Mensuration 3D - Study Mode
[#101] From a solid cylindrical wooden block of height 18 cm and radius 7.5 cm, a conical cavity of the same height and same radius is taken out. What is total surface area (in cm 2 ) of the remaining solid?
Correct Answer
(D) 472.5π
Explanation
Solution: $$eqalign{
& {l^2} = {r^2} + {h^2} cr
& {l^2} = {left( {7.5}
ight)^2} + {left( {18}
ight)^2} cr
& {l^2} = frac{{225}}{4} + 324 cr
& {l^2} = frac{{1521}}{4} cr
& l = frac{{39}}{2} cr
& { ext{Total surface area}} = pi rl + 2pi rh + pi {r^2} cr
& = pi rleft[ {l + 2h + r}
ight] cr
& = pi imes frac{{15}}{2}left[ {frac{{39}}{2} + 36 + frac{{15}}{2}}
ight] cr
& = pi imes frac{{15}}{2}left[ {frac{{126 imes 15}}{2}}
ight] cr
& = left[ {frac{{945}}{2}}
ight]pi cr
& = 472.5pi cr} $$
[#102] The volume of a cylinder is 4312 cm 3 . Its curved surface area is one-third of its total surface area. Its curved surface area (in cm 2 ) is: $$left( {{ ext{Take }}pi = frac{{22}}{7}}
ight)$$
Correct Answer
(C) 616
Explanation
Solution: Curved surface area = $$frac{1}{3}$$ × total surface area 2πrh = $$frac{1}{3}$$ × [2πrh + 2πr 2 ] 6πrh = 2πrh + 2πr 2 4πrh = 2πr 2 4h = 2r h : r = 1x : 2x Volume of cylinder = πr 2 h = 4312 $$frac{{22}}{7}$$ × (2x) 2 × x = 4312 $$frac{{22}}{7}$$ × 4x 3 = 4312 x 3 = 49 × 7 x = 7 Curved surface area = 2πrh = 2 × $$frac{{22}}{7}$$ × (2 × 7) × 7 = 22 × 28 = 616 cm 2
[#103] A right triangular pyramid XYZB is cut from cube as shown in figure. The side of cube is 16 cm. X, Y and Z are mid points of the edges of the cube. What is the total surface area (in cm 2 ) of the pyramid?
Correct Answer
(D) $$32left[ {3 + left( {sqrt 3 }
ight)}
ight]$$
Explanation
Solution: BX = BY = 8 cm ∴ XY = YZ = XZ = 8√2 $$l$$ 2 = 8 2 - (4√2) 2 $$l$$ 2 = 32 $$l$$ = 4√2 Total surface area = $$frac{1}{2}$$ × Perimeter of base × $$l$$ + Area of base = $$frac{1}{2} imes 3 imes 8sqrt 2 imes 4sqrt 2 + frac{{sqrt 3 }}{4}{left( {8sqrt 2 }
ight)^2}$$ = 96 + 32√3 = 32(3 + √3) cm 2
[#104] A spherical ball of radius 1 cm is dropped into a conical vessel of radius 3 cm and slant height 6 cm. The volume of water (in cm 3 ), that can just immerse the ball, is
Correct Answer
(A) $$frac{{5pi }}{3}$$
Explanation
Solution: $$eqalign{
& Delta ABC = { ext{equilateral }}Delta cr
& herefore angle ACB = {60^ circ } cr
& & angle BCP = {30^ circ } cr
& Delta CDO,,angle CDO = {90^ circ } cr
& left( {{ ext{Angle between radius and tangent is }}{{90}^ circ }}
ight) cr} $$ $$eqalign{
& OD = 1P = 1{ ext{ cm}} cr
& OC = 2P = 2left( 1
ight) = 2{ ext{ cm}} cr
& { ext{Then, }}CZ = OC + OZ = 2 + 1 = 3{ ext{ cm}} cr
& Delta CZY,,angle CZY = {90^ circ } cr
& CZ = sqrt 3 P = 3{ ext{ cm}} cr
& YZ = 1P = sqrt 3 { ext{ cm}} cr
& { ext{Now, in cone }}XYC cr
& r = ZY = sqrt 3 { ext{ cm}} cr
& h = CZ = 3{ ext{ cm}} cr
& { ext{Volume of cone}} = frac{1}{3}pi {r^2}h cr
& = frac{1}{3}pi {left( {sqrt 3 }
ight)^2}left( 3
ight) cr
& = 3pi { ext{ c}}{{ ext{m}}^2} cr
& { ext{Volume of sphere}} = frac{4}{3}pi r_s^3 cr
& left( { herefore {r_s} = 1{ ext{ cm}}}
ight) cr
& = frac{4}{3}pi { ext{ c}}{{ ext{m}}^3} cr
& { ext{Volume of water that can immerse the ball}} cr
& = left( {3pi - frac{{4pi }}{3}}
ight){ ext{c}}{{ ext{m}}^3} cr
& = frac{{5pi }}{3}{ ext{ c}}{{ ext{m}}^3} cr} $$
[#105] If the radius of the base of a cone is doubled, and the volume of the new cone is three times the volume of the original cone, then what will be the ratio of the height of the original cone to that of the new cone?
Correct Answer
(B) 4 : 3
Explanation
Solution: $$eqalign{
& R o 1:2 cr
& V o 1:3 cr
& { ext{Height}} o h:H cr
& frac{1}{3} = frac{{1 imes h}}{{4 imes H}} cr
& left{ {frac{1}{3},,pi { ext{ is constant}}}
ight. cr
& frac{h}{H} = frac{4}{3} cr} $$