Volume of Sphere Solved Questions

In this post we will discuss the questions of volume of sphere. All the questions are fully solved step by step for your full understanding but in case if you have any doubt, feel free to contact us for your query.

Some formulas that we have covered in this post are:
1. Volume of sphere
2. Surface area of sphere

These two formulas are important for solving sphere related problems, so make sure you invest some time to remember it.

Questions on Sphere

(01) If three metallic sphere of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere. Find the diameter of new sphere

Volume of sphere solved question
volume\quad of\quad sphere\quad =\frac { 4 }{ 3 } \pi { \quad r }^{ 3 }

Its given in the question,
r1 = 6 cm
r2 = 8 cm
r3 = 10 cm

Volume of New Sphere = Volume of sphere 1 + Volume of sphere 2 + volume of sphere 3

=\frac { 4 }{ 3 } \pi \quad [{ r1 }^{ 3 }\quad +\quad { r }2^{ 3 }\quad +{ \quad r }3^{ 3 }]\\\ \\ =\frac { 4 }{ 3 } \pi \quad [{ 6 }^{ 3 }\quad +\quad 8^{ 3 }\quad +10^{ 3 }]\\\ \\ =\frac { 4 }{ 3 } \pi \quad [{ 216 }\quad +\quad 512\quad +1000]\\\ \\ =\frac { 4 }{ 3 } \pi \quad [{ 12 }^{ 3 }]\

Hence after the simple calculation, we can easily found the volume of big sphere

(02) The surface area of a sphere is same as the curved surface area of right circular cylinder, whose height and diameter is 12 cm each. What is the radius of sphere?

How to calculate surface area of sphere
\ SOLUTION:\ \ Given,\ diameter\quad =\quad 12\quad cm\\\ \\radius\quad =\quad \frac { 12 }{ 2 } \quad =\quad 6\quad cm\\\ \\ height\quad =\quad 12\quad cm\\\ \\ \ According\quad to\quad question,\\\ \\ surface\quad area\quad of\quad sphere\quad =\quad curved\quad surface\quad area\quad of\quad a\quad right\quad circular\quad cylinder\\\ \\ \ 4\pi { r }^{ 2 }\quad \quad =\quad 2\pi rh\\\ \\ \ 2r\quad =\quad h\ r\quad =\quad \frac { h }{ 2 } \quad =\quad \frac { 12 }{ 2 } =6\quad cm\\\ \

Hence the radius of the sphere is 6 cm

(03) How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 * 11 * 12 cm

what is the formula of volume of sphere


Given,\\\ \\diameter\quad =\quad 3mm\quad =0.3\\\ \\quad cm\ \ radius\quad =\quad \frac { 0.3 }{ 2 } cm\\\ \\ number\quad of\quad lead\quad shots\quad \quad =\frac { volume\quad of\quad cuboid }{ volume\quad of\quad one\quad lead\quad shot }\\\ \\ \ volume\quad of\quad cuboid\quad =\quad (9\quad \quad \times \quad 11\quad \quad \times \quad 12\quad ){ cm }^{ 3 }\\\ \\ volume\quad of\quad one\quad lead\quad =\frac { 4 }{ 3 } \pi { r }^{ 3 }\\\ \\ \qquad \qquad \qquad \qquad \qquad \quad \quad =\quad \frac { 4\quad \times \quad 22\quad \times \quad 0.3\quad \times \quad 0.3\quad \times \quad 0.3 }{ 3\quad \times \quad 7\quad \times \quad 2\quad \times \quad 2\quad \times \quad 2 } =\frac { 11\quad \times \quad 0.1\quad \times \quad 0.3\quad \times \quad 0.3 }{ 7 } \\\ \\ \ number\quad of\quad leads\quad =\frac { 9\quad \times \quad 11\quad \times \quad 12\quad \times \quad 7\quad }{ 11\quad \times \quad 0.1\quad \times \quad 0.3\quad \times \quad 0.3 } \\\ \\ \qquad \qquad \qquad \qquad \qquad =\quad 12\quad \times \quad 7\quad \times \quad 10\quad \times \quad 10\quad \times \quad 10\\\ \\ \qquad \qquad \qquad \qquad \qquad =\quad 84000.\

(04) A solid piece of iron of dimensions 49 * 33 * 24 cm is molded into a sphere. What is the radius of sphere

volume of sphere formula
volume\quad of\quad solid\quad =\quad (49\quad \times \quad 33\quad \times \quad 24)\quad { cm }^{ 3 }\\\ \\ volume\quad of\quad sphere\quad =\frac { 4 }{ 3 } \pi { r }^{ 3 }\\\ \\ As\quad the\quad solid\quad is\quad melted\quad into\quad sphere,\quad we\quad know\quad that\\\ \\ volume\quad of\quad sphere\quad =\quad volume\quad of\quad solid\\\ \\ \frac { 4 }{ 3 } \pi { r }^{ 3 }\quad =\quad (49\quad \times \quad 33\quad \times \quad 24)\quad { cm }^{ 3 }\\\ \\ { r }^{ 3 }\quad \quad =\frac { 3 }{ 4 } \times \frac { 7 }{ 22 } \times 49\quad \times \quad 33\quad \times \quad 24\\\ \\ { r }^{ 3 }\quad \quad =3\quad \times \quad 7\quad \times \quad 49\quad \times \quad 3\quad \times \quad 3\\\ \\ { r }^{ 3 }\quad \quad =\quad (3\quad \times \quad 7)^{ 3 }\\\ \\quad \ \ Therefore\quad r\quad =\quad 21\quad cm.\

(05) If a solid sphere of radius 10 cm is molded into 8 spherical solid balls of equal radius, then what is the radius of each balls

volume of sphere question for quantitative aptitude syllabus
Given,\quad radius\quad of\quad big\quad sphere,\quad R\quad =\quad 10\quad cm\\\ \\ volume\quad of\quad big\quad sphere\quad =\frac { 4 }{ 3 } \pi { R }^{ 3 }\\\ \\ number\quad of\quad small\quad balls\quad =8\\\ \\ Let\quad radius\quad of\quad small\quad sphere\quad =\quad r\quad cm\\\ \\ We\quad know\quad that\quad ,\\\ \\ number\quad of\quad balls\quad =\quad \frac { \frac { 4 }{ 3 } \pi { R }^{ 3 } }{ \frac { 4 }{ 3 } \pi { r }^{ 3 } } \\\ \\ \quad \quad \quad \quad 8\quad \quad =\quad \frac { { R }^{ 3 } }{ { r }^{ 3 } } \\\ \\ \quad \quad \quad \quad 8\quad \quad =\quad \frac { { 10 }^{ 3 } }{ { r }^{ 3 } } \\\ \\ \quad \quad \quad \quad \quad { r }^{ 3 }\quad =\quad \frac { { 10 }^{ 3 } }{ 8 } =\frac { { 10 }^{ 3 } }{ { 2 }^{ 3 } } =\left( \frac { 10 }{ 2 } \right) ^{ 3 }={ 5 }^{ 3 }\\\ \\ Therefore\quad ,\quad r=\quad 5\quad cm\

(06) A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls that can be obtained through this process

volume of sphere question for competition exam like GMAT, GRE, Maths olympiad, SSC, SSC-CHSL, SSC=-CGL, Banking , IBPS, SBI PO, NDA
Given,\\\ \\radius\quad of\quad big\quad sphere,\quad R=\quad 8\quad cm\\\ \\radius\quad of\quad small\quad sphere,\quad r\quad =\quad 2\quad cm\\\ \\ The\quad number\quad of\quad spherical\quad balls\quad \quad =\frac { volume\quad of\quad big\quad sphere }{ volume\quad of\quad small\quad sphere } \\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { \frac { 4 }{ 3\quad } \pi \quad { R }^{ 3 } }{ \frac { 4 }{ 3 } \pi { r }^{ 3 } } \\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { { R }^{ 3 } }{ { r }^{ 3 } } \quad =\frac { { 8 }^{ 3 } }{ { 2 }^{ 3 } } \quad =\frac { 512 }{ 8 } =64\ \

(07) A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. What is the height of cone?

what is the volume of sphere
Given,\\\ \\ internal\quad diameter\quad of\quad sphere=\quad 4\quad cm\\\ \\ internal\quad radius\quad of\quad sphere,r\quad =\quad 2\quad cm\\\ \\ external\quad diameter\quad of\quad sphere\quad =8\quad cm\\\ \\ external\quad radius\quad of\quad sphere,R\quad =\quad 4\quad cm\\\ \\ base\quad diameter\quad of\quad cone\quad =8\quad cm\\\ \\ base\quad radius\quad of\quad cone\quad =4\quad cm\\\ \\ \\\ \\ volume\quad of\quad sphere\quad =\frac { 4 }{ 3 } \pi { (R }^{ 3 }-{ r }^{ 3 })\quad cm^{ 3 }\quad =\frac { 4 }{ 3 } \pi { (4 }^{ 3 }-{ 2 }^{ 3 })\quad =\frac { 4 }{ 3 } \pi (64-8)\\\ \\ \qquad \qquad \qquad \qquad \qquad =\frac { 4 }{ 3 } \pi \times \quad 56\quad cm^{ 3 }\\\ \\ \\\ \\volume\quad of\quad cone\quad =\quad \frac { 1 }{ 3 } \pi { r }^{ 2 }h\\\ \\ \qquad \qquad \qquad \qquad \quad \quad =\quad \frac { 1 }{ 3 } \pi \times \quad 4\quad \times \quad 4\quad \times \quad h\quad cm^{ 3 }\\\ \\ according\quad to\quad question,\\\ \\ volume\quad of\quad cone\quad =\quad volume\quad of\quad sphere\\\ \\ \frac { 1 }{ 3 } \pi \times \quad 4\quad \times \quad 4\quad \times \quad h\quad cm^{ 3 }\ =\quad \frac { 4 }{ 3 } \pi \times \quad 56\quad cm^{ 3 }\\\ \\ h\quad \quad =\quad \frac { 56 }{ 4 } =14\quad cm.\ \

Hence the height of the cone is 14 cm

(08) A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. Find the percentage of wood wastage

what is the formula of surface area of sphere
Given,\\\ \\ height\quad of\quad cone\quad =\quad 9\quad cm\\\ \\ base\quad diameter\quad of\quad cone\quad =\quad 18\quad cm\\\ \\ base\quad radius\quad of\quad cone\quad =\quad 9\quad cm\\\ \\ radius\quad of\quad sphere\quad =\quad 9\quad cm\\\ \\ \\\ \\volume\quad of\quad cone\quad =\frac { 1 }{ 3 } \pi { r }^{ 2 }h\\\ \\ \qquad \qquad \qquad \quad \quad =\quad \frac { 1 }{ 3 } \pi \times \quad 9\quad \times \quad 9\quad \times \quad 9\quad cm^{ 3 }\\\ \\ \\\ \\volume\quad of\quad sphere\quad =\frac { 4 }{ 3 } \pi { r }^{ 3 }\\\ \\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 4 }{ 3 } \pi \quad \times \quad 9\quad \times \quad 9\quad \times \quad 9\quad cm^{ 3 }\\\ \\ \\\ \\ volume\quad of\quad wood\quad wasted\quad =\quad volume\quad of\quad sphere\quad -\quad volume\quad of\quad cone\\\ \\ \qquad \quad \ \qquad \qquad =\quad (\frac { 4 }{ 3 } \pi \quad \times \quad 9\quad \times \quad 9\quad \times \quad 9)\quad cm^{ 3 }\quad -(\frac { 1 }{ 3 } \pi \times \quad 9\quad \times \quad 9\quad \times \quad 9)cm^{ 3 }\\\ \\ \qquad \qquad \ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\quad (\pi \quad \times \quad 9\quad \times \quad 9\quad \times \quad 9)\quad cm^{ 3 }\ \ %ge\quad of\quad wood\quad wasted\quad =\frac { volume\quad of\quad wood\quad wasted }{ volume\quad of\quad sphere } \times \quad 100\ \ \qquad \qquad \qquad \qquad \qquad \qquad =\frac { (\pi \quad \times \quad 9\quad \times \quad 9\quad \times \quad 9)\quad cm^{ 3 } }{ \quad (\frac { 4 }{ 3 } \pi \quad \times \quad 9\quad \times \quad 9\quad \times \quad 9)\quad cm^{ 3 } } \times \quad 100\quad =\quad \frac { 3 }{ 4 } \times \quad 100\quad =3\quad \times 25\quad =\quad 75%\


Hence 75% of the wood gets wasted

(09)A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into water until it is completely immersed. Find the rise of water level in the vessel

quantitative aptitude for competition exams like GMAT, GRE, Maths olympiad, SSC, SSC-CGL,, RBI, NDA, AFCAT, Banking, SBI Po
Given\\\ \\ quad \ radius\quad of\quad cylinder\quad =\quad 4\quad cm\\\ \\radius\quad of\quad sphere\quad =\quad 3\quad cm\\\ \\ \\\ \\according\quad to\quad question,\\\ \\ \ volume\quad of\quad cylinder\quad =\quad volume\quad of\quad sphere\\\ \\ \qquad \qquad \qquad \qquad \pi { R }^{ 2 }h\qquad \quad =\quad \frac { 4 }{ 3 } \pi { r }^{ 3 }\\\ \\ \qquad \qquad 4\quad \times \quad 4\quad \times \quad h\quad \quad =\quad \frac { 4 }{ 3 } \quad \times \quad 3\quad \times \quad 3\quad \times \quad 3\\\ \\ \qquad \qquad h\quad \quad =\quad \frac { 9 }{ 4 } \quad cm\ \

(10) A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of cylinder. Find the number of spherical balls

volume of sphere calculation
Let\quad radius\quad of\quad cylinder\quad =\quad r\quad cm\\\ \\ height\quad of\quad cylinder\quad =\quad 8r\quad cm\\\ \\ radius\quad of\quad spherical\quad ball\quad =\quad \frac { r }{ 2 } cm\\\ \\ \\\ \\ therefore\quad number\quad of\quad balls\quad =\quad \frac { volume\quad of\quad cylinder }{ volume\quad of\quad sphere } \quad =\quad \frac { \pi { R }^{ 2 }h }{ \frac { 4 }{ 3 } \pi { r }^{ 3 } } \\\ \\ \\\ \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\frac { r\quad \times \quad r\quad \times \quad 8r\quad }{ \frac { 4 }{ 3 } \times \quad \frac { r }{ 2 } \times \frac { r }{ 2 } \times \frac { r }{ 2 } } \quad =\quad \frac { 8\quad \times \quad 3\quad \times \quad 2\quad \times \quad 2\quad \times \quad 2 }{ 4 } =48\

11) The volume of the greatest sphere that can be cutoff from a cylindrical log of wood of radius 1 cm and height 5 cm is

how to calculate surface area of sphere
Given,\\\ \\ radius\quad of\quad cylinder\quad =\quad 1\quad cm\\ height\quad of\quad cylinder\quad =\quad 5\quad cm\\ therefore\quad radius\quad of\quad sphere\quad =\quad 1\quad cm\\\ \\ according\quad to\quad question,\quad we\quad have\quad to\quad find\quad volumr\quad of\quad sphere\quad having\quad radius\quad 1\quad cm\\ volume\quad of\quad sphere\quad =\frac { 4 }{ 3 } \pi { r }^{ 3 }\\\ \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 4 }{ 3 } \pi \quad \times \quad 1\quad \times \quad 1\quad \times \quad 1\\\ \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 4 }{ 3 } \pi \quad cm^{ 3 }\\ 

Leave a Comment

Your email address will not be published. Required fields are marked *

You cannot copy content of this page