Question

The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. Find the cost of painting the vessel all over, at the rate of 5 paise per sq.cm.

No spam.​

Answer 1

$\huge\underline\mathfrak{Answer\:\:-}$

Given :

• Internal diameter of hollow hemispherical vessel = 24 cm

• External diameter of hollow hemispherical vessel = 25 cm.

• Let r be the internal radius and R be the external radius respectively.

• r = 12 cm
• R = 12.5 cm

________________________

$\rm\underline{\:\:\:\:Total\:surface\:area\:to\:be\:painted\:\::-}$

$\: \quad \sf \: External \: curved \: surface \: area + Internal \: curved \\ \sf\: surface \: area + Area \: of \: ring \\ \\ \\ \colon\implies \tt \: 2\pi {R}^{2} + 2\pi {r}^{2} + \pi \: ( {R}^{2} - {r}^{2} ) \\ \\ \\ \colon \implies \tt\pi \: (2 {R}^{2} + 2\pi {r}^{2} + R - r) \\ \\ \\ \colon \implies \tt\pi \: (3 {R}^{2} + {r}^{2} ) \\ \\ \\ \colon\implies \tt \frac{22}{7} \: (3 \times {12.5}^{2} + {12}^{2} ) \\ \\ \\ \colon \implies \tt \: \frac{22}{7} \: (468.75 + 144) \\ \\ \\ \colon\implies \tt \frac{22}{7} \times 612.75 \\ \\ \\ \colon\implies \tt \blue{1925.78 \: sq.cm}$

____________________

Rate of painting = 5 paise per sq.cm.

Thus,

$\qquad \sf \: Area \times Rate \\ \\ \implies \sf \frac{1925.78 \times \cancel5}{ \cancel{10}0} \\ \\ \\ \implies \sf \green{Rs.96.29}$

Answer 2

$\large{\underline{\mathfrak{\red{Answer:}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large{\boxed{\rm{\blue{Cost\:of\:painting\:=\:Rs\:96.29}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\large{\underline{\mathfrak{\red{Step\:-\:by\:-\:step\:-\:explanation:}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\begin{lgathered}\bold{Given} \begin{cases}\sf{Internal\:diameter\:of\:spherical\:vessel\:=\:24\:cm} \\ \sf{External\:diameter\:of\:hemispherical\:vessel\:25\:cm}\\ \sf{Cost\:of\:painting\:vessel\:at\:the\:rate\:of\:5\:paise\:per\:cm^2\:=?}\end{cases}\end{lgathered}$

⠀⠀⠀⠀⠀⠀⠀

★ Internal radius = $\sf{\dfrac{24}{2}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ Internal radius (r) = 12 cm

⠀⠀⠀⠀⠀⠀⠀

★ External radius = $\sf{\dfrac{25}{2}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ External radius (R) = 12.5 cm

⠀⠀⠀⠀⠀⠀⠀

$\rule{200}2$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\small{\underline{\bold{\sf{\purple{Total\:surface\:area\:to\:be\:painted:}}}}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\boxed{\sf{External\:CSA\:+\:Internal\:CSA\:+\:Area\:of\:ring}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{2 \pi {R}^{2}\:+\:2 \pi {r}^{2}\:+\: \pi\:(R^2\:-\:r^2)}$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\small{\green{\sf{Taking\:\pi\:as\:common-}}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{\pi (2R^2\:+\:2r^2\:+\:R^2\:-\:r^2)}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{\pi (3R^2\:+\:r^2)}$

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\small{\pink{\sf{Putting\:the\:values-}}}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{\dfrac{22}{7}\:[3\:\times\:(12.5)^2\:+\:(12)^2]}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{\dfrac{22}{7}\:\times\:612.75}$

⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{\red{1925.78\:cm^2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\rule{200}2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large{\orange{\sf{Rate\:of\:painting\:per\:cm^2\:=\:5\:paise}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\sf{Cost\:of\:painting\:1925.78\:cm^2\:=\:\dfrac{5\:\times\:1925.78}{100}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

: $\implies$ $\large{\boxed{\rm{\blue{Cost\:of\:painting\:=\:Rs\:96.29}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\therefore{\large{\bold{\sf{Cost\:of\:painting\:vessel\:at\:the\:rate\:of\:5\:paise\:per\:cm^2\:is\:Rs\:96.29}}}}$