NCERT Solutions of class 10 maths Exercise 13.3 of chapter 13-Surface areas and Volumes - Future Study Point

# NCERT Solutions of class 10 maths Exercise 13.3-Surface areas and Volumes

NCERT Solutions of class 10 maths Exercise 13.3 of the chapter 13-Surface areas and Volumes-Here NCERT Solutions of the exercise 13.3 are the Solutions of the chapter ‘Surface areas and Volumes’ of the NCERT maths coursebook of class 10 endorsed by CBSE a reputed Board of India for school education.All questions of exercise 13.3 are solved by an expert of maths in a step-by-step way, so everyone would comprehend the profundity of the idea of maths.

There are a variety of questions in chapter- 13 based on surface areas and volumes of compound figures made of cubes, cuboids, cylinders, cones, and spheres. In exercise 13.3, the questions based on evaluation of the volumes, height, and radius of solids when one solid is melted and recast into another solid. In this exercise for the evaluation of physical quantities like height, radius, surface areas, and volume, you need to keep in mind the following points.

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• In the case when one metallic object is melted and recast into another type of article then the volume of both will be the same.
• In the case when the earth is taken out from aย  well to be dug out and something is made by the earth taken out from the well then the volume of the well is equivalent to the thing to be made.
• In a question, if you are given the speed of the water (distance/unit time)then the volume of water streamed in per unit time =depth รheight รspeed.

NCERT Solutions Class 10 Science from chapter 1 to 16

The questions of exercise 13.3 are very important for your board examination of 2020-21. We likewise have highlighted the questions which are important and most important, so that you could pay attention to those questions.

### NCERT Solutions of class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

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## NCERT Solutions of class 10 maths Exercise 13.3 -Surface areas and Volumes

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Q1.A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.(Imp)

Ans.We are given the radius of the sphere = 4.2 cm

Radius of the cylinder = 6 cm

Let the height of cylinder = h

The volume of sphere = Volume of cylinder

Volume of sphere =ย  ย  ย  ย and Volume of cylinder is =

h = 2.74

Hence the height of the cylinder = 2.74 cm

Q2.Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.(Imp)

Ans. We are given the radii of three metalic spheres 6 cm, 8 cm and 10 cm respectively

Sum of the volumes of metalic spheres = Volume of a single solid sphere

The volume of sphere =ย  ย

Let the radius of the single solid sphere = r

r ยณ = 6ยณ + 8ยณ + 10ยณ

rยณ = 216 + 512+1000

rยณ = 1728

Hence the radius of single solid sphere is 12 cm

Q3.A 20 m deep well with a diameter of 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.(Very Imp)

Ans. We are given the depth of the well, h =ย  20 m, radius, r of the well = 7/2= 3.5 cm

Let the height of the platform made by the earth taken out from the well =H

The shape of well is the shape of a cylinder

The volume of well = The volume of the earth taken out from the well= Volume of the platform = ฯrยฒh =22/7 ร3.5ยฒร20 =22ร0.5ร3.5ร20 = 770 mยณ

The volume of plateform = Length รBreadth ร Height = 22 ร 14 ร H

Therefore

22 ร14 H = 770

H =ย  2.5 m

Hence the height of the plateform = 2.5 mโ

Q4.A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.(Most Imp)

Ans.Let the height of the embankment= H

Radius of well = 3/2 = 1.5 m, height, h =14 m

The volume of well = The volume of the earth taken out from the well= Volume of the embankment

Volume of well =ฯrยฒh =ฯ(1.5)ยฒ14

Outer radius of embankment =1.5 +4 = 5.5 m

Inner radius of embankment = 1.5 m

Volume of embankment =ฯ(5.5)ยฒH – ฯ(1.5)ยฒH=ฯH(5.5ยฒ- 1.5ยฒ)=ฯH(30.25 -2.25)=28ฯH

28ฯH = ฯ(1.5)ยฒ14

28H =2.25ร14

H = 2.25/2 =1.125

Hence the height of embankment is 1.125 m

Q5. A container shaped like right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream .The ice-cream is to be filled into cones of height 12 cm and diameter 6 cm ,having hemisperical shape on the top .Find the number of such cones which can be filled with ice-cream.(Most Imp)

Ans.

We are given diameter of right circular cylinder= 12 cm,so radius,r = 12/2 = 6 cm

Height of the cylinder ,h =15 cm

Volume of cylinder = ฯrยฒh = ฯ(6)ยฒ15 =540ฯ cmยณ

The volume of ice-cream = Volume of cone + Volume of hemisphere

The height,h of the cone =12 cm and radius of cone and hemisphere,r= 6/2= 3 cm

Volume of cone =ย  ย

= 3ฯ ร12 = 36ฯ

Volume of hemisphere =ย  ย

= 2ฯร 9 = 18ฯ

โด The volume of ice-cream = 36ฯ + 18ฯ =54 ฯ cmยณ

Number of ice-cream cones to be filled

Hence the number of cones to be filled are 10

Q6.How many silver coins, 1.75 cm in diameter and of thickness 2 mm,must be meted to form a cuboid of dimensions 5.5 cm ร 10 cm ร 3.5 cm ?(Imp)

Ans.Let number of coins to be formed = n

Radius of coins,r = 1.75/2 = 0.875 cm and thickness,h=2 mm = 0.2 cm

Volume of coin = ฯrยฒh

=4.4 ร 0.125ร0.875=0.48125 cmยณ

Volume of cuboid = Lengthร Breadth รHeight =5.5 cm ร 10 cm ร 3.5 cm =192.5 cmยณ

Number of coins to be melted

= 400

Hence the number of coins to be melted are 400

Q7. A cylindrical bucket ,32 cm high and with radius of base 18 cm ,is filled with sand .This bucket is emptied on the ground and a conical heap of sand is formed .If the height of conical heap is 24 cm,find the radius and slant height of the heap. (Very Imp)

Ans. The volume of the bucket = Volume of conical heap

We are given the height of the cylindrical bucket,h = 32 cm and radius of base, = 18 cm

Volume of cylndrical bucket = ฯrยฒh = ฯ(18)ยฒ32

The height of conical heap =24 cm

Volume of conical heap =ย  ย

The volume of the bucket = Volume of conical heap

rยฒ = (18 ร18ร 4)

rยฒ=9 ร2ร9ร2ร4

r = 9ร2ร2= 36 cm

Hence radius of the conical heap = 36 cm

Slant height of the conical heap,l =โ(rยฒ +hยฒ) = โ(36ยฒ + 24ยฒ)=โ(1296 +576)=โ(1872)=12โ13 cm

Q8.Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?(Most Imp)

Ans.Let area to be irrigated A in time 30 minutes=30 ร60=1800 seconds

Standing water in the area= 8 cm= o.08 m

We are given width of canal is =6 m and depth=1.5 m

The speed of water =10 km/h = (10ร1000)/(60ร60) =25/9 m/s

Volume of water flowed in the area in per second =Width รDepth รSpeed of water=6ร1.5ร25/9=2ร0.5ร25=25 mยณ/s

In 30 minutes (1800 s) the water flowed in the area =25 ร1800=45000 mยณ

The volume of standing water in 30 minutes = Aรh =0.08 A

0.08 A = 45000

A = 45000/0.08= 562500 mยฒ

Q9.A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?(Most Imp)

Ans. We are given the diameter of the pipe = 20 cm = 0.20 m, so its radius =.20/2=0.10 m

The speed of water in the pipe = 3 km/h = 3000/(60ร60) =3000/3600=(5/6) m/s

The volume of water flowed per second through the pipe=ฯrยฒh=ฯ(0.10)ยฒร5/6=0.05ฯ/6 mยณ/s

The diameter of cylindrical tank = 10 m,so its radius =10/2= 5m

Depth(h) of the tank = 2 m

Volume of the tank =ฯrยฒh =ฯ(5ยฒ)ร2= 50ฯ mยณ

Time taken by the pipe in filling the tank =(Volume of tank)/(Volume of water flowed per second inside the pipe) = 50ฯ /(0.05ฯ/6) =300/0.05= 6000 seconds=100 minutes= 1h 40 minutes

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