RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2

Other Exercises

Question 1.
Find, in terms of π the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4 cm.
Solution:
Radius of the circle (r) = 4 cm
Angle at the centre subtended an arc = 30°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 1

Question 2.
Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length \((\frac { 5\pi }{ 3 } \) cm.
Solution:
Radius of the circle (r) = 5 cm 571
Length of arc = \(\frac { 5\pi }{ 3 }\) cm
Let θ be the angle subtended by the arc, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 2

Question 3.
An arc of length 20tc cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.
Solution:
Length of an arc = 20π cm
Angle subtended by the arc = 144°
Let r be the radius of the circle, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 3

Question 4.
An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π ; the radius of the circle.
Solution:
Length of arc = 15 cm
Angle subtended at the centre = 45°
Let r be the radius of the circle, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 4
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 5

Question 5.
Find the angle subtended at the centre of a circle of radius ‘a’  by an arc of length \((\frac { a\pi }{ 4 } )\)  cm.
Solution:
Radius of the circle (r) = a cm
Length of arc = \(\frac { a\pi }{ 4 }\)   cm
Let θ be the angle subtended by the arc at the centre, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 6

Question 6.
A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector.
Solution:
Radius of the sector of a circle (r) = 4 cm
Angle at the centre (θ) = 30°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 7
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 8

Question 7.
A sector of a circle of radius 8 cm contains an angle of 135°. Find the area of the sector.
Solution:
Radius of the sector of the circle (r) = 8 cm
Angle at the centre (θ) = 135°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 9

Question 8.
The area of a sector of a circle of radius 2 cm is 7 is cm2. Find the angle contained by the sector.
Solution:
Area of the sector of a circle =π cm2
Radius of the circle (r) = 2 cm
Let 0 be the angle at the centre, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 10

Question 9.
The area of a sector of a circle of radius 5 cm is 5π cm2. Find the angle contained by the sector.
Solution:
Area of the sector of a circle = 5π cm2
Radius of the circle (r) = 5 cm
Let 9 be the angle at the centre, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 11

Question 10.
Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm. [NCERT Exemplar]
Solution:
Let the central angle of the sector be θ.
Given that, radius of the sector of a circle (r) = 5 cm
and arc length (l) = 3.5 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 12
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 13

Question 11.
In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.
Solution:
Radius of the circle (r) = 25 cm
Angle at the centre (θ) = 72°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 14

Question 12.
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
Solution:
Radius of the circle (r) = 5.7 m
Perimeter of the sector = 27.2 m
Length of the arc = Perimeter – 2r
= (27.2 – 2 x 5.7) m
= 27.2 – 11.4 = 15.8 m
Let θ be the central angle, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 15
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 16

Question 13.
The perimeter of a certain sector of a circle of radius 5.6 cm is 27.2 m. Find the area of the sector.
Solution:
Radius of the sector (r) = 5.6 cm
and perimeter of the sector = 27.2 cm
∴ Length of arc = Perimeter – 2r
= 27.2 – 2 x 5.6
= 27.2- 11.2= 16.0 cm
∴ θ be the angle at the centre, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 17

Question 14.
A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120°. Find the length of its arc and the area.
Solution:
Radius of the sector of a circle (r) = 21 cm
Angle at the centre = 120°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 18

Question 15.
The minute hand of a clock is \(\sqrt { 21 } \)  cm long, Find the area described by the minute hand on the face of the clock between 7.00 A.M. and 7.05 A.M.
Solution:
Length of minute hand of a clock (r) = \(\sqrt { 21 }\)  cm
Period = 7 a.m. to 7.05 a.m. 5 minutes
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 19

Question 16.
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8.00 A.M. and 8.25 A.M.
Solution:
Length of minute hand of a clock (r) = 10 cm
Period = 8 A.M. to 8.25 A.M. = 25 minutes
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 20

Question 17.
The sector of 56° cut out from a circle contains area 4.4 cm2, Find the radius of the circle.
Solution:
Area of a sector = 4.4 cm2
Central angle = 56°                                        ‘
Let r be the radius of the sector of the circle, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 21

Question 18.
Area of a sector of central angle 200° of a circle s 770²cm. Find the length of the corresponding are of this sector.
Solution:
Let the radius of the sector AOBA be r.
Given that, Central angle of sector AOBA = θ = 200°
and area of the sector AOBA = 770 cm2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 22
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 23

Question 19.
The length of minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.           [NCERT Exemplar]
Solution:
We know that, in 60 min, minute hand revolving = 360°
In 1 min, minute hand revolving = \(\frac { 360\circ }{ 60\circ }\)

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 24
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 25

Question 20.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. [CBSE 2013]
Solution:
Length of minute hand (r)= 14 cm
Area swept by the minute hand in 5 minutes
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 26

Question 21.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (0 the length of the arc (ii) area of the secter formed by the arc. (Use π = 22/7) [CBSE 2013]
Solution:
Radius of a circle (r) = 21 cm
Angle at the centre = 60°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 27

Question 22.
From a circular piece of cardboard of radius 3 cm two sectors of 90° have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters (Take π = 22/7).
Solution:
Radius of the circular piece of cardboard (r) = 3 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 28
∴ Two sectors of 90° each have been cut off
∴ We get a semicular cardboard piece
∴ Perimeter of arc ACB
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 29

Question 23.
The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector.
Solution:
Let r be the radius of the circle and 0 be the central angle of the sector of the circle Then area of circle = πr2

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 30

Question 24.
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm. Find the area of the sector of the circle formed by the chord AB.
Solution:
Radius of the circle with centre O (r) = 4 cm
Length of chord AB = 4 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 31

Question 25.
In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find 
(i)  the circumference of the circle,
(ii) the area of the circle,
(iii) the length of the arc AB,
(iv) the area of the sector OAB.
Solution:
Radius of the circle (r) = 6 cm
Length of chord = 10 cm
and central angle (θ) =110°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 32
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 33

Question 26.
Figure, shows a sector of a circle, centre O, containing an angle θ°. Prove that :
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 34
Solution:
Radius of the circle = r
Arc AC subtends ∠θ at the centre of the
circle. OAB is a right triangle
In the right ΔOAB,
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 35
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 36
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 37

Question 27.
Figure, shows a sector of a circle of radius r cm containing an angle θ°. The area of the sector is A cm2 and perimeter of the sector is 50 cm. Prove that
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 38
Solution:
Radius of the sector of the circle = r cm
and angle at the centre = 0
Area of sector OAB = A cm2
and perimeter of sector OAB = 50 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 39
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 40

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1

Other Exercises

Question 1.
On which axis do the following points lie?
(i) P (5, 0)
(ii) Q (0 – 2)
(iii) R (-4, 0)
(iv) S (0, 5)
Solution:
(i) P (5, 0)
Its ordinate or y-axis is 0. It lies on x-axis
(ii) Q (0 – 2)
Its abscissa or x-axis is 0. It lies on y-axis
(iii) R (-4, 0)
Its ordinate is 0 It lies on x-axis
(iv) S (0, 5)
Its abscissa is 0. It lies on y-axis

Question 2.
Let ABCD be a square of side 2a. Find the coordinates of the vertices of this square when
(i) A coincides with the origin and AB and AD are along OX and OY respectively.
(ii) The centre of the square is at the-origin and coordinate axes are parallel to the sides AB and AD respectively.
Solution:
ABCD is a square whose side is 2a
(i) A coincides with origin (0, 0)
AB and AD are along OX and OY respectively
Co-ordinates of A are (0, 0), of B are (2a, 0) of C are (2a, 2a) and of D are (0, 2a)
(ii) The centre of the square is at the origin (0, 0) and co-ordinates axes are parallel to the sides AB and AD respectively.
Then the co-ordinates of A are (a, a) of B are (-a, a), of C are (-a, -a) and of D are (a, -a) as shown in the figure given below :
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1 1

Question 3.
The base PQ of two equilateral triangles PQR and PQR’ with side 2a lies along y- axis such.that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R’ of the triangles.
Solution:
∆PQR and PQR’ are equilateral triangles with side 2a each and base PQ and mid of point of PQ is 0 (0, 0) and PQ lies along y-axis
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1 2

 

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1

Other Exercises

Question 1.
Find the circumference and area of a circle of radius 4.2 cm.
Solution:
Radius of a circle = 4.2 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 1

Question 2.
Find the circumference of a circle whose area is 301.84 cm2.
Solution:
Area of a circle = 301.84 cm2
Let r be the radius, then πr2 = 301.84
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 2

Question 3.
Find the area of a circle whose circum­ference is 44 cm.
Solution:
Circumference of a circle = 44 cm
Let r be the radius,
then 2πr = circumference
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 3

Question 4.
The circumference of a circle exceeds the diameter by 16.8 cm. Find the circum­ference of the circle. (C.B.S.E. 1996)
Solution:
Let r be the radius of the circle
∴  Circumference = 2r + 16.8 cm
⇒  2πr = 2r + 16.8
⇒  2πr – 2r = 16.8
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 4

Question 5.
A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7)
Solution:
Radius of the circle (r) = Length of the rope = 28 m .
Area of the place where the horse can graze
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 5

Question 6.
A steel wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.  (C.B.S.E. 1997)
Solution:
Area of square formed by a wire =121 cm2
∴ Side of square (a) = \(\sqrt { Area } \)  = \(\sqrt { 121 } \)  = 11 cm Perimeter of the square = 4 x side = 4 x 11 = 44 cm
∴Circumference of the circle formed by the wire = 44cm
Let r be the radius
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 6

Question 7.
The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.
Solution:
Let R and r be the radii of two circles and their ratio between them circumference = 2 : 3
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 7

Question 8.
The sum of radii of two circles is 140 cm and the difference of their circum­ferences is 88 cm. Find the diameters of the circles.
Solution:
Let R and r be the radii of two circles Then R + r = 140 cm  …….(i)
and difference of their circumferences
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 8
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 9

Question 9.
Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm. [NCERT Exemplar]
Solution:
Let the radius of a circle be r.
Circumference of a circle = 2πr
Let the radii of two circles are r1 and r2 whose
values are 15 cm and 18 cm respectively.
i.e., r1 = 15 cm and r2 = 18 cm
Now, by given condition,
Circumference of circle = Circumference of first circle + Circumference of second circle
⇒   2πr = 2π r1 + 2πr2 =
⇒  r = r1 + r2
⇒   r = 15 + 18
∴ r = 33 cm
Hence, required radius of a circle is 33 cm.

Question 10.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Solution:
Radius of first circle (r1) = 8 cm
and radius of second circle (r2) = 6 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 10
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 11

Question 11.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.
Solution:
Radius of the first circle (r1) = 19 cm
and radius of the second circle (r2) = 9 cm S
um of their circumferences = 2πr1 + 2πr2
= 2π (r+ r2) = 2π (19 + 9) cm
= 2π x 28 = 56π cm
Let R be the radius of the circle whose circumference is the sum of the circum­ferences of given two circles, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 12

Question 12.
The area of a circular playground is 22176 m2. Find the cost of fencing this ground at the rate of ₹50 per metre.  [NCERT Exempiar]
Solution:
Given, area of a circular playground  = 22176 m2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 13
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 14

Question 13.
The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.
Solution:
ABCD is a square whose each side is 10 cm
∴  AB = BC = CD = DA = 10 cm
AC and BD are its diagonals
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 15
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 16

Question 14.
If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
Solution:
Let r be the radius of the circle a be the side of the square
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 17
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 18

Question 15.
The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle. (Use π = 22/7 and \(\sqrt { 3 } \)  = 1.73)
Solution:
Area of the inscribed circle of ΔABC = 154 cm2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 19
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 20

Question 16.
A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be ₹2640 at the rate of ₹12 per metre. Then, the field is to be thoroughly ploughed at the cost of ₹0.50 per m2. What is the amount required to plough the field ? (Take π = 22/7)
Solution:
Cost of the fencing the circular field = ₹2640
Rate = ₹12 per metre 2640
∴ Circumference = \((\frac { 2640 }{ 12 } )\) = 220 m
Let r be the radius of the field, then = 2πr = 220
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 21

Question 17.
A park is in the form of a rectangle 120 m x 100 m. At the centre of the park there is a circular lawn. The area of park excluding lawn is 8700 m2. Find the radius of the circular lawn. (Use π = 22/7).
Solution:
Area of the park excluding lawn = 8700 m2
Length of rectangular park = 120 m
and width = 100 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 22
∴ Area of lawn = l x b
= 120 x 100 m2 = 12000 m2
Let r be the radius of the circular lawn, then area of lawn = πr2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 23

Question 18.
A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.
Solution:
Distance covered by the car in 450 revolutions = 1 km = 1000 m
∴ Distance covered in 1 revolution = \((\frac { 1000 }{ 450 } )\)
= \((\frac { 20 }{ 9 } )\) m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 24

Question 19.
The area of enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Solution:
Area of enclosed between two concentric circles = 770 cm2
Radius of the outer circle (R) = 21 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 25

Question 20.
An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.[NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 26
Solution:
Let the diameters of concentric circles be k, 2k , 3k
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 27

Question 21.
The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/hr? [NCERT Exemplar]
Solution:
Given, radius of wheel, r = 35 cm
Circumference of the wheel = 2πr
= 2 x \((\frac { 22 }{ 7 } )\) x 35 = 220 cm
But speed of the wheel = 66 kmh-1
= \((\frac { 66 x 1000 }{ 60 } )\) m/ mm
= 1100 x 100 cm min-1
= 110000 cm min-1
∴ Number of revolutions in 1 min
= \((\frac { 110000 }{ 220 } )\)= 500 revolution
Hence, required number of revolutions per minute is 500.

Question 22.
A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per m2. [NCERT Exemplar]
Solution:
Given that, a circular pond is surrounded by a wide path.
The diameter of circular pond = 17.5 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 28

Question 23.
A circular park is surrounded by a rod 21 m wide. If the radius of the park is 105 m, find the area of the road. [NCERT Exemplar]
Solution:
Given that, a circular park is surrounded by a road.
Width of the road = 21 m
Radius of the park (r1) = 105 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 29
.’. Radius of whole circular portion (park + road),
re = 105 + 21 = 126 m
Now, area of road = Area of whole circular portion – Area of circular park
= πr2 – πr2             [∵ area of circle = πr2]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 30

Question 24.
A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying outside the circle and inside the square.  [NCERT Exemplar]
Solution:
Let the side of a square be a and the radius of circle be r.
Given that, length of diagonal of square = 8 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 31

Question 25.
A path of 4 m width runs round a semi­circular grassy plot whose circumference is 81 \((\frac { 5 }{ 7 } )\)m. Find:
(i) the area of the path
(ii) the cost of gravelling the path at the rate of ₹1.50 per square metre
(iii) the cost of turfing the plot at the rate of 45 paise per m2.
Solution:
Width of path around the semicircular grassy plot = 4 m
Circumference of the plot = 81 \((\frac { 5 }{ 7 } )\)m
= \((\frac { 572 }{ 7 } )\) m
Let r be the radius of the plot, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 32
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 33

Question 26.
Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.
Solution:
Radius of first circle (r1) = 3.5 cm
Radius of second circle (r2) = 7 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 34
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 35

Question 27.
A path of width 3.5 m runs around a semi­circular grassy plot whose perimeter is 72 m. Find the area of the path. (Use π = 22/7)                   [CBSE 2015]
Solution:
Perimeter of semicircle grassy plot = 72 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 36
Let r be the radius of the plot
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 37
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 37.1

Question 28.
A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14)               [CBSE 2014]
Solution:
Diameter of circular pond (d) = 17.5 m
Radius (r) =\((\frac { 1725 }{ 2 } )\) = 8.75 m
Width of path = 2m
∴  Radius of outer cirlce (R) = 8.75 + 2 = 10.75 m
Area of path = (R2 – r2
= [(10.75)2 – (8.75)2](3.14)
= 3.14(10.75 + 8.75) (10.75 – 8.75)
= 3.14 x 19.5 x 2 = 122.46 m2
Cost of 1 m2 for constructing the path ₹25 m2
∴  Total cost = ₹ 122.46 x 25 = ₹3061.50

Question 29.
The outer circumference of a circular race-track is 528 m. The track is every­where 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use π= 22/7).
Solution:
Let R and r be the radii of the outer and inner of track.
Outer circumference of the race track = 528 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 38
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 39

Question 30.
A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.
Solution:
Width of the road = 7 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 40
Circumference of the park = 352 m
Let r be the radius, then 2πr = 352
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 41
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 42

Question 31.
Prove that the area of a circular path of uniform width surrounding a circular region of radius r is πh(2r + h).
Solution:
Radius of inner circle = r
Width of path = h
∴ Outer radius (R) = (r + h)
∴ Area of path = πR2 – πr2
= π {(r + h)2 – r2}
= π {r2 + h2 + 2rh – r2}
= π {2rh + h2}
= πh (2r + h) Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90
Solution:
(c) 7th term (a7) = a + 6d = 34
13th term (a13) = a + 12d = 64
Subtracting, 6d = 30 => d = 5
and a + 12 x 5 = 64 => a + 60 = 64 => a = 64 – 60 = 4
18th term (a18) = a + 17d = 4 + 17 x 5 = 4 + 85 = 89

Question 2.
If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be
(a) 0
(b) p – q
(c) p + q
(d) – (p + q)
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 1

Question 3.
If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 2

Question 4.
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(b) First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be the number of terms and d be the common difference, then
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 3

Question 5.
If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164 ?
(a) 26th
(b) 27th
(c) 28th
(d) none of these
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 4

Question 6.
If the sum of it terms of an A.P. is 2n2 + 5n, then its nth term is
(a) 4n – 3
(b) 3n – 4
(c) 4n + 3
(d) 3n + 4
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 5

Question 7.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :
(a) 13
(b) 9
(c) 21
(d) 17
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 6

Question 8.
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) None of these
Solution:
(a)
4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
=> 3a = 15d
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 7

Question 9.
Let S denotes the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn – k Sn-1 + Sn-2 then k =
(a) 1
(b) 2
(c) 3
(d) None of these
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 8

Question 10.
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 9
(a) S
(b) 2S
(c) 3S
(d) None of these
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 10

Question 11.
If the sum of first n even natural number is equal to k times the sum of first n odd natural numbers, then k =
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 11
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 12

Question 12.
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 13
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 14

Question 13.
If S1 is the sum of an arithmetic progression of ‘n’ odd number of terms and S2 is the sum of the terms of the
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 15
Solution:
(a)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 16
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 17

Question 14.
If in an A.P., Sn = n2p and Sm = m2p, where S denotes the sum of r terms of the A.P., then Sp is equal to
(a) \(\frac { 1 }{ 2 }\) p3
(b) mnp
(c) p3
(d) (m + n) p2
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 18

Question 15.
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn , then S3n : Sn is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 19

Question 16.
In an AP, Sp = q, Sq = p and S denotes the sum of first r terms. Then, Sp+q is equal to
(a) 0
(b) – (p + q)
(c) p + q
(d) pq
Solution:
(c) In an A.P. Sp = q, Sq = p
Sp+q = Sum of (p + q) terms = Sum of p term + Sum of q terms = q + p

Question 17.
If Sn denotes the sum of the first r terms of an A.P. Then, S3n : (S2n – Sn) is
(a) n
(b) 3n
(b) 3
(d) None of these
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 20

Question 18.
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 term is
(a) 3200
(b) 1600
(c) 200
(d) 2800
Solution:
(a)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 21

Question 19.
The number of terms of the A.P. 3, 7,11, 15, … to be taken so that the sum is 406 is
(a) 5
(b) 10
(c) 12
(d) 14
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 22

Question 20.
Sum of n terms of the series
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 23
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 24
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 25

Question 21.
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
(a) 50th
(b) 502th
(c) 508th
(d) None of these
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 26

Question 22.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 27
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 28
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 29

Question 23.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 30
Solution:
(b) Sn is the sum of first n terms
Last term nth term = Sn – Sn-1

Question 24.
The common difference of an A.P., the sum of whose n terms is Sn, is
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 31
Solution:
(a)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 32

Question 25.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 33
Solution:
(b)
In first A.P. let its first term be a1 and common difference d1
and in second A.P., first term be a2 and common difference d2, then
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 34

Question 26.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 35
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 36
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 37

Question 27.
If the first term of an A.P. is a and nth term is b, then its common difference is
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 38
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 39

Question 28.
The sum of first n odd natural numbers is
(a) 2n – 1
(b) 2n + 1
(c) n2
(d) n2 – 1
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 40

Question 29.
Two A.P.’s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th terms is
(a) 11
(b) 3
(c) 8
(d) 5
Solution:
(d) In two A.P.’s common-difference is same
Let A and a are two A.P. ’s
First term of A is 8 and first term of a is 3
A30 – a30 = 8 + (30 – 1) d – 3 – (30 – 1) d
= 5 + 29d – 29d = 5

Question 30.
If 18, a, b – 3 are in A.P., the a + b =
(a) 19
(b) 7
(c) 11
(d) 15
Solution:
(d) 18, a, b – 3 are in A.P., then a – 18 = -3 – b
=> a + b = -3 + 18 = 15

Question 31.
The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18th term is
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 41
Solution:
(a)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 42
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 43

Question 32.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 44
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 45
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 46

Question 33.
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
(a) 24th term
(b) 27th term
(c) 26th term
(d) 25th term
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 47

Question 34.
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
(a) n (n – 2)
(b) n (n + 2)
(c) n (n + 1)
(d) n (n – 1)
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 48

Question 35.
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
(a) 3 : 2
(b) 3 : 1
(c) 1 : 3
(d) 2 : 3
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 49

Question 36.
The sum of first 20 odd natural numbers is
(a) 100
(b) 210
(c) 400
(d) 420 [CBSE 2012]
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 50

Question 37.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 51
Solution:
(a)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 52

Question 38.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 53
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 54

Question 39.
The common difference of the A.P. \(\frac { 1 }{ 2b }\) ,
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 55
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 56

Question 40.
If k, 2k – 1 and 2k + 1 are three consecutive terms of an AP, the value of k is
(a) -2
(b) 3
(c) -3
(d) 6 [CBSE 2014]
Solution:
(b) (2k – 1) – k = (2k + 1) – (2k- 1)
2k – 1 – k = 2
=> k = 3

Question 41.
The next term of the A.P. , √7 , √28, √63, …………
(a) √70
(b) √84
(c) √97
(d) √112 [CBSE 2014]
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS 57
= √(l6 x 7)= √112

Question 42.
The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals
(a) -3
(b) 4
(c) 5
(d) 2 [CBSE 2014]
Solution:
(c) 2 (3y + 5) = 3y – 1 + 5y + 1
(If a, b, c are in A.P., b – a = c – b=> 2b = a + c)
=> 6y + 10 = 8y
=> 10 = 2y
=> y = 5

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Define an arithmetic progression.
Solution:
A sequence a1, a2, a3, …, an is called an arithmetic progression of then exists a constant d
Such that a2 – a1 = d, a3 – a2 = d, ………… an – an-1 = d
and so on and d is called common difference

Question 2.
Write the common difference of an A.P. whose nth term is an = 3n + 7.
Solution:
an = 3n + 7
a1 = 3 x 1 + 7 = 3 + 7 = 10
a2 = 3 x 2 + 7 = 6 + 7 = 13
a3 = 3 x 3 + 7 = 9 + 7 = 16
d = a3 – a2 or a2 – a1 = 16 – 13 = 3 or 13 – 10 = 3

Question 3.
Which term of the sequence 114, 109, 104, … is the first negative term ?
Solution:
Sequence is 114, 109, 104, …..
Let an term be negative
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 1

Question 4.
Write the value of a30 – a10 for the A.P. 4, 9, 14, 19, …………
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 2

Question 5.
Write 5th term from the end of the A.P. 3, 5, 7, 9,…, 201.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 3
= 3 + 190 = 193
5th term from the end = 193

Question 6.
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 4

Question 7.
Write the nth term of an A.P. the sum of whose n terms is Sn.
Solution:
Sum of n terms = Sn
Let a be the first term and d be the common difference an =Sn – Sn-1

Question 8.
Write the sum of first n odd natural numbers.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 5

Question 9.
Write the sum of first n even natural numbers.
Solution:
First n even natural numbers are
2, 4, 6, 8, ……….
Here a = 2, d = 2
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 6

Question 10.
If the sum of n terms of an A.P. is Sn = 3n² + 5n. Write its common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 7

Question 11.
Write the expression for the common difference of an A.P. Whose first term is a and nth term is b.
Solution:
First term of an A.P. = a
and an = a + (n – 1) d = b .
Subtracting, b – a = (n – 1) d
d = \(\frac { b – a }{ n – 1 }\)

Question 12.
The first term of an A.P. is p and its common difference is q. Find its 10th term. [CBSE 2008]
Solution:
First term of an A.P. (a) = p
and common difference (d) = q
a10 = a + (n – 1) d
= p + (10 – 1) q = p + 9q

Question 13.
For what value of p are 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P.? [CBSE 2009]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 8

Question 14.
If \(\frac { 4 }{ 5 }\), a, 2 are three consecutive terms of an A.P., then find the value of a.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 9

Question 15.
If the sum of first p term of an A.P. is ap² + bp, find its common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 10

Question 16.
Find the 9th term from the end of the A.P. 5, 9, 13, …, 185. [CBSE 2016]
Solution:
Here first term, a = 5
Common difference, d = 9 – 5 = 4
Last term, l = 185
nth term from the end = l – (n – 1) d
9th term from the end = 185 – (9 – 1) 4 = 185 – 8 x 4 = 185 – 32 = 153

Question 17.
For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form on A.P.? [CBSE 2016]
Solution:
(3k + 3) – (2k + 1) = (5k – 1) – (3k + 3)
3k + 3 – 2k – 1 = 5k – 1 – 3k – 3
k + 2 = 2k – 4
2k – k = 2 + 4
k = 6

Question 18.
Write the nth term of the A.P.
\(\frac { 1 }{ m }\) , \(\frac { 1 + m }{ m }\) , \(\frac { 1 + 2m }{ m }\) , ……… [CBSE 2017]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS 11

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS

Other Exercises

Question 1.
The radii of the bases 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 ?
Solution:
Radii of the bases of a cylinder and a cone = 3:4
and ratio in their heights = 2:3
Let r1, r2 be the radii and h1 and h2 be their heights
heights of the cylinder and cone respectively,
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 1

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 ?
Solution:
Ratio in the heights of two cones =1:2 and ratio in the perimeter of their bases = 3:4
Let r1, r2 be the radii of two cones and ht and h2 be their heights
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 2

Question 3.
If a cone and sphere have equal radii and equal volumes what is the ratio of the diameter of the sphere to the height of the cone ?
Solution:
Let r be the radius of a cone, then
radius of sphere = r
Let h be the height of cone
Now volume of cone = volume of sphere
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 3

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?
Solution:
Let r and h be the radius and heights of a cone, a hemisphere and a cylinder
∴ Volume of cone =  \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 4

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 ?
Solution:
Radii of two cylinders are in the ratio = 3:5
and ratio in their heights = 2:3
Let r1, r2 be the radii and h1, h2 be the heights of the two cylinders respectively, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 5

Question 6.
Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas ?
Solution:
Ratio in the volumes of two cubes = 1 : 27
Let a1 and a2 be the sides of the two cubes respectively then volume of the first area = a1³
and volume of second cube = a
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 6

Question 7.
Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii ?
Solution:
Ratio the heights of two right circular cylinders = 1:2
Let r1,r2 be their radii and h1, hbe their
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 7

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.
Solution:
Volumes of two cones are in the ratio =1:4 and their diameter are in the ratio = 4:5
Let r1 and r2 be the radii and h,h2 be their
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 8

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 ?
Solution:
Surface areas of a sphere and a cube are equal
Let r be the radius of sphere and a be the edge of cube,
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 9

Question 10.
What is the ratio of the volume of a cube to that of a sphere which will fit inside it?
Solution:
A sphere is fit inside the cube
Side of a cube = diameter of sphere
Let a be the side of cube and r be the radius of the sphere, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 10

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 ?
Solution:
Diameters (or radii), and heights of a cylinder a cone and a sphere are equal,
Let r and h be the radius and height be the cone cylinder, cone and sphere respectively, thus their volumes will be
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 11

Question 12.
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?
Solution:
r is the radius of a hemisphere, then
the diameter of the sphere which is cut out of the hemisphere will be r
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 12

Question 13.
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 value of \((\frac { H }{ R } )\).
Solution:
R is the radius of a hemisphere 2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 13

Question 14.
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.
Solution:
Radius and height of a cone and a cylinder be r and h respectively
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 14

Question 15.
A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ?
Solution:
Let r and h be the radii and heights of the cylinder cone and hemisphere respectively, then
Volume of cylinder = πr²h
Volume of cone = \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 15

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 ?
Solution:
Radii of two cones are in the ratio = 2:1
Let r1, r2 be the radii of two cones and h1, h2 be their heights respectively,
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 16

Question 17.
Two cones have their heights in the ratio 1 : 3 and radii 3:1. What is the ratio of their volumes ?
Solution:
Ratio in heights of two cones = 1:3
and ratio in their ratio = 3:1
Let r1, r2 be their radii and h1, h2 be their
heights, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 17

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 ?
Solution:
Bases of a hemisphere and a cone are equal
and their heights are also equal
Let r and h be their radii and heights
respectively
∴ r = h1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 18

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 of the frustum.
Solution:
r1 , r2 are the radii of the bases of a frustum and r1 > r2
Let h1 be the height of cone and h2 be the height of smaller cone
∴ Height of frustum = h1 – h2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 19

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 ?
Solution:
Slant height of a frustum (l) = 6 cm
Perimeter of upper base (P1) = 24 cm
and perimeter of lower base (P2) = 12 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 20

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 ?
Solution:
In a frustum,
Area of upper base (A1) = 4 cm²
and area of lower base (A2) = 9 cm²
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 21

Question 22.
The surface area of a sphere is 616 cm². Find its radius.
Solution:
Surface area of a sphere = 616 cm²
Let r be the radius, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 22

Question 23.
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. [CBSE 2009]
Solution:
Let r be the radius of the base of the cylinder
small as of cone
and let height of the cylinder = h
Then height of cone = h
∴ Volume of cylinder =  πr²h
and volume of cone = \((\frac { 1 }{ 3 } )\)  πr²h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 23

Question 24.
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. [CBSE 2010]
Solution:
Slant height of frustum (l) = 5 cm
Difference between the upper and lower radii = 4 cm
Let h be height and upper radius r1 and lower radius = r2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 24

Question 25.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Solution:
Volume of hemisphere = Surface area of hemisphere (given)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS 25

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

Other Exercises

Question 1.
A metallic sphere 1 dm in diameter is beaten into a circular sheet of uniform thickness equal to 1 mm. Find the radius of the sheet.
Solution:
Diameter of a sphere = 1 dm = 10 cm
∴  Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 cm
Volume of metal used in the sphere = \((\frac { 4 }{ 3 } )\) πr³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 1

Question 2.
Three solid spheres of radii 3,4 and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.
Solution:
Radius of first sphere (r1) = 3 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 2

Question 3.
A spherical shell of lead, whose external diameter is 18 cm, is melted and recast into a right circular cylinder, whose height is 8 cm and diameter 12 cm. Determine the internal diameter of the shell.
Solution:
Diameter of the cylinder = 12 cm
∴  Radius (r1) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h) = 8 cm
∴ Volume = πr1²h = π(6)² x 8 cm³
= π x 36 x 8 = 288π cm³
Now volume of metal used in spherical shell = 288π cm
External diameter = 18 cm 18
∴  External radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm
Let r be the internal radius, then
Volume of the metal = \((\frac { 4 }{ 3 } )\) π (R³ – r³ )
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 3

Question 4.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of two well = 10 m
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 m
Depth (h) = 8.4 m
∴ Volume of earth dug out = πr²h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 4

Question 5.
 In the middle of a rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = \((\frac { 7 }{ 2 } )\) m
Depth (h) = 10 m
∴ Volume of earth dug out = πr²h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 5

Question 6.
The inner and outer radii of a hollow cylinder are 15 cm and 20 cm, respectively. The cylinder is melted and recast into a solid cylinder of the same height. Find the radius of the base of new cylinder.
Solution:
Inner radius of hollow cylinder (r) = 15 cm
Outer radius (R) = 20 cm
Let h be the height of the hollow cylinder,
Then volume of metal used = πR (R² – r²)
= πh (20²- 15²) cm³
=  πh (400 – 225) cm³
= 175 πh cm³
Volume of the new cylinder = 175nh cm³
Height = h
Let R be the radius of new cylinder,
then πR²h = 175 πh
⇒ R²= 175
⇒ R = \(\sqrt { 175 } \)
= 13.2
∴ Radius = 13.2 cm

Question 7.
Two cylindrical vessels are filled with oil. Their radii are 15 cm, 12 cm and heights 20 cm, 16 cm respectively. Find the radius of a cylindrical vessel 21 cm in height, which will just contain the oil of the two given vessels.
Solution:
Radius of first cylinder (r1) = 15 cm
and radius of second cylinder (r2) = 12 cm
Height of the first cylinder (h1) = 20 cm
and height of second cylinder (h2) = 16 cm
∴ Volume of both of cylinders
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 6

Question 8.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:
Diameter of cylindrical bucket = 28 cm
∴ Radius (r) = \((\frac { 28 }{ 2 } )\) = 14 cm
Height (h) = 72 cm
∴ Volume of water filled in it = πr²h
= \((\frac { 22 }{ 7 } )\) x 14 x 14 x 72 cm³ = 44352 cm³
∴ Volume of water in rectangular tank = 44352 cm³
Length of tank (l) = 66 cm
and breadth (b) = 28 cm
Let h1 be its height
∴  Ibh1 = 44352
⇒  66 x 28 h1= 44352
⇒ h1 = \((\frac { 44352 }{ 66×28 } )\) = 24
∴Height of water in the tank = 24 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 7

Question 9.
A cubic cm of gold is drawn into a wire 0.1 mm in diameter, find the length of the wire.
Solution:
Volume of solid gold = 1 cm³
Diameter of cylinderical wire = 0.1 mm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 8

Question 10.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Radius (r) = \((\frac { 3 }{ 2 } )\) m
Depth (h) = 14 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 9
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 10

Question 11.
A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, find the height to which the water level rises in it.
Solution:
Internal radius of the conical vessel (r1) = 10 cm
Height (h1) = 48 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 11

Question 12.
The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm.
Solution:
Vertical height of conical tent (h) = 42 dm
and diameter base (b) = 5.4 dm
∴ Radius (r) = \((\frac { 5.4 }{ 2 } )\) = 2.7 dm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 12

Question 13.
A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:
Let r and h be the radius and height of a circular cylinder and also of a cone, then curved surface area of the cylinder = 2πrh
and curved surface area of cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 13

Question 14.
A sphere of diameter 5 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise ?
Solution:
Diameter of sphere = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 14
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 15

Question 15.
A spherical ball of iron has been melted – and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made ?
Solution:
Let the radius of larger ball = r
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 16

Question 16.
Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m x 16 m x 11 m.
Solution:
Dimensions of a rectangular tank = 28m x 16m x 11m
∴ Volume = 28 x 16 x 11 m³ = 4928 m³
∴  Volume of cylindrical tank = 4928 m³
Radius of the cylindrical tank = 28 m
Let h be depth of the tank, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 17

Question 17.
A hemispherical bowl of internal radius IS cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter S cm and height 6 cm. How many bottles are necessary to empty the bowl? (C.B.S.E. 2001C)
Solution:
Internal radius of hemispherical bowl (r) = 15 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 18

Question 18.
In a cylindrical vessel of diameter 24 cm, filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Find the increase in height of water level.
Solution:
Diameter of the cylindrical vessel = 24 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 19

Question 19.
 A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere of lead (r1) = 7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 20

Question 20.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4\((\frac { 2 }{ 3 } )\)cm and height 3 cm. Find the number of cones so formed. (C.B.S.E. 2004)
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 21

Question 21.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the wire is 108 m, find its diameter. (C.B.S.E. 1994)
Solution:
Diameter of copper sphere – 18 cm 18
Radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm 4
Volume = \((\frac { 4 }{ 3 } )\) π (R³)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 22

Question 22.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (R) = 7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 23

Question 23.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. (C.B.S.E. 2004)
Solution:
Radius of sphere (R) = 10.5 cm
∴ Volume of sphere =\((\frac { 4 }{ 3 } )\) πR³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 24

Question 24.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1:2:3.
Solution:
Let radius of a cone, a hemisphere and a cylinder be r
and height in each case = h
∴ h = r
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 25

Question 25.
A hollow sphere of internal and external diameters 4 and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Outer diameter of a hollow sphere = 8 cm
∴ Outer radius (R) = \((\frac { 8 }{ 2 } )\) = 4 cm
and inner diameter = 4 cm
∴ Inner radius (r)=\((\frac { 4 }{ 2 } )\) =2 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 26

Question 26.
The largest sphere is carved out of a cube of the side 10.5 cm. Find the volume of the sphere.
Solution:
Side of a cube = 10.5 cm
∵ The largest sphere is carved out of the cube,
∴ Diameter of the cube = side of the cube = 10.5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 27

Question 27.
Find the weight of a hollow sphere of metal having internal and external diameters as 20 cm and 22 cm, respectively if 1 cm³ of metal weighs 21 g.
Solution:
Internal diameter of a hollow sphere = 20 cm
and external diameter = 22 cm
∴ Outer radius (R) = \((\frac { 22 }{ 2 } )\) = 11 cm 20
and inner radius = \((\frac { 20 }{ 2 } )\) = 10 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 28

Question 28.
A solid sphere of radius ‘r’ is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thinkness 2 cm, find the value of ‘r’.
Solution:
Radius of solid sphere = r
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 29

Question 29.
Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm find the number of lead spheres dropped in the water.
Solution:
Diameter of cylindrical diameter = 18 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 30

Question 30.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm arc cut off. Find the volume of the remaining solid.
Solution:
Diameter of right solid cylinder = 7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 31

Question 31.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Use TC = 3.1416)
Solution:
Total height of the solid =108 cm
Each diameter of base of hemispherical part = 36 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 32

Question 32.
The surface area of a sphere is the same as the curved surface area of a cone having the radius of the bases as 120 cm and height 160 cm. Find the radius of the sphere.
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 34

Question 33.
 A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 35

Question 34.
A rectangular vessel of dimensions 20 cm x 16 cm x 11 cm is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10 cm. If the conical vessel is filled completely, determine its height. (Use π = 22/7)
Solution:
Dimension of rectangular vessel are 20 cm x 16 cm x 11 cm
Volume of vessel = 20 x 16 x 11 cm³= 3520 cm³
∴ Volume of water in conical vessel = 3520 cm³
Radius of the top of vessel = 10 cm
Let h be its height, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 36

Question 35.
If r1 and r2 be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is (r1³ + r1³ )1/3.
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 37

Question 36.
A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared. Determine the diameter of the sheet.
Solution:
Diameter of solid sphere = 6 cm 6
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 38

Question 37.
A hemispherical tank full of water is \((\frac { 25 }{ 7 } )\) emptied by a pipe at the rate of  litres per second. How much time will it take to half-empty the tank, if the tank is 3 metres in diameter ?
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 39

Question 38.
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 40

Question 39.
The radius of the base of a right circular cone of semi-vertical angle a is r. Show that its volume is \((\frac { 1 }{ 3 } )\) πr³ cot a and curved surface area is πr² cosec α.
Solution:
Radius of circular cone = r
and semi vertical angle = α
Let AO = h and slant height AC = l
In ΔAOC, AO ⊥ BC
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 41

Question 40.
An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm.
Solution:
Diameter of cylindrical portion = 20 cm
∴ Radius (r) = \((\frac { 20 }{ 2 } )\)  = 10 cm
Height of (h1) = 2.8 m = 280 cm
and height of cone (h2) = 42 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 42

Question 41.
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
Solution:
Diameter of the tent = 105 m
∴ Radius (r) = \((\frac { 105 }{ 2 } )\)  m
Height of cylindrical part (h1) = 3m
Slant height of conical part (h2) – 53 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 43
∴ Total surface area of the tent = curved surface area of the conical part + curved surface area of the cylindrical area
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 44

Question 42.
Height of a solid cylinder is 10 cm and diameter 8 cm. Two equal conical hole have been made from its both ends. If the diameter of the holes is 6 cm and height 4 cm, find (i) volume of the cylinder, (ii) volume of one conical hole, (iii) volume of the remaining solid.
Solution:
Height of the solid cylinder (h1) = 10 cm
Diameter = 8 cm
∴Radius (r1) = \((\frac { 8 }{ 2 } )\) = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 45

Question 43.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm, and height 4 cm are cut off. Find the volume of the remaining solid.
Solution:
Diameter of the base of a cylinder = 7 cm
∴ Radius (r1) = \((\frac { 7 }{ 2 } )\) cm
Height of cylinder (h1) = 15 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 46

Question 44.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Useπ = 3.1416)
Solution:
Total height of the solid =108 cm
Diameter of base of each hemisphere = 36 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 47

Question 45.
The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
Solution:
Radius of cylinder (r) = 7 cm
and height (h) = 14 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 48
The diameter of the largest sphere curved out of the given cylinder = diameter of the cylinder
= 2 x 7 = 14 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 49

Question 46.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is 24 m. The height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas required for making the tent. (Use π = 22/7)
Solution:
Diameter of the base of the cone = 24 m
∴  Radius (r) = \((\frac { 24 }{ 2 } )\) = 12 m
Height of the cylindrical part (h1) = 11 m
Total height of the tent = 16 m
Height of the conical part (h2)
= 16- 11 = 5 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 50

Question 47.
Area of the canvas required for the tent = 1320 m2 47. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm find the total surface area and volume of the toy. (C.B.S.E. 2000, 2002)
Solution:
Radius of the toy (r) = 3.5 cm
Total height of the toy = 15.5 cm
∴ Height of the conical part = 15.5 – 3.5 = 12 cm
Slant height of the conical part (l)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 51
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 52

Question 48.
A cylindrical container is filled with ice-cream, whose diameter is 12 cm and height is 15 cm. The whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.
Solution:
Diameter of the cylindrical container = 12 cm
Radius (r1) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h1) = 15 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 53
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 54

Question 49.
Find the volume of a solid in the form of a right circular cylinder with hemi-spherical ends whose total length is 2.7 m and the diameter of each hemispherical end is 0.7 m.
Solution:
Total length of solid = 2.7 m
Diameter of each hemisphere at the ends = 0.7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 55
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 56

Question 50.
A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs. 45 per m².
Solution:
Total height of the tent = 8.25 m
Height of cylindrical part (h1) = 5.5 m
∴ Height of conical part (h2) = 8.25 – 5.5 = 2.75m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 57
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 58

Question 51.
An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gram mass approximately. (Use π = 355/115)
Solution:
Diameter of the base of the cylindrical pole = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of cylindrical portion (h1) = 110 cm
and height of conical portion (h2) = 9 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 59
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 60

Question 52.
The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.
Solution:
Radius of the building (r) = 12m
Height of the cylindrical portion (h1) = 3.5 m and
slant height of conical portion (l) = 12.5 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 61
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 62

Question 53.
A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.
Solution:
In right angled ΔABC, ∠B = 90°
AB = 3 cm and BC = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 63
Now revolving the triangle along CA,
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 64
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 65

Question 54.
A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\) = 3 cm
Height of conical portion (h) = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 66

Question 55.
Find the mass of a 3.5 m long lead pipe, if the external diameter of the pipe is 2.4 cm, thickness of the metal is 2 mm and the mass of 1 cm3 of lead is 11.4 grams.
Solution:
External diameter of a cylindrical pipe = 2.4 cm
Radius (R) = \((\frac { 2.4 }{ 2 } )\) = 1.2 cm
Thickness of the pipe = 2 mm =\((\frac { 2 }{ 10 } )\) = 0.2 cm
∴ Inner radius (r) = 1.2 – 0.2 = 1.0 cm
Height (length) of the pipe (h) = 3.5 m
= 350 cm
Volume of the mass of the pipe = πh (R² – r²)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 67

Question 56.
A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid.
Solution:
Total height of the solid = 19 cm
Diameter of the cylinder = 7 cm
Radius (r) = \((\frac { 7 }{ 2 } )\) cm
Height of the cylinder = 19 – 2 x \((\frac { 7 }{ 2 } )\) cm
= 19-7 =12cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 68
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 69

Question 57.
A golf ball has diameter equal to 4.2 cm. Its surface has 200 dimples each of radius 2 mm. Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemi-spherical.
Solution:
Diameter of the golf ball = 4.2 cm
∴  Radius (R) = \((\frac { 4.2 }{ 2 } )\) =2.1 cm
Radius of each hemispherical dimples (r)  = 2 mm = \((\frac { 2 }{ 10 } )\) = \((\frac { 2 }{ 5 } )\) cm
Curved surface area of one dimple = 2πr²
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 70

Question 58.
The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity. (Take π = 22/7).
Solution:
Height of the bucket (frustum) (h) = 24 cm
Upper radius (r1) = 15 cm
and lower radius (r2) = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 71

Question 59.
The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheet required to make this bucket.
Solution:
Height of the bucket (frustum) (h) = 30 cm
Upper radius (r1) = 21 cm
and lower radius (r2) = 7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 72
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 73

Question 60.
The radii of the ends of a frustum of a right circular cone are 5 metres and 8 metres and its lateral height is 5 metres. Find the lateral surface and volume of the frustum.
Solution:
Upper radius of a frustum (r1) = 8 m
and lower radius (r2) = 5 m
Lateral height (l) = 5m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 74

Question 61.
A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum. (Take π = 22/7)
Solution:
Upper diameter = 28 cm
and lower diameter = 4 cm
Height (h) = 9 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 75
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 76

Question 62.
A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction.
Solution:
Water in a bucket (frustum) = 15.25l
Upper diameter = 25 cm
and lower diameter = 20 cm
∴ Upper radius (r1) = \((\frac { 25 }{ 2 } )\) cm
and lower radius (r2) = \((\frac { 20 }{ 2 } )\) cm =10 cm
Volume = 15.25 /= 1525 x 10 cm³ = 15250 cm³
Let h be its height
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 77

Question 63.
If a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts. (C.B.S.E. 2000C)
Solution:
Radius of the cone (r1) = 10 cm
Cone is divided into 2 parts Such that PQ || AB
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 78
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 79

Question 64.
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 pnetres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres
Solution:
Radius of the cylinder (r) = 14 m
and total height of the tent = 13.5 m
Height of the cylindrical part (h1) = 3 m
Height of conical part (h2) = 13.5-3.0 = 10.5m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 80

Question 65.
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required. (Use : π = 22/7).
Solution:
Upper diameter of the frustum = 18 cm
and  lower diameter = 8m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 81
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 82

Question 66.
A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the shape of a right circular cone mounted on a hemisphere of radius 3 cm. If the height of the toy is 12 cm, find the number of toys so formed. (C.B.S.E. 2006C)
Solution:
Diameter of solid cylinder = 12 cm
and height (h1) = 15 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 83
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 84

Question 67.
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 ₹21 per litre. (Use π = 22/7)
Solution:
Upper radius (R) = 20 cm
Lower radius (r) = 8 cm
Height (h) = 24 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 85

Question 68.
 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. [NCERT Exemplar]
Solution:
The cone of maximum size that is carved out from a cube of edge 14 cm will be of base radius 7 cm and the height 14 cm.
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 86

Question 69.
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. [NCERT Exemplar]
Solution:
Let h be the height of the given cone. One dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following figure:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 87

Question 70.
A wall 24 m, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies \((\frac { 1 }{ 10 } )\) th of the volume of the wall, then find the number of bricks used in constructing the wall. [NCERT Exemplar]
Solution:
Given that, a wall is constructed with the help of bricks and mortar.
∴ Number of bricks
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 88
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 89

Question 71.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket. [NCERT Exemplar] 
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 90

Question 72.
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. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 91

Question 73.
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. [NCERT Exemplar]
Solution:
If two cones with same base and height are joined together along their bases, then the shape so formed is look like as figure shown.
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 92

Question 74.
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. [NCERT Exemplar]
Solution:
Given that, side of a solid cube (a) = 1 cm
Height of conical cavity i.e., cone, h = 7 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 93
Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.
Radius of conical cavity i. e., cone, r = 3 cm
⇒ Diameter = 2 x r = 2 x 3 = 6 cm
Since, the diameter is less than the side of a cube that means the base of a conical cavity is not fit inhorizontal face of cube.
Now, volume of cube = (side)³ = a³ = (7)³ = 34³ cm³
and volume of conical cavity i.e., cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 94

Question 75.
Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacitites are 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 95
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 96
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 97

Question 76.
An icecream cone full of icecream having radius 5 cm and height 10 cm as shown’in the figure. Calculate the volume of icecream, provided that its 1/6 parts is left unfilled with icecream.
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 98
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 99

 

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Other Exercises

Question 1.
Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 1

Question 2.
Show that the sequence defined by an = 3n² – 5 is not an A.P.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 2

Question 3.
The general term of a sequence is given by an = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.
Solution:
General term of a sequence
an = -4n + 15
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 3

Question 4.
Write the sequence with nth term :
(i) an = 3 + 4n
(ii) an = 5 + 2n
(iii) an = 6 – n
(iv) an = 9 – 5n
Show that all of the above sequences form A.P.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 4
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 5

Question 5.
The nth term of an A.P. is 6n + 2. Find the common difference. [CBSE 2008]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 6

Question 6.
Justify whether it is true to say that the sequence, having following nth term is an A.P.
(i) an = 2n – 1
(ii) an = 3n² + 5
(iii) an = 1 + n + n²
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 7
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 8
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 9

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Other Exercises

Question 1.
Write the first five terms of each of the following sequences whose nth terms are:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 1
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 2
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 3
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 4
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 5
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 6

Question 2.
Find the indicated terms in each of the following sequences whose nth terms are:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 7
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 8

Question 3.
Find the next five terms of each of the following sequences given by :
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 9
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 10
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 11
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 12
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 13

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

Other Exercises

Question 1.
A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at the rate of ?1.20 per dm2. (Use % = 3.14)
Solution:
Upper diameter = 40 cm
and lower diameter = 20 cm
∴  Upper radius (r1) = \((\frac { 40 }{ 2 } )\) = 20 cm
and lower radius (r2) = \((\frac { 20 }{ 2 } )\) = 10 cm
Depth or height (h) = 12 cm
Volume of the bucket
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 1

Question 2.
A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.
Solution:
Base diameter of frustum = 20 cm 20
∴  Radius (r1) = \((\frac { 20 }{ 2 } )\) = 10 cm
and diameter of top = 12 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 2

Question 3.
The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.
Solution:
Perimeter of the top of frustum = 18 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 3

Question 4.
The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.
Solution:
Perimeter of the top of frustum = 44 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 4
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 5
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 6

Question 5.
If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use π = 22/7). (C.B.S.E. 2000)
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 7

Question 6.
The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be \((\frac { 1 }{ 125 } )\)  of the volume of the original cone, determine at what height above the base the section is made.
Solution:
Total height of the cone (h1) = 20 cm
Let a cone whose height is h2 is cut off Then height of the remaining portion (frustum)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 8
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 9

Question 7.
If the radii of the circular ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the surface area of the bucket.
Solution:
Height of the bucket (frustum) (A) = 24 cm
Radius of the top (r1) = 15 cm 1
and radius of the bottom (r2) = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 10

Question 8.
The radii of the circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum.
Solution:
Height of the frustum (A) = 12 cm
Radius of the top (r1) = 12 cm
and radius of the bottom (r2) = 3 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 11

Question 9.
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 frutum 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)
Solution:
Radius of the bottom of the tent (r1) = 13 m
and radius of the top (r2) = 7 m
Height of frustum portion (h1) = 8 m
Slant height of the conical cap (l2) = 12 m
Let l1 be the slant height of the frustum portion, then
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 12

Question 10.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹44 per litre which the container can hold. [NCERT Exemplar]
Solution:
Given that, height of milk container (h) = 16 cm
Radius of lower end of milk container (r) = 8 cm
and radius of upper end of milk container (R) = 20 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 13
∴ Volume of the milk container made of metal sheet in the form of a frustum of a cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 14

Question 11.
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 the rate of ₹25 per litre. (Use π = 3.14) [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 15

Question 12.
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). [CBSE 2006C]
Solution:
Volume of frustum (bucket) = 12308.8 cm³
Upper radius (r1) = 20 cm
and lower radius (r2) = 12 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 16
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 17

Question 13.
A bucket made of aluminium sheet is of height 20 cm and its upper and lower ends are of radius 25 cm and 10 cm respectively. Find the cost of making the bucket if the aluminium sheet costs Rs. 70 per 100 cm². (Use π = 3.14) (C.B.S.E. 2006C)
Solution:
Height of bucket (frustum) (h) = 20 cm
Upper radius (r1) = 25 cm
and lower radius (r2) = 10
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 18

Question 14.
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. (C.B.S.E. 2005)
Solution:
Upper radius of frustum (r1) = 3.3 cm
and lower radius (r2) = 27 cm
Slant height (l) = 10 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 19

Question 15.
A bucket made up of a 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 volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is Rs. 20 per 100 cm². (Use π = 3.14) (CBSE 2008)
Solution:
Lower radius of bucket (r) = \((\frac { 16 }{ 2 } )\) = 8 cm
and upper radius (R) = \((\frac { 40 }{ 2 } )\) = 20 cm
Height (h) = 16
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 20

Question 16.
A solid is in the shape of a frustum of a cone. The diameter of the two circular ends are 60 cm and 36 cm and the leight is 9 ³cm. Find the area of its whole surface and the volume. [CBSE 2010]
Solution:
In a solid frustum upper diameter = 60 cm
∴ Radius (r1) = \((\frac { 60 }{ 2 } )\) = 30 cm
Lower diameter = 36 cm 36
∴ Radius (r2) = \((\frac { 36 }{ 2 } )\) = 18 cm
Height (h) = 9 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 21

Question 17.
A milk container is made of metal sheet in the shape of frustum of a cone whose volucrn is 10459\((\frac { 3 }{ 7 } )\) 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 Rs. 1.40 per cm². (Use π = 22.7) [CBSE 2010]
Solution:
Volume of frustum = 10459\((\frac { 3 }{ 7 } )\) cm3 73216
= \((\frac { 73216 }{ 7 } )\) cm³
Lower radius (r2) = 8 cm
and upper radius (r1) = 20 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 22
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 23

Question 18.
A solid cone of base radius 10 cm is cut into two parts through the mid-pint of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone. [CBSE 2013]
Solution:
Radius of solid cone (r) = 10 cm
Let total height = h
In ΔAOB,
C is mid point of AO and CD || OB
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 24

Question 19.
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 ₹10 per 100 cm². (Use π = 3.14). [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 25

Question 20.
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 = 22/7 and \(\sqrt { 5 } \) = 2.236).  [ CBSE 2015]
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 26
Solution:
Total height of cone = 12 cm
Radius of its base = 6 cm
A cone of height 4 cm is cut out
Height of the so formed frustum = 12 – 4 = 8 cm
Let r be the radius of the cone cut out
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 27

Question 21.
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.
Solution:
Let the height and radius of the given cone be H and R respectively.
The cone is divided into two parts by drawing a plane through the mid point of its axis and parallel to the base.
Upper part is a smaller cone and the bottom part is the frustum of the cone.
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 28
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 29

Question 22.
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 ₹40 per litre, find the amount received by the person. [CBSE 2017]
Solution:
Radii of the bucket in the form of frustum of cone = 30 cm
and 12 cm Depth of the bucket = 35 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 30

Question 23.
A reservoir in the form of the frustum of a right circular cone contains 44 x 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)
Solution:
A reservoir is a frustum in shape and its upper radius (r1) = 100 m
Lower radius (r2) = 50 m
and capacity of water in it = 44 x 107 litres
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 31
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 32

P.Q. A metallic right circular cone 20 cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire  \((\frac { 1 }{ 16 } )\)cm, find the length of the wire.

Solution: In the cone ABC, ∠A = 90°, AL ⊥ BC and = 20 cm
It is cut into two parts at the middle point M on the axis AL
AL bisects ∠A also
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 33
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 34

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

Other Exercises

Question 1.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.
Solution:
Diameter of the base of the tent = 24 m
∴ Radius (r)= \((\frac { 24 }{ 2 } )\)  = 12m
Total height of the tent = 16 m
Height of the cylindrical portion (h1) = 11 m
Height of the conical portion (h2) =16-11 = 5 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 1

Question 2.
A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.
Solution:
Radius of the base of the rocket (r) = 2.5 m
Height of cylindrical portion (h1) = 21 m
Slant height of the conical portion (l) = 8 m
Let height of conical portion = h2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 3

Question 3.
A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44
dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m2 . (Use π = 22/7).

Solution:
Total height of the tent = 77 dm
Height of cylindrical part (h1) = 44 dm
= 4.4 m
Height of conical part (h2) = 7.7 – 4.4 = 3.3 m
Diameter of the base of the tent = 36 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 4
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 5

Question 4.
A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy. (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\)  = 3 cm
Height (h) = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 6
Total surface area of the toy = curved surface area of the conical part + surface area of the hemispherical part
= πrl + 2πr² = πr (l + 2r)
= 3.14 x 3 (5 + 6) = 3.14 x 3 x 11 cm²
= 3.14 x 33 = 103.62 cm²

Question 5.
A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the ~ heights of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7)
Solution:
Radius of the common base (r) = 3.5 m
Height of cylindrical part (h1) = 10 cm
Height of conical part (h2) = 6 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 7
Now total surface area of the solid = curved surface of conical part + curved surface of cylindrical part + curved surface of hemispherical part
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 8

Question 6.
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. (C.B.S.E. 2002)
Solution:
Radius of the base of the cylindrical part (r) = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 9
Height of cylindrical part (h1) = 13 cm
Height of the conical part (h2) = 30 – (13 + 5) = 30- 18 = 12 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 10

Question 7.
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) (C.B.S.E. 2000C)
Solution:
Radius of the cylindrical tub (R) = 5 cm
and height (h1) = 9.8 cm
Radius of the solid (r) = 3.5 cm
and height of cone (h1) = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 11
= 38.5 x 4 = 154 cm3
∴ Water flowed out of the tub = 154 cm³
Remaining water in the tub = 770 – 154
= 616 cm³

Question 8.
A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 m. The heights of the cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the volume of the tent.
Solution:
Radius of the tent (r) = 20 m
Height of the conical part (h1) = 2.1 m
and height of the cylindrical part (h2) = 4.2 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 12

Question 9.
A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.
Solution:
Diameter of cylindrical part = 21 cm 21
∴ Radius (r) = \((\frac { 21 }{ 2 } )\) cm
Height of cylindrical part (h1) = 18 cm
and height of each conical part (h2) = 9 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 13

Question 10.
A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder.
Solution:
Base radius of the cylinder (r) = 5 cm
and height (h) = 12 cm
∴ Volume = πr2h = n (5)2 x 12 cm3 = 300π cm3
∵ The base and height of the cone drilled are the same as those of the cylinder
∴ Volume of cone = \((\frac { 1 }{ 3 } )\)πr2h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 14

Question 11.
A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at Rs. 100 per square metre.
Solution:
Diameter of the base of the tent = 20 m
∴ Radius (r) =\((\frac { 20 }{ 2 } )\) = 10 m
Height of cylindrical part (h1) = 2.5 m
and height of conical part (h2) = 7.5 m
Slant height of the conical part (l)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 15
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 16

Question 12.
A boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter. Find the volume of the boiler.
Solution:
Diameter of the cylinder = 2 m
∴ Radius (r) = \((\frac { 1 }{ 3 } )\) = 1 m
Height (length) of cylindrical part (h) = 2 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 17

Question 13.
A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is \((\frac { 14 }{ 3 } )\) m and the diameter of hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.
Solution:
Diameter of the cylindrical part = 3.5 m
∴Radius (r) = \((\frac { 3.5 }{ 2 } )\) = 1.75 = \((\frac { 7 }{ 4 } )\) m
and height (h) = \((\frac { 14 }{ 3 } )\) m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 18
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 19

Question 14.
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 . (C.B.S.E. 2006C)
Solution:
Total height = 104 cm
Radius of hemispherical part (r) = 7 cm
Height of cylinder (h) = 104 cm – 2 x 7 cm = 104- 14 = 90 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 20
Total outer surface area = curved surface area of the cylindrical part + 2 x curved surface area of each hemispherical part
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 21

Question 15.
A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required ?
Solution:
Diameter of inner cylinder = 14 cm
∴Radius (r) = \((\frac { 14 }{ 2 } )\) = 7 cm
Diameter of outer cylinder = 16 cm
∴ Radius (R) = \((\frac { 16 }{ 2 } )\) = 8 cm
Height (h) = 42cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 22

Question 16.
A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm³ of iron has 7.8 gm mass. (Use π = 3.14)
Solution:
Length of roller (h) = 1 m = 100 cm
Inner diameter = 54 cm
Thickness of iron sheet = 9 cm
∴ Inner radius (r) = \((\frac { 52 }{ 2 } )\) = 27 cm
and outer radus (R) = 27 + 9 = 36 cm
∴ Volume of the mass = πR²h – πr²h
= πh (R²- r²)
= 3.14 x 100 (36² – 27²) cm3
= 314 x (36 + 27) (36- 27) cm3
= 314 x 63 x 9 cmc
= 178038 cm3
Weight of 1 cm3 of iron = 7.8 gm
∴ Total weight = 178038 x 7.8 gm
= 1388696.4 gm
= 1388.6964 kg
= 1388.7 kg

Question 17.
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.
Solution:
Diameter of hemisphere = 14 cm
Total height =13 cm
Radius of hemisphere = \((\frac { 14 }{ 2 } )\) = 7 cm
∴ Height of cylindrical part =13-7 = 6 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 23
∴  Inner surface area of the vessel = inner surface area of cylindrical part + inner surface area of hemispherical part
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 24

Question 18.
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.
Solution:
Radius of cone (r) – 3.5 cm
Total height of the toy = 15.5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 25
Height of the conical part (h) = 15.5 – 3.5 = 12 cm
∴ Slant height of the cone (l)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 26

Question 19.
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.
Solution:
In cylindrical metallic pipe,
length of pipe = 14 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 27
Difference between outside and inside
surface area = 44 m2
Volume of pipe material = 99 cm3
Let R and r be the outer and inner radii of the pipe respectively, then Outer surface area – inner surface area = 44 cm2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 28

Question 20.
A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Solution:
Height of cylinder (H) = 15 cm
and diameter =12 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 29
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 30

Question 21.
A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that the mass of 1 cm3 of iron is 8 gm.
Solution:
Diamter of the base = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of the cylindrical portion (h1)= 110 cm
and height of conical portion (h2) = 9 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 31
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 32

Question 22.
A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.
Solution:
Height of conical part (h) = 2 cm
Diameter of base = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 33
Now volume of the cylinder which circum scribes the toy = πr²h
= π (2)² x 4 = 16π cm³
∴  Difference of their volumes = 16π – 8π = 8πcm³

Question 23.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of conical part = 60 cm
and height (h) = 120 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 34
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 35

Question 24.
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. (C.B.S.E. 2009)
Solution:
Internal diameters of cylindrical vessel = 10 cm
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) =5 cm
and height (h) = 10.5 cm
∴ Volume of water filled in it
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 36

Question 25.
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. [CBSE 2010]
Solution:
Edge of cube = 21 cm
∴ Diameter of the hemisphere curved out of
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 37

Question 26.
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 the hemisphere, calculate the height of the cone and the surface area of the toy. (Use π = 22/7).
Solution:
Radius of base of the conical part (r) = 21 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 38
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 39

Question 27.
A solid is in the shape of a cone surmounted on a hemi-sphere, 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. [CBSE 2012]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 40
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 41

Question 28.
An wooden toy is made by scooping out a hemisphere 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. (Use π = 22/7). [CBSE 2013]
Solution:
Height of cylindrical part (h) = 10 cm
Radius of the base (r) = 3.5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 42

Question 29.
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 = 22/7). [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 43

Question 30.
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid, (take π = 22/7). [CBSE 2014]
Solution:
Diameter of solid cylinder= 4.2 cm
∴ Radius (r) =\((\frac { 4.2 }{ 2 } )\) = 2.1 cm
Height (h) = 2.8 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 44

Question 31.
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]
Solution:
Side of a solid cube (a) = 21 cm
∴ Volume = a3, = (21 )3 cm3
= 9261 cm3
Diameter of the base of cone = 21 cm
Now radius of cone curved from it (r) =\((\frac { 21 }{ 2 } )\) cm
and height (h) = 21 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 45

Question 32.
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 the making of toy is 166 \((\frac { 5 }{ 6 } )\) cm3 . Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹10 per cm2. (Take π  = 22/7). [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 46
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 47

Question 33.
In the given figure, from a cuboidal solid metalic 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. (Take  π = 22/7) [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 48
Radius of hole = \((\frac { 7 }{ 2 } )\)cm and height = 5 cm
Length of block (l) = 15 cm
Breadth (b) = 10 cm and height = 5 cm
∴ Surface area = 2(lb + bh + hl)
= 2(15 x 10 + 10 x 5 + 5 x 15) cm2
= 2(150 + 50 + 75) = 2 x 275 = 550 cm2
Area of circular holes of both sides = 2 x πr2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 49

Question 34.
 A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and
contains 41 \((\frac { 19 }{ 21 } )\) m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building? [NCERT Exemplar]

Solution:
Let total height of the building = Internal diameter of the dome = 2rm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 50
Radius of building (or dome) = \((\frac { 2r }{ 2 } )\) = r m
Height of cylinder = 2r-r = rm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 51

Question 35.
A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm x 5 cm x 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. [NCERT Exemplar]
Solution:
Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cuboid pen stand (b) = 5 cm
and height of cuboid pen stand (h) = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 52
∴ Volume of cuboid pend stand = l x b x h= 10 x 5 x 4 = 200 cm3
Also, radius of conical depression (r) = 0.5 cm
and height (depth) of a conical depression (h1) = 2.1 cm
∴ Volume of a conical depression = πrh1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 53

Question 36.
 A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height of the building. Find the height of the building, if it contains 67 \((\frac { 1 }{ 21 } )\) m3 of air.
Solution:
Let the radius of the hemispherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height, therefore
2 r = \((\frac { 2 }{ 3 } )\)h. This implies r = \((\frac { h }{ 3 } )\). Let H metres be the height of the cylindrical portion.
Therefore, H = h – \((\frac { h }{ 3 } )\) = \((\frac { 2 }{ 3 } )\)h metres.
Volume of the air inside the building = Volume of air inside the dome + Volume of the air inside the
cylinder = \((\frac { 2 }{ 3 } )\) πr3 + πr2H, where H is the height of the cylindrical portion
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 54

Question 37.
A solid toy is in the form of a hemisphere surmounted 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. Also, find the total surface area of the toy. [NCERT Exemplar]
Solution:
Let r be the radius of the hemisphere and the cone and h be the height of the cone.
Volume of the toy=Volume of the hemisphere + Volume of the cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 55

= \((\frac { 1408 }{ 7 } )\) cm3
A cube circumsrcibes the given solid. Therefore, edge of the cube should be 8 cm. Volume of the cube = 83 cm3 = 512 cm3 Difference in the volume of the cube and
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 56

Question 38.
A circus tent is in the shape of a 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. [ICBSE 2017]
Solution:
We have, diameter of base of cylinder = d = 56 m
Radius of base of cylinder = r=  \((\frac { d }{ 2 } )\)=  \((\frac { 52 }{ 2 } )\)= 28 m
Height of tent = 27 m
Height of cylinder = 6 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 57
Height of conical portion = 27 – 6 = 21 m
Radius of conical portion, r = 28 m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 58

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.