Question

5. A vessel is in the form of an inverted cone. Its
height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim.
When lead shots, each of which is a sphere of radius
0.5 cm are dropped into the vessel, one-fourth of
the water flows out. Find the number of lead shots
dropped in the vessel.​

Answer 1

$\huge{\bf{\red{Solution:-}}}$

$\bf{For \: cone}$

$\sf \: Radius \: of \: the \: top \: (r) = 5 \: cm$

$\sf \: Height \: (h) = 8 \: cm$

$\therefore \: \: \: \: \: \sf \: \: Volume \: of \: the \: cone$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{3}\pi r {}^{2}h = \frac{1}{3}\pi(5 ) {}^{2} 8 \\ \end{gathered}$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{3}\pi \times 25 \times 8 = \frac{200}{3}\pi \: cm {}^{3} \\ \end{gathered}$

$\therefore \: \: \: \: \: \sf \: Volume \: of \: water \: in \: cone$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \frac{200\pi}{3} \: cm {}^{3} \\ \end{gathered}$

$\sf \: [∵ \: the \:cone \:is \: filled \: \:with \: water\: upto\: the\:brim \:(i.e., \: top)]$

$\: \: \: \bf{For \: spherical \: lead \: shot}$

$\begin{gathered} \sf \: Radius \: (R) = 0.5 \: cm = \frac{5}{10} = \frac{1}{2} \: cm \\ \end{gathered}$

$\therefore \: \: \: \: \: \: \: \sf \: Volume \: of \: a \: spherical \: lead \: shot$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{4}{3}\pi R {}^{3} = \frac{4}{3}\pi \bigg( \frac{1}{2} \bigg) {}^{2} = \frac{4\pi}{3} \times \frac{1}{8} \\ \end{gathered}$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{\pi}6 \: cm {}^{3} \: \: \: \: \: \: \: \: \: \: ...(1) \\ \end{gathered}$

$\bf{Given :} \: \sf \: Volume \: of \: water \: that \: flows \: out$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \frac{1}{4 } \: volume \: of \: the \: (water) \: cone \\ \end{gathered}$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{4} \bigg( \frac{200\pi}{3} \bigg) \: cm {}^{3} = \frac{50\pi}{3} \: cm {}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(2) \\ \end{gathered}$

$\sf \: Let \: the \: number \: of \: lead \: shots \: dropped \: in \: the \: vessel \: be \: \red{n}..$

$\sf \: Then,$

$\sf \: Volume \: of \: n \: lead \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: n \times volume \: of \: one \: lead \: shot$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{n\pi}{6} \: cm {}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [by(1)] \: \: ...(3) \\ \end{gathered}$

$\sf \: According \: to \: the \: question,$

$\begin{gathered} \sf \: These \: n \: lead \: shots \: make \: \frac{1}{4} \: of \: the \: water \: \\ \sf \: flow \: out \: of \: the \: cone.\end{gathered}$

$\therefore \: \: \: \: \: \sf \: by \: (3) \: and \: (2)$

$\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \sf \frac{n\pi}{6} = \frac{50\pi}{3} \\ \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf3n\pi = 300\pi \: \: \: \: \: \: \: \: \: \: \: (Cross - multiplying)$

$\begin{gathered}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: n \: = \frac{300\pi}{3\pi} = 100 \\ \end{gathered}$

$\sf \: Hence, \: the \: number \: of \: lead \: shots \: dropped \: in \: the \: vessel \: is \: 100.$