A cubical block of side 0.5 m floats on water with 30 % of its volume under water . What is the maximum weight that can be put on the block without fully submerging it under water ? [Take ,density of water =10^{3}Kg/m^{3}]

  • Option 1)

    46.3Kg

  • Option 2)

    87.5Kg

  • Option 3)

    65.4Kg

  • Option 4)

    30.1Kg

Answers (1)

 initial cond: 

B= mg

\left ( \rho \right )\left ( \frac{3}{10}Vg \right )=mg

\left ( 1000Kg/m^{3} \right )\left ( \frac{3}{10} \times (0.5)^3\right )=m

m=37.5Kg

finally \, \, \rho vg =\left ( m+M \right )g

\frac{\left ( m+M \right )g}{mg}=\frac{\rho Vg }{\rho \times \frac{3}{10}Vg}

1+\frac{M}{m}=\frac{10}{3}\Rightarrow \frac{M}{m}=\frac{7}{3}

M=\frac{7}{3}\left ( 37.5 \right )=87.3 Kg

=87.3 Kg

 


Option 1)

46.3Kg

Option 2)

87.5Kg

Option 3)

65.4Kg

Option 4)

30.1Kg

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