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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 = 103 kg/m³)
A 46. 3 kg
B 87. 5 kg
C 30. 1 kg
D 65. 4 kg

Answer :

The maximum weight that can be put on the is 87.5 kg

How is it so?

Density:

The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.

Given volume and density so by using the above formula, mass will be calculated. i.e. Mass = Volume × Density

Calculation:

Given, Side of the cubical block = 0.5 m

Density of water = 103 kg/m³

We know that, Density = Mass/Volume

Mass = Volume × Density

When a body floats then the weight of the body = up thrust

As a result, [tex](50) ^ 3 * 30/100 * (1) * g =M_cube g ----(1)[/tex]

Let m mass should be placed, then

[tex]50 ^ 3 \times (1) \times g = (M_cube +m)g ----(2)[/tex]

Subtracting equation (i) from equation (ii),

[tex]((50)^ 3 *(1)* g)) - ((50) ^ 3 * 30/100 * (1) * g) = M_cube g + m)g - M_cube g[/tex]

[tex](50) ^ 3 * g - ((50) ^ 3 * 0.3g) = M_cube g + mg - M_cube g[/tex]

[tex]\Rightarrow mg = (50) ^ 3 * g(1 - 0.3)[/tex]

[tex]Mg = 125 * 10 ^ 3 * 0.7g[/tex]

[tex]\Rightarrow m = 87.5kg[/tex]

Therefore, the maximum weight that can be put on the is 87.5 kg

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Rewritten by : Barada