Mathematics · Physics

Sphere Geometry

157 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

A shpere has a mass of 12.2 kg $\pm $ 0.1 kg and radius 10 cm $\pm $ 0.1 cm, h=the maximum % error in density is 

  1. 10%

  2. 2.4%

  3. 3.83%

  4. 4.2%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Density rho = m / V = m / ((4/3)*pi*r^3). Relative error in rho = dm/m + 3*dr/r. dm/m = 0.1/12.2 = 0.0082. dr/r = 0.1/10 = 0.01. Total error = 0.0082 + 3(0.01) = 0.0382 or 3.82%.

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

The error in the measurement of the radius of a sphere is $0.5$ %. Find the permissible error in the measurement of surface area?

  1. $0.1$ %
  2. $10$ %
  3. $5$ %
  4. $1$ %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The percentage error in measurement of radius is given, $\dfrac{\Delta r}{r}\times 100 =0.5$%
Thus, $\dfrac{\Delta r}{r}=0.5/100=0.005$
The surface area of sphere is $S=4\pi r^2$
Take ln and differentiate, $\dfrac{\Delta S}{S}=2\dfrac{\Delta r}{r}=2\times 0.005=0.01 $
The permissible or % error in the measurement of surface area $=\dfrac{\Delta S}{S}\times 100=0.01\times 100=1$%

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 a sphere is measured as  $ (10 \pm 0.02) $ cm. The error in the measurement of its volume is:

  1. 251 cc

  2. 25.1cc

  3. 2.51 cc

  4. 251.2cc
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

$r = 10$

$\Delta r=0.02$


 We know that,

The volume of sphere is

  $ V=\dfrac{4}{3}\pi {{r}^{3}} $

 $ V=\dfrac{4}{3}\times 3.14\times {{\left( 10 \right)}^{3}} $

 $ V=4186.7\,cc $

Now, taking log

$\log V=\log \dfrac{4}{3}+3\log r$

Differentiating on both sides

$\dfrac{\Delta V}{V}=0+3\dfrac{\Delta r}{r}$

Now, the error is

  $ \dfrac{\Delta V}{V}=3\times \dfrac{\Delta r}{r} $

 $ \Delta V=V\times 3\times \dfrac{\Delta r}{r} $

 $ \Delta V=4186.7\times 3\times \dfrac{0.02}{10} $

 $ \Delta V=25.1\,cc $

Hence, the error in the measurement of its volume is $25.1$ cc

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 the measurement of radius of a sphere is 1%, then the error in the measurement of volume will be :

  1. 8%

  2. 5%

  3. 3%

  4. 1%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that a sphere has volume $V=\dfrac{4}{3}\pi r^3$ meaning $V \alpha r^{3}$.
Thus, error in measurement of volume will be $ 3 \times$ 1% = 3%.

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

If error in measurement of radius of a sphere is 1%, what will be the error in measurement of volume?

  1. 1%
  2. $ \dfrac{1}{3}$ %

  3. 3%

  4. 10%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The volume is given by $V=\dfrac{4}{3} \pi R^3$, where $R$ is radius of sphere.


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

$\dfrac{\delta V}{V}= 3 \times 1 $ % $=3 $ %

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 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)