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 physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The external and internal radius of a hollow cylinder are to be measured to be (4.23 $\pm$ 0.01)cm and (3.89 $\pm$ 0.01)cm. The thickness of the wall of the cylinder is :

  1. (0.34 $\pm$ 0.02) cm
  2. (0.17 $\pm$ 0.02)cm
  3. (0.17 $\pm$ 0.00)cm
  4. (0.34 $\pm$ 0.00)cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Thickness $= R _{ext}-R _{int}=4.23-3.89=0.34$

Now, $\Delta Thickness = (\Delta R _{ext}/R _{ext}+\Delta R _{int}/R _{int})\times Thickness=0.02$

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The radius of curvature of a concave mirror measured by a spherometer is given by $R=\dfrac{l^2}{6h}+\dfrac{h}{2} $. The measured value of $l$ is $3 cm$ using a meter scale with least count $0.1 cm $ and measured value of  $h $ is $ 0.045 cm$ using spherometer with least count $0.005 cm$. Compute the relative error in measurement of radius of curvature. 

  1. $3$
  2. $0.3$
  3. $0.2$
  4. $0.6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $l=3 cm, \Delta l=0.1 cm,  h=0.045 cm, \Delta h=0.005 cm$ 
Now, $R=\dfrac{l^2}{6h}+\dfrac{h}{2} $

Take $ln$ and differentiate (only be taken magnitude)
so, relative error , $\dfrac{\Delta R}{R}=2\dfrac{\Delta l}{l}+\dfrac{\Delta h}{h}+\dfrac{\Delta h}{h}=2\dfrac{0.1}{3}+2\dfrac{0.005}{0.045}=0.3$

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is :

  1. $\cfrac { 2 }{ 3 } \alpha $
  2. $\cfrac { 5 }{ 2 } \alpha $
  3. $\cfrac { 3 }{ 2 } \alpha $
  4. $\alpha $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

S=$4{\pi}R^2$

$\ln { S }$= $\ln ({ 4 {\pi} }) + \ln( { R^2 })$
$\ln { S } = 2\ln { R }$
$\dfrac{\Delta S}{S} = 2 \dfrac{\Delta R}{R} = \alpha$
$\dfrac{\Delta R}{R} = \dfrac {\alpha}{2}$ ------------(1)

V= $\dfrac {4}{3} \pi R^3$
$\ln {V}$ = $\ln ({\dfrac {4}{3} \pi}) + \ln {R^3}$
$\ln {V} = 3 \ln {R}$

$\dfrac{\Delta V}{V} = 3 \dfrac{\Delta R}{R}$

$\dfrac {\Delta V}{V} =3 (\dfrac {\alpha}{2})$

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

If the error in measuring the radius of a sphere is 2%, then the error in the measurement of volume is:

  1. 8%

  2. 6%

  3. 2%

  4. 9%.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Percentage error in radius is given as $2$% i.e.  $\dfrac{\Delta r}{r}\times 100 = 2$ %
Volume of sphere   $V = \dfrac{4\pi}{3}r^3$
Percentage error in volume   $\dfrac{\Delta V}{V}\times 100 = 3\times \dfrac{\Delta r}{r}\times 100 = 3\times 2 = 6$ %
Multiple choice physics static electricity properties of charges charge properties of charge

Two metallic spheres, one hollow and the other solid, have same diameter. The hollow sphere will hold charge

  1. Same as the solid sphere

  2. 2 times as the solid sphere
  3. $\displaystyle\frac{1}{2}$ times as the solid sphere

  4. Zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The capacity of the metallic sphere is proportional to the radius. It does not matter whether it is solid or hollow. Therefore the hollow sphere will hold same charge as the solid sphere.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

  1. 11.73 $cm^3$
  2. 11.736 $cm^3$
  3. 11.7 $cm^3$
  4. 117 $cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Radius of the sphere, $r = 1.41 cm$ ($3$ significant figures)
Volume of the sphere,
$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$
Rounded off upto $3$ significant figures $= 11.7 cm^3$.
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of a sphere is $5$ cm. Its volume will be given by (according to the theory of significant figures) :

  1. $523.33\ { cm }^{ 3 }$
  2. $5.23\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
  3. $5.0\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
  4. $5\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume$=\dfrac{4}{3}{\pi r}^{3}=\dfrac{4}{3}\times \pi \left ( 5 \right )^{3}=523.33\ {cm}^{3}$$=5.2333\times 10^{2}\ cm^{3}$
                                                                                   $\downarrow $
                                                                          5 significant figures
Since radius has single significant figure, so, volume should also have single significant figure.

For single significant figure, we have to drop all after decimal.
$\Rightarrow$ Volume $= 5\times 10^{2}\ cm^{2}$ (if the digit to be dropped is less than 5, preceding digit is left unchanged)

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The diameter of a sphere is $4.24\ m$. Its surface area with due regard to significant figures is :

  1. 5.65 ${ m }^{ 2 }$
  2. 56.5 ${ m }^{ 2 }$
  3. 565 ${ m }^{ 2 }$
  4. 5650 ${ m }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$d=4.24\;m$
$SA=$ Surface area $=4\pi r^{2}=4\pi \left ( \dfrac{d}{2} \right )^{2}=\pi d^{2}$
$=\pi \left ( 4.24 \right )^{2}$
$=56.47\ m^{2}$
Since significant figure in 4.24 is 3, so we express the answer in 3 significant figures.
$SA=56.47\ m^2 \approx 56.5\ m^2$
(Rounding off to 3 significant figures - if digit to be dropped is more than 5, the preceding digit is raised by 1)

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The volume of a sphere is $ 1.76\;{cm }^{ 3 }$. The volume of 25 such spheres according to the idea of significant figures in ${cm}^{ 3 }$ is 

  1. 44.00

  2. 44.0

  3. 44

  4. 4.4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of 1 sphere $=1.76 cm^{3}$
Volume of 25 spheres $=25\times 1.76$
                                      $=44 {cm}^{3}$
                                      $=44.0 {cm}^{3}$
                                 (Since volume is reported in 3 significant figure)

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of sphere is measured to be $\displaystyle \left ( 2.1\pm 0.5 \right )$cm.
 Calculate its surface area with error limits.

  1. $\displaystyle S=\left ( 55.4\pm 26.4 \right )cm^{2}$
  2. $\displaystyle S=\left ( 55.4\pm 26.4 \right )mm^{2}$
  3. $\displaystyle S=\left ( 55.4\pm 13.2 \right )cm^{2}$

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

Surface area, $\displaystyle S=4\pi r^{2}=\left ( 4 \right )\left ( \frac{22}{7} \right )\left ( 2.1 \right )^{2}$$\displaystyle =55.44=55.4cm^{2}$
Further, $\displaystyle \frac{\Delta S}{S}=2.\frac{\Delta r}{r}$
              $\displaystyle \Delta S=2\left ( \frac{\Delta r}{r} \right )\left ( S \right )=\frac{2\times 0.5\times 55.4}{2.1}$$\displaystyle =26.38=26.4 cm^{2}$

$\displaystyle \therefore S=\left ( 55.4\pm 26.4 \right )cm^{2}$

Multiple choice

What is the size range of particles in a colloid?

  1. 1-10 nm

  2. 10-100 nm

  3. 100-1000 nm

  4. All of the above
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The size range of particles in a colloid is typically between 1 and 1000 nanometers.

Multiple choice

What is the formula for the volume of a sphere?

  1. \(V = \frac{4}{3} \pi r^3\)
  2. \(V = \pi r^2\)
  3. \(V = 2\pi r\)
  4. \(V = \frac{1}{3} \pi r^2\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The formula for the volume of a sphere is (V = \frac{4}{3} \pi r^3), where (r) is the radius of the sphere.

Multiple choice

The ancient Indian mathematical text, the 'Ganita Sara Samgraha', contains a formula for calculating the volume of a sphere. What is the formula?

  1. Volume = (4/3) * π * radius^3

  2. Volume = π * radius^2

  3. Volume = (1/3) * π * radius^3

  4. Volume = (2/3) * π * radius^3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Ganita Sara Samgraha provides a formula for calculating the volume of a sphere, which is Volume = (4/3) * π * radius^3.