Verified answer

Given that :

• Diameter of a wooden sphere is 21cm.

Therefore, Radius of a wooden sphere (R) = 21/2cm.

• Radius of the cylinder (r) = 1.5 cm
• Height of the cylinder (h) = 7 cm

To find :

The cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution :

To find the surface area of Silver painted, we have to subtract Surface Area of sphere with Upper part of cylinder for support.

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★ Formula for the surface area of Sphere:

=> 4πR²

★ Formula for the Upper part of Cylinder:

=> πr²

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So, In Equation:

Surface Area of Silver painted = (Surface Area of sphere - Upper part of Cylinder)

=> 4πR² - πr²

=> π(4R² - r²)

[tex] \mapsto \: \frac{22}{7} \times [4 \times ( { \frac{21}{2} })^{2} - ( { \frac{15}{10} })^{2} ][/tex]

[tex] \mapsto \: \frac{22}{7} \times [ \frac{4 \times 441}{4} - \frac{9}{4} ][/tex]

[tex] \mapsto \: \frac{22}{7} [ \frac{1764 - 9}{4} ][/tex]

[tex] \mapsto \: \frac{22}{7} \times \frac{1755}{4} = 1378.928{cm}^{2}[/tex]

Hence, Surface Area of 8 such type of Spherical part :

=> 8 × 1378.928

=> 11031.424cm²

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★ 1p = 1/100 = Rs.0.01

=> 25p = 25/100 = Rs.0.25

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Now, Cost of Silver paint over 1cm² = Rs.0.25

Therefore, Cost of Silver paint over 11031.424cm²:

=> 0.25 × 11031.424

=> Rs. 2757.856

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Then, Curved surface area of a Cylindrical support:- 2πrh

==> 2×22/7 × 15/10 × 7 = 66cm²

Hence, Curved surface area of 8 such Cylindrical support :

=> 66 × 8 = 528cm²

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★ 1p = 1/100 = Rs.0.01

=> 5p = 5/100 = Rs.0.05

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Cost of black paint over 1cm² of Cylindrical support = Rs.0.05

»» Therefore, Cost of black paint over 528cm² :

=> 0.05 × 528

=> Rs.26.4

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Thus, Total Cost of Paint required:-

=> Rs.2757.856 + Rs.26.4

=> Rs. 2784.256

Step-by-step explanation:

Question :-

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find he cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

[tex] \rule{300pt}{0.1pt}[/tex]

[tex]\mathbb\color{darkcyan}{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex]

[tex] \rule{300pt}{0.1pt}[/tex]

Here, diameter of a sphere =21 cm

[tex] \text{Radius of a sphere \( \rm (r)=\dfrac{21}{2} cm \)}[/tex]

[tex] \text{Surface area of a sphere \( \rm=4 \pi r^{2} \)}[/tex]

[tex] \text{ \( \therefore \) Surface area of 8 spheres}[/tex]

[tex] \[ \begin{array}{l} \displaystyle\rm=8 \times 4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2} cm ^{2} \\ \\ \displaystyle\rm =8 \times 22 \times 3 \times 21 cm ^{2} \\ \\ \displaystyle\rm =11088 cm ^{2} \end{array} \][/tex]

Now, radius of a cylinder (r)=1.5 cm

height of a cylinder (h)=7 cm

[tex] \text{Curved surface area of a cylinder \( \rm =2 \pi r h \)}[/tex]

[tex] \text{\( \therefore \quad \) Curved surface area of \( 8 \) cylinders}[/tex]

[tex]\[ \begin{array}{l} \displaystyle \rm=8 \times 2 \times \frac{22}{7} \times 1.5 \times 7 cm ^{2} \\ \\ \rm=528 cm ^{2} \end{array} \][/tex]

Surface area of the top of 8 cylinders

[tex]\[ \begin{array}{l} \rm =8 \pi r^{2} \\ \\ \rm=8 \times \dfrac{22}{7} \times(1.5)^{2} cm ^{2} \\ \\ \rm =56.57 cm ^{2} \end{array} \][/tex]

[tex] \text{\( \therefore \quad \) Cost of painting}[/tex]

[tex]\[ \begin{array}{l} \displaystyle\rm =₹\left[(11088-56.57) \times \frac{25}{100}+528 \times \frac{5}{100}\right] \\ \\ \displaystyle\rm=₹\left[(11031.43) \times \frac{25}{100}+\frac{2640}{100}\right] \\ \\ \displaystyle\rm=₹\left[\frac{275785.75}{100}+\frac{2640}{100}\right]\\ \\ \displaystyle\rm=₹\left[\frac{278425.75}{100}\right]\\ \\ \boxed{\color{blue} \displaystyle\rm=₹ \: \: 2784.25} \end{array} \][/tex]

Hence, the cost of paint required = ₹ 2784.25