Question

a vessel is in the form of an inverted cone. its height is 8 cm and radius of its top, which is open, is 5 cm. it is filled with water upto the brim. when 100 lead shots, each of which is a sphere are dropped into the vessel, one fourth of the water flows out. find the radius of the lead shot​

Answer 1

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

Given that,

• A vessel is in the form of an inverted cone. its height is 8 cm and radius of its top, which is open, is 5 cm.

So,

• Radius of cone, r = 5 cm

• Height of cone, h = 8 cm

We know,

$\rm \: Volume_{(Cone)} = \dfrac{1}{3} \pi \: {r}^{2}h - - - (1)$

Further given that,

• It is filled with water upto the brim. When 100 lead shots, each of which is a sphere are dropped into the vessel, one fourth of the water flows out.

Let assume that radius of spherical lead shot = R cm

So,

$\rm \: Volume_{(Water\:overflow)} = \dfrac{1}{4} \times Volume_{(Cone)}$

$\rm \: Volume_{(Water\:overflow)} = \dfrac{1}{4} \times \frac{1}{3} \times \pi \: {r}^{2}h$

$\rm\implies \:\rm \: Volume_{(Water\:overflow)} = \dfrac{1}{12} \pi \: {r}^{2}h - - - (2)$

Also,

$\rm \: Volume_{(Water\:overflow)} = 100 \times Volume_{(lead \: shot)}$

$\rm \: Volume_{(Water\:overflow)} = 100 \times \frac{4}{3} \: \pi \: {R}^{3}$

$\rm\implies \:\rm \: Volume_{(Water\:overflow)} = \frac{400}{3} \: \pi \: {R}^{3} - - - (3)$

So, on equating equation (2) and (3), we get

$\rm \: \dfrac{1}{12}\pi {r}^{2}h = \dfrac{400}{3}\pi {R}^{3}$

$\rm \: {r}^{2}h \: = \: 1600 {R}^{3}$

$\rm \: {5}^{2} \times 8 \: = \: 1600 {R}^{3}$

$\rm \: 25 \: = \: 200 {R}^{3}$

$\rm \: {R}^{3} = \dfrac{25}{200}$

$\rm \: {R}^{3} = \dfrac{1}{8}$

$\rm \: {R}^{3} = \dfrac{1}{2 \times 2 \times 2}$

$\rm \: {R}^{3} = \dfrac{1}{ {2}^{3} }$

$\rm\implies \:R = \dfrac{1}{2} \: = \: 0.5\: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

As the water is filled up to the brim in the vessel

Volume of water in the vessel = Volume of the conical vessel

On dropping a certain number of lead shots (sphere) into the vessel one-fourth of the water flows out.

Volume of all lead shots dropped into the vessel = 1/4 × Volume of the water in the vessel

Hence,

Number of lead shots = 1/4 × volume of the water in the vessel ÷ volume of each lead shot

We will find the volume of the water in the vessel and lead shot by using formulae;

Volume of the sphere = 4/3 πr3

where r is the radius of the sphere

Volume of the cone = 1/3 πR2h

where R and h are the radius and height of the cone respectively

Height of the conical vessel, h = 8 cm

Radius of the conical vessel, R = 5 cm

Radius of the spherical lead shot, r = 0.5 cm

Number of lead shots = 1/4 × volume of the water in the vessel ÷ volume of each lead shot

= 1 /4 × (1/3) πR2h × 3/4 πr3

= πR2h/12 × 3/4πr3

= R2h / 16r3

= (5cm × 5 cm × 8 cm) / (16 × 0.5 cm × 0.5 cm × 0.5 cm)

= 100

Thus, the number of lead shots dropped in the vessel is 100.