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.

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