To find the tension in the cotton, we need to consider the forces acting on the beeswax.
Given:
Density of beeswax (ρ_beeswax) = 0.95 g/cm^3 = 950 kg/m^3 (converting to kg/m^3)
Mass of beeswax (m_beeswax) = 190 g = 0.19 kg
Length of cotton (L) = 5 cm = 0.05 m
Density of brine (ρ_brine) = 1.05 g/cm^3 = 1050 kg/m^3 (converting to kg/m^3)
Let's analyze the forces acting on the beeswax when it is completely immersed in the brine.
The upward buoyant force on the beeswax is equal to the weight of the displaced brine. The weight of the beeswax is acting downwards.
Weight of the beeswax (F_weight_beeswax) = mass of beeswax * acceleration due to gravity
F_weight_beeswax = 0.19 kg * 9.8 m/s^2 = 1.862 N
The volume of the beeswax (V_beeswax) can be calculated using the formula:
V_beeswax = m_beeswax / ρ_beeswax
V_beeswax = 0.19 kg / 950 kg/m^3 = 0.0002 m^3
The volume of the displaced brine is equal to the volume of the beeswax.
The buoyant force (F_buoyant) on the beeswax is given by:
F_buoyant = ρ_brine * g * V_beeswax
F_buoyant = 1050 kg/m^3 * 9.8 m/s^2 * 0.0002 m^3 = 2.058 N
Since the cotton is anchored to the lead weight, the tension in the cotton (T_cotton) is the difference between the weight of the beeswax and the buoyant force acting on it:
T_cotton = F_weight_beeswax - F_buoyant
T_cotton = 1.862 N - 2.058 N
T_cotton ≈ -0.196 N
The negative value indicates that the tension in the cotton is acting upward (opposite to the weight of the beeswax). However, it is important to note that a negative tension value does not have a physical interpretation in this context. It implies that the cotton is not under tension and may be slack in this scenario.
Therefore, the tension in the cotton is approximately 0 N (or negligible tension).