Volume And Surface Area - Study Mode

[#61] If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?
Correct Answer

(B) 1 : 4

Explanation

Solution: Let original radius = R Then, new radius = $$frac{{ ext{R}}}{2}$$ $$eqalign{
& herefore frac{{{ ext{Volume of reduced cylinder }}}}{{{ ext{Volume of original cylinder}}}} cr
& = frac{{pi imes {{left( {frac{R}{2}}
ight)}^2} imes h}}{{pi imes {R^2} imes h}} cr
& = frac{1}{4},Or,1:4 cr} $$

[#62] The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is :
Correct Answer

(C) $$36$$

Explanation

Solution: Let the length of each pipe be $$l$$ inches Then, volume of water in thinner pipe : $$eqalign{
& = left[ {pi imes {{left( {frac{1}{2}}
ight)}^2} imes 1}
ight] ext{cu.inch} cr
& = left( {frac{{pi l}}{4}}
ight) ext{cu.inch} cr} $$ Volume of water in thinker pipe : $$eqalign{
& = left( {pi imes {3^2} imes l}
ight) ext{cu.inch} cr
& = left( {9pi l}
ight) ext{cu.inch} cr} $$ ∴ Required number of pipes : $$eqalign{
& = frac{{9pi l}}{{left( {frac{{pi l}}{4}}
ight)}} cr
& = 36 cr} $$

[#63] A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :
Correct Answer

(A) 12π cm 3

Explanation

Solution: Clearly, we have r = 3 cm and h = 4 cm ∴ Volume : $$eqalign{
& = frac{1}{3}pi {r^2}h cr
& = left( {frac{1}{3} imes pi imes {3^2} imes 4}
ight)pi {r^3} cr
& = 12pi { ext{ c}}{{ ext{m}}^3} cr} $$

[#64] The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :
Correct Answer

(B) 18 cm

Explanation

Solution: Let the height of the cone be h cm Then, $$eqalign{
& pi imes {r^2} imes 6 = frac{1}{3} imes pi imes {r^2} imes h cr
& Rightarrow h = 18,cm cr} $$

[#65] For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?
Correct Answer

(C) 30%

Explanation

Solution: Volume of the sphere : $$ = left[ {frac{4}{3}pi {{left( {10}
ight)}^3}}
ight]{ ext{ c}}{{ ext{m}}^3}$$ Surface area of the sphere : $$ = left[ {4pi {{left( {10}
ight)}^2}}
ight]{ ext{ c}}{{ ext{m}}^2}$$ ∴ Required percentage : $$eqalign{
& = left[ {frac{{4pi {{left( {10}
ight)}^2}}}{{frac{4}{3}pi {{left( {10}
ight)}^3}}}} imes 100
ight]\% cr
& = 30\% cr} $$