Question

A cubical block of side 10cm is surmounted by a hemisphere. What is the largest diameter that hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of rupees 5 per sq. cm. [use $\pi = \dfrac{{22}}{7}$]

Answer 1

Hint: To solve this question, at first we will obtain the surface area of the cubical block and the curved surface area of the hemisphere and thus will obtain the total surface area. Then by multiplying the rate of the painting per sq. cm. with the total surface area we will get the cost of painting.

Complete step-by-step answer:
A cubical block of side 10cm is surmounted by a hemisphere.
Representing the above statement in a diagram as follows:

Here, the red portion represents the cube and the pink one represents the hemisphere.
The side of the cube is a = 10 cm.
Hence the largest possible diameter of the hemisphere can be d = 10 cm.
Hence, radius of the hemisphere is $r = \dfrac{d}{2} = \dfrac{{10}}{2} = 5cm$ ………(1)
We need to paint the solid we got above. To paint that, we need to paint the whole cube and curved surface area of the hemisphere except the base of the hemisphere.
Now, we will find the surface area of cube by using the formula:
Surface area of a cube is given by $6{a^2}$, where a is the length of side of the cube.
Hence, surface area of this cube is $6{a^2} = 6 \times {\left( {10} \right)^2} = 600c{m^2}$. ………….(2)
The curved surface area of the hemisphere is given by $2\pi {r^2}$, where r is the radius of the hemisphere.
So, the CSA of the given hemisphere will be $2\pi {r^2} = 2\pi \times {5^2}$ (Using 1)
Now, putting in the value of $\pi $, we will get:-
CSA of hemisphere = $2\pi \times {5^2} = 2 \times \dfrac{{22}}{7} \times 25$
On simplifying it, we will get:
CSA of hemisphere = $2\pi \times {5^2} = 2 \times \dfrac{{22}}{7} \times 25$
CSA of hemisphere =$\dfrac{{1110}}{7}c{m^2}$ ………….(3)
Area of the base of the hemisphere will be given by $\pi {r^2}$, where r is the radius of the hemisphere. (Because base is circle only)
Now, putting in the values, we will get:
Area of base = $\pi {r^2} = \dfrac{{22}}{7}{\left( 5 \right)^2} = \dfrac{{22}}{7} \times 25$
On simplifying it, we will get:
Area of base = $\dfrac{{555}}{7}c{m^2}$ ………………(4)
Hence, total surface area to be painted = surface area of the cube + curved surface area of the hemisphere – area of the base of the hemisphere
Now, using (2), (3) and (4), we will get:-
The total surface area to be painted $ = 600 + \dfrac{{1100}}{7} - \dfrac{{550}}{7} = 600 + \dfrac{{550}}{7}$
On simplifying it by taking LCM in RHS, we will get:-
The total surface area to be painted $ = \dfrac{{4200 + 550}}{7}$
On simplifying it further, we will get:-
The total surface area to be painted $ = \dfrac{{4750}}{7}c{m^2}$
We know that the cost of painting per square cm is Rs.5.
Hence the cost of painting 678.57 sq. cm $ = 678.57 \times 5 = 3392.85$rupees

Therefore the cost of painting the total surface area of the solid is Rs.3392.85.

Note: We subtract the area of the base of the hemisphere because that much part of the cube is covered by the base of the hemisphere and we would never be able to paint it.
A cube is a three dimensional solid object bounded by six square faces. It has 6 faces, 12 edges and 8 vertices.
The surface area of the cube is $6{a^2}$. Where a is the side of the cube.
A hemisphere is three dimensional solid which is exactly half of the sphere.
The surface area of the hemisphere is the sum of the curved surface area and surface area of its base i.e. $2\pi {r^2} + \pi {r^2} = 3\pi {r^2}$. Where r is the radius of the hemisphere.