Question

${\LARGE{\underline{\underline{\pmb{\red{\frak{Question}}}}}}}$
A hemispherical bowl made of brass has inner diameter of10.5 cm. \\ \\ Find the cost of tin-plating it on inside at rate of Rs.16 per 100 $\sf{cm^2}$
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Answer 1

Given : Inner diameter of hemispherical bowl is 10.5 cm . Rate of tin-plating is Rs.16 per 100 cm² .

To Find : Cost of tin-plating

$\\ \qquad{\rule{200pt}{3pt}}$

SolutioN : For finding the Cost we know that the tin-plating will be done within the Surfa toce of the bowl .So, first le us Calculate the Surface Area and than we can calculate the Cost . Let's Solve :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Surface \; Area{\small_{(HemiSphere)}} = 2 \pi {r}^{2} }}}}}$

Where :

• $\sf{ \pi = \dfrac{22}{7} }$

• r = Radius

$\dag$ Calculating the Surface Area :

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \pi {r}^{2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times { \bigg( \dfrac{Diameter}{2} \bigg)}^{2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times { \bigg( \dfrac{10.5}{2} \bigg)}^{2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times { \bigg( \dfrac{105}{20} \bigg)}^{2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times \dfrac{105}{20} \times \dfrac{105}{20} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times \cancel\dfrac{105}{20} \times \cancel\dfrac{105}{20} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{7} \times \dfrac{21}{4} \times \dfrac{21}{4} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 2 \times \dfrac{22}{\cancel7} \times \dfrac{\cancel{21}}{4} \times \dfrac{21}{4} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = \cancel2 \times 22 \times \dfrac{3}{4} \times \dfrac{21}{\cancel4} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 22 \times \dfrac{3}{4} \times \dfrac{21}{2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = \cancel{22} \times \dfrac{3}{4} \times \dfrac{21}{\cancel2} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = 11 \times \dfrac{3}{4} \times 21 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = \dfrac{33}{4} \times 21 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = \dfrac{693}{4} } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { Surface \; Area = \cancel\dfrac{693}{4} } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; {\dashrightarrow{\underline{\boxed{\pmb{\sf{ Surface \; Area = 173.25 \; {cm}^{2} }}}}}} \; {\red{\bigstar}} \end{gathered}$

$\dag$ Calculating the Cost :

$\begin{gathered} \implies \; \; \sf { Cost = Surface \; Area \times Rate } \\ \\ \end{gathered}$

$\begin{gathered} \implies \; \; \sf { Cost = 173.25 \times \dfrac{16}{100} } \\ \\ \end{gathered}$

$\begin{gathered} \implies \; \; \sf { Cost = \dfrac{17325}{100} \times \dfrac{16}{100} } \\ \\ \end{gathered}$

$\begin{gathered} \implies \; \; \sf { Cost = \dfrac{277200}{10000} } \\ \\ \end{gathered}$

$\begin{gathered} \implies \; \; \sf { Cost = \cancel\dfrac{277200}{10000} } \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; {\implies{\underline{\boxed{\pmb{\sf{ Cost = ₹ \; 27.72 }}}}}} \; {\purple{\bigstar}} \end{gathered}$

$\therefore \;$ Cost of tin-plating the bowl is ₹ 27.72 .

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Answer 2

Solution

Given That,

• Inner Diameter of Hemishpherical Bowl is 10.5cm
• Rate of tin-plating = Rs.16 per 100 cm² .

Now,

$\displaystyle \rm{ \R= \dfrac{Diameter}{2}}$

$\implies \displaystyle \rm{ r=\frac{10.5}{2}}$

${ \implies \displaystyle \rm{r= \cancel{\frac{ 10.5}{2} } = 5.25cm}}$

Surface Area:

${ \implies \displaystyle \rm{ = 2\pi r ^{2} }}$

${ \implies \displaystyle \rm{ = 2 \times \frac{22}{7} \times 5.25 \times 5.25 }}$

${ \implies \displaystyle \rm{ = 173.25 {cm}^{2} }}$

Cost of Panting:

${ \implies \displaystyle \rm{ = \frac{16 \times 173.25}{100} }}$

${ \implies \displaystyle \rm{ \red{ =27.72 }}}$

${ \qquad{ \rule{500pt}{2pt}}}$

Cost of Tin-Panting is Rs. 27.72

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