Math, asked by smriti2004, 1 year ago

Two solids of right circular cylindrical shape hi 49 cm and 35 cm high and their bases diameter are 16 cm 14 cm respectively. Both are melted and moulded into a single cylinder 56 cm. Find its base diameter.

Answers

Answered by mino9472852163
4

Answer:

Step-by-step explana

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Answered by Anonymous
5

Answer :-

Given for Cylinder 1 ( C_1

Height = 49 cm

Diameter = 16 cm

then radius = 8 cm

Now volume of C_1

 = \pi {r}^{2} h \\  \\  =  \frac{22}{7}  \times 8 \times 8 \times 49 \\  \\  = 64 \times 22 \times 7 \\  \\  = 9856 \:  {cm}^{3}

Given for Cylinder 2 ( C_2)

Height = 35 cm

Diameter = 14 cm

then radius = 7 cm

Now volume of  C_2

 = \pi {r}^{2} h \\  \\  =  \frac{22}{7}  \times 7 \times 7 \times 35 \\  \\  = 22 \times 49 \times 5 \\  \\  = 5390 \:  {cm}^{3}

Now for Cylinder 3 ( C_3)

Volume = Volume of  C_1 + Volume of  C_2

= 9856 + 5390

=  15246 \: cm^3

Height = 56 cm

Now by putting values in volume equation

 15246 = \pi {r}^{2} h \\  \\  =  >  {r}^{2}  =  \frac{15246}{\pi \: h}  \\  \\  =  >  {r}^{2}  =  \frac{15246}{ \frac{22}{7}  \times 56}  \\  \\  =  >  {r}^{2}  =  \frac{272.25 \times 7}{22}  \\  \\  =  >  {r}^{2}  = 86.625 \\  \\  =  >  r \:  =  \sqrt{86.625}  \\  \\  =  > r \:  = 9.307255 \\  \\ or \:   \\ \\  r \:  = 9.31 \: cm \:

Hope it helps you

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