# NCERT Solutions for Class 10 Maths

Find 100% accurate solutions for NCERT Class X Math. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Abbreviation:     CSA = Curved Surface Area, TSA = Total Surface Area and V = Volume

1.    Volume of cube = (Side)3

According to question,    (Side)3 = 64     ⇒   (Side)3 = 43       ⇒      Side = 4 cm

For the resulting cubiod, length (1) = 4 + 4 = 8 cm, breadth (b) = 4 cm and height (h) = 4 cm Surface area of resulting cuboid = 2 (lb + bh+hl) = 2 (8 x 4 + 4 x 4 + 4 x 8)

= 2 (32 + 16 + 32) = 2 x 80 = 160 cm2

2.   ∵ Diameter of the hollow hemisphere = 14 cm ∴ Radius of the hollow hemisphere = 7 cm

Total height of the vessel = 13 cm

∴ Height of the hollow cylinder = 13 – 7 = 6 cm

∴ Inner surface area of the vessel

= Inner surface area of the hollow hemisphere + Inner surface area of the hollow cylinder 3.    Radius of the cone = 3.5 cm ∴ Radius of the hemisphere = 3.5 cm

Total height of the toy = 15.5 cm

∴ Height of the cone = 15.5 – 3.5 = 12 cm

Slant height of the cone  ∴ TSA of the toy = CSA of hemisphere + CSA of cone 4.     Greatest diameter of the hemisphere = Side of the cubical block = 7 cm

∴ TSA of the solid = External surface area of the cubical block + CSA of hemisphere 5.     ∵ Diameter of the hemisphere = l, therefore radius of the hemisphere = Also, length of the edge of the cube = l

∴ Surface area of the remaining alt=”” width=”231″ height=”47″ solid =  6.    Radius of the hemisphere Let            radius = r = 2.5 mm

Cylindrical height = Total height – Diameter of sphere = h = 14 – (2.5 + 2.5) = 9 mm Surface area of the capsule = CSA of cylinder + Surface area of the hemisphere 7.    Diameter of the cylindrical part = 4 cm ∴ Radius of the cylindrical part = 2 cm

TSA of the tent = CSA of the cylindrical part + CSA of conical cap ∴ Cost of the canvas of the tent at the rate of Rs. 500 per m2

= 44 x 500 = Rs. 22000

8.    Diameter of the solid cylinder = 1.4 cm ∴ Radius of the solid cylinder = 0.7 cm

∴ Radius of the base of the conical cavity = 0.7 cm

Height of the solid cylinder = 2.4 cm

∴ Height of the conical cavity = 2.4 cm

∴ Slant height of the conical cavity  ∴ TSA of remaining solid 9.     TSA of the article = 2πrH + 2 (27πr2) 10.  For hemisphere, Radius (r) = 1 cm  For cone, Radius of the base (r) = 1 cm

Height (h) = 1 cm ∴ Volume of the solid = V of hemisphere + V of cone 11.   For upper conical portion, Radius of the base (r) = 1.5 cm Height (h1) = 2 cm For lowe
r conical portion, Volume = 1.57π cm3 For central cylindrical portion

Radius of the base (r) = 1.5 cm

Height (h2) = 12 – (2 + 2) = 8 cm

Volume = πr2h2 = π(1.5)2 x 8 = 18π cm3

∴ Volume of the model = 1.5π + 1.5π + 18π

= 21π

12.   Volume of a gulab jamun  ∴ Volume of 45 gulab jamuns = 1127.28 cm3

Volume of conical depression ∴ Volume of four conical depressions ∴ Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm3

14.   For cone, Radius of the top (r) = 5 cm and height (h) = 8 cm  For spherical lead shot, Radius (R) = 0.5 cm

Volume of spherical lead shot = Volume of water that flows out = Volume of the cone Let the number of lead shots dropped in the vessel be n. 15.   For lower cylinder, Base radius (r) = = 12 cm And Height (h) = 220 cm

Volume = πr2h = π(12)x 220 = 31680π cm3

For upper cylinder, Base Radius (R) = 8 cm And Height (H) = 60 cm

Volume = πr2H = π(8)2 x 60 = 3840π cm3

∴ Volume of the solid Iron pole = V of lower cylinder + V of upper cylinder

= 31680π + 3840π = 35520π

= 35520 x 3.14 = 111532.8 cm3

16.   For right circular cone, Radius of the base (r) = 60 cm And Height (h1) ) = 120 cm For Hemisphere, Radius of the base (r) = 60 cm For right circular cylinder, Radius of the base (r) = 60 cm

And Height (h2) = 180 cm

Volume = πr2h2 = π(60)2 x 180 = 648000πr cm3

Now, V of water left in the cylinder

= V of right circular cylinder – (V of right circular cone + V of hemisphere)

= 648000π – (144000π + 144000π)

= 360000π cm3 17.   Amount of water it holds  Hence, she is not correct. The correct volume is 346.51 cm3.

18.   For sphere, Radius (r) = 4.2 cm For cylinder, Radius (R) = 6 cm

Let the height of the cylinder be H cm.

Then, Volume = πR2H = (6)2 H cm3

According to question, Volume of sphere = Volume of cylinder 19.   Let the volume of resulting sphere be cm.

According to question, 20.  Diameter of well = 7 m

And         Depth of earth dug (h) = 20 m

Length of platform (l) = 22 m, Breadth of platform (b) = 14 m

Let height of the platform be h’ m

According to question,

Volume of earth dug = Volume of platform 21.   Diameter of well = 3 m

∴ Radius of well (r) = m and Depth of earth dug (h) = 14 m

Width of the embankment = 4 m

∴ Radius of the well with embankment Let the height of the embankment be h’ m

According to the question,

Volume of embankment = Volume of the earth dug 22.  For right circular cylinder, Diameter = 12 cm

∴ Radius (r) = = 6 cm and height (h) = 15 cm

For cone, Diameter = 6 cm

∴ Radius (r1) = = 3 cm and height (h1) = 12 cm

Let n cones be filled with ice cream.

Then, According to question,

Volume of n cones = Volume of right circular cylinder 23.  For silver coin, Diameter = 1.75 cm For cuboid, Length (l) = 5.5 cm, Breadth (b) = 10 cm and Height (h’) = 3.5 cm Let n coins be melted.

Then, According to question,

Volume of n coins = Volume of cuboid 24.   For cylindrical bucket, Radius of the base (r) = 18 cm and height (h) = 32 cm

Volume = πr2 h = π (18)2 x 32 = 10368π cm3

For conical heap, Height (h’) = 24 cm

Let the radius be r2 cm.

According to question, Volume of bucket = Volume of conical heap 25.  For canal, Width = 6 m and Depth = 1.5 m = Speed of flow of water = 10 km/h = 10 x 1000 m/h = 10000 m/h ∴ Speed of flow of water in 30 minutes ∴ Volume of water that flows in 30 minutes ∴ The area it will irrigate 26.   For cylindrical tank, Diameter = 10 m

∴ Radius (r) = = 5 m and Depth (h) = 2 m

∴ Volume = πr2h = π(5)2 x 2 = 50π m3

Rate of flow of water (h’) = 3 km/h = 3000 m/h = m/min = 50 m/min

For pipe, Internal diameter = 20 cm, therefore radius (r1) = 10 cm = 0.1 m

∴ Volume of water that flows per minute = (r)2 h’ = π(0.1)2 x 50 = m3

∴ Required time = = 100 minutes  28.   Let r1 cm and r2 cm be the radii of the ends (r1 > r2) of the frustum of the cone.

Then, 1 = 4 cm

r1 =18 cm                     ⇒                            πr1 = 9 cm

2πr2 = 6 cm                      ⇒                     πr2 = 3 cm

Now, CSA of the frustum = πr1(r1 + r2)

= r1 + πr2)1 = (9 + 3) x 4 = 48 cm2

29.   Here, r1 = 10 cm, r2 = 4 cm and l = 15 cm 30.   Here, r1 = 20 cm, r2 = 8 cm and h = 16 cm ∴ Area of the metal sheet used = 1959.36 cm2

Cost of metal sheet = 1959.36 = 156.7488 = Rs. 156.75

Diameter of the wire = cm

∴ Radius of the wire = cm

Let the length of the wire be l cm.

Then, Volume of the wire According to the question, 32.  Number of rounds to cover 12 cm, i.e. 120 mm Here, Diameter = 10 cm, Radius (r) = Length of the wire in completing one round = 2πr = 2π x 5 = 10π cm

Length of the wire in completing 40 rounds = 10π x 40 = 400π cm

Radius of the copper wire Volume of wire = = 9π cm3

Mass of the wire = 9 x (3.142) x 8.88 = 787.98 g

Volume of the double cone Surface area of the double cone = 52.75 cm2

34.  Volume of cistern = 150 x 120 x 110 = 1980000 cm3

Volume of water = 129600 cm3

Volume of cistern to be filled = 1980000 – 129600 = 1850400 cm3

Volume of a brick = 22.5 x 7.5 x 6.5 = 1096.875 cm3

Let n bricks be needed.

Then, water absorbed by n bricks  Hence, the two are not approximately equivalent.

36.  Slant height of the frustum of the cone (l) = Area of the tin sheet required = CSA of cylinder + CSA of the frustum 37.  According to the question, the frustum if difference of the two cones OAB and OCD (in figure).

For frustum, height = h,  slant height = l  and radii of the bases = r1   and  r2    (r1 > r2 )

OP =  h1,  OA = OB = l

∴ Height of the cone =  h1 – h MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams!

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