 45 =n × 4/3 × π × 3³

That is 5 Solid Spheres of diameter 6 cm can be formed by the Solid Cylinder of height 45 cm and diameter 4 cm.

Given: Surface area ratio of two Spheres is: 16:9
Volume of the Sphere is: 4/3 × π × r³ (where r is radius of sphere)
Surface area of the sphere is: 4 × π × r² (where r is radius of sphere)
Let S₁ and S₂ be two different spheres.
(Surface area of) S₁: (Surface area of) S₂ = 16:9
4 × π × (r₁)²: 4 × π × (r₂)² = 16:9 (here r₁ and r₂ are the radii of S₁ and S₂ respectively)
(r₁)²: (r₂)² = 16:9
r₁: r₂ = √16:√9
r₁: r₂ = 4:3
Now,
Let V₁ and V₂ be the volumes of the spheres S₁ and S₂ respectively.
∴ V₁:V₂ = 4/3 × π × (r₁)³:4/3 × π × (r₂)³ (here r₁ and r₂ are the radii of S₁ and S₂ respectively)
⇒ V₁:V₂ = (r₁)³: (r₂)³
⇒ V₁:V₂ = (4)³: (3)³
⇒ V₁:V₂ = 64:27
∴ The ratios of the volumes is: 64:27

Given: surface area of a sphere is 616 cm²
Surface area of the sphere is: 4 × π × r² (where r is radius of sphere)
∴ 4 × π × r² = 616 cm²

r = √49 = 7 cm
∴ Diameter = 2 × r = 2 × 7 = 14cm

Let r₁ be the initial radius of the sphere.
∴ r₁= r
Let r₂ be the radius after increasing it 3 times the size of initial radius.
∴ r₂ = 3r
Let V₁ be the initial volume of the sphere
∴ V₁ = 4/3 × π × (r₁)³ = 4/3 × π × r³
Let V₂ be the volume of the sphere after its radius is increased by 3 times.
∴ V₂ = 4/3 × π × (r₂)³ = 4/3 × π × (3r)³
⇒ V₂ = 27 × 4/3 × π × r³
∴ If radius is increased by 3 times, its volume will be increased by 27 times.

Given: Height of the frustum of a cone: 16 cm
Diameters of the Circular ends: 40cm and 16 cm.
Radius of the Circular ends: 40/2 = 20cm and 16/2 = 8cm

Here slant height h can be found by using Pythagoras theorem.
∴ s² = h² + (R-r)² (here R is 20cm and r is 8cm)
⇒ s² = 16² + (20—8)²
⇒ s² = 16² + (12)²
⇒ s² = 256 + 144
⇒ s² = 400
⇒ s = √400 = 20
∴ Slant height 0f the Frustum is 20cm

Given: Diameter of a sphere: 18cm ⇒ radius = 18/2 = 9cm
Diameter of Cylindrical vessel: 36cm ⇒ radius = 36/2 = 18cm
It is given that Sphere is dropped into the cylindrical vessel containing some water.
∴ Volume of sphere = Volume of water in Cylinder displaced (raised)
Let V₁ be the volume of the Sphere
∴ V₁ = 4/3 × π × (r₁)³
V₁ = 4/3 × π × 9³
Let V₂ be the volume of the water displaced in the cylindrical vessel
∴ V₂ = π × (r₂)² × h (here r₂ is the radius of the Cylinder and h is the level of water raised in the vessel after dropping the sphere into the cylindrical vessel)
V₂ = π × 18² × h
Since V₁=V₂
4/3 × π × 9³ = π × 18² × h

∴ The water level rises by 3cm when the dropped sphere is completely submerged in the cylindrical vessel.

Given: A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base
Let ‘H’ be the height of the cone.
Let ‘R’ be the Radius of the complete cone.
Volume of a cone is given by: 1/3πr²h
Here,
AB = BD = H/2
Let r be the radius of the smaller cone.
∴ In ΔABC and ΔADE
∠ABC = ADE (90°)
∠CAB = ∠EAB (common)
∴ ΔABC ~ ΔADE (AA similarity criterion)
⇒ OA/OC = AB/CD (Corresponding sides are proportional)

⇒ R = 2r
Volume of smaller cone = 
Volume of whole cone = 

∴ The ratio of the volume of the smaller cone to the whole cone is 1:8.

Bucket is in the shape of a frustum of a cone
Therefore,
Given: Height of the frustum: 40 cm
Radius of the Circular ends: 24cm and 15cm

Here slant height h can be found by using Pythagoras theorem.
∴ s² = h² + (R-r)² (here R is 20cm and r is 8cm)
⇒ s² = 40² + (24—15)²
⇒ s² = 40² + (9)²
⇒ s² = 1600 + 81
⇒ s² = 1681
⇒ s = √1681 = 41
∴ Slant height 0f the Frustum is 41cm.

Given: Bottom of a solid is hemispherical and conical above it, both have same radius and same surface areas.
∴ CSA of hemisphere = CSA of Cone
⇒ 2 × π × r² = πrl (where r is the radius and l is the slant height)


∴ r:l = 1:2
That is ratio of radius and the slant height of the given solid is 1:2.

Given: Radius of the base of a right circular cylinder is halved, keeping the height the same.
Let initial Radius of Right Circular Cylinder be ‘r’.
∴ Radius of the Cylinder after its radius is halved is ‘r/2’
Let ‘h’ be the height of the both the cylinders.
Let V₁ be the volume of the initial Cylinder.
Let V₂ be the volume of the Cylinder after the initial Cylinders base radius is halved.
Volume of the Cylinder is given by: πr²h
∴ V₁:V₂ = π(r₁)²h : π(r₂)²h
∴ V₁:V₂ = π(r)²h : π(r_/2)²h
⇒ V₁:V₂ = 1: (1/4)
⇒ V₁:V₂ = 4:1
Thus, ratio of the volume of the cylinder thus obtained to the volume of original is 4:1.

Given: Cubical ice-cream brick of edge 22.
Ice-cream cone of radius 2 cm and height 7 cm.
Let ‘n’ be the number of students who get ice-cream cones.
Let V₁ be the volume of the Cubical Ice-cream brick.
Volume of Cube is given by: a³ (where a is the edge length)
∴ V₁ = a³ = 22³
Let V₂ be the Volume of the Ice-cream Cone.
Volume of Cone is given by: 1/3 × π × r² × h (where r is the radius of the base and h is the edge height of the cone)
∴ V₂ = 1/3 × π × r² × h = 1/3 × π × 2² × 7
Here,
V₁ = n × V₂
∴ 22³ = n × 1/3 × π × 2² × 7

∴ 363 Children can get ice cream cones.

Given: A wall of dimension (270 cm × 300cm × 350cm).
A wall of dimension (22.5cm × 11.25cm × 8.75cm).
1/8 space is covered by the mortar.
Here 1/8 volume of the wall is covered by mortar,
∴ 1 – 1/8 = 7/8 volume of the wall is covered by bricks.
Volume of the cuboid is given by: lbh (here l, b, h are length , breadth, height respectively).
Let V₁ be the volume of the wall.
∴ V₁ = l × b × h = 270 × 300 × 350
⇒ V₁ = 270 × 300 × 350 = 28350000 cm²
Let V₂ be the volume of the brick.
∴ V₂ = l × b × h = 22.5 × 11.25 × 8.75
⇒ V₂ = 22.5 × 11.25 × 8.75 = 70875/32 cm²
Let ‘n’ be the number of bricks required to occupy 7/8 volume of the wall.
∴ n × 70875/32 = 28350000 × 7/8
⇒ n = 28350000 × 7/8 × 32/70875 = 11200
∴ n = 11200
That is 11200 bricks required to occupy 7/8 volume of the wall.

Given: A solid metallic cylinder of base diameter 2 cm and height 16 cm.
Twelve solid sphere of the same size are made by melting a solid metallic cylinder .
Let V₁ be the volume of the cylinder
Volume of the cylinder is given by: π × r² × h
∴ V₁ = π × 1² × 16 (diameter is 2cm, ∴ radius is 1cm)
Let V₂ be the volume of the each Sphere.
Volume of the Sphere is given by: 4/3 × π × r³ (where r is the radius of the sphere)
∴ V₂ = 4/3 × π × r³
Here,
V₁ = 12 × V₂
∴ π × 1² × 16 = 12 × 4/3 × π × r³

∴ r = ∛1 = 1 cm
∴ Diameter of the Sphere is 2 × r = 2 × 1 = 2cm

Given: Diameter of two circular ends of a bucket are 44 cm and 24 cm, and the height of the bucket is 35 cm.
Bucket is in the shape of frustum.
Let V be the Volume of the Bucket(Frustum)
Volume of the frustum is given by: π/3 × h × (R² + r² + Rr) (here r and R are the radii of smaller and larger circular ends respectively)
∴ V = π/3 × h × (R² + r² + R × r)
⇒ V = π/3 × 35 × (22² + 12² + 22 × 12) (diameters are 44 and 24 cm, ∴ their radii are 22cm and 12cm respectively)
⇒ V = π/3 × 35 × (484 + 144 + 264) = π/3 × 35 × (892)
⇒ V = π/3 × 35 × (892) = 32706.6 cm³ = 32.7 litres (∵ 1000cm³ = 1 litre)
∴ The capacity of the bucket is: 32.7 litres

Given: slant height of a bucket is 45
Radii of its top and bottom are 28 cm and 7 cm respectively.
Bucket is in the shape of a frustum.
Let A be the CSA of the Frustum
CSA of the frustum is given by: πl(r₁ + r₂) (here l is the slant height and r1 and r2 are the radii of top and bottom of the bucket(frustum))
∴ A = πl(r₁ + r₂)
⇒ A = π × 45 × (28 + 7) = π × 45 × (35) = 4950 cm²
∴ A = 4950 cm²
That is , curved surface area of the bucket is 4950 cm²

Given: Volume ratio of two Spheres is: 64:27
Volume of the Sphere is: 4/3 × π × r³ (where r is radius of sphere)
Surface area of the sphere is: 4 × π × r² (where r is radius of sphere)
Let S₁ and S₂ be two different spheres.
(Volume of) S₁: (Volume of) S₂ = 64:27
4/3 × π × (r₁)³: 4/3 × π × (r₂)³ = 64:27 (here r₁ and r₂ are the radii of S₁ and S₂ respectively)
(r₁)³: (r₂)³ = 64:27
r₁: r₂ = ∛64:∛27
r₁: r₂ = 4:3
Now,
Let SA₁ and SA₂ be the surface areas of the spheres S₁ and S₂ respectively.
∴ SA₁:SA₂ = 4 × π × (r₁)²:4 × π × (r₂)² (here r₁ and r₂ are the radii of S₁ and S₂ respectively)
⇒ SA₁:SA₂ = (r₁)²: (r₂)²
⇒ SA₁:SA₂ = (4)²: (3)²
⇒ SA₁:SA₂ = 16:9
∴ The ratio of the Surface area of spheres is: 16:9.

Given: hollow cube of internal edge 22 cm
Spherical marbles of diameter 0.5 cm ⇒ radius = 0.5/2 = 0.25 cm
1/8 space of the cube remains unfilled.
Here, 1/8 space of the cube is unfilled, that is 7/8 of the cube is filled with marbles of radius 0.25cm.
Volume of the sphere is : 4/3 × π × r³ (where ‘r’ is the radius of the cube)
Let V₁ be the volume of the Sphere.
∴ V₁ = 4/3 × π × (0.25)³ = 11/168 cm³
Volume of the Cube : a³ (where ‘a’ is the edge length of the cube)
Let V₂ be the volume of the cube.
∴ V₁ = (22)³ = 10648 cm³
Let ‘n’ be the number of sphere required to fill the 7/8 space of the cube.
∴ n × V₁= V₂ × 7/8
⇒ n × 11/168 = 10648 × 7/8
⇒ n × 11/168 = 9317
⇒ n = 9317 × 168/11 = 142296
∴ Number of marbles required to fill 7/8 of the cube id 142296.

Given: Metallic spherical shell of internal and external diameter 4 cm and 8 cm respectively
Recast into the form of a cone of base diameter 8 cm.
Volume of spherical shell is : 4/3 × π × [(r₁)³ — (r₂)³] (here r₁ an r₂ are External and internal radii respectively)
Let v₁ be the volume of given Spherical shell.
∴ v₁ = 4/3 × π × [(8/2)³ — (4/2)³] = 4/3 × π × [(4)³ — (2)³] = 4/3 × π × [64 — 8] = 4/3 × π × 56 cm³
Volume of the cone is given by : 1/3 × π × r² × h
Let v₂ be the volume of cone.
∴ V₂ = 1/3 × π × (8/2)² × h = 1/3 × π × 16 × h cm³
We know that when a object is moulded from one shape to other its volume does not change.
∴ v₁=v₂
That is, 4/3 × π × 56 = 1/3 × π × 16 × h

∴ h = 14cm
That is, height of the cone = 14cm

Given: A cylinder of diameter 0.5 cm
The length of the entire capsule is 2 cm
Here, Radius of the Cylinder and Hemisphere is 0.5/2 = 0.25cm
Height of the Cylinder = length of capsule – 2 × (radius of hemisphere) = 2 – 0.5 = 1.5cm
Volume of Cylinder is: πr²h (here r, h are radius and height of the cylinder respectively)
Volume of hemisphere is: 4/3πr³ (here r is the radius of the hemisphere)
Let V₁ be the Cylindrical part of the capsule
∴ V₁ = πr²h = π × (0.25)² × (1.5) = 0.09375π cm³
Let V₂ be the Cylindrical part of the capsule
∴ V₁ = 4/3πr³ = 4/3π(0.25)³ = 0.0625/3π cm³
∴ Volume of the capsule = V₁ + 2V₂ = 0.09375π + 2 × 0.0625/3π = 0.3599 ≈ 0.36cm³
∴ Volume of the capsule is 0.36cm³

Given: A room with dimensions (12 m × 9 m × 8 m)
Room is in the shape of a Cuboid.
Longest rod that can be placed in a room is nothing but its diagonal.
Let length of the rod be ‘L’.
Length of diagonal of a Cuboid = √(l² + b² + h²)
∴ L = √(l² + b² + h²)
L = √(12² + 9² + 8²) m
⇒ L= √(144 + 81 + 64) m
⇒ L = √289 m
⇒ L = 17 m
∴ The length of the longest rod is 17 m

Given: Length of the diagonal of a cube is 6 √ 3 cm.
We know that the Length of a diagonal of a cube is given by: a√ 3 (here ‘a’ is edge length of cube)
∴ a√ 3 = 6√ 3
⇒ a = 6cm
∴ Edge length of the cube is 6cm.
Surface area of a cube is given by: 6a² (here ‘a’ is the edge length)
Let ‘S’ be the surface area of the given Cube.
∴ S = 6a²
⇒ S = 6 × (6)²
⇒ S = 216 cm²
∴ Surface area of the given cube is 216 cm²

Given: volume of a cube is 2744 cm³.
We know that the Volume of a cube is given by: a³ (here ‘a’ is edge length of cube)
∴ a³ = 2744
⇒ a = ∛ 2744
⇒ a =14cm
∴ Edge length of the cube is 14cm.
Surface area of a cube is given by: 6a² (here ‘a’ is the edge length)
Let ‘S’ be the surface area of the given Cube.
∴ S = 6a²
⇒ S = 6 × (14)²
⇒ S = 1176 cm²
∴ Surface area of the given cube is 1176 cm²

Given: The total surface area of a cube is 864 cm².
We know that the Volume of a cube is given by: 6a² (here ‘a’ is edge length of cube)
∴ 6a² = 864
⇒ a² = 864/6 = 144
⇒ a =√ 144
⇒ a =12cm
∴ Edge length of the cube is 12cm.
Volume of a cube is given by: a³ (here ‘a’ is the edge length)
Let ‘V’ be the Volume of the given Cube.
∴ V = a³
⇒ V = (12)³
⇒ V = 1728 cm³
∴ Volume of the given cube is 1728 cm³

Given: Bricks each measuring (25 cm × 11.25 cm × 6 cm)
A wall with dimensions (8 m × 6 m × 22.5 cm)
We know that the Brick and wall are in the shape of a Cuboid.
∴ Volume of a Brick and wall is given by: l × b × h (here l,b,h are the length , breadth, height of the wall and brick)
Let V₁ be the volume of the Wall.
∴ V₁ = l × b × h = 800 × 600 × 22.5 (∵ 8 m = 800cm and 6 m = 600cm)
⇒ V₁ = 10800000 cm³
Let V₂ be the volume of a Brick.
∴ V₂ = l × b × h = 25 × 11.25 × 6
⇒ V₂ = 1687.5 cm³
Let ‘n’ be the number of Bricks required to build the Wall.
∴ V₁ = n × V₂
10800000 = n × 1687.5
⇒ n = 10800000/1687.5 = 6400
∴ n = 6400
That is, 6400 bricks are required to build the wall.

Given: The area of the base of a rectangular tank is 6500 cm².
Volume of water contained in it is 2.6 m³.
We know that Rectangular tank is in the shape of a cuboid.
And also area of the base of a cuboid is given by : l × b (here l and b are length and breadth respectively)
∴ l × b = 6500 cm² ….1
Volume of a cuboid is given by: l × b × h (here l, b, h are length, breadth, height respectively)
∴ l × b × h = 2.6 m³ = 2.6 × 1000000 cm³ = 2600000 cm³ …2
From —1 and —2
6500 × h = 2600000

∴ h = 4m
That is, depth of water in the tank is 4m.

Given: The volume of a wall is 12.8 m³.
Height is 5 times the breadth and Length is 8 times the height.
Let breadth = x
∴ Height = 5x
⇒ Length = 8 × (5x) = 40x
Volume of a Cuboid is given by: l × b × h (here l, b, h are, length, breadth, height respectively)
∴ l × b × h = 12.8
⇒ x × 5x × 40x = 12.8
⇒ 200x³ = 12.8
⇒ x³ = 12.8/200 m³ = 8/125
⇒ x= ∛8/125 m = 2/5 m = 0.4 m
∴ x = 0.4m = 0.4 × 100 = 40cm
That is Breadth is: 40cm

Given: areas of three adjacent faces of a cuboid are x, y, z respectively.
Let l, b, h be the length , breadth, height of the cuboid respectively.
∴ l × b = x – 1
b × h = y – 2
h × l = z – 3
Volume of the cuboid is: l × b × h
multiply eq’s –1, –2, –3
That is ,
l × b × b × h × l × h = x × y × z l² × b² × h²
⇒ l² × b² × h² = xyz
⇒ (l × b × h)² = xyz
⇒ (V)² = xyz (∵Volume of the cuboid is: l × b × h)
⇒ V = √xyz
∴ V = √xys
That is volume of the given Cuboid is : √xyz

Given: Sum of length, breadth and height of a cuboid is 19 cm.
Length of diagonal is 5 √ 5 cm.
Let l, b, h be the length , breadth, height of the cuboid respectively.
∴ l + b + h = 19cm – 1
Length of a diagonal in a cuboid is given by : √(l² + b² + h²)
∴ √(l² + b² + h²) = 5√5 cm ⇒ (l² + b² + h²) = (5√5)² = 125 cm² – 2
Surface area of Cuboid is: 2(lb + bh + hl)
On squaring eq – 1 on both sides
We get
(l + b + h)² = 19²
⇒ l² + b² + h² + 2(lb + bh + hl) = 361 cm²
⇒ 125+ 2(lb + bh + hl) = 361 ( from eq – 2)
⇒ 2(lb + bh + hl) = 361 – 125 = 236 cm²
∴ 2(lb + bh + hl) = 236 cm²
That is, Surface area of the given Cuboid is 236cm².

Given: Edge of a cube is increased by 50%
Let ‘a’ be the Edge of the cube
Area of the cube is : 6l² (where ‘l’ is the edge of a cube)
Let A₁ be the initial surface area of the cube.
∴ A₁ = 6a²
Now,
50% of the edge is: a × 50/100 = a/2
∴ Edge = a + a/2 after increasing edge by 50%
That is edge of the cube after increasing it by 50% is 3a/2
Let A₂ be the surface area of the cube after increasing the edge by 50%
∴ A₂= 6 × (3a/2)² = 9/4 × 6a²
Here increase in area = A₂ –A₁
⇒ increase in area = 9/4 × 6a² – 6a² = 5/4 × 6a²
Now, increase in percentage is:
Increase in % =   × 100 = 125%
∴ If each edge of a cube is increased by 50%, then the percentage increase in the surface area is 125%.

Given: cuboidal granary with dimensions (8 m × 6 m × 3 m)
Volume of the bag : 0.64 m³
Volume of a Cuboid is given by: l × b × h (here l, b, h are length, breadth, height of the Cuboid respectively)
Let ‘V’ be the Volume of the Cuboidal granary.
∴ V = l × b × h = 8 × 6 × 3 = 144
Let ‘n’ be the number of bags that can fit in cuboidal granary.
∴ n × 0.64 = 144
⇒ n = 144/0.64 = 225
∴ n = 225
That is a total of 225 bags, each of volume 0.64 m³ can fit in Cuboidal granary.

Given: A cube of side 6 cm.
It is cut into a number of cubes each of side 2 cm.
Volume of a cube is given by: a³ (here ‘a’ is the side of a cube).
Let V₁ be the volume of a cube with side 6cm
∴ V₁ = a³
⇒ V₁ = 6³ = 216 cm³
Let V₂ be the volume of a cube with side 2cm
∴ V₁ = a³
⇒ V₁ = 2³ = 8 cm³
Let ‘n’ be the number of cubes of side 2cm which are cut from cube of side 6cm.
∴ n × V₂ = v₁
⇒ n × 8 = 216
⇒ n = 216/8 = 27
∴ Total of 27 cubes each of side 2cm can be cut from a cube of side 6cm.

Given: 5 cm of rain falls.
2 hectares of ground.
2 hectares = 20000 m³ (∵ 1 hectares = 10000 m³)
∴ Area of land which is filled with rain upto 5cm high is 20000 m³
Let V be the volume of the water on the land.
The volume of the water on the land is: area(of land) × height of the water on the land.
∴ V = Area(of land) × height
⇒ V = 20000 × 5/100 (∵ 1cm = 1/100 m )
⇒ V = 1000 m³
∴ The volume of the water that falls on 2 hectares of ground, is: 1000m³

Given: 5 cm of rain falls.
2 hectares of ground.
2 hectares = 20000 m³ (∵ 1 hectares = 10000 m³)
∴ Area of land which is filled with rain upto 5cm high is 20000 m³
Let V be the volume of the water on the land.
The volume of the water on the land is: area(of land) × height of the water on the land.
∴ V = Area(of land) × height
⇒ V = 20000 × 5/100 (∵ 1cm = 1/100 m )
⇒ V = 1000 m³
∴ The volume of the water that falls on 2 hectares of ground, is: 1000m³

Given: Volumes of the cubes are in the ratio 1:27.
Volume of a cube is given by: a³ ( here ‘a’ is the side of the cube).
Surface are of a cube is given by: a² ( here ‘a’ is the side of the cube).
Let a₁, a₂ be the side of first cube and second cube respectively.
∴ (a₁)³: (a₂)³ = 1:27
⇒ a₁: a₂ = ∛1 : ∛27
⇒ a₁: a₂ = 1 : 3
⇒ (a₁)²: (a₂)² = 1²: 3²
⇒ (a₁)²: (a₂)² = 1 : 9
∴ Ratio of the surface areas of the given cubes is 1:9.

Given: Diameter of the base of a cylinder is 4 cm.
Its height is 14 cm.
Volume of the Cylinder is given by: πr²h. (here r and h are radius and diameter respectively)
Let V be the volume of the cylinder.
∴ V = πr²h
⇒ V = π × (4/2)² × 14 (∵ 4 is the diameter and 4/2is the radius)
⇒ V = π × (2)² × 14 = π × 4 × 14
⇒ V = 176 cm³
∴ The volume of the cylinder is 176 cm³

Given: Diameter of the base of a cylinder is 28 cm.
Its height is 20 cm.
Volume of the Cylinder is given by: 2πr(r + h). (here r and h are radius and diameter respectively)
Let S be the Surface area of the cylinder.
∴ V = 2πr(r + h)
⇒ S = 2 × π × (28/2) × [(28/2) + 20] (∵ 28 is the diameter and 28/2is the radius)
⇒ V = 2 × π × 14 × (14 + 20) = 2 × π × 14 × 24
⇒ V = 2992 cm²
∴ The Surface area of the cylinder is 2992 cm²

Given: The radius of a cylinder is 14 cm.
Its curved surface area is 1760 cm².
Curved surface area of a Cylinder is : 2πrh (here r and h are radius and height respectively)
∴ 2πrh = 1760
⇒ 2 × π × 14 × h = 1760

∴ h = 20cm
That is height of the given Cylinder is 20cm.

Given: Radius and height of the Cylinder are 80cm and 20cm respectively.
Lateral Surface area of a Cylinder is : 2πrh
Let S₁ be the Lateral surface area of the Cylinder
∴ S₁ = 2πrh
⇒ S₁ = 2 × π × 80 × 20 = 3200π
Total Surface area of a Cylinder is : 2πr(r + h)
Let S₂ be the Total surface area of the Cylinder
∴ S₂ = 2πr(r + h)
⇒ S₂ = 2 × π × 80 × (80 + 20) = 16000π
Ratio of Total surface area to Lateral surface area id S₂ : S₁
∴ S₂ : S₁ = 16000π : 3200π
⇒ S₂ : S₁ = 5:1
∴ The ratio of the total surface area to the lateral surface area of a cylinder is 5:1

Given: CSA of a cylindrical pillar is 264 m²
Volume of cylindrical pillar id 924 m³
CSA of a cylinder is : 2πrh (here r and h are radius and height respectively)
∴ 2πrh = 264 …. 1
Volume of the Cylinder is : πr²h (here r and h are radius and height respectively)
∴ πr²h = 924 …. 2
Divide eq –2 by eq –1
We get,


∴ r = 7
Substitute ‘r’ in eq –1
∴ 2 × π × 7 × h = 264
⇒ 2 × (22/7) × 7 × h = 264
⇒ 44 × h=264
⇒ h = 264/44 = 6 cm
∴ height of the cylindrical pillar is 6 cm

Given: The ratio between the radius of the base and the height of the cylinder is 2: 3.
Volume of the Cylinder is 1617cm³
Let 2x and 3x be radius and height of the Cylinder respectively.
Volume of the Cylinder is given as: πr²h
∴ πr²h = 1617
⇒ π × (2x)² × (3x) = 1617
⇒ 12π × x³ = 1617


∴ r = 2 × (7/2) = 7cm
and, h = 3 × (7/2) = 10.5 cm
Total surface area of a cylinder is: 2πrh(r + h)
Let S be the TSA of a cylinder
∴ S = 2πrh(r + h)
⇒ S = 2 × π × (7) × (7 + 10.5) = 2 × π × (7) × 17.5 = 770
∴ S = 770cm²
That is, Total surface area of the cylinder is 770 cm²

Given: The radii of two cylinders are in the ratio 2:3.
Heights of the cylinders are in the ratio 5:3.
Volume of cylinder is: πr²h (here r and h are radius and height of the cylinder respectively)
Let V₁ be the volume of first cylinder
∴ V₁ = π(r₁)²h₁
Let V₂ be the volume of second cylinder
∴ V₂ = π(r₂)²h₂
∴ V₁ : V₂ = π(r₁)²h₁ : π(r₂)²h₂
⇒ V₁ : V₂ = π × (2)² × 5 : π × (3)² × 3
⇒ V₁ : V₂ = 20π : 27π = 20:27
∴ V₁ : V₂ = 20:27
That is the ratio of their volume is 20:27.

Given: Two cylinders of equal volume.
Heights of the cylinders are in the ratio 1:2
Volume of cylinder is: πr²h (here r and h are radius and height of the cylinder respectively)
Let V₁ be the volume of first cylinder
∴ V₁ = π(r₁)²h₁
Let V₂ be the volume of second cylinder
∴ V₂ = π(r₂)²h₂
Here,
V₁=V₂
⇒ π(r₁)²h₁ = π(r₂)²h₂
⇒ (r₁)²h₁ = (r₂)²h₂
⇒ (r₁)² : (r₂)² = h₂ : h₁
⇒ (r₁)² : (r₂)² = 2 : 1
⇒ r₁ : r₂= √2 :1
∴ ratio of the radii of given cylinders is √2 :1

Given: The height of the cone is 12 cm.
The radius of the cone is 5 cm.
Curved surface area of a cone is : πrl (here r and l are radius and slant height respectively)
l = √(r² + h²)
∴ πrl = πr√(r² + h²)
Let S be the CSA of the cone.
∴ S = πr√(r² + h²)
⇒ S = π × (5) × √((5)² + (12)²)
⇒ S = π × (5) × √(25 + 144) = π × (5) × √(169)
⇒ S = π × (5) × 13 = 65π
∴ S = 65π cm²
That is CSA of the cone is 65π cm²

Given: Diameter of the base of a cone is 42 cm.
Volume of the cone is 12936 cm³.
Volume of the cone is given by: (1/3) × π × r² × h
∴ (1/3) × π × r² × h = 12936
⇒ (1/3) × π × (42/2)² × h = 12936 (here diameter = 42cm, ∴ r = (42/2) = 21)
⇒ (1/3) × π × (21)² × h = 12936

∴ Height of the given cone is 28cm

Given: Area of the base of a right circular cone is 154 cm².
Height of the cone is 14 cm.
Curved surface area of a cone is : πrl (here r and l are radius and slant height respectively)
Area of the base is given by: πr²
∴ π × r² = 154
⇒ r² = 154/π = 49
⇒ r = √49 = 7cm
∴ radius of the base of the cone is 7cm
Now,
l = √(r² + h²)
∴ πrl = πr√(r² + h²)
Let S be the CSA of the cone.
∴ S = πr√(r² + h²)
⇒ S = π × (7) × √((7)² + (14)²)
⇒ S = π × (7) × √(49 + 196) = π × (7) × √(245)
⇒ S = π × (7) × 7√5 = (22/7) × 7 × 7√5 = 154√5
∴ S = 154√5 cm²
That is CSA of the cone is 154√5 cm²

Given: Radius of the base and the height of a cone is increased by 20%
The volume of the cone is : 1/3πr²h
New radius = r + 20/100r = 120/100r = 6/5r
Similarly, new radius = 6/5h
New Volume = 1/3 × π × (6/5r)² × (6/5h) = 216/125 × 1/3πr²h
Increase in volume = 216/125 × 1/3πr²h – 1/3πr²h =91/125 × 1/3πr²h
∴ Increase in % = 
∴ Volume will be increased by 72.8%

Given: The radii of the base of a cylinder and a cone are in the ratio 3:4.
Heights of the base of a cylinder and a cone are in the ratio 2:3.
Volume of cylinder is: πr²h (here r and h are radius and height of the cylinder respectively)
Volume of cylinder is: πr²h
Let V₁ be the volume of first cylinder
∴ V₁ = π(r₁)²h₁
Let V₂ be the volume of the cone.
∴ V₂ = 1/3π(r₂)²h₂
∴ V₁ : V₂ = π(r₁)²h₁ :1/3 π(r₂)²h₂
⇒ V₁ : V₂ = π × (3)² × 2 :1/3 × π × (4)² × 3
⇒ V₁ : V₂ = 18π : 1/3 × 48π = 18:16 = 9:8
∴ V₁ : V₂ = 9:8
That is the ratio of their volume is 9:8.

Given: A metallic Cylinder of radius 8cm and height 2cm
Right circular cone of height 6cm
Volume of a cylinder is given by: π × r² × h
Volume of a cone is given by: 1/3 × π × r² × h
Let V₁ be the volume of the Cylinder
∴ V₁ = π × r² × h
⇒ V₁ = π × (8)² × (2) = 128π
Let V₂ be the volume of the cone
∴ V₂ = 1/3 × π × r² × h
⇒ V₂ = 1/3 × π × (r)² × (6) = 2πr²
Here, Solid Cylinder is melted and made into a Solid cone.
∴ V₁ = V₂
⇒ 128π = 2πr²
⇒ 2πr² = 128π
⇒ r² = 128π/2π = 64
⇒ r² = 64
⇒ r = √64 = 8cm
∴ Radius of the base of the cone is 8cm.

Given: The height of the conical tent is 14m and its floor area is 346.5m².
Canvas of width 1.1m
Surface area of the tent is: πrl
Let be the Surface area of the cone.
Area of the floor is : πr²= 346.5 m²
⇒ πr²= 346.5
⇒ r² = 346.5/π = 441/4
⇒ r = √(441/4 ) = 21/2
⇒ l = √(h² + r²) (here l is the slant height and h , r are height and radius respectively of cone)
⇒ l = √(14² + (21/2)²) = √(196 + (441/4)) = √1225/4 = 35/2
∴ S = πrl = π ×(21/2 x 35/2) = (33x35)/2
∴ Area of the canvas required to cover the tent is: (33 x 35)/2  m²
∴ l × b = (33 x 35)/2 m² (here l is length and b is breadth or width of the canvas)
⇒ l × 1.1 = (33 x 35)/2 ( width of the canvas is 1.1m)
⇒ l = (33 x 35)/(2 x 1.1) = 525m
∴ Length of the canvas required to cover the tent is 525m

Given: Diameter of the sphere is 14cm
Radius of the sphere is 14/2 = 7cm
Volume of the Sphere is given by: 4/3πr³ (here r is the radius of the sphere)
Let V be the volume of the sphere)
∴ V = 4/3πr³
⇒ V = 4/3π(7)³ = (4/3) x (22/7) × 343
⇒ V = (4/3) × 22 × 49 = 4312/3 = 1437(1/3)
∴ Volume of the sphere is 1437(1/3)

Given: Volumes of the spheres are in the ratio 8:27.
Volume of a sphere is given by: 4/3πr³ (here ‘r’ is the radius of the sphere).
Surface are of a sphere is given by: 4πr² (here ‘r’ is the radius of the sphere).
Let r₁, r₂ be the radii of first sphere and second sphere respectively.
∴ 4/3π(r₁)³: 4/3π(r₂)³ = 2:27
⇒ (r₁)³: (r₂)³ = 8 : 27
⇒ r₁: r₂ = ∛8 : ∛27
⇒ (r₁): (r₂) = 2: 3
Now , here
4π(r₁)²: 4π(r₂)² = (r₁)² : (r₁)²
⇒ (r₁)² : (r₁)² = (2)² : (3)² = 4:9
∴ Ratio of the surface areas of the given spheres is 4:9.

Given: A hollow metallic Sphere with external diameter8cm and internal diameter 4cm
A cone having base diameter 8cm
Radius of the cone is: 8/2 = 4cm
External Radius of the sphere is 8/2 = 4cm
Internal Radius of the sphere is 4/2 = 2cm
Volume of a Hollow Sphere is given by: 4/3π((r₁)³ – (r₂)³) (here r₁ and r₂ are External and internal radii of the hollow sphere respectively)
Let V₁ be the Volume of the Hollow sphere.
∴ V₁ = 4/3π((r₁)³ – (r₂)³)
⇒ V₁ = 4/3π((4)³ – (2)³) = 4/3π(64 – 8) = 4/3π(56)
Volume of the cone is given by 1/3 × π × r² × h
Let V₂ be the volume of the cone
∴ V₂ = 1/3 × π × r² × h
⇒ V₂ = 1/3 × π × (4)² × h
Here, Hollow sphere is melted and moulded into a cone.
∴ V₁ = V₂
⇒ 4/3π(56) = 1/3 × π × (8)² × h
⇒ 1/3 × π × 16 × h = 4/3 × π × (56)

∴ Height of the cone is 14cm

Given: A metallic cone having base radius 2.1 cm and height 8.4 cm
Volume of a cone is given by: 1/3 × π × r² × h
Let V₁ be the volume of the cone
∴ V₁ = 1/3 × π × r² × h
⇒ V₁ = 1/3 × π × (2.1)² × (8.4)
Volume of a sphere is given by: 4/3 × π × r³
Let V₂ be the volume of the cone
∴ V₁ = 4/3 × π × r³
Here,
V₁ = V₂
∴ 1/3 × π × (2.1)² × (8.4) = 4/3 × π × r³
⇒ r³ = 1/3 × π × (2.1)² × (8.4) × 3/4π = (2.1)³
⇒ r³ = (2.1)³
⇒ r = ∛(2.1)³ = 2.1 cm
∴ Radius of the sphere is 2.1 cm

Given: The volume of a hemisphere is 19404 cm³.
Volume of the hemisphere is given by: 2/3 × π × r³
∴ 2/3 × π × r³ = 19404
⇒ r³ = 19404 × 3/2π = 9261
⇒ r³ = 9261 ⇒ r = ∛9261
⇒ r = 21
Now, Total surface area of hemisphere is given by: 3πr²
Let S be the TSA
∴ S = 3 × π × r²
⇒ S = 3 × π × (21)² = 4158 cm²
∴ S = 4158 cm²
That is TSA of the given sphere is 4158 cm²

Given: The surface area of a sphere is 154 cm².
TSA of the sphere is given by: 4 × π × r²
∴ 4 × π × r² = 154
⇒ r² = 154 × 1/4π = 49/4
⇒ r² = 49/4 ⇒ r = √49/4
⇒ r = 7/2
Now, Volume of hemisphere is given by: (4/3)πr³
Let V be the Volume of the hemisphere
∴ V = (4/3) × π × r³
⇒ V = (4/3) × π × (7/2)³ = (539/3) cm³ = 179 (2/3)cm³
∴ V = 179 (2/3)cm³
That is Volume of the given sphere is 179 (2/3)cm³

Given: Radius of the hemisphere: 7cm .
TSA of the hemisphere is given by: 3πr²
Let S be the TSA of the hemisphere.
∴ S = 3πr²
⇒ S = 3 × π × (7)² = 147π cm²
∴ TSA of the hemisphere is 147π cm²

Given: The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm.
Bucket is in the shape of frustum.
Let V be the Volume of the Bucket(Frustum)
Volume of the frustum is given by: (π/3) × h × (R² + r² + Rr) (here r and R are the radii of smaller and larger circular ends respectively)
∴ V = (π/3) × h × (R² + r² + R × r)
⇒ V = (π/3) × 40 × (35² + 14² + 35 × 14)
⇒ V = (π/3) × 40 × (1225 + 196 + 490) = (π/3) × 40 × (1911)
⇒ V = (π/3) × 40 × (1911) = 80080 cm³
∴ The volume of the bucket is: 80080 cm³

Given: The radii of the end of a bucket are 5 cm and 15 cm and it is 24 cm high
Bucket is in the shape of frustum.
TSA of a frustum of a cone = πl(r₁ + r₂) + πr₁² + πr₂² (here l , r₁, r₂ are the slant height, radii of the frustum)
Let S be the TSA of the bucket
∴ S = πl(r₁ + r₂) + π(r₂)² (here , top of the bucket is not closed but bottom is closed, ∴ π(r₂)² = 0 )
l = √(h² + (R-r)²)
⇒ S = π × √(h² + (R-r)²) × (r₁ + r₂) + π(r₂)²
⇒ S = π × √(24² + (15-5)²) × (5 + 15) + π × (5)²
⇒ S = π × √(576 + 100) × (20) + π × 25
⇒ S = π × √(676) × (20) + π × 25
⇒ S = π × 26 × (20) + π × 25
⇒ S = π × 520 + π × 25
⇒ S = π × (520 + 25)
⇒ S= 3.14 × 545 = 1711.3 cm²
∴ The surface area of the bucket is 1711.3 cm²

Given: A circus tent is cylindrical to a height of 4 m and conical above it.
Its diameter is 105 m and its slant height is 40 m.
CSA of the Cylinder is given by: 2πrh (here, r and h are radius and height respectively)
Let V₁ be the CSA of the cylindrical part of tent
∴ V₁ = 2πrh
⇒ V₁ = 2 × π × (105/2) × 4 = 1320cm² (here diameter is 105 cm, ∴ radius = (105/2) cm)
CSA of the cone is given by: πrl (here, r and l are radius and slant height respectively)
Let V₂ be the CSA of the conical part of tent
∴ V₂ = πrl
⇒ V₂ = π × (105/2) × 40 = 6600 cm² (here diameter is 105 cm, ∴ radius = (105/2) cm)
∴ Total area of the canvas is V₁ + V₂
∴ V₁ + V₂ = 1320 + 6600 = 7920cm²
∴ Total area of the canvas is: 7920 cm²

Assertion is wrong and Reason is Wrong.
Assertion (A):
Given: The radii of the end of a bucket are 5 cm and 15 cm and it is 24 cm high
Bucket is in the shape of frustum.
TSA of a frustum of a cone = πl(r₁ + r₂) + πr₁² + πr₂² (here l , r₁, r₂ are the slant height, radii of the frustum)
Let S be the TSA of the bucket
∴ S = πl(r₁ + r₂) + π(r₂)² (here , top of the bucket is not closed but bottom is closed, ∴ π(r₂)² = 0 )
l = √(h² + (R-r)²)
⇒ S = π × √(h² + (R-r)²) × (r₁ + r₂) + π(r₂)²
⇒ S = π × √(24² + (15-5)²) × (5 + 15) + π × (5)²
⇒ S = π × √(576 + 100) × (20) + π × 25
⇒ S = π × √(676) × (20) + π × 25
⇒ S = π × 26 × (20) + π × 25
⇒ S = π × 520 + π × 25
⇒ S = π × (520 + 25)
⇒ S= 3.14 × 545 = 1711.3 cm²
∴ The surface area of the bucket is 1711.3 cm²
Reason(R):
Here,
Surface area is π {R² + r² + l(R + r)},where l² = h² + (R –r)².
Assertion is wrong and Reason is Wrong.

Assertion is wrong and Reason is correct.
Assertion (A):
Given: A hemisphere of radius 7 cm.
Total surface area of the hemisphere is: 3πr²
Let S be the TSA of the hemisphere.
∴ S = 3πr² = 3π(7)²
∴ S = 462cm²
The total cost to pain it: 462 × 5 = Rs 2310
Reason(R):
The total surface area of a hemisphere is 3πr².
∴ Assertion is wrong and Reason is correct.

Assertion is correct and Reason is correct explanation of the given assertion.
Assertion (A):
Given: A coin which is 1.75 cm in diameter and 2 mm thick.
A cuboid with dimensions (10cm × 5.5cm × 3.5cm)
A coin is in the form of a cylinder.
Volume of the Cylinder is given by: πr²h
Let V₁ be the volume of the coin.
∴ V₁ = πr²h
⇒ V₁ = π × (1.75/2)² × (2/10) (here, 1.75 is the diameter,∴ radius = 1.75/2 , and 1mm = 1/10 cm)
⇒ V₁ = π × (0.875)² × 0.2 = 77/160 cm²
Volume of a cuboid is given by: l × b × h
Let V₂ be the volume of the of the cuboid.
∴ V₂ = l × b × h
⇒ V₂ = 10 × 5.5 × 3.5 = 192.5 cm²
Let ‘n’ be the number of coin melted .
∴ n × V₁ = V₂
⇒ n × (77/160) = 192.5
⇒ n = 192.5 × (160/77) = 400
∴ n = 400
That is 400 coin when melted can be moulded into a cuboid of given dimensions.
Reason(R):
Volume of a cylinder of base radius r and height h is given by V= (πr²h) cubic units. And, area of a cuboid = (l × b × h) cubic units.
∴ Assertion is correct and Reason is correct explanation of the given assertion.

Assertion is wrong and Reason is Wrong.
Assertion (A):
Given: volumes of two sphere are in the ratio 27:8.
Volume of the sphere is given by: 4/3 πr³
Let V₁ be the volume of the first sphere.
Let V₂ be the volume of the first sphere.
∴ V₁:V₂ = 4/3 π(r₁)³ : 4/3 π(r₂)³
⇒ 27:8 = (r₁)³ : (r₂)³
⇒ r₁ : r₂ = 3:2
Surface area of the sphere is given by: 4πr²
Let S₁ be the Surface area of the sphere.
Let S₂ be the Surface area of the sphere.
∴ S₁ : S₂ = 4π(r₁)²:4π(r₂)²
⇒ S₁ : S₂ = (r₁)²: (r₂)²
⇒S₁ : S₂ = (3)²: (2)²
⇒S₁ : S₂ = 9:4
Reason(R):
Volume of a sphere = 4/3 πr³.
Surface area of a sphere = 4πR².
Assertion is wrong and Reason is Wrong.

Assertion is correct and Reason is Wrong.
Assertion (A):
Given: A cone of radius 3cm and height 4cm.
CSA of the cone is given by: πrl (here r is radius and l is slant height)
l = √(h² + r²)
Let S be the CSA of the cone
∴ S = πrl = π × r × √(h² + r²)
⇒ S = π × 4 × √(4² + 3²) = π × 3 × √(16 + 9) = π × 3 × √(25) = π × 3 × 5 = 15π
∴ S =15π
Reason (R):
Volume of a cone is : 1/3 πr²h
∴ Assertion is correct and Reason is Wrong.