Question

How many silver coins 1.75cm in diameter and of thickness 2mm must be melted to form a cuboid of dimensions 5.5cm10cm 3.5?

Answer 1

Answer:

Step-by-step explanation:

Given

The dimensions of cuboid.

5.5 /10 /3.5

The dimensions of cylinderical coin

Diameter = 1.75cm and Height = .2 cm

To find = no. Of coins to fir a cuboid of dimensions 5.5/10/3.5

Vol of cuboid = n× vol of coins

Here (n) is the no. Of coins

L×b×h = πr^2h ×n

l×b×h/πr^2h = n

Substitute the value

7×5.5×10×3.5/22×[1.75/2]^2×.2 = n

7×55×3.5/22×[3.0625/4] ×.2 = n

7×5×3.5/2×[3.0625/4]×.2 = n

35×3.5/[3.0625/10] = n

35×35/3.0625 = n

1225/3.0625 = n

400 = n

Therefore

N = 400

Hence the no. of coins required for making a cuboid is 400

Answer 2

For a coin : $\sf{r=\frac{175}{200}=\frac{7}{8}cm,h=2mm=\frac{2}{10}cm}$

Let n be the number of coins. Then,

n × Volume of one coin = Volume of a cuboid

= $\sf{n×π{r}^{2}h=l×b×h}$

= $\sf{n \times \frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times \frac{2}{10} = 5.5 \times 10 \times 3.5}$

Hence, $\sf{n = \frac{55 \times 10 \times 35 \times 7 \times 8 \times 8 \times 10}{22 \times 7 \times 7 \times 2 \times 10 \times 10} = 400}$