Question

plz solve 1,3,4and6.​

Answer 1

♛ QUESTION 1 :-

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6cm. Find height of the cylinder.

Solution :-

${Radius}_{hemisphere}$ = 4.2cm

${Radius}_{cylinder}$ = 6cm

Height (h) = ?

The object fromed by recasting the hemisphere will be

same in volume.

So, Volume of sphere = Volume of cylinder

$\frac{4}{3}\pi {{r}_{1} }^{3} = \pi {{r}_{2}}^{2} h$

$\implies \: \frac{4}{3} \times {(4.2)}^{3} = \pi {(6)}^{2} h$

$\implies \: \frac{4}{3} \times \frac{4.2 \times 4.2 \times 4.2}{36} = h$

$\implies \: h = {(1.4)}^{3} = 2.74 \: cm$

Height of cylinder = 2.74 cm

♛ QUESTION 3 :-

A 20 m deep well with diameter 7 cm is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution :-

Diameter = 7 cm

So, radius = $\dfrac{D}{2} = \dfrac{7}{2} = 3.5cm$

Earth dug out from (cylindrical) well

= π × r × r × h

= $\frac{22}{7} \times 3.5 \times 3.5 \times 20$

= 770m³

to from platform ...... (cuboidal)

volume of cuboid = l × b × h

770 = 22 × 14 × h

So, height = $\frac{770}{22} \times 14 = 2.5m$

♛ QUESTION 4 :-

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution :-

The shape of the well will be cylindrical as shown in the figure below:

Given:

Depth (h) = 14m

$({R}_{1})$ = $\dfrac{3}{2}m$

Width of embankment = 4 m

From the figure, it can be observed that our embankment will be in a cylindrical shape having outer radius, $({R}_{2})$ = $4+ \frac{3}{2} = \frac{11}{2}m$

Let height of embankment be ${h}_{2}$

Volume of soil dug from well = Volume of earth used to form the embankment

$π \times {{r}_{1}}^{2} \times {h}_{1} = [tex] π \times ({{r}_{2}}^{2} - {{r}_{1}}^{2}) \times [tex]{h}_{2}$

$\pi \times { (\frac{3}{2}) }^{2} \times 14 = \pi \times ( { \frac{11}{2} }^{2} - { \frac{3}{2} }^{2} ) \times h2$

${h}_{2} = \frac{9}{8} = 1.125m$