A test tube has lower part as hemisphere and upper part cylindrical with the same radius. If 6cm³ water is added to the test tube,it will be completely filled. But if 2000/3 cm³ of water is added, 5 cm height will remain empty. Find the radius of the tube and the height of the cylindrical part. (π = 22/7)
Answers
Radius of the tube is 6.48 cm and height of the cylindrical part is 4.5203 cm
Step-by-step explanation:
Given data
A test tube has a lower part as hemisphere and upper part as cylinder with same radius
Volume of the container = 6 cm³
If 2000 /3 cm³ of water is added 5 cm height will remain empty
π = 22 ÷ 7
Find the radius and height of the cylindrical part
Total volume = Volume of the cylinder + volume of the hemisphere
Total volume = πr²h + (2÷3) πr³ -----------> 1
Total volume = πr² (h + (2÷3 r) ------------> 2
Substitute the values in above equation
6 cm³ = (22÷7) r² (h + (2÷3)r ) ------------> 3
6 cm³ = 3.14 r² (h+0.67r) ------------->4
6÷3.14 = r² (h+0.67r)
1.909 = r²h+0.67r³) ------------> 5
If (2000÷3) cm³ of water is added , 5 cm height will remain empty and that can be written as
(2000 ÷ 3) = (volume of the cylinder - 5 cm ) + volume of hemisphere
(2000 ÷ 3) = (πr²h - 5) + (2÷3) πr³
(2000 ÷ 3) = πr²(h-5+ (2÷3)r) -------------->6
(2000 ÷ 3) = (πr² ÷ 3)( 3h-15+2r)
2000 = (πr²)( 3h-15+2r)
2000 = 3.14 r²(3h-15+2r)
2000÷3.14 = r²(3h-15+2r)
636.61 = r²(3h-15+2r)
636.61 = 3hr² -15 r² +2r³ ----------------> 7
From equation ( 5 )
h = (1.909-0.67 r³) ÷r² ----------------> 8
Substitute the value of h in equation 7
636.61 = 3 r² ((1.909-0.67 r³) ÷r²) -15 r² + 2 r³
636.61 = 3(1.909-0.67 r³) -15 r² + 2 r³
636.61 = 5.727 - 2.0 r³ - 15 r² +2 r³
636.61 = 5.727 +2 r³
636.61 - 5.727 = 2 r³
630.883 = 2 r³
r³ = 630.883 ÷ 2
r³ = 315.441
r = 6.48 cm
substitute the value of r in equation 8
h = (1.909 -0.67(6.807)³) ÷ (6.807)²
h = 4.5203 cm
Radius of the tube is 6.48 cm and height of the cylindrical part is 4.5203 cm
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Answer:333
Step-by-step explanation: