Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes with solutions includes all the important topics with detailed explanation that aims to help students to score more marks in Board Exams 2020. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 13 Surface Areas and Volumes.
Important Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes
Expert teachers at CBSETuts.com collected and solved 2 Marks and 4 mark important questions for Class 10 Maths Chapter 13 Surface Areas and Volumes.
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2016
Short Answer Type Questions II [3 Marks]
Question 1.
In Figure, a decorative block, made up of two solids a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block.(use π=22/7)
Solution:
Side of the cube = 6 cm
Question 2.
A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment
Solution:
Question 3.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder, (use π=22/7)
Solution:
Question 4.
A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find the volume of the cone, (use π = 3.14)
Solution:
Question 5.
A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of base diameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy(use π =22/7)
Solution:
Question 6.
In figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at therate of rupees 500/sq. metre. (use π=22/7)
Solution:
Question 7.
A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel
Solution:
Question 8.
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 3×5/9 cm. Find the diameter of the cylindrical vessel
Solution:
Question 9.
A hemispherical tank, of diameter 3 m, is full of water. It is being emptied by a pipe 3×4/7 litre per second. How much time will it take to make the tank half empty?
Solution:
Question 10.
A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice-cream. The whole ice-cream is to be divided into 10 children in equal ice-cream cones, with conical base surmounted by hemispherical top. If the height of conical portion is twice the diameter of base, find the diameter of conical part of ice-cream cone.
Solution:
Question 11.
A metal container, open from the top, is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper circular ends are as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at of rupees 35/litre
Solution:
=1/3 x 22/7 x 21[400+64+160]=22 x 624/1000 liters[1 cm³=1/1000 l]
cost of 1l milk=35rupees
cost of milk in container=35 22 x 624/1000=480.48 rupees
Long Answer Type Questions [4 Marks]
Question 12.
A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm³. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket, (use π=3.14)
Solution:
Let R, r and V be the upper radius, lower radius and volume of the frustum respectively then
R = 20 cm, r = 12 cm and V = 12508.8 cm³
12308.8 = 1/3x 3.14 [400 + 144 +240]h
h=12308.8 x 3/3.14 x 784 h=15 cm
Now, l (slant height) =√(20-12)²+15²=√64+225=√289=17 cm²
Total area of the metal sheet=curved surface area of the cone+area of the base= π(R+r)l+πr²
=π(20+12)x 17+π x 12 x 12
=π x 32 x 17 +144π =688π
=688 x 3.14 =2160.32 cm²
Question 13.
The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle
Solution:
Question 14.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs ? 120 per sq.m, find the amount shared by each school to set up the tents.What value is genetated by above problem?
Solution:
Question 15.
In figure, shown a right circular cone of height 30 cm. A small cone is cut off from the top by a plane parallel to the base. If the volume of the small cone is 1/27 of the volume of given cone, find at what height above the base is the section made.
Solution:
2015
Short Answer Type Questions II [3 Marks]
Question 16.
In fig, from the top of a solid cone of a height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid
Solution:
Question 17.
A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in making of toy is 166 x 5/6 cm3 Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of 10 rupees per cm2
Solution:
Question 18.
In figure, from a cuboidal solid metallic block of dimensions 15 cm X 10 cm X 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.
Solution:
Cuboid: Length of cuboid, l = 15 cm
Breadth cuboid, b = 10 cm
Height of cuboid, h = 5 cm
Cylinder: Diameter of cylinder = 7 cm
Radius of cylinder, r = 7/2 cm
Height of cylinder, h’ = 5 cm
Surface area of remaining block = Total surface area of cuboidal block + Curved Surface Area of cylinder – Area of 2 circles
= 2(15 x 10 + 10 x 5 + 5 x 15) + 2 x 22/7 x 7/2 x 5- 2 x 22/7 x 7/2 x 7/2
= 2(150 + 50 + 75) + 110 – 77 = 550 + 33 = 583 cm².
Question 19.
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs 100 rupees per sq. m, find the amount, the associations will have to pay.What values are shown by these associations ?
Solution:
Height of cylinder = 4 m
Radius =2.1 m
Curved surface area of cylinder = 2πrh = 2 x 22/7 x 2.1 x 4 = 52.8 m²
Radius of cone = 2.1 m, height of cone = 2.8 m.
Let slant height = l
l = √r²+h² [ Pythagoras theorem]
Slant height of cone, l = √(2.1)² + (2.8)² = 3.5 m
Curved surface area of cone =πrl = 22/7 x 2.1 x 3.5 = 23.1 m²
Area of canvas required for one tent = 52.8 + 23.1 = 75.9 m²
Canvas required for 100 tents = 100 x 75.9 = 7590 m²
Total cost = rupees (7590 x 100) = rupees 759000
Amount to be paid by association = 50/100 x 759000 = 379500 rupees
Care for the society, values shown by welfare associations.
Question 20.
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.
Solution:
Radius of hemispherical bowl = 18 cm
Volume of liquid in bowl = 2/3πr³
=2/3 x π x 18 x 18 x 18 =3888π cm³
liquid wasted=10/100 x 3888π cm³ = 3888π/10 cm³
liquid transferred into bottles =3888π-3888π/10
=34992π/10 cm³
radius of bottle=3cm
let height = X cm
volume of bottle =πr²h
=π x 3 x 3 x X =9πX cm³
volume of 72 bottles =72 x 9πX cm³
=648πX cm³
648πX = 34992π/10
X=34992/10 x 648 =5.4 cm
Height of each bottle = 5.4 cm
Question 21.
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of 5 rupee per 100 sq. cm
Solution:
Largest diameter of hemisphere = 10 cm = Side of cube
.’. radius = 5 cm
Total surface area of the solid = Surface area of cube +Curved surface area of hemisphere – Area of base of hemisphere
= 6(side)² + 2πr² – πr²
= 6 x 10² + 2 x 3.14 x 5 x 5 – 3.14 x 5 x 5
= 600 + 78.5 = 678.5 cm²
Total cost = 678.5 x 5 /100 =
33.92 rupees
Question 22.
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area
Solution:
Volume of the cone =1/3 x πr²h
=1/3π x 3.5/2 x 3.5/2 x 3 cm³ =12.25/4 πcm³
volume of 504 =504 x 12.25/4 πcm³ =1543.5 πcm³
volume of the sphere=1543.5 π³=volume of 504 cones
4/3πr³=1543.5π
r³1543.5×3/4=1157.625
r=3√1157.625=10.5cm
Diameter of sphere=21cm
surface area of sphere=4πr²
=4×22/7×10.5×10.5=1386cm²
Question 23.
Two spheres of 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.
Solution:
Radius of smaller sphere = 3 cm
Volume of smaller sphere =4/3πx3x3x3cm³=36π cm³
When Mass=1kg,then volume=36π cm³
When Mass = 7 kg
volume=7×36π cm³=252π cm³
total value of two spheres=36π+252π =288π cm³
let the radius of sphere so formed=R cm
volume of big sphere=total volume of 2 sphere
4/3πR³=288π
R³=288 x 3/4=72 x 3 =216
R=6 cm
Diameter =12 cm
Question 24.
A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth 8/9 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape
Solution:
Given, Radius of the cylinder = 3 cm
Height = 5 cm
Volume of cylinder = πr²h =π x 3 x 3 x 5 cm³=45π cm³
Radius of cone = 3/2 cm
Height of the cone =8/9 cm
Volume of the cone =1/3 πr²h
=1/3π x 9/4 x 8/9 = 2π/3cm³
Volume of metal left =45π-2π/3=133π/3 cm³
Ratio=volume of metal left in cylinder/volume of the metal taken out=133π/3:2π/3=133:2
Question 25.
A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right- circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres
Solution:
Radius of cone = 30 cm
height of cone = 60 cm
Volume of cone = 1/3πr² h =1/3π x 30 x 30 x 60 =18000 π cm³
Radius of cylinder = 60 cm
Height of cylinder = 180 cm
Volume of cylinder =πr²h
=π x 60 x 60 x 180 cm³ = 648000 π cm³
volume of water left=volume of cylinder-volume of cone
=648000π-18000π
=630000 π cm³
=630000 x 22/7 =1980000 cm³ =1.98 m³
Question 26.
The rain water from a 22 m x 20 m roof drains into a cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills 4/5th of the cylindrical vessel, then find the rainfall in cm
Solution:
Let the rainfall=x m
volume of rain water=lbh
=22 x 20 x X =440X m³
radius of cylindrical vessel=1 m
height =3.5m
volume of vessel=πr²h
=22/7 x 1² x 3.5=22 x 0.5 =11 m³
now,it is given that vessel is filled upto 4/5 of the volume by rain water
4/5 x 11=440X
4×11/5×440=X
X=1/50
X=1/50×100=2 cm
rainfall=2cm
Long Answer Type Questions [4 Marks]
Question 27.
A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to form a platform 27 m x 11 m. Find the height of the platform.
Solution:
cylinder:Diameter of cylinder =6 m
radius of cylinder,r=6/2=3 m
height of cylinder,h=21 m
cuboid of platform:Lenght=21m,breadth=11m,height=h
according to question,
volume of cuboid platform=volume of cylindrical well
lbh’=πr²h
27x11xh’=22/7x3x3x21
h’=22x3x3x21/7x27x11=2m
height of platform =2m
Question 28.
A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment.
Solution:
Given, Diameter of the well = 4 m
Radius of the well = 2 m
Depth of the well = 14 m = Height
Volume of earth taken out from the well = πr2h
= πx2x2x14 = 56π m3
Earth taken out from the well evenly spread to form an embankment having height 40 cm = 0.4 m
Let external radius of embankment be R.
Internal radius = 2 m = radius of well
Volume of embankment (cylindrical)=π(R²-r²)h
=π(R²-4) x 4/10 =56π
R²-4=56×10/4=140
R²=140+4=144
R=12m
Width of embankment=R-r=12-2=10m
Question 29.
Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm, If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe
Solution:
Let the radius of the pipe=r cm
Speed of water=2.52 km/hr=2520 m/h
Volume of the water that flows in half an hour=1/2πr²h
=1/2π x r/100 x r/100 x 2520 =126πr²/1000 m³
volume of the water in cylindrical tank=π x 40/100 x 40/100 x 3.15 m³
volume of the water flowing in 1/2hr=volume of water in cylindrical tank
126πr²/1000=π x 40/100 x 40/100 x 3.15
r²=40/100 x 40/100 x 3.15 x 1000/126
r²=4⇒r=2 cm
internal diameter of pipe=4cm
Question 30.
From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire
Solution:
Given, Radius of hemisphere=4.2cm
Volume of hemisphere=2/3πr³=2/3π x (4.2)³ cm³=49.392π cm³
Volume of 2 hemispheres =π x 49.392π cm³
=98.784π cm³
Height of cylinder =10cm
Radius =4.2cm
Volume of cylinder =πr²h
=π x (4.2)² x 10=176.4π
Volume of metal left =176.4π-98.784π
=77.616π cm³
Radius of wire =0.7 cm
Let length of wire =X cm
Volume of cylindrical wire=π x 0.7 x 0.7 x X
=0.49πX cm³
Volume of cylindrical wire =volume of metal left from cylinder
0.49πX=77.616π
X=77.616/0.49=158.4 cm
Length of wire=158.4 cm
Question 31.
A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top, which is open, is 5 cm. 100 spherical lead balls are dropped into the vessel. One-fourth of the water flows out of the vessel. Find the radius of a spherical ball.
Solution:
Radius of cone = 5 cm and height of cone = 8 cm
.’. Volume of the cone=1/3πr²h
=1/3π x 5 x 5 x 8 =200/3π cm³
Let radius of spherical lead ball =r cm
Volume of lead ball=4/3πr³ cm³
A.T.Q., volume of 100 lead balls=1/4 x volume of cone
100 x 4/3πr³=1/4 x 200/3π
400r³=50
r³=50/400=1/8
r=1/2=0.5 cm
Radius of spherical lead ball=0.5 cm
Question 32.
Milk in a container, which is in the form of a frustum of a cone of height 30 cm and the radii of whose lower and upper circular ends are 20 cm and 40 cm respectively, is to be distributed in a camp for flood victims. If this milk is available at the rate of 35 rupees per litre and 880 litres of milk is needed daily for a camp, find how many such containers of milk are needed for a camp and what cost will it put on the donor agency for this. What value is indicated through this by the donor agency
Solution:
Given, R1 = 20 cm, R2 = 40 cm, height = 30 cm, where R1 = radius of lower end,
R2 = radius of bigger end at top
Volume of the container open at top=1/3πh[R1²+R2²+R1R2]
=1/3π x 30(20²+40²+20 x 40)
=28000π=28000 x 22/7= 88000 cm³
Now, 880 liters=880 x 1000 cm³
=880000 cm³
Number of container of milk=880000/88000=10
Cost of 880 l of milk= 880 x 35 =30800 rupees
The moral value depicted is kindness
2014
Very Short Answer Type Question [1 Mark]
Question 33.
If the total surface area of a solid hemisphere is 462 cm², find its volume.
Solution:
Total surface area of solid hemisphere=2πr²+πr²=3πr²
Now given 3πr²=462
r²=462 x 7/3 x 22 =49
r=7cm
volume of hemisphere=2/3πr³=2/3 x 22/7 x (7)³=2/3 x 22/3 x 22 x 49 cm³
=718.67 cm³
Short Answer Type Questions II [3 Marks]
Question 34.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled competely
Solution:
Internal diameter of pipe = 20 cm
So,internal radius of pipe=10cm=1/10m
In 1 hour, 4 km = 4000 m length of water which flows in the tank.
So, volume of water in 1 hour which flows out =πr²h=π(1/10)² x 4000=40πm³
Diameter of cylindrical tank = 10 m
So, radius of cylindrical tank = 5m ,
Height of cylindrical tank = 2 m
.’. Volume of cylinder tank =πr²h=π x 5 x 5 x 2 =50π m³
Time taken to fill the tank=volume of the tank/volume of the water in 1hr
=50π/40π=5/4 hrs=1 hr 15 minutes
Question 35.
A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/12 cm, find the length of wire
Solution:
Question 36.
The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left
Solution:
Question 37.
Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/h. How much area will it irrigate in 10 minutes, if 8 cm of standing water is needed for irrigation?
Solution:
Given; canal is 6 m wide, 1.5 m deep
and in 1 hour, 4 km length of water flows out.
.’. Volume of water flows out in 1 hour = l x b x h = 6 x 1.5 x 4000 m³ = 36000 m³
.’. Volume of water flows out in 10 minutes=36000/60 x 10 = 6000 m³—(1)
Suppose this water irrigates X m² of area and we require 8 cm of standing water.
.’. Volume of water required = Area of cross-section x Length= X x 8/100 m³—-(2)
from 1& 2,we get
X x 8/100 =6000⇒X=6000 x 100/8 =75000 m²
∴75000 m² of area is irrigated
Question 38.
In Figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter; a semicircle is added on outside the region. Find the area of the shaded region
Solution:
Triangle AED is right-angled at E.
∴AD²=AE²+ED²
AD=√9²+12²=√81+144=√225=15 cm
BC=AD=15 cm
Now, Area of shaded portion = Area of rectangle + Area of Semicircle – Area of triangle AED
= l x b x 1/2πr²-1/2 x base x height
=[20 x 15 +1/2π(15/2)²-1/2 x 9 x 12] cm²
=[300+1/2 x 3.14 x (15/2)²-54] cm²
=[246 + 88.31] cm²= 334.31 cm²
Question 39.
A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m.Find the cost of cloth used at the rate 25 rupees /metre.
Solution:
Base diameter of conical tent = 14 m and height = 24 m.
∴Slant height(l)=√(7)²+(24)² m
=√49+576 m =√625 m =25 m
Area of the cloth required for conical tent=πrl =22/7 x 7 x 25 =550 m²
Width of cloth=5 m
Length of cloth required =550/5 =110 m
Cost of cloth = 25 rupee x 110 =2750 rupees
Question 40.
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 24 cm, then find its slant height correct upto one place of decimal
Solution:
For cylindrical bucket:
Base radius = 18 cm and height = 32 cm.
For cone: Height = 24 cm.
Let base radius = r cm
When cylinder is converted into cone, then their volumes are equal.
.’. Volume of cone = Volume of cylinder
=1/3πr² x 24 = π(18)² x 32
=> r² = (18)² x 4
r = 18 x 2 = 36 cm
Radius of the base of cone = 36 cm
Slant height (l) =√(24)² + (36)²cm = √576+1296 cm
[∴Using Pythagoras theorem l = √h² + r²]
= √1872 cm = 43.26 cm = 43.3 cm
Long Answer Type Questions [4 Marks]
Question 41.
Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which 2/5 th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant
Solution:
Question 42.
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid
Solution:
Question 43.
150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.
Solution:
Let rise in water level in cylinder be h cm when 150 spherical marbles,
each of diameter 1.4 cm are dropped and fully immersed.
Then volume of 150 spherical marbles=volume of water raised in cylinder
150 x 4/3π(0.7)³ = π(7/2)²h
150 x 4/3 x 7 x 7 x 7/1000 =7 x 7/4 x h
h= 4 x 4 x 7/20=28/5 =5.6 cm
water level rise by 5.6 cm
Question 44.
A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of rupee 21 per litre. 22
Solution:
Radius of lower end (r1) = 8 cm Radius of upper end (r2) = 20 cm
Height of frustum = 24 cm
Volume of the container, in form of frustum of cone,
=πh/3[r1²+r2²+ r1r2]
= 22/7 x 24/3[(8)2 + (20)2 + 8 x 20]
=22/7 x 8 [64 + 400 + 160] =22/7 x 8 x 624 cm³= 15689.14 cm3
= 15.68914 L
.’. Cost of milk which can completely fill the container = 21 x 15.68914 = 329.47 rupees.
Question 45.
A hemispherical depression is cut out from one face of a cubical block of side 7 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of remaining solid
Solution:
Edge of the cube=7 cm
Diameter of hemisphere =7 cm
Now,surface area of remaining solid=surface area of 6 faces of cube+surface area of hemisphere-surface area of circular top of diameter 7 cm
6(edge)²+2πr²-πr²
=6(7)²+2πr(7/2)²-π(7/2)²
=[6 x 49+22/7 x 49/4] cm² =[294 +38.5]cm² =332.5 cm²
Question 46.
A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find:
- the volume of water which can completely fill the bucket
- the area of the metal sheet used to make the bucket.
Solution:
2013
Short Answer Type Questions II [3 Marks]
Question 47.
A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter, the diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel.
Solution:
Diameter of hemispherical bowl = 14 cm
Radius of hemispherical bowl = 7 cm
Height of cylinder = 13 cm – 7 cm = 6 cm
Radius of cylinder = 7 cm Now,
Total surface area of vessel
= Curved Surface Area of hemispherical bowl +
curved surface area of cylinder
= 2πr² + 2πrh = 2π x 7 x 7 + 2π x 7 x 6
= 98π + 84π = 182π cm²
=26 x 22 = 572 cm²
Question 48.
A wooden toy was made by scooping out a hemispherical of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy.
Solution:
Radius of cylinder r = 3.5 cm ; Height of cylinder h = 10 cm Radius of hemisphere = 3.5 cm
Remaining volume of wood in toy
= Volume of cylinder – Volume of two hemispherical scoops
=πr²h-2 x 2/3πr³
=π x (3.5)² x [10-4/3 x 3.5]
=22/7 x 35/10 x 35/10 x 16/3 =1232/6=205.33 cm³
Question 49.
A toy is in the form of a cone mounted on a hemisphere of same radius 7 cm. If the total height of the toy is 31 cm, find its total surface area.
Solution:
Radius of hemisphere = height of hemisphere = 7cm
Height of cone = total height – (height of hemisphere)
= 31 – 7 = 24 cm
Slant height of cone =l =√h²+r²=√(24)²(7)²
=√576+49=√625=25 cm
Curved surface area of cone = πrl =22/7 x 7 x 25 = 550 cm²
Curved surface area of hemisphere = 2πr²
= 2 x 22/7 x 7 x 7 = 308 cm²
Total surface area = 550 + 308 = 858 cm²
Question 50.
A solid cone of base radius 10 cm is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone
Solution:
Question 51.
A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, find its width
Solution:
Diameter of sphere radius of sphere=8 cm
Volume of sphere=4 cm
Volume of sphere=4/3π(4)³=4/3π(64) cm³
Length of cylindrical wire, h=12 m=1200 cm
Radius of cylindrical wire=r
Volume of cylindrical wire =πr²h=π(r)² x 1200 cm³
volume of sphere= volume of cylinder
4/3π(64)=π(r)² x 1200
4 x 64/3 x 1200 =r²
64/(3 x 3) x (100)=r²
r=8/3 x 10=8/30=0.26 cm
Width of wire = 0.26 cm
Question 52.
The total surface area of a solid cylinder is 231 cm². If the curved surface area of this solid cylinder is 2/3 of its total surface area, find its radius and height
Solution:
Let radius of the base of cyclinder = r cm and height = h cm.
Total surface area = 2πr (r + h ) = 231 cm².
Curved surface area = 2πrh
Curved surface area =2/3 Total surface area
2πrh=2/3 x 2πr(r+h)⇒2πrh=2/3 x 231
2πrh=154 cm²—-(1)
2πr(r+h)=231
2πr+2πrh=231
2πr²+154=231
2πr²=231-154=77
2 x 22/7 x r² =77⇒r²=77 x 7/22 x 2⇒r=7/2 cm=3.5 cm =radius
From (1)
2πrh=154
2 x 22/7 x 7/2 x h =154
Height=154/22=7 cm
Long Answer Type Questions [4 Marks]
Question 53.
Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Solution:
Diameter of cylindrical pipe = 2 cm
Radius of cylindrical pipe, r = 2/2 = 1 cm
Rate of flowing of water in 1 sec = 0.4 m/s = 40 cm/sec.
So, volume of water in 1 sec = πr²h
= π x 1 x 1 x 40 = 40π cm³
volume of the water in 1/2 hr=40π x 1800 =7200π cm³
Radius of cylindrical tank = 40 cm
Let h’ be the rise in level of water in the tank.
So, A.T.Q.
Volume of tank = Volume of flowing water
π x 40 x 40 x h’ = 72000π
h’=72000/40 x 40=45 cm
Hence, rise in level of water = 45 cm
Question 54.
A bucket open at the top and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of rupee 10 per 100 cm2,
Solution:
Question 55.
Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/hr
Solution:
Diameter of cylindrical pipe = 7 cm
Radius of cylindrical pipe = 7/2 cm
Volume of flowing water in 1 min = 192.5 litre = 192.5 x 1000 cm³ = 192500 cm³
Volume of flowing water in 1 hour = 192500 x 60 cm³
Let ‘h’ be the length covered by flowing water in cylindrical pipe in 1 hour
Volume of water flowing through cylindrical pipe in 1 hour=πr²h=22/7(7/2)² x h
192500 x 60 x 7 x 4/22 x 49 =h
h=300000 cm=300000/100000=3 km
Rate of flowing water in 1hr =3km/hr
Question 56.
A container open at the top and made up of metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the cost of metal sheet used to make the container, if it costs ? 10 per 100 cm².
Solution:
Radius of upper end of frustum r1 = 20 cm
Radius of lower end of frustum r2 = 8 cm
Height of frustum = 16 cm
Slant height of frustum =l = √h² + (r1 – r2)²
=√(16)2+ (20-8)² = √256 +144 = √400 = 20 cm
Curved surface area of frustum open at top
= πt(r1 + r2)l + πr2²
=22/20 (20 + 8)20 + 22/7 (8)²
=22/7[28 x 20 + 64]
= 22/7[560 + 64] = 22/7 x 624 cm²
Total cost of metal sheet used = 10/100 x 22/7 x 624 = 196.141 rupees.
Question 57.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled ?
Solution:
Question 58.
A bucket open at the top is of the form of a frustum of a cone. The diameters of its upper and lower circular ends are 40 cm and 20 cm respectively. If a total of 17600 cm3 of water can be filled in the bucket, find its total surface area
Solution:
2012
Short Answer Type Questions I [2 Marks]
Question 59.
The volume of a hemisphere is 2425 1/2 cm ³ . Find its curved surface area.
Solution:
Question 60.
A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.
Solution:
Question 61.
A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal,then find the ratio of the radius and the height of the conical part
Solution:
Short Answer Type Questions II [3 Marks]
Question 62.
From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid.
Solution:
Question 63.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then find the radius and slant height of the heap.
Solution:
Let,Height of cylinder, H=32 cm
Radius of cylinder, R=18 cm
Height of cone, h =24 cm
Radius of cone, r =?
Volume of bucket=Volume of heap
πR²H=1/3πr²h
18 x 18 x 32=1/3 x r² x 24
18 x 18 x 32/8=r²
1296=r²
Radius of heap, r =√1296=36 cm
slant height of heap,l=√h²+r²=√24²+36²=√576+1296=√1872=43.27 cm
Question 64.
A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied in a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.
Solution:
Hemispherical Bowl:
Radius, R = 9 cm
Cylindrical Vessel:
Radius, r = 6 cm
Let, height = h cm
Content of hemispherical bowl is completely transferred to cylindrical vessel.
So, Volume of cylindrical vessel = Volume of hemispherical bowl
=> πr²h = 2/3πR³
6 x 6 x h =2/3 x 9 x 9 x 9 => h = 2 x 9 x 9 x 9/3 x 6 x 6= 13.5 cm
.’. Height of water in cylindrical vessel = 13.5 cm
Question 65.
A sphere of diameter 6 cm is dropped into a cylindrical vessel, partly filled with water, whose diameter is 12 cm. If the sphere is completely submerged in water, by how much will the surface of water be raised in the cylindrical vessel?
Solution:
Diameter of sphere is 6 cm, so radius of sphere i.e. r =6/2 = 3 cm Diameter of the cylinder is 12 cm.
.’. Radius (R) =12/2 = 6 cm
Let the height of cylinder is H after increasing the water level.
Now, A.T.Q., Volume of sphere = Volume of cylinder
4/3πr³=πR²H
4/3 x (3)³=(6)².H
H=4x3x3x3/3x6x6=1cm
Water level will be raised by 1 cm.
Long Answer Type Questions [4 Marks]
Question 66.
A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid
Solution:
Let r be the common radius of each of them and h be the height of the cone.
.’. r = 3.5 cm, h = 6 cm
Volume of the solid = Volume of the cone +Volume of the hemisphere
=1/3πr²h+2/3πr³
=1/3×22/7×3.5×3.5×6+2/3×22/7×3.5×3.5×3.5
=1/3×22/7x 3.5 x 3.5 [6 + 2 x 3.5]
= 1/3 x 22 x 0.5 x 3.5 [6 + 7]
=1/3x 22 x 0.5 x 3.5 x 13 = 166.83 cm³
Question 67.
A container shaped like a right circular cylinder having base radius 6 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm and radius 3 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Solution:
Let radius of cylinder, R = 6 cm
Height, H = 15 cm
Radius of cone, r = 3 cm
Height of cone, h = 12 cm
Number of cones=Volume of cylinder/Volume of cone
πR²H/1/3πr²h=3 x 6 x 6 x 15/3 x 3 x 12
n = 15
Number of cones filled with ice-cream = 15
Question 68.
A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket.
Solution:
r1=28 cm,r2=21 cm,h=?
volume=28.49l
=28.49 x 1000 =28490 cm³
v=1/3π(r1²+r2²+r1r2 )h
28490=22/3 x 7[(28)²+(21)²+28 x 21]h
28490 x 3 x 7/22=(784+441+588)h
27195=1813h
h=27195/1813=15 cm
Height of bucket=15 cm
Question 69.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid
Solution:
Let r be the common radius of each of them and h be the height of the cone.
r = 7 cm, h = 2r = 2×7 = 14cm
Volume of the solid = Volume of the cone +Volume of the hemisphere
=1/3πr²h+2/3πr³=1/3xπxr²[h+2r]
=1/3×22/7×7[14+14]
=1/3x22x7x28=1437.33 cm³
Question 70.
A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas use in making the tent, if the breadth of the canvas is 1.5 m.
Solution:
Question 71.
A hemispherical tank, full of water, is emptied by a pipe at the rate of litres per sec. How much time will it take to empty half the tank if the diameter of the base ofthe tank is 3 m?
Solution:
Question 72.
A drinking glass is in the shape of the frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass
Solution:
Question 73.
A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm3 and its diameter is 7 cm, then find the height of the toy
Solution:
Question 74.
The radii of internal and external surface of a hollow spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder
Solution:
It is given that r = 3 cm and R = 5 cm, where R–>radius of external surface, r–>radius of internal surface,
Question 75.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass
Solution:
2011
Short Answer Type Questions I [2 Marks]
Question 76.
Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid
Solution:
Length of resulting cuboid, l = 4 cm + 4 cm = 8 cm
Breadth, b = 4 cm, Height, h = 4 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2 (8 x 4 + 4 x 4 + 8 x 4) = 160 cm²
Question 77.
A toy is in the shape of a solid cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 21 cm and 40 cm respectively, and the height of cone is 15 cm, then find the total surface area of the toy
Solution:
Question 78.
From a solid cylinder of height 20 cm and diameter 12 cm, a conical cavity of height 8 cm and radius 6 cm is hollowed out. Find the total surface area of the remaining
Solution:
Question 79.
A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere
Solution:
Height of the cone, H = 20 cm
Radius of base, R = 5 cm Let radius of the sphere be ‘r’.
A.T.Q. Volume of the cone = Volume of the sphere
1/3πR²H=4/3πr³
5 x 5 x 20 =4xr³
=>5x5x20/4 = r³
=> r³ = 5x5x5
=> r = 5 cm
Diameter of the sphere = 2 x 5 = 10 cm
Question 80.
Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid
Solution:
Let ‘a’ be the side of each cube of volume 27 cm³ then a³ = 27
=> a = 3 cm
When two cubes are joined end to end to form a solid then cuboid will be formed.
length of cuboid, l = 6 cm = 3 cm + 3 cm
breadth of cuboid, b = 3 cm
height of cuboid, h = 3 cm
Total surface area of the cuboid = 2 (lb + bh + hL) = 2(6 x3 + 3×3 + 3×6)
= 2(18 + 9 + 18) = 2 x 45 = 90 cm²
Question 81.
The dimensions of a metallic cuboid are 100 cm x 80 cm x 64 cm. It is melted and recast into a cube. Find the surface area of the cube
Solution:
Dimensions of the metallic cuboid are 100 cm x 80 cm x 64 cm
Metallic cuboid is recasted into a cube.
Then, Volume of cuboid = Volume of cube
=> 100 x 80 x 64 = a³ (where a is the side of cube)
=>³√100 x 80 x 64 = a
³√512000 = a
> a = 80 cm
Now, Surface area of the cube = 6a² = 6(80)² = 6 x 80 x 80 = 38400 cm²
Question 82.
A wooden article was made by scooping out a hemisphere of radius 7 cm, from each end of a solid cylinder of height 10 cm and diameter 14 cm. Find the total surface area of the article
Solution:
Short Answer Type Questions II [3 Marks]
Question 83.
The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm3, then find the value of r
Solution:
Question 84.
An open metal bucket is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the cost ofmilk which can completely fill the bucket at ? 30 per litre
Solution:
Question 85.
From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid
Solution:
Cylinder:
Height, H = 14 cm; Diameter = 7 cm; Radius, R = 7/2 cm
Cone:
height, h = 4 cm; radius, r = 2.1 cm
Volume of remaining solid = Volume of cylinder – Volume of 2 cones
= πR²H – 2 x 1/3πr²
=22/7×7/2×7/2×14-2×1/3×22/7×2.1×2.1×4
= 539 – 36.96 = 502.04 cm³
Question 86.
The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm respectively and its height is 8 cm. Find its total surface area
Solution:
Long Answer Type Questions [4 Marks]
Question 87.
From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid
Solution:
Question 88.
Water is flowing at the rate of 10 km/hour through a pipe of diameter 16 cm into a cuboidal tank of dimensions 22 m x 20 m x 16 m. How long will it take to fill the empty tank
Solution:
Question 89.
Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
Solution:
Question 90.
A farmer connects a pipe of internal diameter 20 cm, from a canal into a cylindrical tank in his field, which is 10 m in diameter and 4 m deep. If water flows through the pipe at the rate of 5 km/hour, in how much time will the tank be filled?
Solution:
Question 91.
Water is flowing at the rate of 6 km/h through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of the water in the tank will rise by 7 cm.
Solution:
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Question 92.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted to form a cone of base diameter 8 cm. Find the height and the slant height of the cone.
Solution:
2010
Very Short Answer Type Questions [1 Mark]
Question 93.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm respectively. Find the curved surface area of the frustum
Solution:
Question 94.
The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, write the height of the frustum
Solution:
Question 95.
The slant height of a frustum of a cone is 10 cm. If the height of the frustum is 8 cm, then find the difference of the radii of its two circular ends
Solution:
Short Answer Type Question II [3 Marks]
Question 96.
The rain-water collected on the roof of a building, of dimensions 22 m X 20 m, is drained into a cylindrical vessel having base diameter 2 m and height 3.5 m. If the vessel is full up to the brim, find the height of rain-water on the roof
Solution:
Long Answer Type Questions [4 Marks]
Question 97.
A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 3/7 cm³ The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of rupee 1.40 per square centimeter.
Solution:
Question 98.
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy.
Solution:
Question 99.
The difference between the outer and inner curved surface areas of a hollow right circular cylinder, 14 cm long, is 88 cm². If the volume of metal used in making the cylinder is 176 cm³, find the outer and inner diameters of the cylinder
Solution:
Question 100.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.
Solution:
Question 101.
A container, open at the top, and made of a metal sheet, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper ends as 7 cm and 14 cm respectively. Find the cost of milk which can completely fill the container at the rate of rupee 25 per litre. Also, find the area of the metal sheet used to make the container.
Solution:
Question 102.
A solid copper sphere of surface area 1386 cm² is melted and drawn into a wire of uniform cross-section. If the length of the wire is 31.5 m, find the diameter of the wire
Solution:
Question 103.
A solid right circular cone of diameter 14 cm and height 4 cm is melted to form a hollow hemisphere. If the external diameter of the hemisphere is 10 cm, find its internal diameter. Also find the total curved surface area of the hemisphere.
Solution:
