Mathematics · Physics

Sphere Geometry

183 Questions

Sphere geometry deals with calculating the volume and surface area of round objects. It includes problems on spherical shells, recasting spheres, and understanding radius variations. These mathematical formulas are essential for competitive exam preparation.

Volume of a sphereSurface area calculationsSpherical shellsRadius ratio variationsRecasting spheres

Sphere Geometry Questions

Multiple choice

The size of a colloidal particle is

  1. About 10-10 m

  2. 10 -9 -10 -7 m

  3. Greater than 10 -7 m

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Colloidal particles are intermediate in size between true solutions and suspensions, typically ranging from 1 nanometer to 100 nanometers (10^-9 to 10^-7 m).

Multiple choice

A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller right circular cones of base radius 3.5 cm and height 3 cm each. Find the number of cones formed?

  1. 125

  2. 126

  3. 130

  4. 120
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of sphere = 4/3$\pi$r3 = 4/3$\pi$ (10.5)3 cm3 Volume of cone of base radius 3.5 cm and height 3 cm = 1/3$\pi$ (3.5)2 x 3 cm3 Number of cones formed =$\frac{Volume of sphere}{Volume of cone} \Rightarrow \frac{4/3\pi \times 10.5 \times 10.5}{{1/3\pi}\times 3.5 \times 3.5 \times 3}$     $\Rightarrow$ 126

Multiple choice

A solid right circular cone of base diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter of the sphere.

  1. 3 cm

  2. 4 cm

  3. 6 cm

  4. 8 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Radius of base of cone is 7 cm and height is 8 cm. Volume of cone = 1/3 $\pi$ r2 h = 1/3$\pi$ (7)2 x 8 cm3 External radius of hollow sphere = 5 cm  Let internal radius be r cm. Volume of hollow sphere = 4/3 $\pi$ (R3 - r3) cm3 = 4/3 $\pi$ (53 - r3) cm3 Since volume of material used for cone will be the same as the volume of material used to make hollow sphere, 1/3 $\pi$ (7)2 8 = 4/3 $\pi$ (125 - r3) $\Rightarrow$ 98 = 125 - r 3 r3 = 125 - 98 = 27 r = 3 cm So, internal diameter of the sphere will be 2r = 2 x 3 = 6 cm.

Multiple choice maths surface area and volume of sphere solids surface area of a prism surface area of a prism and a pyramid volume of a sphere surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

From a solid sphere of radius $R$, a concentric solid sphere of radius $\dfrac{R}{2}$ is removed. The total surface area increases by

  1. $0\%$
  2. $25\%$
  3. $50\%$
  4. $75\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Solution:- (B) $25 \%$
Initial area of sphere $ \left( {A} _{1} \right) = 4 \pi {R}^{2}$
New area of sphere $\left( {A} _{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2} = 5 \pi {R}^{2}$
$\therefore$ Increase in area $= \cfrac{{A} _{2} - {A} _{1}}{{A} _{1}} \times 100$
$\Rightarrow$ Increase in area $= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100 = 25 \%$
Hence the area will be increased by $25 \%$.
Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

The dimension of sphere of influence of an molecules is

  1. $1 A^o$
  2. $10 A^o$
  3. $100 A^o$
  4. $0.1 A^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation


The Molecular range is the radius around a molecule in a substance to which the molecule can exert attraction force on another molecule$.$  This is distance$,$
Sphere of influence is the spherical space/volume around a molecule in a substance$,$ inside which the molecule at center can exert attractive force on another molecule$.$
The molecular range is very small like $1 nanometer.$
Surface tension is the surface energy $-$ additional energy required to increase surface by a unit area$.$  This is also the tension force exerted $($along the line$)$ on the boundary of a surface per unit length$.$
Hence,
option $(A)$ is correct answer.
Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

If $2$ bubble of radius $r _1$ & $r _2$ are combined then find radius of common surface.

  1. $\dfrac{r _1r _2}{r _1+r _2}$
  2. $\dfrac{r _1r _2}{r _1-r _2}$
  3. $\sqrt{r _1r _2}$
  4. $\dfrac{r _1+r _2}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When two bubbles of radii r1 and r2 are joined, the radius of the common interface r is given by 1/r = 1/r1 - 1/r2 (assuming r1 < r2). Thus, r = (r1 * r2) / (r2 - r1).

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

If the circumference of the inner edge of a hemispherical bowl is $\displaystyle\frac{132}{7}:cm$, then what is the capacity?

  1. $12\pi\:cm^3$
  2. $18\pi\:cm^3$
  3. $24\pi\:cm^3$
  4. $36\pi\:cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle2\pi r=\frac{132}{7}\implies r=3$

$\therefore$ Capacity $\displaystyle=\frac{2}{3}\pi r^3=18\pi$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A hollow spherical shell is made of metal of density $4.8$ g/cm$^3$. If its internal and external radii are $10$ cm and $12$ cm respectively, find the weight of the shell

  1. $15.24 $ kg
  2. $12.84 $ kg
  3. $14.64 $ kg

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of spherical shell
$= \displaystyle \frac{4 \pi}{3} (R^3 - r^3) = \frac{4 \pi}{3} (12^3 - 10^3)$
$=\displaystyle \frac{4}{3} \times \pi \times (12 -10) (12^2 +12 \times 10 +10^2)$
$=\displaystyle \frac{4}{3} \times \pi \times 2 \times 264 cm^3$
$Weight = volume \times density = \displaystyle \frac{4}{3}\times \pi \times 364 \times 4.8 = 14.64 kg$