Ex.13.4 Q5 Surface Areas and Volumes Solution - NCERT Maths Class 9


Question

A hemispherical bowl made of brass has inner diameter \(10.5\rm\, cm.\) Find the cost of Tin-plating it on the inside at the rate of \(16 \text{per} \,100 \, \rm{m^2} \).

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-4 | Question 5

Text Solution

Reasoning:

Hemisphere is half of a sphere so surface area is \(\begin{align}\frac{{4\pi {r^2}}}{2} = 2\pi {r^2} \end{align}\).

What is known?

Ratio of tin plating per \(100\,\rm\,{m^2} \) and inner diameter.

What is unknown?

Cost of tin planting

Steps:

Inner Diameter \((2r) = 10.5\rm\, cm\)

Inner Radius \(r = \frac{{10.5}}{2} \)

Surface area

\[\begin{align} &=2\pi {r^2} \\&= 2 \times \frac{{22}}{7} \times \frac{{10.5}}{2} \times \frac{{10.5}}{2}\\ &= 173.25\,\rm{cm^2} \end{align}\]

Cost of plating is \(16\) per \(100\)\(\,\rm{cm^2}\)

\(∴ \)Cost of plating for \(173.25 \,\rm{cm^2}\)

\[\begin{align} & = \frac{{173.25 \times 16}}{{100}}\\& = \rm{Rs}\,\,{{27}}.{\rm{ }}{{72}} \end{align}\]

Answer:

Cost of tin plating \( = \rm Rs 27.72 \)

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