the hemispherical bowl of internal and external diameter 6cm and 10 cm is melted and formed in to a cylinder of dianeter 14 cm fond the height of tge cylinder
Answers
Answered by
92
if any solid particle is melted and formed new particle by it then
volume of first particle=volume of 2nd particle
now,
volume of bowl=2/3π{R^3-r^3}
where R is external and r is internal radius of bowl
volume of bowl =2/3 π{(5)^3-(3)^3}
=2/3 x πx{125-27}
=2/3 x π x 98
=196/3π cm^3
let length of cylinder =h
volume of cylinder=πR1^2.h
=π(7)^2.h
now,
49π.h=196π/3
h=4/3 cm
volume of first particle=volume of 2nd particle
now,
volume of bowl=2/3π{R^3-r^3}
where R is external and r is internal radius of bowl
volume of bowl =2/3 π{(5)^3-(3)^3}
=2/3 x πx{125-27}
=2/3 x π x 98
=196/3π cm^3
let length of cylinder =h
volume of cylinder=πR1^2.h
=π(7)^2.h
now,
49π.h=196π/3
h=4/3 cm
Answered by
17
Answer:
Step-by-step explanation:
R = 10/2 = 5cm
r = 6/2 = 3 cm
volume of hemisphere = volume of cylinder
2/3*π(R3-r3) = πr'2h
2/3*98 = 196h
h = 2/3 * 98/196 = 1/3cm = 0.33cm
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