The document Previous Year Questions - Surface Area and Volumes (part-1) Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

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**Short Answer Type Questions**

**Q.1. A juice seller was serving his customers using glasses as shown in Fig, The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14). [CBSE 2019 (30/5/1)] Ans.** Since, the inner diameter of the glass = 5 cm and height =10 cm, the apparent capacity of the glass = πr

= (3.14 x 2.5 x 2.5 x 10) cm

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.

= 32.71 cm

So, the actual capacity of the glass

= apparent capacity of glass - volume of the hemisphere

= (196.25 - 32.71) cm

Ans.

Ratio of their volumes =

⇒

Ratio of their surface area

Ans.

Volume of the conical heap of rice

Canvas cloth required to just cover the heap

Ans.

Let r be the radius of both cylinder and cone and h be the height of cylinder and cone

Therefore, slant height of cone

Ratio of their curved surface

⇒

Put,

∴

Squaring both sides, we get

Ans.

Radius of the conical part (r) = 28 m

Height of conical part (h) = 21 m

Curved surface area of conical part = πrl

= π(28)35 m^{2} = 980π m^{2}

Radius of cylindrical part = 28 m

Height of cylindrical part = 6 m

Curved surface area of cylindrical part = 2πrh

= 2π(28)6 = 336π m^{2}

Total curved surface area = 980π + 336π

∴ Area of canvas used = 4136 m^{2}**Q.6.The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe. [AI 2017] Ans.** Let the length of cylindrical pipe = 4 marks

Its inner radius r = 30 cm = 3.0 m

Length of cuboid = 4.4 m

Breadth of cuboid = 2.6 m

and height of cuboid = 1.0 m

Inner radius ‘r’ of pipe = 30 cm = 0.3 m

Width of pipe = 5 cm = 0.05 m

∴ Outer radius ‘R’ of pipe = 0.3 m + 0.05 m = 0.35 m

Let the length of pipe formed = H metres.

A.T.Q.

Solid cuboid is melted to form a hollow cylindrical pipe with thickness.

∴ Volume of cuboid = Volume of outer cylinder - Volume of inner cylinder

H = 112 m, Hence the length of the pipe is 112 m.

Ans.

r + h = 37 cm [Given]

⇒ h = (37 - r) cm

Total surface area = 2πr (r + h)

⇒

⇒

⇒ h = (37 - 7) cm = 30 cm

Volume o f the cylinder =

= 22 x 7 x 30 cm

= 4620 cm

Ans.

Let R and r be the radii of two circular ends respectively. Therefore, we have

∴ Curved surface area of the frustum = (πR + πr)l

= (9 + 3) x 4 = 12 x 4 = 48 cm

Ans.

Let r be the radius of the base of cone and h be the height of conical part of the toy.

Then, r = OB = 3.5 cm

h = OC = CD - OD = (15.5 - 3.5) cm = 12 cm

Also, radius of the hemisphere, r = 3.5 cm

∴ Total surface area of the toy

= surface area of cone + surface area of hemisphere

∴ Speed = 5 km/h

∴ Length in t hours = 5000 t m.

According to the question,

Volume of water flown through pipe = Volume of water in tank

πr

Hence required time is 2 hours.

Ans.

A.T.Q.

Volume of solid hemisphere = Surface area of a solid hemisphere

⇒

∴ Diameter of hemisphere = 2 x r = 2 x 9/2 = 9 unit

Ans.

Let r = 5x and h = 12x

Volume of cone = 1/3 πr

⇒

⇒ x

⇒ x = 1

So, the value of r = 5 cm and h = 12 cm

TSA of cone = πr(l + r) = 3.14 x 5(13 + 5)

= 3.14 x 90 = 282.6 cm

Let length of wire = h cm and radius of wire = 3/20cm

According to question

CSA of wire = CSA of cylinder

⇒

⇒ h = 400 cm

Ans.

Volume of water in the hemispherical tank = 2/3πr

Ans.

Volume of ice-cream in one ice-cream cone =

∴ Volume of ice-cream in 10 such cones = 10 x 2πr

20πr

Diameter of conical ice-cream cup = 6 cm

**Q.16. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by Find the diameter of the cylindrical vessel. [CBSE (AI) 2016]Ans.** Volume of sphere = 4/3 π(6)^{3} cm^{3 }

Volume of water rise in cylinder = ∴ ⇒ r = 9 cm**Q.17. Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere. [CBSE (F) 2015]Ans.** Volume of the smaller sphere Volume of smaller sphere x density = mass∴ 36π (density of metal) = 1

Density of metal = 1/36π

∴ Volume of bigger sphere x density = mass

Volume of new sphere = volume of smaller sphere + volume of bigger sphere(R')

(R')

(R')

(R')

R' =3 x 2

R' = 6 cm

∴ Diameter of new sphere =12 cm.

Ans.

Radius of cylindrical bottle, r = 6/2 = 3 cm

Let height of cylindrical bottle = h

Since 10% liquid is wasted, therefore only 90% liquid is filled into 72 cylindrical bottles.

∴ The volume of 72 cylindrical bottles = 90% of the volume in the bowl

⇒

**Long Answer Type Questions**

**Q.1. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid, (use π = 22/7) [Allahabad 2019]Ans.** Diameter of cylinder

= diameter of the hemisphere = 7 cm

∴ Radius of cylinder = 7/2 cm

Total height of the solid = 20 cm

Height of the cylinder

Volume of the solid= volume of the cylinder + 2 x volume of one hemisphere

Ans.

Area of metal sheet used = π(R + r)l + πr

= π(20 + 12) x 17 + π x 144

Hence, the height of the bucket is 15 cm and the area of the metal sheet used is 2160.32 cm

Ans.

Therefore, the slant height of the frustum,

So, Area of the metallic sheet usedCurved surface area of frustum of cone + Area of circular base + Curved surface area of cylinder

= [π x 35.44 (22.5 + 12.5) + π x (12.5)

= 4860.9 cm

Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)

OR

A girl empties a cylindrical bucket, full of sand, of base radius 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct up to one place of decimal.

**[CBSE (E) 2014, 2019 (30/5/1)] ****Ans. **We have,

Radius of cylindrical bucket =18 cm

Height of cylindrical bucket = 32 cm

And the height of conical heap = 24 cm

Let the radius of the conical heap be r cm

Volume of the sand = volume of the cylindrical bucket

= πr^{2}h = π x (18)^{2} x 32

Now, volume of conical heap Here, volume of the conical heap will be equal to the volume of sand∴ 8πr^{2} = π x (18)^{2} x 32

⇒ r^{2} = 18 x 18 x 4 = (18)^{2} x (2)^{2}

⇒ r^{2} = (36)^{2} or r = 36 cm**Q.5. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find the area of the metal sheet used to make the bucket. [Use π = 3.14] [CBSE 2018]Ans. **For bucket,

Upper diameter (D) = 30 cm

∴ Lower diameter (d) = 10 cmHeight of bucket (h) = 24 cm∴ The area of the metal sheet used = CSA of the frustum (bucket) + Area of bottom part (base)

Ans.

Let r

Now, in ΔVOA and VO' A', we have

⇒ ⇒ ∴ Volume of the frustum = Let the length of wire of diameter 1/16 cm be l cm. Then,Volume of metal used in wire Since the frustum is recast into a wire of length l cm and diameter 1/16 cm.∴ Volume of the metal used in wire = volume of the frustum

⇒ ⇒

∴ Radius of metallic cylinder (r) = 6 cm

Height (h) = 15 cm

Volume of cylinder = πr

Radius of cone = 3 cm

Height of cone = 9 cm

Volume of cone = 1/3 π(3)

Number of toys so formed = If the question is “A tent is in the form of a cylinder surmounted by a cone. Find the capacity of the tent and the cost of canvas for making the tent at Rs 100 per sq.m.”, then the solution is given as “Tent is a combination of a cylinder and a cone. For capacity∴ Volume (capacity) of the tent = Volume of the cylindrical part + Volume of the conical part

For the cost of the canvas, we find the total surface area.

Total surface area = Curved area of the cylindrical part + Curved surface area of the conical part

Now, proceed to find its cost.”

Note: Don’t solve as “Total surface area of canvas = Total surface area of cylinder + Total surface area of a cone and proceed further"

This is the wrong solution.

Ans.

Height of conical vessel (h) = 24 cm

Radius of cylindrical vessel (R) = 10 cm

Let H be the height of water in the cylindrical vessel Now, the total volume of the conical vesselAccording to the question,3/4 of the volume of water from the conical vessel is emptied into the cylindrical vessel.

⇒ 3/4 x Volume of conical vessel =Volume of water in the cylindrical vessel.

⇒ 25 x 6 = 10 x 10 x H⇒ ⇒ H = 1.5 cm∴ Height of water in the cylindrical vessel = 1.5 cm

Q.9. In a rain-water harvesting system, the rainwater from a roof of 22 m x 20 m drains into a cylindrical tank having a diameter of base 2 m and height of 3.5 m.

Ans.

Breadth of roof (b) = 20 m

Let the height of water column collected on roof= hm

∴ Volume of standing water on rooftop = lbh

= (22 x 20 x h) m

This water is taken into a cylindrical tank of diameter of base 2 m and height 3.5 m.

Tank gets completely filled with this amount of water.

⇒ Volume of water from roof-top = Volume of cylinder

Diameter of tank = 2 m

∴ Radius of tank = 2/2 = 1 m

Hence rainfall is 2.5 cm.

Ans.

Radius of the cylinder = Height of the cylinder = 2.4 cmAlso, radius of the cone = 0.7 cm and height of the cone = 2.4 cm

Now, slant height of the cone =∴ Total surface area of the remaining solid= curved surface area of cylinder + curved surface area of the cone + area of upper circular base of the cylinder.

Ans.

Since B is the mid-point of AD and BC is parallel to DE, therefore C is the mid-point of AE.

∴ AC = CE

Also, ΔABC ~ ΔADE

∴ The required ratio = 1 : 7.

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