Question

two solid of right cylindrical shape of 49 cm and 35 cm high and their base diameter are 16 cm, 14 cm respectively both are melted and molded into a single cylinder 56 cm find its base diameter ​

Answer 1

Answer:

18.6 cm

Step-by-step explanation:

Let the radious of 2 cylinders be r cm and Radious of 3rd cylinders be R cm

$\textbf{\underline{\underline{According\:To\:Question:}}}$

Volume of Cylinder 1 + Volume of Cylinder 2 = Volume of Cylinder 3

$\implies \sf\pi {r}^{2} h + \pi {r}^{2} h = \pi \: (R) ^{2} h$

$\implies\sf \pi( {r}^{2} + {r}^{2} h) = \pi \: R ^{2} h$

$\implies \sf( {8}^{2} \times 49 + {7}^{2} \times 35) = R ^{2} \times 56$

$\implies \sf(64 \times 49 + 49 \times 35) = R ^{2} \times \: 56$

$\implies \sf3136 + 1715 = R ^{2} \times 56$

$\implies \sf4851 = R ^{2} \times 56$

$\implies \sf \: R ^{2} = \dfrac{4851}{56}$

$\implies \sf \: R ^{2} = 86.625$

$\implies \sf \: R = \sqrt{86.625}$

$\implies \sf \: R = 9.3\: cm$

Diameter = 2 × Radious

Diameter = 2 × 9.3 cm

$\huge{\boxed{\tt{18.6\:cm}}}$