Question

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form
a cuboid o​

Answer 1

Answer:

Hope it helps..

Mark as brainliest is proves to be helpful for u

Answer 2

Appropriate Question:

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to forma cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm.

Solution :

Let the required number of couns be n .

Now, volume of n coins = volume of the cuboid.

Therefore ,

$\sf{}n \times (\pi \times ( \dfrac{1.75}{2} ) ^{2} \times {(0.2)}^{2} ) = 5.5 \times 10 \times 3.5 \\ \\ \sf{}n \times \frac{22}{7} \times \frac{175}{200} \times \frac{175}{ 200} \times \frac{2}{10} = \frac{55}{ 10} \times 10 \times \frac{35}{10} \\ \\ \sf{}n \times \frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times 2 = 55 \times 35 \\ \\ \sf{} \frac{n \times 11 \times 7}{16} = 55 \times 35 \\ \\ \sf{}n = \frac{55 \times 35 \times 16}{7 \times 11} \\ \\ \sf{} \implies 5 \times 5 \times 16 \\ \\ \boxed{ \sf{}n = 400}$

Therefore, number of coins is equals to 400.

$\rule{200pt}{4pt}$

$\underline{ \underline{ \sf{ \red{ \bold{ more \: Formulas}}}}} \\ \\ \: \sf{}volume \: of \: cube \: = {a}^{3} \\ \\ \sf{}volume \: of \: cylinder = \pi {r}^{2} h \\ \\ \sf{}volume \: of \: cone = \frac{1}{3} \pi \: {r}^{2} h \\ \\ \sf{}volume \: of \: sphere \: = \frac{4}{3} \pi \: {r}^{3} \\ \\ \sf{} volume \: of \: hemisphere = \frac{2}{3} \pi {r}^{3}$