Here, l = 26 cm, upper radius = 18 cm,
lower radius = 8 cm
d = difference in radius = 18- 8 = 10cm.
Let h be the height of bucket
\(\begin{aligned} & \therefore \quad h=\sqrt{l^2-d^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{(26)^2-(10)^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{676-100} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{576}=24 \mathrm{~cm} \end{aligned}\)

Let radius of bigger cylinder be R = 4 cm and length = 30 cm
Let radius of smaller cylinders be r = 2 cm and length = 8 cm
Now, volume of wood used to make rolling pin = volume of bigger cylinder + 2 x volume of smaller cylinder
= 3.14 x 4 x 4 x 30 + 2 x 3.14 x 2 x 2 x 8
= 1507.2 + 200.96
= 1708.16 cm³
Since mass of 1 cm³ wood = \({{1}\over{2}}\) g
= \({\pi}{r}^{2}h+2\times{{2}\over{3}}{\pi}{r}^{3}\)
= \({\pi}({1.4})^{2}\times2.2+2\times{{2}\over{3}}{\pi}({1.4})^{3}\)
\(={{22}\over{7}}\times(1.4)^{2}\left[2.2+{{4}\over{3}}\times1.4 \right]\)
= \({{22}\over{7}}\times1.96\times{{12.2}\over{3}}={{75.152}\over{3}}\) cm²

Radius of cylinder and hemisphere (r)=\(\frac{5}{2}\)cm
Height of cylinder (h)=10 cm
(i) Volume of the measuring cane=πr²h-\(\frac{2}{3}\)πr³
=\(\frac{22}{7}\times{\frac{5}{2}}\times{\frac{5}{2}}\times{10}-\frac{2}{3}\times{\frac{22}{7}}\times{\frac{5}{2}}\times{\frac{5}{2}}\times{\frac{5}{2}}\)
=196.43-32.74=163.69 cm³
(ii) Volumes of solid figures (mensuration)

Volume of one cube=125 cm³
(edge)³=125
edge=\(\sqrt [ 3 ]{ 125 } \)=5 cm
Length of the cuboid = 20 cm and
height = 5 cm
Surface area of cuboid =2(lb+bh+lh)
=2[20x5+5x5+5x20]
= 2 x 225 = 450 cm²
Volume of the cuboid =lxbxh
= 20 x 5 x 5
= 500 cm³ 

Water displaced by a person = 0.04 m³ 
Water displaced by 500 persons = 500 x 0.04 = 20 m³
Volume of cuboid = Volume of water displaced
Ibh=20
80 x 50 x h=20 
h=\(\frac{20}{80\times{50}}\)=\(\frac{1}{200}\)m
⇒ h=0.5 cm

Dimensions of the metallic cuboid are 100cm x 80cm x 64cm
100 x 80 x 64=a³(where a=side of cube)
\(\Rightarrow\)10 x 2 x 4=a \(\Rightarrow\) 80
cmSurface area of the cube=6a²
=6(80)²=6 x 80 x 80=38400 cm²

Let the internal diameter of the pipe be r m.
Water flows in 1 hour = 2.52 km.
Water flows in \(\frac {1}{2}\)  hour = 2.52 = 1.26 km
= 1260m
Volume of water flows in \(\frac {1}{2}\) hour = \(\pi\)r²h.
=\(\pi\)r² x 1260
Volume of the water in cylindrical tank
\(=\pi \times { \left( \frac { 40 }{ 100 } \right) }^{ 2 }\times 3.15\)
Volume of water flow = Volume of increased water
\(\pi { r }^{ 2 }\times 1260=\pi { \left( \frac { 2 }{ 5 } \right) }^{ 2 }\times 3.15\)
\(\Rightarrow \quad 1260{ r }^{ 2 }=\frac { 2 }{ 5 } \times 3.15\)
\(\Rightarrow \quad { r }^{ 2 }=\frac { 4 }{ 25 } \times \frac { 315 }{ 100 } \times \frac { 1 }{ 1260 } =\frac { 1 }{ 1260 } \)
\(\Rightarrow \quad r=\frac { 1 }{ 50 } m=2cm\)
Internal diameter of pipe = 4 cm.

Volume of bowl =\(=\frac { 2 }{ 3 } \pi { r }^{ 3 }\)
Volume of liquid in bowl
\(=\frac { 2 }{ 3 } \pi \times { (18) }^{ 3 }{ cm }^{ 3 }\)
Volume of liquid after wastage
\(=\frac { 2 }{ 3 } \pi \times { (18) }^{ 3 }\times \frac { 90 }{ 100 } { cm }^{ 3 }\)
Volume of bottle = \(\pi\)r²h
Volume of liquid in 72 bottles = \(\pi\) x (3)²X h x 72 cm²
Volume of liquid in bottles = volume after wastage
\(\pi \times { (3) }^{ 2 }\times h\times 72=\frac { 2 }{ 3 } \pi \times { (18) }^{ 3 }\times \frac { 90 }{ 100 } \)
\(\Rightarrow \quad h=\frac { \frac { 2 }{ 3 } \pi \times { (18) }^{ 3 }\times \frac { 90 }{ 100 } }{ \pi \times { (3) }^{ 2 }\times 72 } \)
= 5.4 cm.

250.5 m²

\(6{ \left( 19 \right) }^{ 1/3 }cm\)

Volume of the bucket = 12308.8 cm³
r₁ = 20 cm and r₂ = 12 cm, h = ?
We have volume of bucket \(=\frac{\pi h}{3}\left(r_{1}^{2_{2}}+r_{2}^{2}+r_{1} r_{2}\right)\)
\(\begin{array}{l} \Rightarrow 12308.8=3.14 \times \frac{h}{3}\left(20^{2}+12^{2}+20 \times 12\right) \\ \Rightarrow \quad h=12308.8 \times \frac{3}{3.14 \times 784}=15 \mathrm{~cm} \end{array}\)
\(\therefore\)Height of the bucket, h = 15 cm
Surface area of the metal sheet used \(=\pi r_{2}^{2}+\pi\left(r_{1}+r_{2}\right) l\)
\(\because \quad l =\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}=\sqrt{15^{2}+8^{2}}\)
\(=\sqrt{225+64}=17 \mathrm{~cm}\)
\(\therefore\) Surface area of the metal sheet used
\(\begin{array}{l} =\pi r_{2}^{2}+\pi\left(r_{1}+r_{2}\right) l \\ =3.14+12^{2}+\frac{22}{7} \times 32 \times 17 \\ =3.14(144+544)=216032 \mathrm{sq} \mathrm{cm} \end{array}\)
Hence, height of the bucket = 15 cm, area of metal, speed used = 216032 sq cm

Diameter of spherical marble = 14 cm
Radius rl = 1.4/2 = 0.7 = 7/10 cm
Diameter of cylindrical vessel = 7 cm
Radius R = 7/2 cm
Let h be the rise in water level then
Volume of 150 spherical marbles =volume of water rises
\(\Rightarrow \quad 100\times \frac { 4 }{ 3 } \times \pi \times \frac { 7 }{ 10 } \times \frac { 7 }{ 10 } \times \frac { 7 }{ 10 } =\pi \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times h\)
\(\Rightarrow \quad h=\frac { 4\times 7 }{ 5 } \)
\(\Rightarrow \quad \frac { 28 }{ 5 } =h\)
\(\Rightarrow\) h = 5.6 cm
The rise is the level of water, h = - 5.6 cm.

(i) 8 cm
(ii) 1386 cm²
(iii) Social justice, caring