Question

A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness .If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder. ​

Answer 1

GIVEN:

• Radius of sphere = 6 cm
• External radius of cylinder = 5 cm
• Height of the cylinder = 32 cm

TO FIND:

• What is the thickness of the cylinder ?

SOLUTION:

Let the internal radius of the cylinder be 'r' cm and external radius be 'R' cm

We know that the formula for finding the Volume of sphere is:-

$\large{\boxed{\bf{\star \: VOLUME = \dfrac{4}{3} \pi r^3 \: \star}}}$

We know that the formula for finding the Volume of hollow cylinder is:-

$\large{\boxed{\bf{\star \: VOLUME = \pi (R^2 - r^2) h \: \star}}}$

According to question:-

$\large\bf{\star \: \dfrac{4}{3} \pi r^3 = \pi (R^2 - r^2 ) h\: \star}$

On putting the given values in the formula, we get

$\sf{\longmapsto \dfrac{4}{3} \cancel{\pi} r^3 = \cancel{\pi }(R^2 - r^2 ) h}$

$\sf{\longmapsto \dfrac{4}{3} \times (6)^3 = (5^2 - r^2 ) \times 32 }$

$\sf{\longmapsto \dfrac{4}{\cancel{3}} \times \cancel{216} = (25-r^2 ) \times 32 }$

$\sf{\longmapsto \dfrac{288}{32} = 25-r^2 }$

$\sf{\longmapsto 9-25 = -r^2 }$

$\sf{\longmapsto \cancel{-}16 = \cancel{-} r^2 }$

$\sf{\longmapsto \sqrt{16} = r }$

$\bf{\longmapsto 4 \: cm = r }$

• Internal radius = r = 4 cm

❍ Thickness of cylinder = (R–r) = (5–4) = 1 cm

❝ Hence, the thickness of the cylinder is 1 cm ❞

______________________

Answer 2

$\rule{300}3$

$\huge{\blue{\fbox{\bold{\red{UR\:QUESTION}}}}}$

$\huge {\mathbf {\purple {Q.}}}}$A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness .If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.

$\rule{300}3$

$\huge{\blue{\fbox{\bold{\red{UR\:ANSWER}}}}}$ ✓

$\Large\underline\bold{given}$

$\sf\dashrightarrow Radius\: of\: sphere\: = 6 cm$

$\sf\dashrightarrow External\: radius\: of \:cylinder = 5 cm$

$\sf\dashrightarrow Height\: of\: the \:cylinder\: = 32 cm$

$\Large\underline\bold{TO FIND,}$

$\undeline{\sf{\red{thickness\: of\: the \:cylinder }}}$

$\Large\underline\bold{diagram}$

$\setlength{\unitlength}{1mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(5,0){20}}\put(25,30){\circle*{1}}\put(29,26){\sf\large{6cm}}\end{picture}$

$\Large\underline\bold{solution,}$

$\sf\underline\bold{let,}$

$\sf\implies internal\:radius='r'cm$

$\sf\implies external\:radius\:= 'R' cm$

$\sf\bold{volume\:of\:sphere=\dfrac{4}{3} \pi r^3 is }$

$\sf\implies volume\:of\:hollow\:cylinder= \pi (R^2 - r^2) h$

$\sf\underline\bold{A.T.Q...........i.e,. according\:to\: question}$

$\sf\implies \dfrac{4}{3} \pi r^3 = \pi (R^2 - r^2 ) h$

$\Large\bold{therefore,}$

$\sf{\implies \dfrac{4}{3} \cancel{\pi} r^3 = \cancel{\pi }(R^2 - r^2 ) h}$

$\sf{\implies \dfrac{4}{3} \times (6)^3 = (5^2 - r^2 ) \times 32 }$

$\sf{\implies \dfrac{288}{32} = 25-r^2 }$

$\sf{\implies 9-25 = -r^2 }$

$\sf{\implies \cancel{-}16 = \cancel{-} r^2 }$

$\sf{\implies \sqrt{16} = r }$

$\sf{\implies 4 \: cm = r }$

$\sf{\fbox{r=4cm}}$

$\sf\therefore Thickness\: of \:cylinder = (R-r) = (5-4) = 1 cm$

$\sf{\fbox{thickness=1cm}}$

$\rule{300}3$

$\tt\huge\red{\text{BE \:BRAINLY}}$

$\rule{300}3$