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Question

Answers

A. 3cm

B. 4cm

C. 4.5cm

D. 5.5cm

Answer

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147.9k+ views

Hint: Volume of hemisphere is given as $\dfrac{2}{3}\pi {{r}^{3}}$and total surface area can be given by $3\pi {{r}^{2}}$where r is the radius of the hemisphere and value of \[\pi \] be$\dfrac{22}{7}$.

Complete step-by-step answer:

As we know that volume of solid hemisphere can be given as $\dfrac{2}{3}\pi {{r}^{3}}$and total surface area can be given as$3\pi {{r}^{2}}$, where r is the radius of the hemisphere and value of \[\pi \] is$\dfrac{22}{7}$.

Now, it is given that both volume and total surface area are equal in magnitude. Hence, $\dfrac{2}{3}\pi {{r}^{3}}$and $3\pi {{r}^{2}}$are equal to each other.

Hence, we get

$\dfrac{2}{3}\pi {{r}^{3}}=\dfrac{3\pi {{r}^{2}}}{1}$

On cross-multiplying, we get$2\pi {{r}^{3}}=9\pi {{r}^{2}}$.

Now subtract $9\pi {{r}^{2}}$from both sides of the above equation, we get

$\begin{align}

& 2\pi {{r}^{3}}-9\pi {{r}^{2}}=9\pi {{r}^{2}}-9\pi {{r}^{2}} \\

& 2\pi {{r}^{3}}-9\pi {{r}^{2}}=0\ldots \ldots (1) \\

\end{align}$

Now, taking $\pi {{r}^{2}}$as common from both the terms i.e. $2\pi {{r}^{2}}$and $9\pi {{r}^{2}}$.

So, we get

$\pi {{r}^{2}}\left( 2r-9 \right)=0\ldots \ldots (2)$

as $\pi =\dfrac{22}{7}$, so it can never be zero.

Hence, the value of ${{r}^{2}}$or \[\left( 2r-9 \right)\] may be zero. So, we can equate ${{r}^{2}}$and \[\left( 2r-9 \right)\]to ‘0’ to get the value of r.

So, we get

${{r}^{2}}=0$

or $r=0$(Not possible)

A hemisphere of zero radius is not possible.

Hence r=0 can be ignored and not possible.

Now, equating \[\left( 2r-9 \right)\] to 0, we get

\[2r-9=0\]

Add 9 to both sides of above equation, we get

\[2r-9+9=0+9\]

or

$2r=9$

Now, dividing by 2 on both sides of the above equation, we get

$\dfrac{2r}{2}=\dfrac{9}{2}$

or

$r=\dfrac{9}{2}=4.5$

As we have given that volume is expressed in $c{{m}^{3}}$and area is expressed in$c{{m}^{2}}$. Hence, radius calculates should be in cm. Therefore, the radius of the hemisphere is 4.5cm.

Note: One can go wrong while writing the total surface area of the hemisphere. He/she may use $2\pi {{r}^{2}}$as total surface which is half of total surface which is half of total surface area of sphere i.e. $\dfrac{4\pi {{r}^{2}}}{2}=2\pi {{r}^{2}}$which us wrong as we are not including area of base. So, total surface area of hemisphere is $2\pi {{r}^{2}}+\pi {{r}^{2}}=3\pi {{r}^{2}}$.

One can get confused, how volume and surface area can be equal as both have different units i.e. $c{{m}^{3}}$and$c{{m}^{2}}$. So, one needs to take care with the statement that we are equating only magnitudes.

Complete step-by-step answer:

As we know that volume of solid hemisphere can be given as $\dfrac{2}{3}\pi {{r}^{3}}$and total surface area can be given as$3\pi {{r}^{2}}$, where r is the radius of the hemisphere and value of \[\pi \] is$\dfrac{22}{7}$.

Now, it is given that both volume and total surface area are equal in magnitude. Hence, $\dfrac{2}{3}\pi {{r}^{3}}$and $3\pi {{r}^{2}}$are equal to each other.

Hence, we get

$\dfrac{2}{3}\pi {{r}^{3}}=\dfrac{3\pi {{r}^{2}}}{1}$

On cross-multiplying, we get$2\pi {{r}^{3}}=9\pi {{r}^{2}}$.

Now subtract $9\pi {{r}^{2}}$from both sides of the above equation, we get

$\begin{align}

& 2\pi {{r}^{3}}-9\pi {{r}^{2}}=9\pi {{r}^{2}}-9\pi {{r}^{2}} \\

& 2\pi {{r}^{3}}-9\pi {{r}^{2}}=0\ldots \ldots (1) \\

\end{align}$

Now, taking $\pi {{r}^{2}}$as common from both the terms i.e. $2\pi {{r}^{2}}$and $9\pi {{r}^{2}}$.

So, we get

$\pi {{r}^{2}}\left( 2r-9 \right)=0\ldots \ldots (2)$

as $\pi =\dfrac{22}{7}$, so it can never be zero.

Hence, the value of ${{r}^{2}}$or \[\left( 2r-9 \right)\] may be zero. So, we can equate ${{r}^{2}}$and \[\left( 2r-9 \right)\]to ‘0’ to get the value of r.

So, we get

${{r}^{2}}=0$

or $r=0$(Not possible)

A hemisphere of zero radius is not possible.

Hence r=0 can be ignored and not possible.

Now, equating \[\left( 2r-9 \right)\] to 0, we get

\[2r-9=0\]

Add 9 to both sides of above equation, we get

\[2r-9+9=0+9\]

or

$2r=9$

Now, dividing by 2 on both sides of the above equation, we get

$\dfrac{2r}{2}=\dfrac{9}{2}$

or

$r=\dfrac{9}{2}=4.5$

As we have given that volume is expressed in $c{{m}^{3}}$and area is expressed in$c{{m}^{2}}$. Hence, radius calculates should be in cm. Therefore, the radius of the hemisphere is 4.5cm.

Note: One can go wrong while writing the total surface area of the hemisphere. He/she may use $2\pi {{r}^{2}}$as total surface which is half of total surface which is half of total surface area of sphere i.e. $\dfrac{4\pi {{r}^{2}}}{2}=2\pi {{r}^{2}}$which us wrong as we are not including area of base. So, total surface area of hemisphere is $2\pi {{r}^{2}}+\pi {{r}^{2}}=3\pi {{r}^{2}}$.

One can get confused, how volume and surface area can be equal as both have different units i.e. $c{{m}^{3}}$and$c{{m}^{2}}$. So, one needs to take care with the statement that we are equating only magnitudes.