Q.1 In fig., sectors of two concentric circles of radii 7 𝑐𝑚 and 3. 5 𝑐𝑚 are given. Find the area of the shaded region. (Use π = 22/7 )

 

(OR)

 

Q.2 A solid sphere of radius 3 𝑐𝑚 is melted and then recast into small spherical balls each of diameter 0. 6 𝑐𝑚. Find the number of balls.

 

We are given the radii of the two concentric sectors, and we have to find the area of the shaded region.

The area of the shaded region can be found by subtracting the area of the smaller sector from the area of the bigger sector.

Now, we know that area of a sector is given by 𝐴 = $\frac{\theta }{360°}\times {\mathrm{\pi r}}^2$ , where θ is the sector's

angle is enclosed, and 𝑟 is the radius of the sector. The radius of the bigger sector, 𝑅 = 7 𝑐𝑚

The radius of the smaller sector, 𝑟 = 3. 5 𝑐𝑚

Both these sectors enclose an angle of 30$°$, i.e.,   θ = 30$°$

Area of bigger circle = $\frac{\theta }{360°}\times \pi R^2$

𝐴 =  $$\begin{gathered} \frac{30°}{360°}\times \frac{22}{7}\times 7\times 7 \\ \frac{1}{12}\times 22\times 7 \\ \frac{154}{12} \end{gathered}$$

A = 12.83 cm²

Area of Smaller circle = $\frac{\theta }{360°}\times \pi r^2$

a= $$\begin{gathered} \frac{30°}{360°}\times \frac{22}{7}\times 3.5\times 3.5 \\ \frac{1}{12}\times 22\times 3.5\times 0.5 \\ \frac{38.5}{12} \\ 3.21 c{m}^2 \end{gathered}$$

So, the area of the shaded region = 𝐴 − 𝑎
= 12. 83 − 3. 21
= 9. 62 𝑐𝑚²
Hence, the area of the shaded region is 9.62 cm²

^(OR)

We are given that a solid sphere is melted and recast into small spherical balls. Next, we have to find the number of balls formed.
The radius of the solid sphere, 𝑅 = 3 𝑐𝑚
And the diameter of each spherical ball, 𝑑 = 0. 6 𝑐𝑚
So, the radius of each spherical ball, 𝑟 = 𝑑/2
⇒ 𝑟 = 0.6/2 
⇒ 𝑟 = 0. 3 𝑐𝑚
When a solid is converted from one shape to another, the volume of the solid remains unaltered.
This is because the volume of a solid is the amount of three-dimensional space occupied by the solid. Therefore, whatever the shape of the solid, it will occupy the same amount of space. Let 𝑥 be the total number of spherical balls formed.
So, the volume of a solid sphere = 𝑛 × volume of each spherical ball The volume of the solid sphere, 𝑉 = 4/3 π𝑅³
⇒ 𝑉 = 4/3 × (22/7) × 3 × 3 × 3
⇒ 𝑉 = 792/7
⇒ 𝑉 = 113. 14 𝑐𝑚³
The volume of each spherical ball, 𝑣 = 4/3 π𝑟³
⇒ 𝑣 = 4/3 × 22/7 × 0. 3 × 0. 3 × 0. 3
⇒ 𝑣 = 0.792 
⇒ 𝑉 = 0. 113 𝑐𝑚³
So, the number of spherical balls formed, 𝑛 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑠𝑝ℎ𝑒𝑟𝑒 /𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑏𝑎𝑙𝑙
n = 113.14 / 0.11314
⇒ 𝑛 = 1000
Hence, 1000 spherical balls are formed during recasting