"Time is a great healer, death is a better one "

- DEBJIT

UNIT-12

PROBLEMS BASED ON CONVERSION OF SOLIDS

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylindrical tub is 5 cm and its height is 22 10.5 cm, find the volume of water left in the cylindrical tub (use π = ] 7 (Ans: 683.83 cm³) Ans: No. of solid = vol of cone + vol of hemisphere 1 2 = π r2 h + π r3 3 3 1 = π r2 [h +2 r] 3 On substituting we get, = 141.17 cm3 vol of cylinder = π r2 h On substituting we get, = 825 cm3 volume of H2O left in the cylinder = 825 – 141.17 = 683.83 cm3 2. A bucket of height 8 cm and made up of copper sheet is in the form of frustum of right circular cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate i) the height of the cone of which the bucket is a part ii) the volume of water which can be filled in the bucket iii) the area of copper sheet required to make the bucket (Leave the answer in terms of π (Ans: 129 π cm2) 1.

Ans: Let total height be h h 3 = (similar ∆’s ) => h+8 9

=> h = 4 cm ∴ ht. of cone which bucket is a part = 4 cm Substitute to get Ans.: for ii) iii)

95

3. A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is 6 : π .

Ans: S.A. of sphere = S.A of cube 4π r2 = 6a2

6a 2 r= 4Π

4 Πγ 3 v1 3 = ∴ ratio of their volume v2 a3 On simplifying & substituting, we get √6 : √π

4.

A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Ans : 3768cu.cm,1318.8 Sq.cm)

Ans: BC = 152 + 202 = 25 cm Apply Py. Th to right ∆ OAB & OAC to get OB = 9cm OA = 12cm Vol of double cone = vol of CAA1 + vol of BAA1 1 1 = π × 122 × 16 + π × 122 × 9 3 3 3 = 3768 cm

SA of double cone = CSA of CAA1 + CSA...

PROBLEMS BASED ON CONVERSION OF SOLIDS

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylindrical tub is 5 cm and its height is 22 10.5 cm, find the volume of water left in the cylindrical tub (use π = ] 7 (Ans: 683.83 cm³) Ans: No. of solid = vol of cone + vol of hemisphere 1 2 = π r2 h + π r3 3 3 1 = π r2 [h +2 r] 3 On substituting we get, = 141.17 cm3 vol of cylinder = π r2 h On substituting we get, = 825 cm3 volume of H2O left in the cylinder = 825 – 141.17 = 683.83 cm3 2. A bucket of height 8 cm and made up of copper sheet is in the form of frustum of right circular cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate i) the height of the cone of which the bucket is a part ii) the volume of water which can be filled in the bucket iii) the area of copper sheet required to make the bucket (Leave the answer in terms of π (Ans: 129 π cm2) 1.

Ans: Let total height be h h 3 = (similar ∆’s ) => h+8 9

=> h = 4 cm ∴ ht. of cone which bucket is a part = 4 cm Substitute to get Ans.: for ii) iii)

95

3. A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is 6 : π .

Ans: S.A. of sphere = S.A of cube 4π r2 = 6a2

6a 2 r= 4Π

4 Πγ 3 v1 3 = ∴ ratio of their volume v2 a3 On simplifying & substituting, we get √6 : √π

4.

A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Ans : 3768cu.cm,1318.8 Sq.cm)

Ans: BC = 152 + 202 = 25 cm Apply Py. Th to right ∆ OAB & OAC to get OB = 9cm OA = 12cm Vol of double cone = vol of CAA1 + vol of BAA1 1 1 = π × 122 × 16 + π × 122 × 9 3 3 3 = 3768 cm

SA of double cone = CSA of CAA1 + CSA...

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