two solids of diet cylindrical shape are 49 cm and 35 cm height and their base diameters are 16 cm 14 cm respectively both are melted and moulded into a single cylinder 56 cm find its base diameter
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According to question
Radius of first cylinder R1 = 16/2 = 8 Cm
Height of first cylinder H1 = 49 Cm
So Volume V1 = pi R1^2 H1
= pi * 64 * 49
Similarly
Radius of second cylinder R2 = 14/2 = 7 Cm
Also Height H2 = 35 Cm
So Volume V2 = pi R2^2 H2
= pi * 49 * 35
When Moulded Total volume V = V1 + V2
V = pi [ 64 * 49 + 49 * 35 ]
Height of New cylinder H = 56 Cm
Let radius be R
So V = pi R^2 H
Pi [ 64 * 49 + 49 * 35 ] = pi R^2 * 56
R = 9.30 Cm
Thus Diameter = 18.60 Cm
Radius of first cylinder R1 = 16/2 = 8 Cm
Height of first cylinder H1 = 49 Cm
So Volume V1 = pi R1^2 H1
= pi * 64 * 49
Similarly
Radius of second cylinder R2 = 14/2 = 7 Cm
Also Height H2 = 35 Cm
So Volume V2 = pi R2^2 H2
= pi * 49 * 35
When Moulded Total volume V = V1 + V2
V = pi [ 64 * 49 + 49 * 35 ]
Height of New cylinder H = 56 Cm
Let radius be R
So V = pi R^2 H
Pi [ 64 * 49 + 49 * 35 ] = pi R^2 * 56
R = 9.30 Cm
Thus Diameter = 18.60 Cm
Rakeshp:
thanks bro
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