Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Figure). The radius BO of the hemisphere (as well as of the cone) \(=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm} .\)
So, volume of the toy = \(\begin{aligned} & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \end{aligned}\)
\(\begin{aligned} & =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3 \end{aligned}\)
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm
So, the colume required = volume of the right circular cylinder - volume of the toy
= 3.14 \(\times\)2² \(\times\)4 - 25.12) cm³
= 25.12 cm³
hence, the required difference of the two volumes = 25.12 cm³.

The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid \(+\frac{1}{2}\) volume of the cylinder 
\(=\left[15 \times 7 \times 8+\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15\right] \mathrm{m}^{3}=1128.75 \mathrm{~m}^{3}\)
Next, the total space occupied by the machinery = 300 m³
And the total space occupied by the workers = 20 × 0.08 m³ = 1.6 m³
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 – (300.00 + 1.60) = 827.15 m³

Denote radius of cone by r, slant height of cone by l, height of cone by h, radius of cylinder by r' and height of cylinder by h'. Then r = 2.5 cm, h = 6 cm, r' = 1.5 cm, h' = 26 – 6 = 20 cm and 
\(l=\sqrt{r^{2}+h^{2}}=\sqrt{2.5^{2}+6^{2}} \mathrm{~cm}=6.5 \mathrm{~cm}\)
Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted.
So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder
\(=\pi r l+\pi r^{2}-\pi\left(r^{\prime}\right)^{2} \)
\(=\pi\left[(2.5 \times 6.5)+(2.5)^{2}-(1.5)^{2}\right] \mathrm{cm}^{2} \)
\(=\pi[20.25] \mathrm{cm}^{2}=3.14 \times 20.25 \mathrm{~cm}^{2} \)
\(=63.585 \mathrm{~cm}^{2}\)
Now, the area to be painted yellow = CSA of the cylinder + area of one base of the cylinder
\(=2 \pi r^{\prime} h^{\prime}+\pi\left(r^{\prime}\right)^{2} \)
\(=\pi r^{\prime}\left(2 h^{\prime}+r^{\prime}\right) \)
\(=(3.14 \times 1.5)(2 \times 20+1.5) \mathrm{cm}^{2} \)
\(=4.71 \times 41.5 \mathrm{~cm}^{2} \)
\(=195.465 \mathrm{~cm}^{2}\)

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

This top is exactly like the object we have discussed. So, we can conveniently use the result we have arrived at there. That is :
TSA of the toy = CSA of hemisphere + CSA of cone
Now, the curved surface area of the hemisphere = \(\frac{1}{2}\left(4 \pi r^{2}\right)=2 \pi r^{2} \)
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}\)
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
\(=\left(5-\frac{3.5}{2}\right) \mathrm{cm}=3.25 \mathrm{~cm}\)
So, the slant height of the cone \((l)=\sqrt{r^{2}+h^{2}}=\sqrt{\left(\frac{3.5}{2}\right)^{2}+(3.25)^{2}} \mathrm{~cm}=3.7 \mathrm{~cm}(\text { approx. })\)
Therefore, CSA of cone \(=\pi r l=\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
This gives the surface area of the top as
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}+\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
\(=\frac{22}{7} \times \frac{3.5}{2}(3.5+3.7) \mathrm{cm}^{2}=\frac{11}{2} \times(3.5+3.7) \mathrm{cm}^{2}=39.6 \mathrm{~cm}^{2}(\text { approx. })\)
You may note that ‘total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere

Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass \(=\pi r^{2} h\)
\(=3.14 \times 2.5 \times 2.5 \times 10 \mathrm{~cm}^{3}=196.25 \mathrm{~cm}^{3}\)
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e., it is less by \(\frac{2}{3} \pi r^{3}=\frac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \mathrm{~cm}^{3}=32.71 \mathrm{~cm}^{3}\)
So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= (196.25 – 32.71) cm³
=163.54 cm³

Given,spherical glass vessel is a combination of a sphere as its base and a cylinder as its neck.

For cylindrical portion,
Height of the cylinder, h₁ = 8 cm
Radius of the cylinder,
\(r_1=\frac{2}{2}=1 \mathrm{~cm}\)
For spherical portion,
Radius of the sphere,
\(r_2=\frac{8.5}{2} \mathrm{~cm}\)
\(\therefore\) Volume of water filled in a spherical glass vessel = Volume of the cylinder + Volume of the sphere
\(\begin{aligned} & =\pi r_1^2 h_1+\frac{4}{3} \pi r_2^{3} \\ \end{aligned}\)
\(\begin{aligned} & =3.14 \times 1 \times 1 \times 8+\frac{4}{3} \times 3.14 \times \frac{8.5}{2} \times \frac{8.5}{2} \times \frac{8.5}{2} \\ \end{aligned}\)
\(\begin{aligned} & =25.12+\frac{1928.3525}{6} \end{aligned}\)
= 25.12 + 321.39 = 346.51
So, the correct answer is 346.51 cm³.

Given, solid is a combination of a cone and a hemisphere and it is placed into a right circular cylinder.

Height of the cylinder, h = 180 cm = 1.8 m
\(\left[\because 1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m}\right]\)
Radius of the cylinder, r = 60 cm = 0.6 m
\(\therefore\) Volume of water filled in a right circular cylinder = \(\pi\) r² h
\(\begin{aligned} & =\frac{22}{7} \times 0.6 \times 0.6 \times 1.8 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{14.256}{7} \mathrm{~m}^3 \end{aligned}\)
For conical portion,
Height, h₁ = 120 cm = 1.2 m
Radius, r₁ = 60 cm = 0.6 m
For hemispherical portion,
Radius, r₂ = 60 cm = 0.6 m
\(\therefore\) Volume of the solid = Volume of the cone + Volume of the hemisphere
\( =\frac{1}{3}\times \pi r_{1}^{2}k_{1}+\frac{2}{3}\pi r_{2}^{1}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times(0.6)^2 \times(1.2)+\frac{2}{3} \times \frac{22}{7} \times(0.6)^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{21} \times(0.6)^2[1.2+2 \times 0.6)=\frac{22}{21} \times 0.36(1.2+1.2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{21} \times 0.36 \times 2.4=\frac{19.008}{21}=\frac{6.336}{7} \mathrm{~m}^3 \end{aligned}\)
Clearly, volume of water left in the cylinder = volume of water filled in a right circular cylinder - volume of the solid
\(=\frac{14.256}{7}-\frac{6.336}{7}=\frac{7.92}{7}=\)1.131429 m³ = 1.131 m³

Height (h₁) of larger cylinder = 220 cm
Radius (r₁) of larger cylinder = \(\frac{24}{2}\)= 12 cm
Height (h₂) of smaller cylinder = 60 cm
Radius (r₂) of smaller cylinder = 8 cm
Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder
\(\begin{aligned} & =\pi r_1^2 h_1+\pi r_2^2 h_2 \\ \end{aligned}\)
\(\begin{aligned} & =\left(\pi(12)^2 \times 220\right)+\left(\pi(8)^2 \times 60\right) \\ \end{aligned}\)
\(\begin{aligned} & =\pi[144 \times 220+64 \times 60] \\ \end{aligned}\)
\(\begin{aligned} & =3.14[31,680+3,840] \\ \end{aligned}\)
\(\begin{aligned} & =3.14 \times 35520=111,532.8 \mathrm{~cm}^3 \end{aligned}\)
Mass of 1 cm³ iron = 8 g
Mass of 111532.8 cm³ iron = 11532.8 \(\times\)8 = 892262.4 g

Given, one gulabjamun is a combination of a cylinder and two hemispheres. Here, total length of one gulabjamun = 5 cm and diameter = 2.8 cm

\(\therefore\) Radius of cylindrical part = Radius of hemispherical part
\(=r=\frac{2.8}{2}=1.4 \mathrm{~cm}\)
Height of cylindrical part,
h = PQ - (PR + SQ) = 5 - (1.4 + 1.4)
= 5 - 2.8 = 2.2 cm
\(\therefore\) Volume of one gulabjamun = 2 \(\times\) Volume of hemispherical part + Volume of cylindrical part
\(\begin{aligned} & =2 \times\left[\frac{2}{3} \pi r^{3}\right]+\pi r^2 h=\frac{4}{3} \pi r^{3}+\pi r^2 h \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2\left[\frac{4 r}{3}+h\right]=\frac{22}{7} \times 1.4 \times 1.4\left[\frac{4}{3} \times 1.4+2.2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}\left[\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}\right] \\ \end{aligned}\)
\(\begin{aligned} & =22 \times \frac{1}{5} \times \frac{7}{5}\left[\frac{28}{15}+\frac{11}{5}\right]=\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 45 gulabjamuns
\(=45 \times \frac{9394}{375}=\frac{3 \times 9394}{25}=\frac{28182}{25}=1127.28 \mathrm{~cm}^3\)
Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, quantity of syrup found in 45 gulabjamuns
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

Here, toy is a combination of a hemisphere and a cone.

Given, total height of toy,
AD = 15.5 cm
For hemispherical portion,
Radius, OC = OD = OB = 3.5 cm
For conical portion,
Height, OA = AD - OD
= 15.5 - 3.5 = 12 cm
and radius = 3.5 cm
Now, total surface area of the toy = Curved surface area of cone + curved surface area of hemisphere
\(\begin{aligned} & =\pi r l+2 \pi r^2=\pi r \sqrt{h^2+r^2}+2 \pi r^2 \quad\left[\because l=\sqrt{h^2+r^2}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 3.5 \times \sqrt{(12)^2+(3.5)^2}+2 \times \frac{22}{7} \times(3.5)^2 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{144+12.25}+22 \times 3.5 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{156.25}+11 \times 7 \end{aligned}\)
= 11(12.5) + 77
=137.5 + 77
= 214.5 cm²

Given, wooden article is a combination of a cylinder and two hemispheres.

Here, height of the cylinder, h = 10 cm
Clearly, Radius of base of the cylinder = Radius of hemisphere say, r = 3.5 cm
Now, required TSA of the wooden article = 2 \(\times\)CSA of one hemisphere + CSA of cylinder
\(\begin{aligned} & =2 \times\left(2 \pi r^2\right)+2 \pi r h \\ \end{aligned}\)
\(\begin{aligned} & =2 \pi r(2 r+h) \\ \end{aligned}\)
\(\begin{aligned} & =2 \times \frac{22}{7} \times 3.5 \times(2 \times 3.5+10) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 7 \times(7+10)=22 \times 17=374 \mathrm{~cm}^2 \end{aligned}\)

Given, diameter of cylinder = Diameter of conical cavity
= 1.4 cm

\(\therefore\) Radius of cylinder = Radius of conical cavity
\(=\frac{\text { Diameter }}{2}=\frac{1.4}{2}=0.7 \mathrm{~cm}\)
Height of the cylinder = height of the conical cavity
= 2.4 cm
\(\therefore\) Slant height of the conical cavity,
\(\begin{aligned} l & =\sqrt{b^2+r^2}=\sqrt{(2.4)^2+(0.7)^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{5.76+0.49} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{6.25}=2.5 \mathrm{~cm} \end{aligned}\)
Now, TSA of remaining solid = CSA of conical cavity + CSA of cylinder + Area of the base of the cylinder
\(\begin{aligned} & =\pi r l+2 \pi r h+\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =\pi r(l+2 h+r) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 0.7 \times(2.5+2 \times 2.4+0.7) \\ \end{aligned}\)
\(\begin{aligned} & =22 \times 0.1 \times(2.5+4.8+0.7) \end{aligned}\)
\(\begin{aligned} & =2.2 \times 8 \\ \end{aligned}\)
\(\begin{aligned} & =17.6 \approx 18 \mathrm{~cm}^2 \end{aligned}\)

Here, a cone and spherical lead shots are given. Since, lead shots are dropped into the vessel, so the water which flows out from the vessel is equal to the volume of lead shots.
Given, height of the vessel, h = 8 cm and radius of the vessel, r = 5 cm.
\(\therefore\) Volume of water filled in a vessel = Volume of cone
\(\begin{aligned} & =\frac{1}{3} \times \pi \times r^2 \times h \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{4400}{21} \mathrm{~cm}^3 \end{aligned}\)
Also, radius of a spherical lead shot = 0.5 cm
\(\therefore\) Volume of one spherical lead shot \(=\frac{4}{3} \times \pi \times r^3\)
\(=\frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{11}{21} \mathrm{~cm}^3\)
Let n be the number of lead shots dropped in the vessel, then according to the question,
n \(\times\) Volume of one spherical lead shot
= \(\frac{1}{4}\)\(\times\)Volume of water filled in a vessel
\(\begin{aligned} \Rightarrow & & n \times \frac{11}{21} & =\frac{4400}{21} \times \frac{1}{4} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n \times 11 & =1100 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n & =100 \end{aligned}\)
Hence, required number of lead shots is 100.

Given that, medicine capsule is a combination of two hemispheres and one cylinder.

Also, given diameter of the capsule = 5 mm
\(\therefore\) Radius = \(\frac{5}{2}\)= 2.5 mm
and length of the capsule = 14 mm
\(\therefore\) Length of the cylindrical portion = 14 - (2.5 + 2.5) = 9 mm
Now, CSA of one hemispherical portion
\(=2 \pi r^2=2 \times \frac{22}{7} \times 2.5 \times 2.5=\frac{275}{7} \mathrm{~mm}^2\)
and CSA of the cylindrical portion
\(=2 \pi r h=2 \times \frac{22}{7} \times 2.5 \times 9=\frac{22}{7} \times 45=\frac{990}{7} \mathrm{~mm}^2\)
Now, required surface area of medicine capsule = 2 \(\times\) CSA of one hemisherical portion + CSA of the cylindrical portion
\(=2 \times \frac{275}{7}+\frac{990}{7}=\frac{550}{7}+\frac{990}{7}=\frac{1540}{7}=220 \mathrm{~mm}^2\)

Given that, tent is a combination of a cylinder and a cone.

For conical portion,
Slant height, l = 2.8 m
Radius, r = Radius of cylinder  \(=\frac{\text { Diameter }}{2}=\frac{4}{2}=2 \mathrm{~m}\)
For cylindrical portion,
Radius, r = \(\frac{4}{2}=2 \mathrm{~cm}\)
Height, h = 2.1 m
\(\therefore\) Required surface area of the tent = CSA of cone + CSA of cylinder
\(\begin{aligned} & =\pi r l+2 \pi r h=\pi r(l+2 h) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 2 \times(2.8+2 \times 2.1) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{44}{7}(2.8+4.2)=\frac{44}{7} \times 7=44 \mathrm{~m}^2 \end{aligned}\)
Now, cost of the canvas of the tent at the rate of Rs 500 per m² = Surface area \(\times\) Cost per m² = 44 \(\times\) 500 = Rs 22000

Given, side of the cube = Diameter of the hemisphere = l units
\(\therefore\) Radius of the hemisphere, \(r=\frac{l}{2}\) units

Now, required surface area of the remaining solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere
\(\begin{aligned} & =6 \times(\text { Edgc })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times l^2+2 \pi \times\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 l^2+2 \pi \times \frac{l^2}{4}-\pi \frac{l^2}{4}=6 l^2+\pi \frac{l^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{l^2}{4}(\pi+24) \text { sq units } \end{aligned}\)

Given, a cubical block is surmounted by a hemisphere. Therefore, diameter of hemisphere must be equal to the side of cubical block and it is the greatest diameter of hemisphere.

For cubical portion,
Edge = 7 cm
For hemispherical portion,
Diameter = 7 cm
\(\therefore\)   Radius, r = \(\frac{7}{2}\)cm
Now, required surface area of solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere.
\(\begin{aligned} & =6 \times(\text { Edge })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times(7)^2+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}-\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+11 \times 7-\frac{11 \times 7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+77-\frac{77}{2} \end{aligned}\)
= 371 - 38.5 = 332.5 cm²