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

1 thought on “5. A vessel is in the form of an inverted cone. Its<br />height is 8 cm and the radius of its top, which is<br />open, is 5 cm. It”

1. [tex]\huge{\bf{\red{Solution:-}}}[/tex]

   [tex]\bf{For \: cone}[/tex]

   [tex]\sf \: Radius \: of \: the \: top \: (r) = 5 \: cm[/tex]

   [tex]\sf \: Height \: (h) = 8 \: cm[/tex]

   [tex]\therefore \: \: \: \: \: \sf \: \: Volume \: of \: the \: cone[/tex]

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

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{3}\pi \times 25 \times 8 = \frac{200}{3}\pi \: cm {}^{3} \\ \end{gathered} [/tex]

   [tex]\therefore \: \: \: \: \: \sf \: Volume \: of \: water \: in \: cone[/tex]

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \frac{200\pi}{3} \: cm {}^{3} \\ \end{gathered} [/tex]

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

   [tex]\: \: \: \bf{For \: spherical \: lead \: shot}[/tex]

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

   [tex]\therefore \: \: \: \: \: \: \: \sf \: Volume \: of \: a \: spherical \: lead \: shot[/tex]

   [tex]\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} [/tex]

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{\pi}6 \: cm {}^{3} \: \: \: \: \: \: \: \: \: \: …(1) \\ \end{gathered} [/tex]

   [tex]\bf{Given :} \: \sf \: Volume \: of \: water \: that \: flows \: out [/tex]

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \frac{1}{4 } \: volume \: of \: the \: (water) \: cone \\ \end{gathered} [/tex]

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{4} \bigg( \frac{200\pi}{3} \bigg) \: cm {}^{3} = \frac{50\pi}{3} \: cm {}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: …(2) \\ \end{gathered} [/tex]

   [tex]\sf \: Let \: the \: number \: of \: lead \: shots \: dropped \: in \: the \: vessel \: be \: \red{n}..[/tex]

   [tex]\sf \: Then,[/tex]

   [tex]\sf \: Volume \: of \: n \: lead \: [/tex]

   [tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: n \times volume \: of \: one \: lead \: shot[/tex]

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

   [tex]\sf \: According \: to \: the \: question, [/tex]

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

   [tex]\therefore \: \: \: \: \: \sf \: by \: (3) \: and \: (2)[/tex]

   [tex]\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \sf \frac{n\pi}{6} = \frac{50\pi}{3} \\ \end{gathered} [/tex]

   [tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf3n\pi = 300\pi \: \: \: \: \: \: \: \: \: \: \: (Cross – multiplying)[/tex]

   [tex]\begin{gathered}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: n \: = \frac{300\pi}{3\pi} = 100 \\ \end{gathered}[/tex]

   [tex]\sf \: Hence, \: the \: number \: of \: lead \: shots \: dropped \: in \: the \: vessel \: is \: 100.[/tex]

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