Question

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):

Answer 1

Answer:

area a shoukd be the duble of man foot area multiplied by volume of cube

Answer 2

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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