Perimeter of 2 inner semicircle = (400 - 2 x 90) m = (400 - 180) m = 220 m
Radius of each semicircle = \(\frac { 220 }{ 2\pi } \) = \(\frac { 220\times 7 }{ 2\times 22 } \) = 35 m
Perimeter of 2 outer semicircle = (488 - 180) m = 308 m
Radius of each outer semicircle = \(\frac { 308 }{ 2\pi } =\frac { 308\times 7 }{ 2\times 22 } =\quad 49m\)
width of the track = outer radius - inner radius = 49 - 35 = 14 m
Area of rectangular tracks = 2 x area of rectangle
= 2 x i x b = 2 x 90 x 14       (i = 90, b = 14 m)
= 28 x 90 = 2520 m²
Area of two semicircle rings = area of one circle ring
\(=\quad \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =\frac { 22 }{ 7 } \left( { 49 }^{ 2 }-{ 35 }^{ 2 } \right) { m }^{ 2 }\)
= \(22\over7\) x (49 - 35) (49 - 35) m²
= \(22\over7\) x 14 x 84 m² = 44 x 84 = 3696 m²
Total area of track = (2520 + 3696) m² = 6216 m²
Cost of developing the track at the rate of Rs.12.50/ m² = Rs. 6216 x 12.50 = Rs.77,700

Radius of solid sphere (R) = 3 cm
Radius of spherical ball (r) = 0.3 cm
Let 'n' be the number of spherical balls.
\(\Rightarrow\)  n x Volume of a ball = Volume of solid sphere
\(\Rightarrow\)  n x \(\frac{4}{3}\pi\) (0.3)³ = \(\frac{4}{3}\pi\) (3)³
\(\Rightarrow\)   n = \(\frac { 3\times 3\times 3 }{ 0.3\times 0.3\times 0.3 } \) = 1000
\(\therefore\)  Required number of balls = 1000

Given, Radius of cylinder, r=8 cm=Radius of conical hole
Height of cylinder, h=15 cm
In FED, FD=\(\sqrt { { \left( FE \right) }^{ 2 }+{ \left( ED \right) }^{ 2 } } \)
\(\sqrt { 225+64 } \)=\(\sqrt { 289 } \) =17 cm=Slant height(l)

Total surface area, after drilling of cone= curved surface area of cylinder + Area of base of a cylinder + Curved surface area of cone
\(2\pi rh+{ \pi r }^{ 2 }+\pi rl\\ =2\times 3.14\times 8\times 15+3.14\times { \left( 8 \right) }^{ 2 }+3.14\times 8\times 17\)
=753.6+200.96+427.04=1381.6 cm²

Radius of upper end = R₁ = \(\frac { 40 }{ 2 } =20cm\)
Radius of lower end = R₂ = \(\frac { 20 }{ 2 } =10cm\)
Let height be h cm
Volume of bucket = 17600 cm³
\(\Rightarrow \frac { 1 }{ 3 } \pi h(R^{ 1 }_{ 2 }+(R^{ 2 }_{ 2 }+R_{ 1 }R_{ 2 })=17600\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times h\left( 20 \right) ^{ 2 }+\left( 10 \right) ^{ 2 }+20\times 10=17600\)
\(\Rightarrow h=\frac { 17600\times 7\times 3 }{ 22\times 700 } =24cm\)
Now slant height of bucket l= \(\sqrt { h^{ 2 }+(R_{ 1 }-R_{ 2 })^{ 2 } } \)
l = \(\sqrt { (24)^{ 2 }+(10)^{ 2 } } =\sqrt { 676 } =26cm\)
Total surface area = \(\pi R^{ 2 }_{ 2 }+\pi l(R_{ 1 }+R_{ 2 })\)
= \(\pi \left( 10 \right) ^{ 2 }+26(20+10)=\frac { 22 }{ 7 } \times 880cm^{ 2 }=2765.714cm^{ 2 }\)

Ans. x= 25

Let the radius of third ball be x
Volume of spherical ball
=\(\frac{4}{3}\pi\left(\frac{3}{2}\right)^{3}\)cm³
Volume of Ist smaller ball
=\(\frac{4}{3}\pi\left(\frac{1.5}{2}\right)^{3}\)
Volume of IInd smaller ball
=\(\frac{4}{3}\pi\left(\frac{2}{2}\right)^{3}\)
Volume of IIIrd smaller ball =\(\frac{4}{3}\pi\left(x\right)^{3}\)
ATQ,
\(\frac{4}{3}\pi\left(\frac{3}{2}\right)^{3}=\frac{4}{3}\pi\left[\left(\frac{1.5}{2}\right)^{3}+\left(1\right)^{3}+\left(x\right)^{3}\right]\)
=\(\left(\frac{3}{2}\right)^{3}=\left(\frac{1.5}{2}\right)^{3}+1+x^{3}\)
⇒ x=1.25 cm
d=2x=2x1.25=2.5 cm

84 cm²

154 cm²

30.17 cm³

Volume of sphere = Volume of cone 
\(\therefore \quad \frac { 4 }{ 3 } \pi { r }_{ 1 }^{ 3 }=\frac { 1 }{ 3 } \pi { r }_{ 2 }^{ 2 }h\)
\(\Rightarrow \quad \frac { 4 }{ 3 } \times { r }_{ 1 }^{ 3 }=\frac { 1 }{ 3 } (6)^{ 2 }\times \frac { 24 }{ 3 } \)
\(\Rightarrow \quad 4{ r }_{ 1 }^{ 3 }=36\times 24\)
\(\Rightarrow \quad { r }_{ 1 }^{ 3 }={ 6 }^{ 3 }\)
\(\Rightarrow \quad { r }_{ 2 }={ 6 }\)
Hence, radius of sphere
= 6 cm.