Question

A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere.

Answer 1

When the load will not wet, then in such an case sphere is fully submerged. This means that the weight of the sphere is exactly equal to the weight of the fluid displaced.

∴ Weight of the sphere with load = weight of the fluid displaced.

Now, weight of the fluid displaced = Upthrust

= Volume of the sphere × density of water × g

= 4/3 π × 5³ × 1 × g.

Now, weight of the sphere with load = 100 g +  4/3 π × 5³ × x × g

where x is the density of the material of the sphere.

∴  100 g +  4/3 π × 5³ × x × g = 4/3 π × 5³ × 1 × g.

∴ 100  +  4/3 π × 5³ × x  = 4/3 π × 5³ × 1

∴ 100 + 4.19 × 125x = 4.19 × 125

523.81x = 523.81 - 100

523.81x = 423.81

∴ x = 423.81/523.81

∴ x = 0.81 g/cm³.

Now, specific gravity = density of the material/density of water at 4° C.

∴ Specific gravity = 0.81/1

∴ Specific Gravity = 0.81

Hope it helps.

Answer 2

$<i>$Considering the equilibrium state after the load is put in the water.

We can say that

Mg (M is the mass of the sphere) + mg (m is the mass of the load)= F(buoyancy)

Mg (M is the mass of the sphere) + mg (m is the mass of the load)= Vpg (V is the part of the system under water=Volume of Sphere)

M+m= Vp

M/V(density of sphere) = (p-m/V)

Specific gravity = M/pV= 1-m/pV

m= 0.1

$\mathsf\:V=4/3\pi(5*10-2)^3$

$\mathsf\:p=10^3$(density of water)

So value of specific gravity can be found