RD Sharma 2022 Solutions for Class 10 Maths Chapter 14 Surface Areas And Volumes are provided here with simple step-by-step explanations. These solutions for Surface Areas And Volumes are extremely popular among class 10 students for Maths Surface Areas And Volumes Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2022 Book of class 10 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2022 Solutions. All RD Sharma 2022 Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 632:

Question 1:

How many balls, each of radius I cm, can be made from a solid sphere of lead of radius 8 cm?

Answer:

We are given a solid sphere with radius cm.
From this sphere we have to make spherical balls of radius cm.

Let the no. of balls that can be formed be.

We know,
Volume of a sphere =.
So, volume of the bigger solid sphere= …… (a)

Volume of one smaller spherical ball = …… (b)

We know, the volume of the solid sphere should be equal to the sum of the volumes of the n spherical balls formed.

So, using (a) and (b), we get,

Therefore,

Hence, the no. of balls of radius that can be formed out of solid sphere of radius is 512.

Page No 632:

Question 2:

How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm × 1 m × 5 dm?

Answer:

We are given a metallic block of dimension

We know that,

So, the volume of the given metallic block is

We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.
We know,

Volume of a sphere

Here,

Let the no. of bullets formed be n.
We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.

Hence the no. of bullets that can be formed is 8400.

Page No 632:

Question 3:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two of the balls are 1.5 cm and 2 cm respectively. Determine the diameter of the third ball.

Answer:

We have one spherical ball of radius 3 cm

So, its volume …… (a)

It is melted and made into 3 balls.
The first ball has radius 1.5 cm

So, its volume …… (b)

The second ball has radius 2 cm

So, its volume …… (c)

We have to find the radius of the third ball.
Let the radius of the third ball be

The volume of this third ball …… (d)

We know that the sum of the volumes of the 3 balls formed should be equal to the volume of the given spherical ball.

Using equations (a), (b), (c) and (d)

Hence the diameter of the third ball should be 5 cm

Page No 632:

Question 4:

Find the number of metallic circular discs with 1.5 cm base diameter and of height  0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm .

Answer:

Given the diameter of the base of the the circular disc = 1.5 cm
Height = 0.2 cm
Volume of the circular disc = πr2h=π×1.522×0.2=π×0.752×0.2              ...(i)
Height of the cylinder = 10 cm
Diameter = 4.5 cm
Volume of the cylinder = πR2H=π4.522×10=π×2.252×10             ...ii
Now since the circular discs are used to make the cylinder so, let n be the number of circular discs required.
n×Volume of circular disc=Volume of cylinderVolume of cylinderVolume of circular disc=nπ×2.252×10π×0.752×0.2=nn=450
Hence, 450 metallic circular discs need to be melted to form the right circular cylinder.


 

Page No 632:

Question 5:

  How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimension 
6cm × 42 cm × 21 cm . 

Answer:

The dimensions of the solid rectangular lead piece is 66 cm×42 cm×21 cm.
Diameter of the spherical lead shots = 4.2 cm
Let n spherical lead shots be obtained from the rectangular piece. 
n×volume of spherical lead shot=Volume of the rectangular lead pieceVolume of the rectangular lead piecevolume of spherical lead shot=n66×42×2143πr3=n66×42×2143π4.223=n5821238.808=nn=1500
Hence, 1500 lead shots can be formed. 

DISCLAIMER: There is some error in the question given. Instead of 6 cm, there should be 66 cm. 
The result obtained is by taking 66 cm as the dimensions of the rectangular piece.
 

Page No 632:

Question 6:

Three cubes  of a metal whose edges are in the ratios 3 : 4 : 5 are melted and converted into a single cube whose diagonal is 123 cm . Find the edges of the three cubes.

Answer:

The three cubes of metal are in the ratio 3 : 4 : 5. 
Let the edges of the cubes be 3x, 4x and 5x.
Volume of the three cubes will be 
V1=3x3V2=4x3V3=5x3
Diagonal of the single cube = 123 cm
We know diagonal of the cube = a3=123
Hence, the side of the cube = 12 cm
Volume of the bigger cube Vb=123
Volume of the three cubes = Volume of the single  
3x3+4x3+5x3=12327x3+64x3+125x3=1728216x3=1728x3=1728216=8x=2
Hence, the edges of the three cubes  will be 3×2, 4×2, 5×2=6,8,10 cm

Page No 632:

Question 7:

A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm.  Find the number of cones so formed.

Answer:

Radius of the solid metallic sphere, r = 10.5 cm
Radius of the cone, R = 3.5 cm
Height of the cone, H = 3 cm
Let the number of smaller cones formed be n. 
Volume of the metallic sphere, Vs=43πr3=43π10.53
Volume of the cone, Vc=13πR2H=13π3.52×3
Let the number of cones thus formed be n.
n×volume of the cone=volume of the spherevolume of the sphere(Vs)volume of the coneVc=n43π10.5313π3.52×3=n126=n
Hence, 126 cones are thus formed.
 

Page No 632:

Question 8:

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Answer:

The radius of the copper rod is 0.5 cm and length is 8 cm. Therefore, the volume of the copper rod is

Let the radius of the wire is r cm. The length of the wire is 18 m=1800 cm. Therefore, the volume of the wire is

Since, the volume of the copper rod is equal to the volume of the wire; we have

V1=Vπr2×1800=π×0.52×8r2=0.25×81800=1900r=130=0.033 cm

Hence, the radius of the wire is 0.033 cm = 0.33 mm.
So, thickness = 0.33×2=0.66 mm

Page No 632:

Question 9:

How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm × 10 cm × 7 cm?

Answer:

The dimension of the cuboid is 11cm10cm7cm. Therefore, the volume of the cuboid is

The radius and thickness of each coin arecm and 2mm = 0.2cm respectively. Therefore, the volume of each coin is

Since, the total volume of the melted coins is same as the volume of the cuboid; the number of required coins is

Page No 632:

Question 10:

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 (Use π = 22/7).

Answer:

The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is r. Therefore, we have

Therefore, the radius of the metallic sphere is 7 cm and the volume of the sphere is

The sphere is melted to recast a cone of height 28 cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is

Since, the volumes of the sphere and the cone are same; we have

Hence, the diameter of the base of the cone so formed is two times its radius, which is cm.

Page No 632:

Question 11:

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer:

Let the radius of the cone by r
Now, Volume cylindrical bucket = Volume of conical heap of sand
π18232=13πr22418232=8r2r2=18×18×4r2=1296r=36 cm
Let the slant height of the cone be l.
Thus , the slant height is given by
l=242+362=576+1296=1872=1213 cm

Disclaimer: The answer given in the book for the slant height is not correct.

Page No 632:

Question 12:

A solid cuboid of iron with dimensions 53 cm ⨯ 40 cm ⨯ 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe. [CBSE 2015]

Answer:

Volume of solid cuboid of iron = Volume of cylindrical pipe
lbh=πhR2-r253×40×15=227×h822-72253×40×15=227×h42-3.5253×40×15=227×h×3.75h=2698.18 cm


Disclaimer: The answer given in the book is not correct.

Page No 632:

Question 13:

The 34th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.

Answer:

Radius of conical vessel r = 5 cm

Height of conical vessel h = 24 cm

The volume of water = 34×volume of conical vessel.
=34×13πr2h=34×13π×25×24=150π

Let h' be the height of cylindrical vessel, which filled by the water of conical vessel,

Radius of cylindrical vessel = 10 cm

Clearly,

Volume of cylindrical vessel = volume of water
π102h=150πh=150π100πh=1.5 cm

Thus, the height of cylindrical vessel is 1.5 cm.

Page No 632:

Question 14:

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. [CBSE 2014]

Answer:

Let the rise in the level of water in the vessel be h cm.
Now, Volume of 150 spherical marbles = Volume of water displaced in the vessel

150×43×227×1.423=227×722×h200×0.73=722×hh=5.6 cm



Page No 633:

Question 15:

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 25th 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?
[CBSE 2014]

Answer:

Let the number of the balls be n.
Volume of water flows out = Volume of n spherical bolls
25×13πR2h=n×43πr325×2.52×11=n×40.58327.5=0.58nn=440

Page No 633:

Question 16:

16 glass spheres each of radius  2 cm are packed into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water . Find the volume of the water filled in the box .

Answer:

Radius of the glass spheres, r = 2 cm
Dimensions of the cuboidal box = 16cm×8cm×8cm
volume of the spheres = Vs=43πr3=43π23
volume of the cuboidal box = Vc=16×8×8=1024
Volume of water in the cuboidal box = Volume of the cuboidal box − Volume of the 16 glass spheres
=1024-16×43π23=1024-536.6=487.6 cm3
Hence, the volume of the water in the cuboidal box = 487.6 cm3

Page No 633:

Question 17:

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 cm2  , find the volume of cylinder. (Use π =22 / 7 )

Answer:

Let the radius of the base of the cylinder be r cm
Let the height be h cm.
Now given that r + h = 37 cm
Total surface area = 1628 cm2             ...(i)
Total surface area of the cylinder = 2πr2+2πrh=2πrr+h=2πr×37=74πr                ...ii
From equation (i) and (ii) we get
74πr=1628r=162874π=7 cm3
Thus, the height will be 37 − 7 = 30 cm
Thus, the volume of the cylinder = πr2h=π72×30=4620 cm3
Hence the volume is 4620 cm3
 

Page No 633:

Question 18:

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 of Rs 25 per metre. [Use π = 22/7] [CBSE 2014]

Answer:

Let the slant height of the cone be l.
Thus , the slant height is given by
l=1422+242=49+576=625= 25 m
Now, the curved surface area of the tent is given by
227×142×25=550 m2
The curved surface area will be equal to the area of the cloth
550=Length×5Length=110 m
Now, the cost of cloth is given by
110 ⨯ 25
= Rs 2750

Page No 633:

Question 19:

The volume of a hemi-sphere is 242512 cm3. Find its curved surface area. (Use π = 22/7)

Answer:

Let the radius of the hemisphere be r cm.
Volume of hemisphere = 242512 cm3
23πr3=4851223×227r3=48512r3=4851×3×72×2×22r3=441×212×2×2r3=2123r=212 cm
Now, the curved surface area of hemisphere is given by
2πr2=2×227×2122=693 cm2

Page No 633:

Question 20:

If the total surface area of a solid hemisphere is 462 cm2, find its volume (Take π = 22/7 ) [CBSE 2014]

Answer:

Let the radius of the hemisphere be r cm.
Total surface area of hemisphere = 462 cm2
3πr2=4623×227×r2=462r2=49r=7 cm
Now, the volume of hemisphere is given by
23πr3=23×22773=21563 =71823 cm3

Page No 633:

Question 21:

A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone. 

Answer:

Height of the cone, h = 120 cm
Radius of the cone, r = 60 cm

Height of the cylinder, H = 180 cm
Radius of the cylinder, R = 60 cm

Now,
Volume of the cylinder = πR2H=π602×180 cm3
Volume of the cone = 13πr2h=13π602×120

Therefore,
Volume of water left in the cylinder = Volume of cylinder − volume of the cone 
                                                         =π602×180-13π602×120=π602180-40=π×3600140=227×3600×140=1584000 cm3=1.584 m3

Page No 633:

Question 22:

A heap of rice in the form of a cone of diameter 9 m  and height 3.5 m. Find the volume of rice. How much canvas cloth is required to cover the heap  ?

Answer:

The heap of rice is in the form of a cone.
Diameter, d = 9 m
radius, r = 92m
height, h = 3.5 m
Volume, V=13πr2h=13π922×3.5=74.25 m3
Thus, volume of rice = 74.25 m3
The canvas cloth required to cover the heap will be the curved surface area of the cone
l=h2+r2l=3.52+922l=12.25+20.25l=5.7 m
CSA=πrl=π×92×5.7=80.62 m2
Hence, the canvas cloth required to cover the heap will be 80.62 m2

Page No 633:

Question 23:

A factory manufactures 120,000 pencils daily . The pencil are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm . Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at â‚¹0.05 per dm2.

Answer:

Length of the pencil, h = 25 cm
circumference of the base = 1.5 cm
Curved surface area of the pencil which needs to be painted will be
CSA=circumference×height=1.5×25 cm2=37.5 cm2
= 0.375 dm2
Pencils manufactured in one day = 120000
So, the total area to be painted will be 120000×0.375 dm2=45000 dm2
Cost of painting this area will be 45000×0.05=Rs 2250
 

Page No 633:

Question 24:

A well of diameter 2 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm. Find the width of the embankment.

Answer:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

The well is 14m deep. Thus, the height of the solid right circular cylinder is m.

Therefore, the volume of the solid right circular cylinder is

Since, the embankment is to form around the right circular cylinder. Let the width of the embankment be x m. The height of the embankment is h = 40 cm = 0.4 m. Therefore, the volume of the platform is

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

Hence, x = 5

Hence,

Page No 633:

Question 25:

Water flows through a cylindrical pipe , whose inner radius is 1 cm , at the rate of 80 cm /sec in an empty cylindrical tank , the radius of whose base is 40 cm . What is the rise of water level in tank in half an hour ? 

Answer:

The inner radius of the cylindrical pipe r =1 cm.

Rate of flow of water = 80 cm/sec

volume of the water that flows through pipe in 1sec is πr2×80=80π cm3 

volume of the water that flows through pipe in half an hour 80π×30×60=144000π cm3 

radius of the base of the cylindrical tank is R = 40 cm

let the water level in the cylinderical tank after half an hour be h

volume of the raised water = πR2h=π402h

volume of the raised water in tank = volume of the water that flows through pipe

π402h=144000πh=1440001600=90 cm

Thus water level will rise by 90 cm in half an hour.

Page No 633:

Question 26:

Water in a canal 1.5 m wide and 6 m deep is flowing with a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

Answer:

The canal is 1.5 m wide and 6 m deep. The water is flowing in the canal at 10 km/hr. Hence, in 30 minutes, the length of the flowing standing water is

Therefore, the volume of the flowing water in 30 min is

Thus, the irrigated area in 30 min of 8 cm=0.08 m standing water is

Page No 633:

Question 27:

A cylindrical tank full of water is emptied by a pipe at the rate of 225 litres per minute. How much time will it take to empty half the tank, if the diameter of its base is 3 m and its height is 3.5 m? [Use π=227]        [CBSE 2014]

Answer:

Given that, the cylindrical tank has the diameter of its base as 3 m and its height as 3.5 m.

Thus,
Volume of cylindrical tank = πr2h
                                            = 227×322×3.5
                                            = 24.75 m3
Now, 1 m3 = 1000 L
∴ 24.75 m3 = 24750 L
Half the capacity of tank = 12375 L

Now,
Time taken by the pipe to empty 225 litres = 1 minute
Time taken by the pipe to empty 1 litre = 1225 minutes

Therefore,
Time taken by the pipe to empty 12375 litres = 1225×12375
                                                                         = 55 minutes

Page No 633:

Question 28:

Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of the 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.                                                        [CBSE 2015]

Answer:

Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = πr2h
                                                                             = π × (40)2 × 315
                                                                             = 5,04,000 π cm3
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 ÷ 2 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = πd22×126000
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour
 
πd22×126000=504000πd22=4 d=4 cm

Thus, the internal diameter of the pipe is 4 cm.

Page No 633:

Question 29:

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. (Use π = 22/7)

Answer:

The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder are r cm and R cm respectively. The difference between the outer and inner surface area of the hollow cylinder is

By the given condition, this difference is 88 square cm. Hence, we have

The volume of the metal used in making the cylinder is

By the given condition, the volume of the metal is 176 cubic cm. Hence, we have

Hence, we have two equations with unknowns R and r

Adding the two equations, we have

Then from the second equation, we have

Therefore, the outer and inner diameters of the hollow cylinder are 5cm and 3cm respectively.



Page No 652:

Question 1:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.

Answer:

We have the following diagram

For cylindrical part, we have

Therefore, the curved surface area of the cylinder is given by

For conical part, we have

Therefore, the curved surface area of the conical part is

For hemisphere, we have

Therefore the surface area of the hemisphere is

The total surface area of the toy is

Hence, total surface area of the toy is

Page No 652:

Question 2:

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub (Take π = 22/7)

Answer:

To find the volume of the water left in the tube, we have to subtract the volume of the hemisphere and cone from volume of the cylinder.

For right circular cylinder, we have

The volume of the cylinder is

For hemisphere and cone, we have

Therefore the total volume of the cone and hemisphere is

The volume of the water left in the tube is

Hence, the volume of the water left in the tube is

Page No 652:

Question 3:

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Rs 10 per dm2 .

Answer:

We have a solid composed of cylinder with hemispherical ends.

Radius of the two curved surfaces

Height of cylinder is h.

Total height of the body

So, total surface area is given by,

Change the units of curved surface area as,

Cost of polishing the surface is Rs 10 per.

So total cost,

Page No 652:

Question 4:

A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer:

We have to find the inner surface area of a vessel which is in the form of a hemisphere mounted by a hollow cylinder.

Radius of hemisphere and cylinder

Total height of vessel

So, the inner surface area of a vessel,

Page No 652:

Question 5:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.  The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:

We have to find the total surface area of a toy which is a cone surmounted on a hemisphere.

Radius of hemisphere and the base of the cone

Height of the cone,

h = 15.5 − 3.5 = 12 cm
slant height (l)=h2+r2=122+3.52=156.25=12.5 cm

So, total surface area of toy,

S=πrl+2πr2=πrl+2r=227×3.512.5+2×3.5=214.5 cm2

Page No 652:

Question 6:

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder (ii) left in the cylinder. (Take π 22/7)

Answer:

We have a cylindrical vessel in which a cone is inserted. We have,

Radius of the cylinder

Radius of cone

Height of cylinder

Height of cone

(i) We have to find the volume of water displaced from the cylinder when cone is inserted.

So,

So volume of water displaced,

(ii) We have to find the volume of water remaining in the cylinder.

So volume of the water left in the cylinder,

Page No 652:

Question 7:

A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block.

Answer:

We have to find the remaining volume and surface area of a cubical box when a hemisphere is cut out from it.

Edge length of cube

Radius of hemisphere

Therefore volume of the remaining block,

So,

So, remaining surface area of the box,

Therefore,

Put the values to get the remaining surface area of the box,

Page No 652:

Question 8:

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 the base of the cone is 21 cm and its volume is 2/3 of the volume of hemisphere, calculate the height of the cone and the surface area of the toy.
(Use π=22/7).

Answer:

Solution:

Let the height of the conical part be h.

Radius of the cone = Radius of the hemisphere = r = 21 cm

The toy can be diagrammatically represented as

 

Volume of the cone =

Volume of the hemisphere =

According to given information:

Volume of the cone × Volume of the hemisphere

Thus, surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

Page No 652:

Question 9:

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them is being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. (Use π = 22/7).

Answer:

Height of cone = 9.5 − 3.5 = 6 cm
Volume of the solid = Volume of cone + Volume of hemisphere
=13πr2h+23πr3=13×227×3.52×6+23×227×3.53=77+89.83=166.83 cm3

Page No 652:

Question 10:

The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. (Use π=227). [CBSE 2014]

Answer:

The radius of the largest possible sphere is carved out of a wooden solid cube is equal to the half of the side of the cube.
Radius of the sphere = 72=3.5
Volume of the wood left = Volume of cube − Volume of sphere
=Side3-43πr3=73-43×227×3.53=343-179.67=163.33 cm3

Page No 652:

Question 11:

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. (Take π=227). [CBSE 2014]

Answer:

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

=2πrh+πrl+πr2=2×227×4.22×2.8+227×4.22×4.222+2.82+227×4.222=2×227×2.1×2.8+227×2.1×2.12+2.82+227×2.12=2×227×2.1×2.8+227×2.1×3.5+227×2.12=36.96+23.1+13.86=73.92 cm2

The total surface area of the remaining solid is 73.92 cm2.

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Question 12:

The largest cone is curved out from one face of solid cube of side 21 cm. Find the volume of the remaining solid. [CBSE 2015]

Answer:

The radius of the largest possible cone is carved out of a solid cube is equal to the half of the side of the cube.
Also, the height of the cone is equal to the side of the cube.
Radius of the cone = 212=10.5 cm
Volume of the remaining solid  = Volume of cube − Volume of cone
=Side3-13πr2h=213-13×227×10.52×21=9261-2425.5=6835.5 cm3

Disclaimer: The answer given in the book is not correct.

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Question 13:

A pen stand made of wood is in the shape of a cuboid with four conical depression and a cubical depression to hold the pens and pins , respectively . The dimension of the cuboid are 10 cm × 5 cm×4 cm .  The radius of each of the conical depression is 0.5 cm  and the depth is 2.1 cm . The edge of the cubical depression is 3 cm . Find the volume of the wood in the entire stand. 

Answer:

The dimensions of the cuboid = 10 cm × 5 cm × 4 cm
Volume of the total cuboid = 10 cm × 5 cm × 4 cm = 200 cm3 
Radius of the conical depressions, r = 0.5 cm
Depth, h = 2.1 cm
Volume of the conical depression = 13πr2h=13π0.522.1=0.5495 cm3
Edge of cubical depression, a = 3 cm
Volume of the cubical depression = a3=33=27 cm3
Volume of wood used to make the entire stand = Volume of the total cuboid − volume of conical depression − volume of cubical depression
=200-4×0.5495-27=170.802 cm3 

 

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Question 14:

A solid toy s in the form of a hemisphere  surrounded by a right circular cone . The height of cone is 4 cm and the diameter of the base is 8 cm . Determine the volume of the toy. If a cube circumscribes the toy , then find the difference of the volumes of cube and the toy .

Answer:

The height of the cone, h = 4 cm
Diameter of the base, d = 8 cm
Radius of the cone, r = 4 cm
lateral side will be
l=h2+r2l=42+42=42
Volume of the toy = volume of hemisphere + volume of cone
V=23πr3+13πr2h=πr232r+h=π4232×4+4=201.14 cm3
When the cube circumscribes the toy, then 
Volume of the cube = a3=83=512 cm3
Volume of cube-volume of toy=512-201.14=310.86 cm3
Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cone
=2πr2+πrl=2π42+π×4×42=π422+2=171.68 cm2


 

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Question 15:

A circus tent is in the shape of cylinder surmounted by a conical top of same diameter. If their common diameter is 56 m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent.

Answer:


Total height of the tent above the ground = 27 m
Height of the cylinderical part, h1= 6 m
Height of the conical part, h2 = 21 m
Diameter = 56 m
Radius = 28 m
Curved surface area of the cylinder, CSA1 = 2πrh1=2π×28×6=336π
Curved surface area of the cylinder, CSA2 will be
πrl=πrh2+r2=π×28×212+282=28π441+784=28π×35=980π
Total curved surface area = CSA of cylinder + CSA of cone
= CSA1 + CSA2
=336π+980π=1316π=4136 m2
Thus, the area of the canvas used in making the tent is 4136 m2 .

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Question 16:

 A cone of maximum size is carved out from a cube of edge 14 cm . Find the surface area of the cone and of the remaining solid left out after the cone carved out .

Answer:

The base of the largest right circular cone will be the circle inscribed in a face of the cube and its height will be equal to an edge of the cube.

Given, Edge of the cube = 14 cm

Radius of base of the cone, r=142=7 cm

Height of the cone, h = 14 cm

Slant height of the cone, l=h2+r2
l=72+142l=49+196=245=75 cm

Surface area of the cone 
=πrr+l=π77+75=1541+5 cm2
Surface area of the remaining solid = Surface area of the cube − surface area of the cone
=6a2-13πr2h=6×142-1541+5=1022+1545 cm2


 

Page No 652:

Question 17:

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water . Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm .

Answer:

Diameter of the marbles = 1.4 cm
Radius of the marbles, r = 1.42=0.7 cm
Diameter of the cylinderical beaker = 7 cm
Radius of the beaker = 72=3.5 cm
Rise in the level of water = 5.6 cm
let the number of marbles be n.
n×volume of the marbles=volume of the water risenn=volume of the water risenvolume of the marblesn=π7225.643π0.73=150
Hence, the number marbles will be 150.
 



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Question 18:

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape formed.

Answer:

Radius, r = 8 cm
Height, h = 15 cm
l=h2+r2l=152+82l=225+64l=289=17 cm
When the two cones are joined through their bases so, the total surface area of the shape formed will be sum of the curved surface areas
of the two cones.
Curved surface area of the cone
 CSA=πrl=π817=427.4 cm2
similarly curved surface area of the other cone will also be the same.
Hence, the total surface area of the solid formed = 427.4 + 427.4 = 854.8 cm2

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Question 19:

From a solid cube of side 7 cm , a conical cavity of height 7 cm and radius 3 cm is hollowed out . Find the volume of the remaining solid.

Answer:

Side length of the cube, a = 7 cm
Height of the cone, h = 7 cm
radius, r = 3 cm
Volume of the remaining solid = Volume of the cube − volume of the cone
V=a3-13πr2hV=73-13π32×7V=343-66=277 cm3

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Question 20:

The difference between outside and inside surface areas of cylindrical metallic pipe 14 cm long is 44 m2. If the pipe is made of 99 cm3 of metal, find the outer and inner radii of the pipe.

Answer:

We have to find the outer and inner radius of a hollow pipe.

Radius of inner pipe be

Radius of outer cylinder be

Length of the cylinder

Difference between the outer and the inner surface area is 44

So,

So,

…… (1)

So, volume of metal used is 99 , so,

Use equation (1) in the above to get,

Therefore,

…… (2)

Solve equation (1) and (2) to get,

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Question 21:

A solid wooden toy is in the form of hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 16656 cm3. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of â‚¹ 10 per cm2Take π =227.

Answer:

Let h be the height of the cone and r be the radius of the base of cone.
r=3.5 cm
Volume of the wooden toy = 13πr2h+23πr3 
                                           = 13πr2(h+2r) 
                                           = 13×227×3.5×3.5(h+7) 
                                           = 776h+7 
According to the question,
776h+7=16656776h+7=10016h=6 

The height of the wooden toy = 6 cm + 3.5 cm
                                                     = 9.5 cm

Now,
Curved surface area of the hemispherical part = 2×227×3.5×3.5 
                                                                            = 77 cm2

Hence, the cost of painting the hemispherical part of the toy = 77×10 
                                                                                                 = Rs 770

Page No 653:

Question 22:

A wall 24 m , 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm × 16 cm × 10 cm . If the mortar occupies  110th of the volume of the wall, then find the number of bricks used in constructing the wall.

Answer:

Dimensions of the wall are 24 m × 0.4 m × 6 m
Volume of the wall = 24 m × 0.4 m × 6 m = 57.6 m3
Dimensions of the bricks are 25 m × 16 m × 10 m
Volume of each brick = 4000 cm3 = 0.004 m
Volume of mortar = 110×Volume of the wall=110×57.6=5.76 m3
Volume of all the bricks = Volume of the wall − Volume of mortar
=57.6 - 5.76= 51.84 m3
Let the number of bricks used in making the wall be n.
Volume of all the bricksVolume of each brick=n51.840.004=nn=12960
Hence, 12960 bricks are used to make the wall.

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Question 23:

 A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and contains 411921 m3  of air. If the internal diameter of dome is equal to its total height  above the floor , find the height of the building ?

Answer:

let the total height of the building be H m.

let the radius of the base be r m. Therefore the radius of the hemispherical dome is r m.

 Now given that internal diameter = total height
2r=H

Total height of the building = height of the cylinder +radius of the dome
⇒ H = h + r

⇒ 2r = h + r
⇒ r = h

Volume of the air inside the building = volume of the cylinder+ volume of the hemisphere
411921=πr2h+23πr388021=πh2h+23πh388021=πh31+2388021=πh353h=2 m

Hence, height of the building H = 2 × 2 = 4m

 

 

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Question 24:

A building is in the form of a cylinder surrounded by a hemispherical dome. The base diameter of the dome is equal to 23 of the total height of the building . Find the height of the building , if it contains  67121m3 of air.

Answer:


Let the radius of the dome be r. 
Diameter be d.
Let the height of the building be H.
Given d=23H
2r=23Hr=H33r=H
Also, h + r = H
3r=h+r2r=hr=h2
Volume of air = Volume of air in the cylinder + Volume of air int he hemispherical dome
πr2h+23πr3=67121πh22h+23πh23=140821h3=11264176=64h=4 m
Hence, the radius will be r=h2=42=2 m
Height of the building, H = 3r = 3×2=6 m

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Question 25:

Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities are 2: 1 . Find the heights and capacities of the cones . Also, find the volume of the remaining portion of the cylinder.

Answer:

V1 : V2 = 2 : 1
Diameter of the cylinder = 6 cm
Radius, r = 3 cm
Height of the cylinder = 21 cm
Let the height of one cone be H. 
So, the height of the other cone will be 21 − H. 
V1V2=π32Hπ3221-H21=H21-H42-2H=HH=14 cm
Height of one of the cones will be 14 cm and of the other will be 21 − H = 21 − 14 = 7 cm
Volume of cone with height 14 cm = V1=π32×14=396 cm3
Volume of cone with height 7 cm = V2=13π32×7=66 cm3
Volume of the remaining portion of the cylinder = Volume of the cylinder-volume of cone 1-volume of cone 2
V=π32×21-396-66=594-396-66=132 cm3 

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Question 26:

An icecream cone full of icecream having radius 5 cm and height 10 cm as shown in the figure. 16.77. Calculate the volume of icecream, provided that its 1/ 6 part is left unfilled with icecream .

Answer:

Ice cream above the cup is in the form of a hemisphere
So, volume of the ice above the cup = 23πr3=23π53 cm3
Volume of the cup = 13πr2h=13π525=13π×125
Now, 1/6 part of the total is left unfilled. So, 5/6 is filled. 
So, the volume of ice cream
 =56Volume of hemispherical cup+volume of cone=562×125π3+125π3=56×125π32+1=327.38 cm3

Page No 653:

Question 27:

In the given figure, from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.

Answer:

Surface area of the remaining block = Surface area of the cuboid + Curved surface area of cylinder − 2 × Area of base of cylinder

= 215×10+10×5+15×5+2×227×72×5-2×227×72×72 
= 583 cm2

Thus, the surface area of the remaining block is 583 cm2.



Page No 665:

Question 1:

A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m , the height of the frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent.
(Take : π = 22/7)
 

Answer:

The height of the frustum cone is h = 8 m. The radii of the end circles of the frustum are r1 = 13m and r2 =7m.

The slant height of the frustum cone is

The curved surface area of the frustum is

The base-radius of the upper cap cone is 7m and the slant height is 12m. Therefore, the curved surface area of the upper cap cone is

Hence, the total canvas required for the tent is

Hence total canvas is

Page No 665:

Question 2:

 A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the  bucket. Also, find the cost of milk which can completely fill the container , at thr rate of â‚¹25 per litre. (Use π = 3.14).

Answer:

Height of the bucket, h = 30 cm
Radii r1 = 10 cm and r2 = 20 cm
Capacity of the bucket,
V=13πhr12+r1r2+r22=13π×30102+10×20+202=21980 cm3=21.980 liters

l=h2+r2-r12l=302+20-102=1010
Surface area of the bucket
S=CSA+area of the baseS=πr1+r2l+πr12S=π10+201010+π102S=2978.86+314=3292.86 cm2
Cost of milk which can completely fill the container at Rs 25/litre
= 21.980 × 25
= Rs 549.50

Page No 665:

Question 3:

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3 of water. 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 the metal sheet used in its making. (Use π = 3.14).

Answer:

Let the depth of the bucket is h cm. The radii of the top and bottom circles of the frustum bucket are r1 =20cm and r2 =12cm respectively.

The volume/capacity of the bucket is

Given that the capacity of the bucket isCubic cm. Thus, we have

Hence, the height of the bucket is

The slant height of the bucket is

The surface area of the used metal sheet to make the bucket is

Hence

Page No 665:

Question 4:

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area.

Answer:

The slant height of the frustum of a cone is l=10cm. The radii of the upper and lower circles of the bucket are r1 =33cm and r2 =27cm respectively.

The total surface area of the frustum of the cone is

Hence total surface area is

Page No 665:

Question 5:

A solid is in the shape of a frustum of a cone, The diameters of the two circular ends are 60 cm and 36 cm and the height is 9 cm. Find the area of its whole surface and the volume.

Answer:

The height of the frustum of a cone is h=9cm. The radii of the upper and lower circles of the frustum of the cone are r1 =30cm and r2 =18cm respectively.

The slant height of the frustum of the cone is

The volume of the frustum of the cone is

The total surface area of the frustum of the cone is

Page No 665:

Question 6:

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 Rs. 10 per 100 cm2 .
(Use π = 3.14).

Answer:

The slant height of the bucket is given by
l=h2+R-r2=242+15-52=576+100=676=26 cm
Surface area of bucket
= Curved surface area of bucket +  Area of the smaller circlular base
=πlR+r+πr2=3.14×26×15+5+3.14×5×5=1632.8+78.5=1711.3 cm2
Cost of metal sheet used = 10100×1711.3=Rs 171.13



Page No 666:

Question 7:

A bucket, made of metal sheet, is in the form of a cone whose height is 35 cm and radii of circular ends are 30 cm and 12 cm. How many litres of milk it contains if it is full to the brim? If the milk is sold at Rs 40 per litre, find the amount received by the person.

Answer:

The given bucket is in the form of the frustum of a cone.
Height, h = 35 cm
r1 = 30 cm
r2 = 12 cm
Volume=π3hr12+r22+r1r2=π3×35302+122+30×12=π3×35900+144+360=π3×35×1404=51433.2 cm3
= 51.4 litres
Selling price of the milk = Rs 40/litre
So, selling price of 51.4 litres of milk will be 51.4×40=Rs 2056

Page No 666:

Question 8:

A milk container is made of metal sheet in the shape of frustum of cone whose volume is 1045937cm3 . 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 Rs. 1.40 per cm2 .
(Use π = 22/7)

Answer:

Let the depth of the container is h cm. The radii of the top and bottom circles of the container are r1 =20cm and r2 =8cm respectively.

The volume/capacity of the container is

Given that the capacity of the bucket is. Thus, we have

cm

Hence, the height of the container is 16 cm.

The slant height of the container is

The surface area of the used metal sheet to make the container is

The cost to make the container is

Page No 666:

Question 9:

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 in the volumes of two parts of the cone.

Answer:



Let the height of the cone be H.
Now, the cone is divided into two parts by the parallel plane
∴ OC = CAH2
Now, In ∆OCD and OAB
∠OCD = OAB          (Corresponding angles)
∠ODC = OBA          (Corresponding angles)
By AA-similarity criterion ∆OCD ∼ ∆OAB
CDAB=OCOACD10=H2×HCD=5 cm
Volume of first partVolume of second part=13πCD2OC13πCAAB2+ABCD+CD2=52102+105+52=25100+50+25=25175=17CDR=H2×HCD=R2
Volume of smaller coneVolume of whole cone=13π(CD)2OC13π(AB)2OA=(R2)H2R2H=18R2HR

Page No 666:

Question 10:

In the given figure, from the top of a solid cone of 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. (Use π=227 and 5=2.236). [CBSE 2015]

Answer:



Now, In ∆OCD and OAB
∠OCD = OAB          (Corresponding angles)
∠ODC = OBA          (Corresponding angles)
By AA-similarity criterion ∆OCD ∼ ∆OAB
CDAB=OCOACD6=412CD=2 cm
The slant height of the bucket is given by
l=h2+R-r2=82+6-22=64+16=80=45 cm
Surface area of the remaining solid
= Curved surface area of figure +  Area of the smaller circle + Area of the larger circle
=πlR+r+πR2+πr2=227×45×6+2+227×6×6+227×2×2=227×45×8+227×6×6+227×2×2=224.88+113.14+12.57=350.59 cm2

Page No 666:

Question 11:

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

Answer:

We have,Radius of the solid cone, R=CPHeight of the solid cone, AP=HRadius of the smaller cone, QD=rHeight of the smaller cone, AQ=hAlso, AQ=AP2 i.e. h=H2 or H=2h          .....iNow, in AQD and APC,QAD=PAC        Common angleAQD=APC=90°So, by AA criteriaAQD~APCAQAP=QDPChH=rRh2h=rR                 Using i12=rRR=2r                .....iiAs,Volume of smaller cone=13πr2hAnd,Volume of solid cone=13πR2H=13π2r2×2h           Using i and ii=83πr2hSo,Volume of frustum=Volume of solid cone-Volume of smaller cone=83πr2h-13πr2h=73πr2hNow, the ratio of the volumes of the two parts=Volume of the smaller coneVolume of the frustum=13πr2h73πr2h=17=1:7

So, the ratio of the volume of the two parts of the cone is 1 : 7.

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Question 12:

A reservoir in the form of the frustum of a right circular cone contains 44 × 107 litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take: π = 22/7)

Answer:

Let the depth of the frustum cone like reservoir is h m. The radii of the top and bottom circles of the frustum cone like reservoir are r1 =100m and r2 =50m respectively.

The volume of the reservoir is

Given that the volume of the reservoir is. Thus, we have

Hence, the depth of water in the reservoir is

The slant height of the reservoir is

The lateral surface area of the reservoir is

Hence, the lateral surface area is

Page No 666:

Question 13:

A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base . Compare the volumes of two parts. 

Answer:

Radius of the cone, R = 4 cm
Let the height be H. 
Since the plane divides the cone into two parts through the mid point so, a small cone and a frustum will be formed.

OC=CA=H2

Let the radius of the smaller cone be r cm.

In ∆OCD and ∆OAB,

OCD = ∠OAB  (90°)

COD = ∠AOB  (Common)

∴∆OCD ∼ ∆OAB  (AA Similarly criterion)
OAOC=ABCD=OBOD      corresponding sides are proportionalHH2=4rr=2 cm

Now volume of the smaller cone = 13πCD2×OC=13π22×H2=2πH3 cm
Height of the frustum of the cone = H2

Volume of frustum of cone = 13πhr12+r1r2+r22
=13πH242+22+4×2=14πH3 cm3
Volume of the smaller coneVolume of the frustum of the cone=2πH314πH3Volume of the smaller coneVolume of the frustum of the cone=17
Hence, the ratio of the volumes of the two parts will be 1 : 7.


 

Page No 666:

Question 1:

The radii of the base of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?

Answer:

Let r1 and r2 be the radii of the base of a cylinder and a cone.

The volume of cylinder …… (i)

The volume of cone …… (ii)

Dividing (i) by (ii), the, we get

Page No 666:

Question 2:

If the heights of two right circular cones are in the ratio 1 : 2 and the perimeters of their bases are in the ratio 3 : 4, what is the ratio of their volumes?

Answer:

Given that,

Therefore,

The ratios of volume of their cones will be

Page No 666:

Question 3:

If a cone and a sphere have equal radii and equal volumes. What is the ratio of the diameter of the sphere to the height of the cone?

Answer:

Given that,

A cone and a sphere have equal radii and equal volume

i.e., volume of cone = volume of sphere

Page No 666:

Question 4:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?

Answer:

Let r be the radius of the base.

And h is the height.

Here, h = r.

Now,

The ratio of their volumes will be

Volume of cone: volume of hemisphere: volume of a cylinder

Page No 666:

Question 5:

The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their curved surface areas?

Answer:

Given that,

Now, the ratio of their curved surface area

Page No 666:

Question 6:

Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas?

Answer:

The rate of the value of cubes = 1:27

…… (i)

Now,

The ratio of their surface area

Hence,

Page No 666:

Question 7:

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?

Answer:

Let r1 and r2 be the radii of two right circular cylinders and h1 and h2 be the heights.

Since,

Both the cylinder has the same volume.

Therefore,



Page No 667:

Question 8:

If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then write the ratio of their weights.

Answer:

The ratio of the volume of cones

And

i.e.,

Hence,

Page No 667:

Question 9:

A sphere and a cube have equal surface areas. What is the ratio of the volume of the sphere to that of the cube?

Answer:

Surface area of sphere = sphere area of cube

i.e., 4πr2 = 6a2

…… (i)

 

Page No 667:

Question 10:

What is the ratio of the volume of a cube to that of a sphere which will fit inside it?

Answer:

Ratio of sphere

Now,

Volume of cube

Volume of sphere

The ratio of their volumes

Page No 667:

Question 11:

What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

Answer:

Given that the diameter and the height of the cylinder, cone and sphere are the same.
The volume of cylinder, =πd22d

The volume of cone, =13πd22d

And the volume of sphere, =43πd23

Therefore,

The ratio of their volumes,

v1=v2=v3πd22d=13πd22d=43πd233:1:2
 

Hence, the ratio is 3 : 1 : 2

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Question 12:

The surface area of a sphere is 616 cm2 . Find its radius.

Answer:

The surface area of sphere = 616k cm2

We know that

Taking squire root both the side

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Question 13:

A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone

Answer:

Since, cylinder and a cone both are have same radius and height.

Therefore,

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Question 14:

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.

Answer:

Slant height of the Frustum = 5 cm

Squaring both sides we get

Height of the Frustum = 3 cm

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Question 15:

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?

Answer:

Let the radius of the hemisphere be r units.
Volume of a hemisphere = Surface area of the hemisphere
23πr3=3πr223r=3r=92d=9 units
Hence, diameter of the hemisphere is equal to 9 units.

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Question 16:

The radii of two cones are in the ratio 2 : 1 and their volumes are equal. What is the ratio of their heights?

Answer:

Let the radius of the first cone = 2x

And height of the first cone = h1

Then,

The radius of the second cone = x

Height of the second cone = h2

Then,

Since,

The volumes of the two cones are equal.

Or

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Question 17:

Two cones have their heights in the ratio 1 : 3 and radii 3 : 1. What is the ratio of their volumes?

Answer:

Let the radius of the cone is 3x and x,

And the height of the cone is y and 3y.

Then,

Volume of the first cone

Volume of the second cone

Then the radius of their volume

Or

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Question 18:

A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces?

Answer:

The base of the cone and hemisphere are equal. So radius of the two is also equal.

and

Height of the hemisphere = height of the cone

Then the slant height of the cone

Now, the curved surface area of

Hemisphere

and

The curved surface area of cone

Putting the value of l from eq. (i)

We get

Now,

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Question 19:

If r1 and r2 denote the radii of the circular bases of the frustum of a cone such that r1 > r2, then write the ratio of the height of the cone of which the frustum is a part to the height fo the frustum.

Answer:

Since,

Therefore,

In

Hence,

The ratio of the height of cone of which the frustum is a part to the height fo the frustum.

OVOO'=hh1=r1r1-r2

Hence,

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Question 20:

If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum?

Answer:

The parameter of upper base

The parameter of lower base

The surface area of frustum

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Question 21:

If the areas of circular bases of a frustum of a cone are 4 cm2 and 9 cm2 respectively and the height of the frustum is 12 cm. What is the volume of the frustum?

Answer:

Area of circular bases of frustum is

The height of frustum h = 12 cm

Now, the volume of frustum

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Question 22:

A sphere of maximum volume is cut-out from a solid hemisphere of radius r, what is the ratio of the volume of the hemisphere to that of the cut-out sphere?

Answer:

Since, a sphere of maximum volume is cut out from a solid hemisphere of radius.

i.e., radius of sphere

Therefore,

The volume of sphere

…… (i)

The volume of hemisphere …… (ii)

Divide (i) by (ii).

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Question 23:

A metallic hemisphere is melted and recast in the shape of a cone with the same base radius R as that of the hemisphere. If H is the height of the cone, then write the values of HR.

Answer:

Given,

Radius of the hemisphere = Radius of the cone.

Now,

Volume of the hemisphere

and

Volume of the cone

Volume of the hemisphere = volume of the cone

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Question 24:

A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.

Answer:

Given that

i.e.

Since,

Right, circular cone and right circular cylinder have equal base and equal right.

Therefore,

The total surface area of cylinder

The total surface area of cone

Now,

Hence,

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Question 25:

A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes?

Answer:

Let the diameter of the base for all three be x cm and height be y cm.

For hemisphere radius

Height

(As height of the hemisphere is equal to the radius of hemisphere)

For cone

Radius

Height

(As height is same for all)

For cylinder

Radius

Height

The ratio of their volume is

= cylinder volume : conic volume : hemispherical volume



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Question 1:

A funnel is the combination of ____________ of a cone and __________.

Answer:




A funnel is the combination of ___frustum___ of a cone and __cylinder__.

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Question 2:

A cylindrical pencil sharpened at one edge is the combination of a _________and  a _________.

Answer:




A cylindrical pencil sharpened at one edge is the combination of a __cone__ and  a __cylinder__.

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Question 3:

A surahi is the combination of a _______ and a ________.

Answer:




A surahi is the combination of a __cylinder__ and a __sphere__.

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Question 4:

A plumb line (sahul) is the combination of a _________ and a _________.

Answer:





A plumb line (sahul) is the combination of a __cone__ and a ___hemisphere___.

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Question 5:

The shape of a glass (tumbler) is usually in the form of __________.

Answer:

The shape of a glass (tumbler) is usually in the form of frustum of a cone.

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Question 6:

The shape of a gilli, in the gilli-danda game, is a combination of _______and a ________.

Answer:

The shape of a gilli, in the gilli-danda game, is a combination of two cones and a cylinder.

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Question 7:

A shuttle cock used for playing badminton has the shape of the combination of ________ and _________.

Answer:

A shuttle cock used for playing badminton has the shape of the combination of hemisphere and frustum of a cone.

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Question 8:

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called a _________.

Answer:

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called a frustum of cone.

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Question 9:

Two identical solid cubes of side a are joined end to end. The total surface area of the resulting cuboid is _________.

Answer:

Length of the resulting cuboid, l = a + a = 2a

Breadth of the resulting cuboid, b = a

Height of the resulting cuboid, h = a

∴ Total surface area of the resulting cuboid

= 2(lb + bh + hl)

= 2(2a × a + a × a + a × 2a)

= 2(2a2 + a2 + 2a2)

= 2 × 5a2

= 10a2

Two identical solid cubes of side a are joined end to end. The total surface area of the resulting cuboid is ____10a2_____.

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Question 10:

A spherical steel ball of radius r is melted to make eight new identical balls. The radius of each new ball is _________.

Answer:

Let the radius of each new ball be R.

∴ Volume of spherical steel ball = 8 × Volume of each new ball

43πr3=8×43πR3R3=r38=r23R=r2

A spherical steel ball of radius r is melted to make eight new identical balls. The radius of each new ball is          r2        .

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Question 11:

If a solid cone of base radius r and height h is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the solid so formed is __________.

Answer:

Slant height of the cone, l = r2+h2

Curved surface area of the solid

= Curved surface area of cone + Curved surface area of cylinder

=πrl+2πrh=πrl+2h=πrr2+h2+2h

If a solid cone of base radius r and height h is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the solid so formed is       πrr2+h2+2h         .

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Question 12:

A solid ball is exactly fitted inside the cuboidal box of side a. The volume of the ball is _________.

Answer:

If a solid ball is exactly fitted inside the cuboidal box, then the diameter of the ball is equal to the side of the cuboidal box.

Let the radius of the ball be r units.

2r=ar=a2

Volume of the ball = =43πr3=43πa23=16πa3

A solid ball is exactly fitted inside the cuboidal box of side a. The volume of the ball is          16πa3      .

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Question 13:

A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is _______.

Answer:

Total height of the new cylinder, H = h + h = 2h

Radius of the new cylinder = r

∴ Total surface area of the shape (or new cylinder)

=2πrH+2πr2=2πrH+r=2πr2h+r

A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is     2πr2h+r    .

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Question 14:

Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is _________.

Answer:

When two identical solid hemispheres of equal base radius are stuck together along their bases, then the new solid so formed is a sphere. The radius of the sphere is same as the base radius of each solid hemisphere.

Radius of the sphere = r cm

∴ Total surface area of the combination = Total surface area of sphere = 4πr2 cm2

Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is      4πr2 cm2    .

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Question 15:

The radii of the ends of the frustum of a cone are r1 and r2 cm and If h cm is the height of the frustum, then the volume of the frustum of the cone in cm3 is _________.

Answer:

The radii of the ends of the frustum of a cone are r1 and r2 cm and if h cm is the height of the frustum, then the volume of the frustum of the cone in cm3 is     13πhr12+r22+r1r2    .

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Question 16:

If A1, and A2, denote areas of circular bases of the frustum of a cone and h is its height, then the volume of the frustum is _________.

Answer:

Let r1 and r2 be the radius of the top and bottom faces of the frustum of cone.

A1=πr12r1=A1π      .....(1)
and
A2=πr22r2=A2π         .....(2)

Volume of frustum of cone

=13πhr12+r22+r1r2=13πhA1π+A2π+A1πA2π                        Using 1 and 2=13hA1+A2+A1A2

If A1, and A2, denote areas of circular bases of the frustum of a cone and h is its height, then the volume of the frustum is 13hA1+A2+A1A2.

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Question 17:

The radii of the ends of a frustum of a cone are r1 and r2 cm. If h cm is the height of the frustum then the height of the cone of which the frustum is a part, is ________.

Answer:


Let the height of the cone be H cm.


Here, ∆OAB ~ ∆OCD

OAOC=ABCDH-hH=r1r2hH=1-r1r2
hH=r2-r1r2H=r2r2-r1h

The radii of the ends of a frustum of a cone are r1 and r2 cm. If h cm is the height of the frustum then the height of the cone of which the frustum is a part, is     r2r2-r1h   .

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Question 18:

The radii of the ends of a frustum of a cone are r1 and r2. If l is the slant height of the frustum, then the slant height of the cone of which the frustum is apart, is ________.

Answer:


Let the slant height of the cone be L units.



Here, ∆OAB ~ ∆OCD

OBOD=ABCDL-lL=r1r21-lL=r1r2
lL=1-r1r2lL=r2-r1r2L=r2r2-r1l
The radii of the ends of a frustum of a cone are r1 and r2. If l is the slant height of the frustum, then the slant height of the cone of which the frustum is apart, is     r2r2-r1l    .

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Question 19:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the following figure is ________.

Answer:

From the given figure, the radius of the hemispherical portion is same as the radius of the cylinder.

Capacity of the cylindrical vessel with a hemispherical portion raised upward at the bottom

= Volume of the cylinder − Volume of the hemisphere

=πr2h-23πr3=πr2h-23r=13πr23h-2r

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the following figure is    13πr23h-2r   .

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Question 20:

If a marble of radius 2.1 cm is put into a cylindrical cup, full of water, of radius 5 cm and height 6 cm, then the volume of water that flows out of the cup is ________.

Answer:

Radius of the marble, r = 2.1 cm

Now,

Volume of water that flows out of the cylindrical cup

= Volume of the marble

=43πr3=43×227×2.13=38.808 cm3

If a marble of radius 2.1 cm is put into a cylindrical cup, full of water, of radius 5 cm and height 6 cm, then the volume of water that flows out of the cup is __38.808 cm3__.



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Question 21:

Three solid cubes have a face diagonal of 42 cm each. Three other solid cubes have a face diagonal of 82 cm each. All the cubes are melted together to form a cube. The side of the cube so formed is of length ___________.

Answer:

We know that each face of a cube is a square.

Let the side of solid cube having face diagonal of 42 cm each be x cm.

Diagonal of a square = 2×Side=2x

2x=42

x = 4 cm

Now, let the side of solid cube having face diagonal of 82 cm each be y cm.

2y=82

y = 8 cm

Suppose the side of the bigger cube obtained on melting the given cubes be a cm.

∴ Volume of the bigger cube = 3 × Volume of cube having side 4 cm + 3 × Volume of cube having side 8 cm

a3=3×43+3×83a3=192+1536=1728a3=123a=12 cm

Three solid cubes have a face diagonal of 42 cm each. Three other solid cubes have a face diagonal of 82 cm each. All the cubes are melted together to form a cube. The side of the cube so formed is of length __12 cm__.

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Question 22:

In Fig. 14.78, a circle is inscribed in a square ABCD and the square is circumscribed by a circle. If the radius of the smaller circle is r сm, then the area of the shaded region in cm2 is _________.

Answer:

Side of the square = Diameter of the smaller circle = 2r

Length of diagonal of square = 2 × Side of the square = 22r

∴ Radius of the bigger circle = Length of diagonal of square2=22r2=2r

Area of the shaded region

= 14(Area of the bigger circle − Area of the square)

=14π2r2-2r2=142πr2-4r2=π-22r2 cm2

In Fig. 14.78, a circle is inscribed in a square ABCD and the square is circumscribed by a circle. If the radius of the smaller circle is r сm, then the area of the shaded region in cm2 is    π-22r2  .

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Question 23:

The volume of the greatest right circular cone, which can be cut from a cube of side 4 cm is __________.

Answer:

The greatest right circular cone will be such that the diameter of the base of the cone will be equal to the side of the cube and the height of the cone will be equal to the side of the cube.

Radius of the greatest right circular cone which can be cut from the given cube, r = 42 = 2 cm

Height of the greatest right circular cone which can be cut from the given cube, h = 4 cm

∴ Volume of the greatest right circular cone which can be cut from the given cube

=13πr2h=13π×22×4=163π cm3

The volume of the greatest right circular cone, which can be cut from a cube of side 4 cm is   163π cm3  .

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Question 24:

The volume of a cuboid is 2442 cm3. Its length is 52 cm, breadth and height are in the ratio 3 :7. The height of the cuboid is ________.

Answer:

Let the breadth and height of the cuboid be 3x and 7x, respectively.

Volume of the cuboid = Length × Breadth × Height

52×3x×7x=2442 cm3542x2=2442x2=245x=245 cm

∴ Height of the cuboid = 7x=7×245=25210 cm

The volume of a cuboid is 2442 cm3. Its length is 52 cm, breadth and height are in the ratio 3 :7. The height of the cuboid is   25210 cm  .

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Question 25:

The sum of the length, breadth and height of a cuboid is 53  cm and length of its diagonal is 35 cm, then its total surface area is ___________.

Answer:

Let the length, breadth and height of the cuboid be l, b and h, respectively.

Given:

 l+b+h=53 cm       .....(1)

Length of diagonal of cuboid = l2+b2+h2=35 cm

l2+b2+h2=45 cm2       .....(2)

Now,

l+b+h2=l2+b2+h2+2lb+2bh+2hl2lb+bh+hl=53 cm2-45 cm2           Using 1 and 22lb+bh+hl=75-45=30 cm2

Total surface area of the cuboid = 2lb+bh+hl=30 cm2

The sum of the length, breadth and height of a cuboid is 53 cm and length of its diagonal is 35 cm, then its total surface area is __30 cm3_.



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