Answer:
The value is [tex]a = 20.4 \%[/tex]
Explanation:
From the question we are told that
The acceleration due to gravity of the planet is [tex]a = 5.40 \ m/s^ 2[/tex]
The percentage volume of the apparatus submerged in the ocean on earth is [tex]k = 25.0 \%[/tex]
Generally the buoyant force acting on the apparatus is equal to the weight of the apparatus
i.e
[tex]\rho_s * V_s * g = m * g[/tex]
Hence the mass of the apparatus is mathematically represented as
[tex]m_s = \rho_s * V_s[/tex]
Here [tex]\rho_s[/tex] is the density of the sea water with a value of [tex]\rho_s = 1030 \ kg/m^3[/tex]
[tex]V_s[/tex] is the volume of the apparatus submerged
So
[tex]m_s = 1030 * 0.25 V[/tex]
Here V is the total volume of the apparatus
=> [tex]m_s = 257.5 V \ kg[/tex]
Generally at the new planet the mass of the apparatus is mathematically represented as
[tex]m_c = \rho_c * V_c[/tex]
Here [tex]\rho_c[/tex] is the density of the glycerin with a value of [tex]\rho_c = 1260 \ kg/m^3[/tex]
Gnerally [tex]m_s = m_c[/tex]
So
[tex]257.5 V = 1260 * V_c[/tex]
=> [tex]V_c = \frac{257.5V }{ 1260 }[/tex]
=> [tex]V_c = 0.204 \ V[/tex]
Hence the percentage volume of the apparatus which is submerged in glycerin at the new planet is
[tex]a = 0.204 * 100 = 20.4 \%[/tex]
