Question

two solids of right cylindrical shape of 49 cm and 35 cm high and their base diameter 16 cm and 14 cm respectively. both are melted and moulded into a single cylinder 56 cm high. find its base diameter

Answer 1

$volume \: of \: two \: cylinders \\ = \frac{\pi}{4} \times {16 }^{2} \times 49 + \frac{\pi}{4} \times {14 }^{2} \times 35 \\ = \frac{\pi}{4} \times 19404 \\ let \: diameter \: of \: single \: cylinder \: be \: \: d \\ then \: \frac{\pi}{4} {d}^{2} \times 56 = \frac{\pi}{4} \times 19404 \\ {d}^{2} = \frac{19404}{56} = 346.5 \\ d = 18.6$

Answer 2

$\underline{\huge{\text{Solution:}}}$

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$\textit{\underline{\text{Volume of two cylinders:}}}$

⇒ $\bold{ \frac{\pi}{4} \times {16}^{2} \times 49 + \frac{\pi}{4} \times {14}^{2} \times 35}$

⇒ $\bold{ \frac{\pi}{4} \times 19404}$

$\textit{\underline{\text{Diameter (d):}}}$

⇒ $\bold{ \frac{\pi}{4} {d}^{2} \times 56 = \frac{\pi}{4} \times 19404}$

⇒ $\bold{ {d}^{2} = \frac{19404}{56} = 346.5}$

$\bold{So,\: the \: base \: diameter \: is}$

⇒ $\bold{d = 18.6}$ ✔️


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$\bold{\fbox{Hope\:It\:Helps\:!!!}}$
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