How many silver coins 1.75 cm in diameter and of thickness 2 mm,

For a circular coin :

Diameter = 1.75 cm

$\Rightarrow$ Radius $(r)=\frac{\mathbf{1 7 5}}{\mathbf{2 0 0}} \mathrm{cm}$

Thickness $(\mathrm{h})=2 \mathrm{~mm}=\frac{2}{10}$

$\therefore$ Volume $=\pi r^{2} h=\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10} \mathrm{~cm}^{3}$

For a cuboid : Length $(\ell)=10 \mathrm{~cm}$,

Breadth (b) = 5.5 cm and Height (h) = 3.5 cm

$\therefore$ Volume $=\ell \times \mathrm{b} \times \mathrm{h}=10 \times \frac{\mathbf{5 5}}{\mathbf{1 0}} \times \frac{\mathbf{3 5}}{\mathbf{1 0}} \mathrm{cm}^{3}$

Number of coins $=\frac{\text { Volme of culoid }}{\text { Volume of one coin }}$

$=\frac{10 \times \frac{55}{10} \times \frac{35}{10}}{\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10}}=400$

Thus, the required number of coins $=400$.