SOLUTION:

Soln. : Volume of each cube = 64 cm³
Let the edge of each cube = 2
∴ x³ = 64 cm³
⇒ x = 4 cm
Now, Length of the resulting cuboid l = 2x cm = 8 cm
Breadth of the resulting cuboid b = x cm = 4 cm
Height of the resulting cuboid h = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm²
= 2 [32 + 16 + 32] cm² = 2 [80] cm² = 160 cm².

SOLUTION:

Soln. : For hemispherical
part, radius (r)
∴ Curved surface area = 2πr²

Total height of vessel = 13 cm
∴ Height of cylinder = (13 - 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh
∴ Inner surface area of vessel = (308 + 264)cm²
= 572 cm²

SOLUTION:

Soln. : Let r and h be the radius of cone, hemisphere and height of cone
and h = (15.5 3.5) cm
= 12.0 cm
Also l² = h² + r² = 12² + (3.5)²
= 156.25
∴ l = 12.5 cm
Curved surface area of the conical part = πrl
Curved surface area of the hemispherical part = 2πr²
Total surface area of the toy = πrl + 2πr² = πr (l + 2r)

= 11 × (12.5 + 7) cm² = 11 × 19.5 cm² = 214.5 cm²

SOLUTION:

Soln. : Let side of the block, l = 7 cm
∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid
= [Total surface area of the cubical block] + [C.S.A. of the hemisphere]
- [Base area of the hemisphere]
= (6 × l²) + 2πr² πr²
[where l = 7 cm and r = cm]


= 332.5 cm²

SOLUTION:

Soln. : Let l be the side of the cube.
∴ The greatest diameter of the hemisphere = l
⇒ Radius of the hemi-sphere
∴ Surface area of hemisphere = 2πr²

Base area of the hemisphere
Surface area of the cube
∴ Surface area of the remaining solid

SOLUTION:

Soln. : Diameter of the hemispherical part
= 2.5 mm
Curved surface area of one hemisphere part = 2πR²
∴ Surface area of both hemispherical parts


Entire length of capsule = 14mm
∴ Length of cylindrical part
= (14 - 2 × 2.5)mm = 9mm
∴ Area of cylindrical part = 2πrh

∴ Total surface area


= 220 mm²

SOLUTION:

Soln. : For cylindrical part :
Radius (r) and height (h) = 2.1 m
∴ Curved surface area
For conical part :
Slant height (l) = 2.8 m and base radius (r) = 2 m
∴ Curved surface area
∴ Total surface area
= [Curved surface area of the cylindrical part] + [Curved surface area of conical part]


Cost of the canvas used :
Cost of 1 m² of canvas = ₹ 500
∴ Cost of 44 m² of canvas = ₹ 500 × 44 = ₹ 22000.

SOLUTION:

Soln. : For cylindrical part:
Height = 2.4 cm and diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
∴ Total surface area of the cylindrical part
= 2πrh + 2πr² = 2πr [h + r]


For conical part :
Base radius (r) = 0.7 cm and height (h) = 2.4 cm
∴ Slant height (l)

∴ Curved surface area of the conical part

Base area of the conical part

Total surface area of the remaining solid
= [(Total surface area of cylindrical part)
+ (Curved surface area of conical part)
- (Base area of the conical part)]

Hence, total surface area to the nearest cm² is 18cm².

SOLUTION:

Soln. : Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Curved surface area = 2πrh

Curved surface area of a hemisphere = 2πr²
∴ Curved surface area of both hemispheres

Total surface area of the remaining solid
= (220 + 154) cm² = 374 cm².

SOLUTION:

Soln. : Here, r = l cm and h = l cm.
Volume of the conical part
and volume of the hemi-spherical part
∴ Volume of the solid shape

SOLUTION:

Soln. : Here, diameter = 3 cm ⇒ Radius (r)
Total height = 12 cm
Height of a cone (h₁) = 2 cm
∴ Height of both cones = 2 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 - 4) cm = 8 cm.
Now, volume of the cylindrical part = πr2h2
Volume of both conical parts
∴ Volume of the whole model

SOLUTION:

Soln. : Since, a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.
Diameter = 2.8 cm ⇒ Radius = 1.4 cm
∴ Length (height) of the cylindrical part
= 5 cm - (1.4 + 1.4) cm = 5 cm - 2.8 cm = 2.2 cm

Now, volume of the cylindrical part = πr²h
and volume of both the hemispherical ends
∴ Volume of a gulab jamun




Volume of 45 gulab jamuns

Since, the quantity of syrup in gulab jamuns = 30% of [volume] = 30% of

= 338 cm² (approx.)

SOLUTION:

Soln. : Dimensions of the cuboid are 15 cm, 10 cm and 3.5 cm.
∴ Volume of the cuboid
= 525 cm³
Since each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,
∴ Volume of each depression

Since there are 4 depressions,
∴ Total volume of 4 depressions
Now, volume of the wood in entire stand
= [Volume of the wooden cuboid] - [Volume of 4 depressions]

= 523.53 cm²

SOLUTION:

Soln. : Height of the conical vessel (h) = 8 cm
Base radius (r) = 5 cm
Volume of water in conical vessel

Now, Total volume of lead shots = of [Volume of water in the cone]

Since, radius of spherical lead shot (r) = 0.5 cm
∴ Volume of 1 lead shot

∴ Number of lead shots
=
= 100
Thus, the required number of lead shots = 100.

SOLUTION:

Soln. : Height of the big cylinder (h) = 220 cm
Base radius (r)
∴ Volume of the big cylinder = πr2h = π(12)² × 220 cm³
Also, height of smaller cylinder (h1) = 60 cm
Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder
= π(8)² × 60 cm³
∴ Volume of iron pole
= [Volume of big cylinder]
+ [Volume of the smaller cylinder]
= π × 220 × 12² + π × 60 × 8² cm³
= 3.14[220 × 12 × 12 + 60 × 8 × 8]cm³


Mass of iron
= 892.2624 kg = 892.26 kg.

SOLUTION:

Soln. : Height of the conical part = 120 cm.
Base radius of the conical part = 60 cm.
∴ Volume of the conical part
Radius of the hemispherical part = 60 cm.
∴ Volume of the hemispherical part
∴ Volume of the solid = [Volume of conical part] + [Volume of hemispherical part]



Volume of the cylinder = πr2h

⇒ Volume of water in the cylinder

∴ Volume of the water left in the cylinder

= 1131428.57142 cm³
= 1.13142857142 m³ = 1.131 m³ (approx).

SOLUTION:

Soln. : Volume of the cylindrical part = πr2h
= 3.14 × 1² × 8 cm³


Radius of spherical part
Volume of the spherical part

Total volume of the glass-vessel




= 346.51 cm³ (approx.)
⇒ Volume of water in the vessel = 346.51 cm³
Since, the child finds the volume as 345 cm³
∴ The child's answer is not correct
The correct answer is 346.51 cm³.

SOLUTION:

Soln. : Radius of the sphere (r1) = 4.2 cm
∴ Volume of the sphere

Radius of the cylinder (r2) = 6 cm
Let h be the height of the cylinder
∴ Volume of the cylinder = πr²h

Since, Volume of the metallic sphere = Volume of the cylinder.
⇒
⇒

Hence, height of the cylinder = 2.744 cm

SOLUTION:

Soln. : Radii of the given spheres are:
r₁ = 6 cm, r₂ = 8 cm, r₃ = 10 cm
⇒ Volume of the given spheres are:

∴ Total volume of the given spheres = V₁ + V₂ + V₃




Let the radius of the new big sphere be R.
∴ Volume of the new sphere

Since, the two volume must be equal.
∴
⇒ R³ = 1728 ⇒ R = 12 cm
Thus, the required radius of the resulting sphere = 12 cm.

SOLUTION:

Soln. : Diameter of the cylindrical well = 7 m
⇒ Radius of the cylinder (r)
Depth of the well (h) = 20 m
∴ Volume
= 22 × 7 × 5m³
⇒ Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14 m
Let h be the height of the platform.
∴ Volume of the platform = 22 × 14 × h m³
∴ 22 × 14 × h = 22 × 7 × 5
⇒
Thus, the required height of the platform is 2.5 m.

SOLUTION:

Soln. : Diameter of cylindrical well (d) = 3 m
⇒ Radius of the cylindrical well
Depth of the well (h) = 14 m
∴ Volume

Let the height of the embankment = H metre.
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment R = (4 + 1.5) m
= 5.5 m.
∴ Volume of the embankment
= πR²H - πr²H = πH [R² - r²] = πH (R + r) (R - r)

Since, Volume of the embankment = Volume of the cylindrical well
∴

Thus, the required height of the embankment = 1.125 m.

SOLUTION:

Soln. : For the circular cylinder:
Diameter = 12 cm
⇒ Radius (r) and height (h) = 15 cm
∴ Volume of total ice cream

For conical and hemispherical part of ice-cream:
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of conical part (H) = 12 cm

Volume = (Volume of the conical part) + (Volume of the hemispherical part)


Let number of ice-cream cones required to fill the total ice cream = n.
∴

Thus, the required number of cones is 10.

SOLUTION:

Soln. : For a circular coin:
Diameter = 1.75 cm
⇒ Radius (r)
Thickness (h)
∴ Volume = πr²h

For a cuboid:
Length (l) = 10 cm,
Breadth (b) = 5.5 cm
and Height (h) = 3.5 cm
∴ Volume =
Number of coins

Thus, the required number of coins = 400.

SOLUTION:

Soln. : For the cylindrical bucket:
Radius (r) = 18 cm and height (h) = 32 cm
Volume = πr²h
⇒ Volume of the sand
For the conical heap:
Height (H) = 24 cm & let radius of the base be (R).
∴ Volume of conical heap

∵ Volume of the conical heap sand = Volume of the sand
∴
⇒
⇒
Let 'l' be the slant height of the conical heap of the sand.

∴ l² = R² + H²
⇒ l² = 24² + 36²
⇒ l² = (12 × 2)² + (12 × 3)²
⇒ l² = 12²[2² + 3²]
⇒ l² = 12² × 13
⇒
Thus, the required radius = 36 cm and slant height

SOLUTION:

Soln. : Width of the canal = 6 m, Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes
∴ Volume of water flown in

Since the above amount (volume) of water is spread in the form of a cuboid of height as 8 cm Let the area of the cuboid = a
∴ Volume of the cuboid = Area × Height

Thus,

= 562500m² = 56.25 hectares
Thus, the required area = 56.25 hectares.

SOLUTION:

Soln. : Diameter of the pipe = 20 cm
⇒ Radius of the pipe (r) = cm = 10 cm
Since, the water flows through the pipe at 3 km/hr.
∴ Length of water column per hour(h) = 3 km
= 3 × 1000 m = 3000 × 100 cm = 300000 cm.
∴ Volume of water = πr₂h = π × 10² × 300000 cm³
= π × 30000000 cm³
Now, for the cylindrical tank,
Diameter = 10 m ⇒ Radius (R) = = 5 × 100 cm = 500 cm
Height (H) = 2 m = 2 × 100 cm = 200 cm
∴ Volume of the cylindrical tank = πr₂H = π × (500)² × 200 cm³
Now, time required to fill the tank

SOLUTION:

Soln. : We have: r₁ = 2 cm, r₂ = 1 cm and h = 14 cm
Volume of the glass

= πh (r₁² + r₂² + r₁r₂)

SOLUTION:

Soln. : We have

Slant height (l) = 4 cm
2πr₁ = 18 cm and 2πr₂ = 6 cm
⇒ and
∴ Curved surface area of the frustum of the cone
= π(r₁ + r₂) l = (πr₁ + πr₂) l = (9 + 3 ) × 4 cm²
= 12 × 4 cm² = 48 cm².

SOLUTION:

Soln. : Here, the radius of the open side (r₁) = 10 cm
The radius of the upper base (r₂) = 4 cm
Slant height (l) = 15 cm
∴ Area of the material required
= [Curved surface area of the frustum] + [Area of the top end]
= π(r₁ + r₂) l + πr₂²

SOLUTION:

Soln. : We have : r₁ = 20 cm, r₂ = 8 cm and h = 16 cm

∴ Volume of the frustum





∴ Cost of milk = ₹ 20
= ₹ 208.998 ≈ ₹ 209
Now, slant height of the given frustum


∴ Curved surface area = π(r₁ + r₂)l


Area of the bottom = πr²

∴ Total area of metal required
= 1758.4 cm² + 200.96 cm² = 1959.36 cm²
Cost of metal required = ₹
= ₹ 156.75.

SOLUTION:

Soln. : Let us consider the frustum DECB of the metallic cone ABC

Here, r₁ = BO and r₂ = DO′
In ∆AOB,
⇒
In ∆ADO′,
⇒
Now, the volume of the frustum DBCE



Let l be the length and D be diameter of the wire drawn from the frustum. Since the wire is in the form of a cylinder,
∴ Volume of the wire = πr²l

∵ [Volume of the frustum] = [Volume of the wire]
∴
⇒
= 7964.44m
Thus, the required length of the wire = 7964.44 m

SOLUTION:

Soln. : Since diameter of the cylinder = 10 cm
∴ Radius of the cylinder (r)
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm
∴ The thickness of cylinder covered in one round
⇒ Number of rounds (turns) of the wire to cover
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds = 40 × 31.4 cm = 1256 cm
Now, radius of the wire
∴ Volume of wire = πr²l
∵ Density of wire = 8.88 gm/cm³
∴ Weight of the wire
= [Volume of the wire] × density


= 787.97 g = 788 g (approx.)

SOLUTION:

Soln. : Let us consider the right ∆BAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse
Obviously, we have obtained two cones on the same base AA' such that radius = DA or DA'
Now,
[∴ ADB ~ ΔCAB]

Also
Since, CD = BC - DB
⇒
Now, volume of the double cone



Surface area of the double cone = πrl1 + πrl2

SOLUTION:

Soln. : ∵ Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Volume of water contained in the cistern = 129600 cm³
∴ Free space (volume) which is not filled with water = (1980000 - 129600) cm³ = 1850400 cm³
Now, volume of one brick = (22.5 × 7.5 × 6.5) cm³ = 1096.875 cm³
∴ Volume of water absorbed by one brick
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
∴ [volume occupied by n bricks] = [(free space in the cistern + volume of water absorbed by n-bricks)]
⇒
⇒
⇒
⇒
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

SOLUTION:

Soln. : Volume of three rivers = 3 {(Surface area of a river) × Depth}


Volume of rainfall = (Surface area) × (Height of rainfall)



Since, 0.7236 km³ ≠ 9.728 km³
∴ The additional water in the three rivers is not equivalent to the rainfall.

SOLUTION:

Soln. : We have,
For the cylindrical part
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm ⇒ Curved Surface area

For the frustum :
Height (H) = 22 - 10 = 12 cm
∴ Slant height (l)


∴ Surface area = π (r₁ + r₂)l

Area of tin required = [Area of the frustum] + [Area of cylindrical portion]

SOLUTION:

Soln. :

We have, Curved surface area of the frustum PQRS

= πr₁l₁ - πr₂l₂ ...(1)
Now, ∆OC₁Q ∼ ∆OC₂S [By AA similarity]
∴
⇒
(∵ l₁ = l + l₂)
⇒ ...(2)
Now, from (1), we get
Curved surface area of the frustum


= πl (r₁ + r₂) [From (2)]
Now, the total surface area of the frustum
= (curved surface area) + (base surface area) + (top surface area)
= πl (r₁ + r₂) + πr₂² + πr₁² = π (r₁ + r₂) l + π (r₁² + r₂²)
= π [(r₁ + r₂) l + r₁² + r₂²]

SOLUTION:

Soln. : We have,
[Volume of the frustum RPQS]
= [Volume of right circular cone OPQ]
- [Volume of right circular cone ORS]

...(1)

Since ∆OC1Q ∼ ∆OC2S [By AA similarity]
∴
⇒ ...(2)
⇒
⇒
⇒ ...(3)
From (1) and (2), we have Volume of the frustum RPQS



[From (3)]