Question

A solid cuboid of iron with dimensions 53 cm × 40 cm × 15 cm is melted and recast into a cylindrical pipe. the outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of the cylinder.

Answer 1

Heya mate, Here is ur answer

Dimension of cuboid=53 cm× 40 cm × 15 cm

Volume of cuboid=lbh

=53×40×15

=31800 cm^3.

Let the height of the cylinder be h.

Outer Diameter=8

Outer Radius (r) =8/2

r=4 cm.

Inner Diameter=7 cm

Inner radius (r2)= 7/2 cm

=3.5 cm
Volume of hollow cylinder = π(r^2-r2^2)×h

$= \frac{22}{7} ( 4 {}^{2} - 3.5 {}^{2} ) \times h$

$= \frac{22}{7} (4 + 3.5)(4 - 3.5) \times h$

$= \frac{22}{7} \times 7.5 \times 0.5 \times h$

$= \frac{82.5}{7} \times h$

When one solid is melted to form another, their volumes remain unchanged.

So,

Volume of cylinder= volume of cuboid

$\frac{82.5}{7} \times h = 31800$

$h = \frac{31800 \times 7}{82.5}$

h=2698.18 (approx)

Warm regards

@Laughterqueen

Be Brainly ✌✌

Answer 2

$\underline \bold {Solution:-}$

Volume of solid cuboid of iron

$= l \times b \times h \\ \\ = 53 \times 40 \times 15 \: {cm}^{3} \\ \\ = 31800 \: {cm}^{3}$

It is melted and recast into a cylindrical pipe of outer diameter 8cm and inner diameter 7cm.

Inner radius (r) = 7/2 = 3.5 cm

Outer radius (R) = 8/2 = 4 cm

Let the length of the pipe be h cm.

Volume of hollow cylindrical pipe

$= \pi( {R}^{2} - {r}^{2} )h \\ \\ = 3.14 \times ( {4}^{2} - {3.5}^{2} )h \\ \\ = 3.14 \times (16 - 12.25)h \\ \\ = 3.14 \times 3.75 \times h \\ \\ = 11.77h \: {cm}^{3}$
Since,

volume of cuboid = Volume of pipe

$31800 \: {cm}^{3} = 11.77h \: {cm}^{3} \\ \\ h = \frac{31800 \: {cm}^{3} }{11.77 \: {cm}^{3} } \\ \\ h = 2717.94$

Therefore, length of pipe = 2717.94 cm.

Hope it may help you.