A block of ice of total area A and thickness 0.5 m is floating in water. In order to just support a man of mass 100 kg, the area A should be (the specific gravity ofice is 0.9):
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Answer:
area a shoukd be the duble of man foot area multiplied by volume of cube
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1.24m21.24m2
4.21m24.21m2
2.41m22.41m2
7.23m27.23m2
Solution :
(c) For equilibrium
(m1+m2)g=ρVg(m1+m2)g=ρVg
Here, m1m1=mass of man =100 kg
m2m2 msss of ice
=0.917×1000V=0.917×1000V
=917V=917V
ρ=1000kg/m2ρ=1000kg/m2
h=0.5 m:. 100g+917Vg=rho=1000Vg`

∴V=100g1000g−917g∴V=100g1000g-917g
∴Ah=10083∴Ah=10083
∴A=10083×0.5∴A=10083×0.5
=2.41m2=2.41m2
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