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 chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

In order to find the density of a solid we have to first know its :

  1. mass and area

  2. weight and area

  3. shape and volume

  4. mass and volume
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The density of a body is defined as the amount of mass contained per unit volume of the body.

$density=mass/volume$
So, we need mass and volume to find the density.
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with  cavity be 8 g per $c{m^3}$ then the volume of cavity is:

  1. zero
  2. 15 $c{m^3}$
  3. 5 $c{m^3}$
  4. 20 $c{m^3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$V _s=5cc$ As given

When it is placed in water net force$=$ new downward force
$\Rightarrow Hog-(V+V _s)g=20g\Rightarrow V+V _s=20\ \therefore V=20-V _s cc\V=20-5cm^2=15cm^2$

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

There are two hollow spheres made of different materials, one with double the radius and one fourth wall thickness of the other are filled with ice. If the time taken for melting the ice completely in larger sphere is $25$ minutes and that for smaller sphere is $16$ minutes, the ratio of the thermal conductivity of the larger sphere to the smaller sphere is :

  1. $4:5$
  2. $5:4$
  3. $25:8$
  4. $8:25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Smaller sphere

Radius $\to \dfrac{R}{2}$. wall thickness $\to 4d$

Larger sphere
Radius $\to R$, wall thickness $\to d$

Vol of ice in smaller sphere $= \dfrac{4}{3} \pi \left(\dfrac{R}{2} - 4d\right)^3$

$m _S = $ Mass of ice in smaller sphere $=\rho _{ice} \times \dfrac{4}{3}\pi \left(\dfrac{R}{2} - 4d\right)^3$

$m _L = $ Mass of ice in larger sphere $= \rho _{ice} \times \dfrac{4}{3} \pi (r-d)^3$

Heat required to melt $m _S = m _sL = \dfrac{4}{3}\pi \left(\dfrac{R}{2} 4d\right)^3\times \rho _{ice} \times L = Q _s$        ...(i)

Heat required to melt $m _L = m _l L = \dfrac{4}{3} \pi (R-d)^3 \times \rho _{ice} \times L = Q _L$          ...(ii)

$\dfrac{Q}{t} = \dfrac{KA}{x} (\Delta T)$

$\dfrac{Q _s}{16} = \dfrac{K _s4\pi\left(\dfrac{R}{2}\right)^2}{4d}\Delta T$      ...(for small sphere)      ...(1)

$\dfrac{A _L}{25} = \dfrac{K _24\pi R^2}{d}\Delta T$        ...(for large sphere)   ....(2)

Dividing eq(1)  by eq(2)
$\therefore \dfrac{Q _L}{25} \times \dfrac{16}{Q _s} = \left(\dfrac{K _L}{K _S}\right)\dfrac{4\pi R^2\Delta T\times 4d}{4\pi\left(\dfrac{R}{2}\right)^2\Delta T \times d}$

$\therefore \dfrac{Q _L}{Q _S} \times \dfrac{16}{25} = \left(\dfrac{K _L}{K _S}\right)\times 16$

but $\dfrac{Q _L}{Q _S} = \dfrac{(R-d)^3}{\left(\dfrac{R}{2}-4d\right)^3}$

here $R > > d$,

$\therefore \dfrac{Q _L}{Q _S} = \dfrac{R^3}{\left(\dfrac{R}{2}\right)^3} = 8$

$\therefore \dfrac{8}{25} = \dfrac{K _L}{K _S}$

Multiple choice chemistry metals physical properties of metals and non-metals some physical properties of metals general characteristics and uses of metals

The mass of steel sphere having density 7850$\;kg :m^{-3}$ and radius 0.15$\;m$ is:

  1. 112$\;kg$.
  2. 290$\;kg$.
  3. 110.9$\;kg$.
  4. 300$\;kg$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass  of the sphere, $M = Density \times Volume$


Volume of the sphere is, $ V= \dfrac {4}{3} \pi r^3$


$ V=  \dfrac {4}{3} \pi (0.15)^3= 0.01412  \ m^{3} $

$M= 7850 \times 0.01412$

$= 110.9 \:kg$

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A sphere of radius $1$ cm has potential of $8000$V. The energy density near the surface of sphere will be?

  1. $64\times 10^5$ $J/m^3$
  2. $8\times 10^3$ $J/m^3$
  3. $32$ $J/m^3$
  4. $2.83$ $J/m^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Energy density = $=(\frac { 1 }{ 2 } )∈0E2$ 
$=(\frac { 1 }{ 2 } )\times 8.86\times 10-12\times \left( \frac { v }{ r }  \right) 2$
$=4.43\times 10-12\times [(8000)/(10-2)]2$
$=283.5\times 10-2$
$=2.83J/m3$
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

In an experimental set up, the density of a small sphere is to be determined. The diameter of the small sphere is measured with the help of a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the sphere has a relative error of 2%, the relative percentage error in the density is

  1. $0.03$%
  2. $3.11$%
  3. $0.08$%
  4. $8.2$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $ %