Question

A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into 12 toys in the shape
of a right circular cone mounted on a hemisphere. Find the total height of the toy if height of the
conical part is 3 times its radius.
(A) 9 cm
(B) 12 cm
(C) 15 cm
(D) 18 cm​

Answer 1

Answer:

cylinder       cone              hemissphere

r=6              r=3                r=3

h=15           h=12-3=>9

volume of cylinder =[ volume of cone +volume of hemisphere]nos. of toys

=>πx36x15=2/3xπx27 + 1/3xπx9x9x n

=>πx36x15=1/3xπ[54+81]x n

=>36x15x3=135x n

=>12=n

Answer 2

$\sf \large \underline{ \underline {GIVEN:-}}$

$\sf \red{•Radius \: of \: the \: cylinder=6cm}$

$\sf \blue{•Height \: of \: the \: cylinder=15cm}$

$\sf \: TO \: FIND \: OUT:-$

$\sf \green{•Total \: height \: of \: the \: toy}=??$

$\sf \underline {SOLUTION}:-$

$\sf \: •Volume \: of \: cylinder=( \pi r²h)</p><p>$

$\sf \: : \implies( \pi ×6×6×15)cm³=(540 \pi)cm³$

$\sf \green{A}/ \orange{Q}$

$\sf•Cylinder \: is \: recasted \: into \: 12 \: toys$

$\sf \therefore Volume \: of \: 12 \: toys=(540 \pi)cm³$

$\sf Volume \: of \: 1 \: toy= {\bigg(} \frac{540 \pi}{12} \bigg)cm³=(45 \pi)cm³$

$\sf •Let \: the \: radius \: of \: the \: hemisphere \:be \: r \: cm.</p><p></p><p>$

$\sf \: Then,the \: height \: of \: the \: cone \: =3r \: cm$

$\sf \blue{Volume \: of \: 1 \: toy}= \purple{Volume \: of \: hemisphere }+ \red{Volume \: of \: cone}$

$\: \: \: \: \: \: \sf = \bigg( \frac{2}{3} \pi r³+ \frac{1}{3} \pi r²×3r \bigg)cm³$

$\sf \: \: \: = \bigg( \frac{5 \pi r³}{3} \bigg)cm³$

$\sf \therefore \frac{5 \pi r³}{3} =45 \pi \implies r³= \bigg(45× \frac{3}{5} \bigg)=27=3³$

$\sf \implies r=3$

$\sf \orange{Radius \: of \: the \: hemisphere }=3cm$

$\therefore \sf \blue {Total \: height \: of \: the \: toy}=(r+3r)cm \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =(4r)cm \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf=(4×3)cm \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf=12cm$

So,Option B is the correct answer