
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
620.4k+ 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
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

