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} = 4^{3} ⇒ 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 cm^{2}

**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

∴ 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 =

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 m^{2}

= 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 = 2*πrH + *2 (27πr^{2})

**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 (*h*_{1}) = 2 cm

For lowe

r conical portion, Volume = 1.57π cm^{3} For central cylindrical portion

Radius of the base *(r) = *1.5 cm

Height (h_{2}) = 12 – (2 + 2) = 8 cm

Volume = π*r ^{2}h_{2} = π*(1.5)

*x 8 = 18π cm*

^{2}^{3}

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

* = *21π

∴ Volume of 45 gulab jamuns = 1127.28 cm^{3}

∴ Volume of four conical depressions

∴ Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm^{3}

**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 = π*r ^{2}h = *π(12)

^{2 }x 220 = 31680π cm

^{3}

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

Volume = π*r ^{2}H* = π(8)

^{2}x 60 = 3840π cm

^{3}

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

= 31680π + 3840π = 35520π

= 35520 x 3.14 = 111532.8 cm^{3}

**16. ** **For right circular cone, **Radius of the base *(r) = *60 cm

And Height (h_{1}) ) = 120 cm

**For Hemisphere, **Radius of the base *(r) = *60 cm

**For right circular cylinder, **Radius of the base *(r) = *60 cm

And Height (h_{2}) = 180 cm

Volume = π*r ^{2}h_{2} = π*(60)

^{2}x 180 = 648000πr cm

^{3}

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 cm^{3}

**17.** Amount of water it holds

Hence, she is not correct. The correct volume is 346.51 cm^{3}.

**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 = πR^{2}H = (6)^{2} H cm^{3}

According to question, Volume of sphere = Volume of cylinder

**19.** Let the volume of resulting sphere be *r *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 (*r _{1}*) =

^{ }= 3 cm and height (h

*) = 12 cm*

_{1}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 = π*r ^{2} h = π *(18)

^{2}x 32 = 10368π cm

^{3}

**For conical heap, **Height *(h’) = *24 cm

Let the radius be r_{2} 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 = π*r ^{2}h = *π(5)

^{2}x 2 = 50π m

^{3}

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 (r_{1}) = 10 cm = 0.1 m

∴ Volume of water that flows per minute = *(r) ^{2 }h’ = *π(0.1)

^{2 }x 50 = m

^{3}

∴ Required time = = 100 minutes

**28. ** Let *r*_{1}* *cm and *r*_{2} cm be the radii of the ends (r* _{1}* >

*r*of the frustum of the cone.

_{2})Then, 1* = *4 cm

2π*r*_{1}* =*18 cm ⇒ π*r*_{1} = 9 cm

*2*π*r*_{2} = 6 cm ⇒ π*r*_{2} = 3 cm

Now, CSA of the frustum = π*r*_{1}(r_{1} + r_{2})

*= *(π*r*_{1} + π*r _{2}*)1 = (9 + 3) x 4 = 48 cm

^{2}

**29. ** Here, r_{1} = 10 cm, r_{2}* = *4 cm and *l = *15 cm

**30. ** Here, r_{1} = 20 cm, r_{2}* = *8 cm and *h = *16 cm

∴ Area of the metal sheet used = 1959.36 cm^{2}

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

Let the length of the wire be *l *cm.

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 = 2π*r = *2π x 5 = 10π cm

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

Mass of the wire = 9 x (3.14^{2}) x 8.88 = 787.98 g

Surface area of the double cone

= 52.75 cm^{2}

**34.** Volume of cistern = 150 x 120 x 110 = 1980000 cm^{3}

Volume of water = 129600 cm^{3}

Volume of cistern to be filled = 1980000 – 129600 = 1850400 cm^{3}

Volume of a brick = 22.5 x 7.5 x 6.5 = 1096.875 cm^{3}

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).

**F****or** **frustum**, height = *h*, slant height = *l** *and radii of the bases = r_{1} and r_{2} (r_{1} > r_{2} )

OP = h_{1}, OA = OB = *l*

∴ Height of the cone = *h _{1}* –

*h*