# Class 10 RD Sharma Solutions – Chapter 16 Surface Areas and Volumes – Exercise 16.2 | Set 3

### Question 25. A hemispherical depression is cut from one face of a cubical wooden block of the edge 21 cm such that the diameter of the hemispherical surface is equal to the edge of the cubical surface. Determine the volume and the total surface area of the remaining block.

**Solution: **

According to the questionEdge of the cubical wooden block (e) = 21 cm,

Diameter of the hemisphere = Edge of the cubical wooden block = 21 cm,

So, the radius of the hemisphere (r) = 10.5 cm

Now, we find the volume of the remaining blockV = volume of the cubical block – volume of the hemisphere

V = e

^{3}− (2/3 πr^{3})= 21

^{3}− (2/3 π10.5^{3}) = 6835.5 cm^{3}

Now, we find the surface area of the blockA

_{S}= 6(e^{2}) = 6(21^{2})

Now, we find the curved surface area of the hemisphereA

_{C}= 2πr^{2 }= 2π10.5^{2}The base area of the hemisphere is

A

_{B}= πr^{2 }= π10.5^{2}So, the remaining surface area of the box

= A

_{S}– (A_{C}+ A_{B})= 6(21

^{2}) – (2π10.5^{2}+ π10.5^{2}) = 2992.5 cm^{2}

Hence, the remaining surface area of the block is 2992.5 cm^{2}and the volume is 6835.5 cm^{3}

### Question 26. A boy is playing with a toy which 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 surface area of the toy.

**Solution:**

According to the questionRadius of the cone (R) = 21 cm,

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

Now, we find thevolume of the coneV1 = 1/3 × πR

^{2}LHere, L is the slant height of the comne

= 1/3 × π21

^{2}L

Now, we find the volume of the HemisphereV2 = 2/3 × πR

^{3}= 2/3 × π x 21

^{3}= 169714.286 cm^{3}Since, volume of the cone = 2/3 of the hemisphere,

So,

V1 = 2/3 V2

V1 = 2/3 ×169714.286

1/3 × π x 21

^{2}x L = 2/3×π x 21^{3}L = 28 cm

So, the height of the cone is 28 cm

Now, we find the curved surface area of the coneA1 = πRL

A1 = π × 21 × 28 cm

^{2}

Now, we find the curved curface area of the hemisphereA2 = 2πR

^{2}A2 = 2π(21

^{2}) cm^{2}Hence, the total surface area of the toy

A = A1 + A2

= π × 21 × 28 + 2π(21

^{2})= 5082 cm

^{2}

Hence, the curved surface area of the toy is 5082 cm^{2}

### Question 27. Consider a solid which is in the form of a cone surmounted on hemisphere. The radius of each of them is being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.

**Solution:**

According to the questionRadius of the cone = Radius of the hemisphere = R = 3.5 cm,

Total height of the solid (H) = 9.5 cm,

Slant height of the cone = H – R

L = 9.5 – 3.5 = 6 cm

Now, we find the volume of the coneV1 = 1/3 × πR

^{2}L= 1/3 × π x 3.5

^{2}x 6 cm^{3}

Now, we find thevolume of the hemisphereV2 = 2/3 × πR

^{3}= 2/3 × π5

^{3}cm^{3}—————(ii)Hence, the total volume of the solid is

V = V1 + V2

= 1/3 × π x 3.5

^{2}x 6 + 2/3 × π x 5^{3}= 166.75 cm

^{3}

Hence, the volume of the solid is 166.75 cm^{3}

### Question 28. A wooden toy is made by scooping out a hemisphere of same radius from each end of the solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of the wood in the toy.

**Solution: **

According to the questionRadius of the cylinder = Radius of the hemisphere (r) = 3.5 cm

Height of the hemisphere (h) = 10 cm

Now, we find the volume of the cylinderV1 = π × r

^{2}× h= π × 3.5

^{2}× 10

Now, we find the volume of the hemisphereV2 = 2/3 × πr

^{3}= 2/3 × π x 3.5

^{3}cm^{3}Hence, the volume of the wood in the toy is

V = V1 – 2(V2)

= π × 3.5

^{2}× 10 – 2(2/3 × π x 3.5^{3})= 205.33 cm

^{3}

Hence, the volume of the wood in the toy is 205.33 cm^{3}

### 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.

**Solution: **

According to the questionDiameter of the wooden solid = 7 cm,

Radius of the wooden solid = 3.5 cm,

Now, we find the volume of the cubeV1 = a

^{3},= 3.5

^{3}

Now, we find the volume of sphereV2 = 4/3 × π × r

^{3}= 4/3 × π × 3.5

^{3}Hence, the volume of the wood left

V = V1 + V2

= 3.5

^{3}− 4/3 × π × 3.5^{3}= 163.33 cm

^{3}

Hence, the volume of the wood left is 163.33 cm^{3}

### Question 30. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.

**Solution: **

According to the questionHeight of the cylinder = Height of the cone = H = 2.8 cm

Diameter of the cylinder = 4.2 cm,

So, the radius of the cylinder = Radius of the cone = R = 2.1 cm

Now, we find the curved surface area of the cylindrical partA1 = 2πRH

= 2π(2.8)(2.1) cm

^{2}

Now, we find the curved surface area the cone = πRLA2 = π × 2.1 × 2.8 cm

^{2}The area of the cylindrical base = AB = πr

^{2}= π(2.1)^{2}Hence, the total surface area of the remaining solid is

A = A1 + A2 + AB

= 2π(2.8)(2.1) + π × 2.1 × 2.8 + π(2.1)

^{2}= 36.96 + 23.1 + 13.86

= 73.92 cm

^{2}

Hence, the total surface area of the remaining solid is 73.92 cm^{2}

### Question 31. The largest cone is carved out from one face of the solid cube of side 21 cm. Find the volume of the remaining solid.

**Solution: **

According to the questionDiameter of the cone = 21 cm,

So, the radius of the cone = 10.5 cm,

The height of the cone is equal to the side of the cone,

Now, we find the volume of the cubeV1 = e

^{3}= 10.5

^{3}

Now, we find the volume of the coneV2 = 1/3 × πr

^{2}L= 1/3 × π10.5221 cm

^{3}Hence, the volume of the remaining solid

V = V1 – V2

= 10.53 – 1/3 × π x 10.5

^{2}x 21= 6835.5 cm

^{3}

Hence, the volume of the remaining solid is 6835.5 cm^{3}

### Question 32. A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of the hemisphere is 3.5 cm and the total wood used in the making of toy is 166 5⁄6 cm^{3}. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of Rs. 10 per square cm.

**Solution: **

According to the questionRadius of the hemisphere = 3.5 cm,

Volume of the solid wooden toy = cm

^{3},Hence, the volume of the solid wooden toy = volume of the cone + volume of the hemisphere

1/3 × πr

^{2}L + 2/3 × πr^{3}= 100161/3 × π3.5

^{2}L + 2/3 × π3.5^{3}= 10016L + 7 = 13

L = 6 cm

Height of the solid wooden toy = Height of the cone + Radius of the hemisphere

= 6 + 3.5 = 9.5 cm

Now, we find thecurved surface area of the hemisphereA = 2πR

^{2}= 2π(3.5

^{2}) = 77 cm^{2}Cost of painting the hemispherical part of the toy = 10 x 77 = Rs. 770

Hence, the cost of painting the hemispherical part of the toy is 770 rupees

### Question 33. How many spherical bullets can be made out of a solid cube of lead whose edge measures 55 cm and each of the bullet being 4 cm in diameter?

**Solution:**

According to the questionDiameter of the bullet = 4 cm,

Radius of the spherical bullet = 2 cm

So, the volume of a spherical bullet is

V = 4/3 × π × r

^{3}= 4/3 × π × 2

^{3}= 4/3 × 22/7 × 2

^{3}= 33.5238 cm^{3}Let’s assume that the total number of bullets be a.

So, the volume of ‘a’ number of the spherical bullets

V1 = V x a

= (33.5238 a) cm

^{3}

and theVolume of the solid cube = (55)^{3}= 166375 cm^{3}

Now, we find the volume of ‘a’ number of the spherical bullets = Volume of the solid cube33.5238 a = 166375

a = 4962.892

Hence, the total number of the spherical bullets is 4963

### Question 34. Consider a children’s toy which is in the form of a cone at the top having a radius of 5 cm mounted on a hemisphere which is the base of the toy having the same radius. The total height of the toy is 20 cm. Find the total surface area of the toy.

**Solution: **

According to the questionRadius of the conical portion of the toy (r) = 5 cm,

Total height of the toy (h) = 20 cm,

Length of the cone = L = 20 – 5 = 15 cm,

Now, we find thecurved surface area of the coneA1 = πrL

= π(5)(15) = 235.7142 cm

^{2}

Now, we find the curved surface area of the hemisphereA2 = 2πr

^{2}= 2π(5)

^{2}= 157.1428 cm^{2}Therefore, The total surface area of the toy is

A = A1 + A2

= 235.7142 + 157.1428

= 392.857 cm

^{2}

Hence, the total surface area of the children’s toy is 392.857 cm^{2}

### Question 35. A boy is playing with a toy conical in shape and is surmounted with hemispherical surface. Consider a cylinder in with the toy is inserted. The diameter of cone is the same as that of the radius of cylinder and hemispherical portion of the toy which is 8 cm. The height of the cylinder is 6 cm and the height of the conical portion of the toy is 3 cm. Assume a condition in which the boy’s toy is inserted in the cylinder, then find the volume of the cylinder left vacant after insertion of the toy.

**Solution: **

According to the questionDiameter of the cone = Diameter of the Cylinder = Diameter of the Hemisphere = 8 cm,

So, Radius of the cone = Radius of the cylinder = Radius of the Hemisphere = r = 4 cm

Height of the conical portion (L) = 3 cm,

Height of the cylinder (H) = 6 cm,

Now, we find the volume of the cylinderV1 = π × r

^{2}× H= π × 4

^{2}× 6 = 301.7142 cm^{3}

Now, we find the vcolume of the Conical part of the toyV2 = 1/3 × πr

^{2}L= 1/3 × π4

^{2}× 3 = 50.2857 cm^{3}

Now, we find the volume of the hemispherical part of the toyV3 = 2/3 × πr

^{3}= 2/3 × π4

^{3}= 134.0952 cm^{3}Hence, the remaining volume the cylinder left vacant after insertion of the toy is

V = V1 – (V2 + V3)

= 301.7142 – (50.2857 + 134.0952) = 301.7142 – 184.3809

= 117.3333 cm

^{3}

Hence, the remaining volume the cylinder left vacant after insertion of the toy is 117.3333 cm3

### Question 36. Consider 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 the right circular cylinder full of water such that it touches bottoms. Find the volume of the water left in the cylinder, if radius of the cylinder is 60 cm and its height is 180 cm.

**Solution: **

According to the questionRadius of the circular cone (r) = 60 cm,

Height of the circular cone (L) = 120 cm,

Radius of the hemisphere (r) = 60 cm,

Radius of the cylinder (R) = 60 cm,

Height of the cylinder (H) = 180 cm,

Now, we find the volume of the circular coneV1 = 1/3 × πr

^{2}L= 1/3 × π60

^{2}× 120= 452571.429 cm

^{3}

Now, we find the volume of the hemisphereV2 = 2/3 × πr

^{3}= 2/3 × π60

^{3}= 452571.429 cm^{3}

Now, we find the volume of the cylinderV3 = π × R

^{2}× H= π × 60

^{2}× 180 = 2036571.43 cm^{3}Hence, the volume of water left in the cylinder is

V = V3 – (V1 + V2)

= 2036571.43 – (452571.429 + 452571.429)

= 2036571.43 – 905142.858 = 1131428.57 cm

^{3}= 1.1314 m

^{3}

Hence, the volume of the water left in the cylinder is 1.1314 m^{3}

### Question 37. Consider a cylindrical vessel with internal diameter 20 cm and height 12 cm is full of water. A solid cone of base diameter 8 cm and height 7 cm is completely immersed in water. Find the value of water when

### (i) Displaced out of the cylinder

### (ii) Left in the cylinder

**Solution: **

According to the questionInternal diameter of the cylindrical vessel = 20 cm,

So, the radius of the cylindrical vessel (r) = 10 cm

Height of the cylindrical vessel (h) = 12 cm,

Base diameter of the solid cone = 8 cm,

So, the radius of the solid cone (R) = 4 cm

Height of the cone (L) = 7 cm,

(i)Now, we find the volume of water displaced out from the cylinder which is equal to the volume of the coneV1 = 1/3 × πR

^{2}L= 1/3 × π4

^{2}× 7 = 117.3333 cm^{3}

Hence, the volume of the water displaced out of the cylinder is 117.3333 cm^{3}

(ii)The volume of the cylindrical vessel isV2 = π × r

^{2}× h= π × 102 × 12 = 3771.4286 cm

^{3}Hence, the volume of the water left in the cylinder is

V = V2 – V1

= 3771.4286 – 117.3333 = 3654.0953 cm

^{3}

Hence, the volume of the water left in cylinder is 3654.0953 cm^{3}