Question

a solid is in the form of a right circular cylinder with a hemisphere shape at one end and a cone at the other end. their common diameter is 4.2 cm and heights of the cylindrical and conical portion are 12 cm and 7cm. respectively find the volume of the solid toy (π = 22/7)​

Answer 1

Step-by-step explanation:

$\textsf {\large{\underline { \underline {SOLUTION:-}}}}$

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

$\bf➠Then, \angle XOA=60⁰ \: and \: \angle \: XOB=30⁰$

$\bf ➠Draw \: AL \perp \: OX \: and \: BM \perp \: OX$

$\bf \:⇰ Let \: AL=BM=h \: meters.$

$\bf \: ⇰ \: Speed \: of \: Jet = 720 \: km/hr$

$\bf = \huge( \small \: 720 \times \frac{5}{18} \huge) \small \: m / s$

$\bf = 200 \: m/s$

$\bf \:➥ Time \: taken \: to \: cover \: the \: distance \: AB=15 \: sec$

$\bf➥ Distance \: covered =(speed×time)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = (200 \times 15)m = 3000m$

$\bf \therefore LM=AB=3000m$

${ \bf \: ⇰ \: Let \: OL} = x \: \bf m$

$\bf➥From \: right \: \triangle OLA, \: we \: have:$

$\bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h} = \frac{1}{ \sqrt{3} }$

$\bf \implies \: x = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: ....(1)$

$\bf \:➥ From \: right \: \triangle \: OMB,we \: have:$

$\bf\frac{OM}{BM} =cot30⁰= \sqrt{3}$

$\bf \implies \: \frac{x + 3000}{h} = \sqrt{3}$

$\bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m \: and \: BM=h \: m]}$

$\bf \implies \: x + 3000 = \sqrt{3} h$

$\bf \implies \: x = ( \sqrt{3} h - 3000) \: \: \: \: \: \: ....(2)$

⇰ Equating the value of x from (1) and (2),we get;

$\bf \: \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} }$

$\implies \bf \: 3h - 3000 \sqrt{3} = h$

$\bf \implies \: 2h = (3000 \times \sqrt{3} ) = (3000 \times 1.762)$

$\bf \implies \: h = (3000 \times 0.866) = 2598$

$\bf \:➥ Hence \: ,the \: required \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}}$

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Answer 2

We have, volume of the solid toy = volume of the conical portion + volume of the cylindrical portion + volume of the hemispherical portion

$= \frac{1}{3}\pi {r}^{2}h \: + \pi {r}^{2} {h}^{ ҅ } \: + \frac{2}{3} \pi {r}^{3} \\ \\ = \frac{1}{3} \times ( {2.1)}^{2} \times 7 \: + \pi \times {(2.1)}^{2}×12 + \\ \frac{2}{3} \times \pi \times {(2.1)}^{2} \\ \\ = \pi \times {(2.1)}^{2}( \frac{1}{3} \times 7 \: + 12 \: + \frac{2}{3} \times 2.1) \\ \\ = \frac{22}{7} \times 4.41( \frac{7}{3} + 12 + \frac{4.2}{3} ) \\ \\ = \frac{97.02}{7} (2.33 + 12 + 1.4) \\ \\ = 13.86 \times 15.73 \\ \\ = 218.01 {cm}^{3}$

I hope this helps.