(i)A block of steel of size 5*5*5 cm ^3 is weighed in water.If relative density of steel if 7,its apparent weight is ?
(ii)When an object of mass 200 kg placed on an ice block floating on water it just sinks.The mass of ice berg is (Relative density of ice = 0.9,relative density of water is 1.02)
please explain the 2 questions with proper explanation and steps.Waiting for answers.
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1. relative density = 7
It means for the same volume, steel weighs 7 times as compared to water. The force of buoyancy will be equal to the weight of water displaced.
weight of steel block in water = weight in air - buoyancy force upwards
7 gm/cm³ * 125 cm³ * g - 1 gm/cm³ * 125 cm³ * g = 0.750 * g Newtons
g = acceleration due to gravity = 9.81 m/sec²
if you want to write weight in kg, then apparent weight = 0.750 kg
2. let mass of ice berg = m. When the object is placed on the ice berg, the total weight is just more than the buoyancy force.
total weight of ice + object = volume of water displaced * density * g
(m + 200 ) g = (m/0.9) * 1.02 * g
m+ 200 = m * 1.02/0.9 = 1.133 * m
m = 200 / 0.133 = 1,500 kg
It means for the same volume, steel weighs 7 times as compared to water. The force of buoyancy will be equal to the weight of water displaced.
weight of steel block in water = weight in air - buoyancy force upwards
7 gm/cm³ * 125 cm³ * g - 1 gm/cm³ * 125 cm³ * g = 0.750 * g Newtons
g = acceleration due to gravity = 9.81 m/sec²
if you want to write weight in kg, then apparent weight = 0.750 kg
2. let mass of ice berg = m. When the object is placed on the ice berg, the total weight is just more than the buoyancy force.
total weight of ice + object = volume of water displaced * density * g
(m + 200 ) g = (m/0.9) * 1.02 * g
m+ 200 = m * 1.02/0.9 = 1.133 * m
m = 200 / 0.133 = 1,500 kg
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