Question

The dimensions of a solid rectangular slab of lead is 66 cm, 42 cm and 21cm respectively. Find by melting this how many sphere of diameter 4.2 cm can be formed ?​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

${\boxed{\sf\:{r=\dfrac{d}{2}}}}$

${\boxed{\sf\:{r=\dfrac{4.2}{2}}}}$

r = 2.1 cm

${\boxed{\sf\:{Dimension\;of\;cuboid}}}$

= (66 × 42 × 21) cm

Assumption

${\boxed{\sf\:{Number\;of\;lead\;be\;p}}}$

Situation,

p × Volume of Sphere = Volume of Cuboid

$\tt{\rightarrow p\times\dfrac{4}{3}\times \pi\times r^3=l\times b\times h}$

Put the values,

$\tt{\rightarrow p\times\dfrac{4}{3}\times\dfrac{22}{7}\times 2.1\times 2.1\times 2.1=66\times 42\times 21}$

$\tt{\rightarrow p\times\dfrac{4}{3}\times\dfrac{1}{7}\times 2.1\times 2.1\times 2.1=3\times 42\times 21}$

$\tt{\rightarrow p\times\dfrac{4}{3}\times\dfrac{1}{7}\times 0.7\times 0.1\times 0.1= 1\times 2\times 1}$

$\tt{\rightarrow p\times\dfrac{2}{3}\times\dfrac{1}{7}\times\dfrac{7}{10}\times\dfrac{1}{10}\times\dfrac{1}{10}=1}$

$\tt{\rightarrow p\times\dfrac{1}{3}\times 1\times\dfrac{1}{5}\times\dfrac{1}{10}\times\dfrac{1}{10}=1}$

p = 3 × 5 × 10 × 10

= 1500

Hence

$\Large{\fbox{Number\;of\;spherical\;lead\;can\;be\;obtained\;is\;1500}}$

Answer 2

Answer :

The Number of spherical lead can be obtained is 1500.

Step-by-step explanation :

Given :

◆ Diameter = 4.2 cm.

As we know,

Radius is the Half of Diameter.

So,

$\sf \implies r=\dfrac{4.2}{2}$

$\sf \implies 2.1 cm$

◆ Radius = 2.1 cm.

◆ Volume of one spherical lead shot = $\sf \dfrac{4}{3}\times (2.1)^3 \times \pi cm^3$

=> (66 × 42 × 21) cm.

Now,

x × Volume of Sphere = Volume of Cuboid.

{ Here - x is the number of lead. }

Values in Equation,

$\sf \\\implies x\times\dfrac{4}{3}\times \pi\times r^3=L\times B\times H}\\ \\ \\\implies x\times\dfrac{4}{3}\times\dfrac{22}{7}\times 2.1\times 2.1\times 2.1=66\times 42\times 21}\\ \\ \\\implies x\times\dfrac{4}{3}\times\dfrac{1}{7}\times 2.1\times 2.1\times 2.1=3\times 42\times 21}\\ \\ \\\implies x\times\dfrac{4}{3}\times\dfrac{1}{7}\times 0.7\times 0.1\times 0.1= 1\times 2\times 1}\\ \\ \\\implies x \times\dfrac{2}{3}\times\dfrac{1}{7}\times\dfrac{7}{10}\times\dfrac{1}{10}\times\dfrac{1}{10}=1}$

$\sf\\\\\implies x\times\dfrac{1}{3}\times 1\times\dfrac{1}{5}\times\dfrac{1}{10}\times\dfrac{1}{10}=1}\\ \\ \\\implies P= 3 \times 5 \times 10 \times 10\\ \\ \\ \implies 1500$

∴ The Number of spherical lead can be obtained is 1500.