Mensuration 3D - Study Mode
[#131] The ratio of total surface area and volume of a sphere is 1 : 7. This sphere is melted to form small spheres of equal size. The radius of each small sphere is $$frac{1}{6}$$ the radius of the large sphere. What is the sum (in cm 2 ) of curved surface areas of small spheres?
Correct Answer
(D) 33264
Explanation
Solution: $$eqalign{
& frac{{{ ext{Total surface area}}}}{{{ ext{Volume}}}} = frac{1}{7} cr
& frac{{4pi {r^2}}}{{frac{4}{3}pi {r^3}}} = frac{1}{7} cr
& r = 21{ ext{ cm}} cr
& { ext{Radius of small sphere}} = frac{1}{6} imes r cr
& = frac{1}{6} imes 21 cr
& = frac{7}{2}{ ext{ cm}} cr
& { ext{Number of small sphere}} cr
& = frac{{{ ext{Volume of large sphere}}}}{{{ ext{Volume of small sphere}}}} cr
& = frac{{frac{4}{3}pi {R^3}}}{{frac{4}{3}pi {r^3}}} cr
& = frac{{21 imes 21 imes 21}}{{frac{7}{2} imes frac{7}{2} imes frac{7}{2}}} cr
& = 27 imes 8 cr
& = 216 cr
& { ext{Curved surface area of small sphere}} cr
& = 216 imes 4pi {r^2} cr
& = 216 imes 4 imes frac{{22}}{7} imes frac{7}{2} imes frac{7}{2} cr
& = 33264{ ext{ c}}{{ ext{m}}^2} cr} $$
[#132] The perimeter of the triangular base of a right prism is 15 cm and radius of the in circle of the triangular base is 3 cm. If the volume of the prism be 270 cm 3 , then the height of the prism is
Correct Answer
(D) 12 cm
Explanation
Solution: $$eqalign{
& r{ ext{ - inradius of incircle of triangle}} cr
& { ext{Perimeter}} = 15{ ext{ cm }}left( {{ ext{given}}}
ight) cr
& herefore { ext{Semiperimeter}}left( S
ight) = frac{{15}}{2}{ ext{cm}} cr
& { ext{Inradius of any triangle}} cr
& r Rightarrow frac{Delta }{S} cr
& r = frac{{{ ext{area}}}}{{{ ext{semiperimeter}}}} cr
& { ext{Where }}Delta { ext{ is the area of triangle }} cr
& herefore r{ ext{ }} = { ext{ }}3{ ext{ cm }}left( {{ ext{given}}}
ight) cr
& Rightarrow 3 = frac{{{ ext{area of triangle}}}}{{frac{{15}}{2}}} cr
& Rightarrow 3 imes frac{{15}}{2} = { ext{area of triangle}} cr
& Rightarrow frac{{45}}{2}{ ext{cm}} = { ext{area of triangle}} cr
& herefore { ext{Volume of prism}} cr
& Rightarrow 270{ ext{ c}}{{ ext{m}}^3},left( {{ ext{given}}}
ight) cr
& herefore 270 = h imes frac{{45}}{2} cr
& h = 12{ ext{ cm}} cr} $$
[#133] 40 men took a dip in a pool 30 m long and 25 m broad. If the average water displaced by a man is 5 m 3 , then what will be the rise (in cm) in level of the pool?
Correct Answer
(B) 26.66
Explanation
Solution: Water displaced by 40 men = 40 × 5 = 200 m 3 Rise in level of pool $$eqalign{
& = frac{{200}}{{30 imes 25}} cr
& = frac{8}{{30}} imes 100 cr
& = 26.66{ ext{ cm}} cr} $$
[#134] The ratio of the of two cones is 5 : 6 and their volumes are in the ratio 8 : 9. The ratio of their height is:
Correct Answer
(A) 32 : 25
Explanation
Solution: $$eqalign{
& {r_1}:{r_2} = 5:6 cr
& {v_1}:{v_2} = 8:9 cr
& frac{{{v_1}}}{{{v_2}}} = frac{{pi r_1^2{h_1}}}{{pi r_2^2{h_2}}} cr
& frac{8}{9} = frac{{25 imes {h_1}}}{{36 imes {h_2}}} cr
& 32:25 = {h_1}:{h_2} cr} $$
[#135] A 15 m deep well with radius 2.8 m is dug and the earth taken out from it is spread evenly to from a platform of breadth 8 m and height 1.5 m. What will be the length of the platform? $$left( {{ ext{Take }}pi = frac{{22}}{7}}
ight)$$
Correct Answer
(B) 30.8 m
Explanation
Solution: $$eqalign{
& { ext{Volume of well}} = pi {r^2}h cr
& = frac{{22}}{7} imes {2.8^2} imes 15 cr
& { ext{Volume of platform}} = 8 imes 1.5 imes x cr
& { ext{Volume of platform}} = { ext{Volume of well}} cr
& 8 imes 1.5 imes x = frac{{22}}{7} imes 2.8 imes 2.8 imes 15 cr
& x = 30.8{ ext{ m}} cr} $$