A hollow sphere of mass 110 kg is chained to the bottom of a lake such that the

Question

A hollow sphere of mass 110 kg is chained to the bottom of a lake such that the sphere floats 30 meters above the bottom of the lake with the chain tight. The total depth of the lake is 122 meters at the location where the sphere is chained. The density of water is 1000kg/m3. The radius of the hollow sphere is 1 meter.

Part A

What is the pressure on the outside of the sphere?

Give your answer in ATMs.

Part B

What is the tension within the chain where it is attached to the sphere? (assume the chain mass is negligible). In newtons

Explanation / Answer

a) depth of sphere = 122 -30 = 92 m

pressure at a depth of 92 m = p*gh

P = 1000*9.8*92 = 902520 Pa

P in ATM = 902520 / 1.0325*10^5

pressure in atmospheres = 8.9 atm

b) weight of the sphere = 110*9.8 = 1078 N

volume = 4/3 pi*r^3 = 4/3*pi*1^3 = 4.188 m3

volume of water displaced = 4.188 m3

weight of water displaced = volume*density = 4.188*1000*9.8 = 41050.4 N

upthrust = 41050.4 N

tension = upthrust - weight

T = 41050.4 - 1078 = 39972.24 N

the tension in the chain = 39972.24 N

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