A rock is suspended by a light string. When the lock is in air, the tension in the string is 39.2 N. When the rock is totally immersed in water, the tension is 28.4 N. When the rock is totally immersed in an unknown liquid, the tension is 18.6 N. What is the density of the unknown liquid?

Solution 33E Step 1 of 7: In the given problem,initially a rock is suspended using a string in air with tension T=39.2 as shown in the figure below Step 2 of 7: The free body diagram for the rock in the above case is as shown below, When rock is suspended in air using string, the tension force will be acting upward with equal magnitude as that of weight of the rock, which acts downward due to gravity. That is T= W= 39.2 N………………...1 Step 3 of 7: Now let us consider the second case , when the rock is immersed inside the liquid as shown in the figure below By archimedes principle, an extra force buoyancy force acts upward on the rock as shown in the free body diagram below, Step 4 of 7: From the free body diagram , when the rock is immersed in liquid buoyancy force is given by B= W-T………………..2 Similarly when the water is immersed in water, the tension in the string ws T = 28.4 N and the weight of rock from the equation 1 , W=39.2 N and density of water = 1000 kg/ m , we need to calculate the volume of liquid(water or unknown liquid) w The buoyancy force in water by equation 2, B w W T w Substituting W=39.2 N and T =w8.4 N B = 39.2 N 28.4 N w B w10.8 N