Question

$\large \pink\star\ \large \mathfrak \pink{QUESTION}$
A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. The height of the conical heap is 12 cm. The radius of the heap at the base is
$\large \blue\star\ \large \bf \blue{With \: Explanation}$
$\large \green\star\ \large \bf \green{No \: Spam}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. The height of the conical heap is 12 cm.

Now, Dimensions of cylindrical bucket.

Radius of cylindrical bucket, r = 21 cm

Height of cylindrical bucket, h = 36 cm

So, Volume of sand in bucket is equals to volume of cylinder of radius = 21 cm and height = 36 cm.

$\boxed{\sf{  \:\rm \: Volume_{(sand \: in \: cylindrical \: bucket)} = \pi \: {r}^{2} \: h \: }}$

Now, Dimensions of conical heap.

Height of conical heap, H = 12 cm

Let assume that radius of conical heap be R cm.

So, Volume of sand in conical heap is equals to volume of cone of radius, R and height = 12 cm.

So,

$\boxed{\sf{  \:\rm \: Volume_{(sand \: in \: conical \: heap)} = \frac{1}{3} \: \pi \: {R}^{2} \: H \: }}$

As, cylindrical bucket full of sand is emptied to form a conical heap.

So,

$\:\rm \: Volume_{(sand \: in \: cylindrical \: bucket)} \: = \: Volume_{(sand \: in \: conical \: heap)} \:$

$\rm \: \pi \: {r}^{2} \: h \: = \: \frac{1}{3} \: \pi \: {R}^{2} \: H$

$\rm \: {r}^{2} \: h \: = \: \frac{1}{3} \: {R}^{2} \: H$

On substituting the values of r, h and H, we get

$\rm \: 21 \times 21 \times 36 = \dfrac{ {R}^{2} }{3} \times 12$

$\rm \: 21 \times 21 \times 36 = {R}^{2} \times 4$

$\rm \: 21 \times 21 \times 9= {R}^{2}$

$\rm \: {R}^{2} = 21 \times 21 \times 3 \times 3$

$\rm \: R = \sqrt{21 \times 21 \times 3 \times 3}$

$\rm \: R = 21 \times 3$

$\rm\implies \:R = 63 \: cm$

Hence, The radius of conical heap, R = 63 cm.

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Additional Information :-

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}$

$\boxed{ \sf{ \: Volume_{(frustum)} = \dfrac{\pi \: h}{3}( {R}^{2} + {r}^{2} + Rr)}}$

$\boxed{ \sf{ \: CSA_{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA_{(cone)} = \pi \: rl}}$

$\boxed{ \sf{ \: TSA_{(cylinder)} = 2\pi \: r(h + r)}}$

$\boxed{ \sf{ \: TSA_{(cone)} = \pi \: r(l + r)}}$

Answer 2

$\huge \mathfrak \green {I \: hope \: this \: answer \: will \: be \: helpful}$