
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
603.6k+ 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
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

