
The surface areas of a solid sphere and a solid hemisphere are equal to S , if their volumes are are ${V_1}$ and ${V_2}$ respectively then \[\dfrac{{{V_1}}}{{{V_2}}}\]
A.$\dfrac{{\sqrt 3 }}{2}$
B.$\dfrac{{3\sqrt 3 }}{8}$
C.$\dfrac{3}{4}$
D.$\dfrac{{3\sqrt 3 }}{4}$
Answer
510.6k+ views
Hint : In such kinds of questions we have to use the basic formulas for volume and surface area of sphere and hemisphere . Also the relation between hemisphere and sphere has to be used to find the ratio between their volumes .
Complete step-by-step answer:
Let R and r be the radii of of the sphere and the hemisphere respectively
It is given that their surface areas S are equal .
We know that the surface areas of the sphere and hemisphere are $4\pi {R^2}$ and $3\pi {r^2}$ respectively .
$ \Rightarrow 4\pi {R^2} = 3\pi {r^2}$
$ \Rightarrow \dfrac{R}{r} = \sqrt {\dfrac{3}{4}} $ ( cancelling out similar terms )
Now , let ${V_1}$ and ${V_2}$ be the volumes of the sphere and hemisphere respectively .
Therefore , $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{4}{3}\pi {R^3}}}{{\dfrac{2}{3}\pi {r^3}}}$
$ = \dfrac{{2{R^3}}}{{{r^3}}}$ $ = 2{\left( {\dfrac{R}{r}} \right)^3}$
Putting value of $\dfrac{R}{r}$ from above
We get
$\dfrac{{{V_1}}}{{{V_2}}} = 2{\left( {\sqrt {\dfrac{3}{4}} } \right)^3}$ $ = \dfrac{{2 \times 3\sqrt 3 }}{8} = \dfrac{{3\sqrt 3 }}{4}$
Note –In such types of questions the key concept we have to remember is that we always recall all the formulas for surface area and volumes of three dimensional shapes . A proper understanding of each and every shape would be beneficial in such questions .
Complete step-by-step answer:
Let R and r be the radii of of the sphere and the hemisphere respectively
It is given that their surface areas S are equal .
We know that the surface areas of the sphere and hemisphere are $4\pi {R^2}$ and $3\pi {r^2}$ respectively .
$ \Rightarrow 4\pi {R^2} = 3\pi {r^2}$
$ \Rightarrow \dfrac{R}{r} = \sqrt {\dfrac{3}{4}} $ ( cancelling out similar terms )
Now , let ${V_1}$ and ${V_2}$ be the volumes of the sphere and hemisphere respectively .
Therefore , $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{4}{3}\pi {R^3}}}{{\dfrac{2}{3}\pi {r^3}}}$
$ = \dfrac{{2{R^3}}}{{{r^3}}}$ $ = 2{\left( {\dfrac{R}{r}} \right)^3}$
Putting value of $\dfrac{R}{r}$ from above
We get
$\dfrac{{{V_1}}}{{{V_2}}} = 2{\left( {\sqrt {\dfrac{3}{4}} } \right)^3}$ $ = \dfrac{{2 \times 3\sqrt 3 }}{8} = \dfrac{{3\sqrt 3 }}{4}$
Note –In such types of questions the key concept we have to remember is that we always recall all the formulas for surface area and volumes of three dimensional shapes . A proper understanding of each and every shape would be beneficial in such questions .
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE
