Abody weighs 10newton in air and 9.5 in water. how will it weigh in alchohol if density 0.8gcm-³

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

The volume of water displaced is equal to the volume of alcohol displaced

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

trig 15.3 cos