Mensuration 3D - Study Mode
[#126] A solid brass sphere of radius 2.1 dm is converted into a right circular cylindrical rod of length 7 cm. The ratio of total surface areas of the rod to the sphere is
Correct Answer
(C) 7 : 3
Explanation
Solution: $$eqalign{
& frac{4}{3}pi {R^3} = pi {r^2}h cr
& frac{4}{3}pi imes 21 imes 21 imes 21 = pi imes {R^2} imes 7 cr
& {R^2} = 21 imes 21 imes 4 cr
& R = 42{ ext{ cm}} cr
& frac{{{ ext{total surface area of rod}}}}{{{ ext{total surface area of sphere}}}} cr
& = frac{{2pi rleft( {h + r}
ight)}}{{4pi {R^2}}} cr
& = frac{{42left( {42 + 7}
ight)}}{{2 imes 21 imes 21}} cr
& = frac{{49}}{{21}} cr
& = frac{7}{3} cr
& = 7:3 cr} $$
[#127] A solid cylinder has radius of base 14 cm and height 15 cm. 4 identical cylinders are cut from each base as shown in the given figure. Height of small cylinder is 5 cm. What is the total surface area (in cm 2 ) of the remaining part?
Correct Answer
(C) 3432
Explanation
Solution: Total surface area of remaining part 6 = 16πrh + 2πR 2 - 8πr 2 + 2πRH + 8πr 2 = 2π(8rh + R 2 - 4r 2 + RH + 4r 2 ) = 2π(8 × 3.5 × 5 + (14) 2 - 4 × 3.5 × 3.5 + 14 × 15 + 4 × (3.5) 2 ) = 3432
[#128] A cylindrical vessel whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the Cylinder exactly. The depth of water in the vessel before the sphere was put, is
Correct Answer
(C) $$frac{7}{3}{ ext{ cm}}$$
Explanation
Solution: Height of water after ball is immersed = 3.5 × 2 = 7 cm $$eqalign{
& Rightarrow { ext{Volume of water}} = pi {r^2}h - frac{4}{3}pi {r^3} cr
& = pi {r^2}left( {h - frac{4}{3}r}
ight) cr
& = frac{{22}}{7} imes 3.5 imes 3.5left( {7 - frac{4}{3} imes 3.5}
ight) cr
& = 11 imes 3.5left( {frac{7}{3}}
ight) cr
& = frac{{269.5}}{3}{ ext{ c}}{{ ext{m}}^3} cr} $$ Volume of water before ball was immersed $$eqalign{
& Rightarrow pi {left( {3.5}
ight)^2} imes h = frac{{269.5}}{3} cr
& h = frac{{269.5 imes 7}}{{3 imes 3.5 imes 3.5 imes 22}} cr
& h = frac{7}{3}{ ext{ cm}} cr} $$
[#129] A hemispherical depression of diameter 4 cm is cut out from each face of a cubical block of sides 10 cm. Find the surface area of the remaining solid (in cm 2 ). $$left( {{ ext{Use }}pi = frac{{22}}{7}}
ight)$$
Correct Answer
(C) $$675frac{3}{7}$$
Explanation
Solution: Side of cubical block = 10 cm ∴ Area of each face of cubical block = 10 2 = 100 cm 2 Radius of Hemisphere = $$frac{4}{2}$$ = 2 cm Surface area of hemisphere $$eqalign{
& = 2pi {r^2} cr
& = 2 imes frac{{22}}{7} imes {2^2}{ ext{ c}}{{ ext{m}}^2} cr
& = frac{{176}}{7}{ ext{ c}}{{ ext{m}}^2} cr} $$ Total surface area of hemisphere $$ = 6 imes frac{{176}}{7} = frac{{1056}}{7}{ ext{ c}}{{ ext{m}}^2}$$ Remaining surface area of each face of cubical block $$eqalign{
& = {10^2} - 2pi r cr
& = 100 - 2 imes frac{{22}}{7} imes 2{ ext{ c}}{{ ext{m}}^2} cr
& = 100 - frac{{88}}{7}{ ext{ c}}{{ ext{m}}^2} cr
& = frac{{612}}{7}{ ext{ c}}{{ ext{m}}^2} cr} $$ ∴ Total surface area of 6 remaining cubical block $$ = 6 imes frac{{612}}{7} = frac{{3672}}{7}{ ext{ c}}{{ ext{m}}^2}$$ ∴ Surface area of remaining solid $$eqalign{
& = left( {frac{{1056}}{7} + frac{{3672}}{7}}
ight){ ext{c}}{{ ext{m}}^2} cr
& = frac{{4728}}{7}{ ext{ c}}{{ ext{m}}^2} cr
& = 675frac{3}{7}{ ext{ c}}{{ ext{m}}^2} cr} $$
[#130] The lateral surface area of frustum of a right circular cone if the area of its base is 16π cm 2 and the diameter of circular upper surface is 4 cm and slant height 6 cm, will be
Correct Answer
(C) 36π cm 2
Explanation
Solution: Base Area = 16π πR 2 = 16π R = 4 Given r = 2 ∵ ΔABC ≅ ΔADE $$eqalign{
& herefore frac{{{ ext{BC}}}}{{{ ext{DE}}}} = frac{{{ ext{AC}}}}{{{ ext{AE}}}} cr
& frac{4}{8} = frac{6}{{{ ext{AE}}}} cr
& { ext{AE}} = 12 = { ext{L}} cr} $$ Surface Area of frustum = πRL - πr$$l$$ = π × 4 × 12 - π × 2 × 6 = 48π - 12π = 36π