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A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is immersed in water without touching the walls of beaker. What will be the balance reading now ?
- a)2 Kg
- b)2.5 Kg
- c)1 KG
- d)3 Kg
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A beaker containing water is placed on the platform of a spring balanc...
Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium,
B=0.5kg
The reading of the spring balance = Weight of water + Buoyant force' reaction pair force downwards
=1.5kg+0.5kg=2kg
Most Upvoted Answer
A beaker containing water is placed on the platform of a spring balanc...
Weight = initial weight + up thrust
and weighing machine measure the normal reaction not the actual mass/weight.
and weighing machine measure the normal reaction not the actual mass/weight.
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Community Answer
A beaker containing water is placed on the platform of a spring balanc...
Given:
- Mass of beaker with water = 1.5 kg
- Mass of stone = 0.5 kg
- Density of stone = 500 kg/m³
To find:
- Balance reading after the stone is immersed in water
Solution:
1. Weight of the beaker with water:
- The balance reading of 1.5 kg indicates the weight of the beaker with water.
2. Weight of the stone:
- Weight = Mass x Acceleration due to gravity
- Mass of the stone = 0.5 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Weight of the stone = 0.5 kg x 9.8 m/s² = 4.9 N
3. Weight of the water displaced by the stone:
- When the stone is immersed in water, it displaces some water. The weight of the displaced water can be calculated using Archimedes' principle.
- According to Archimedes' principle, the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
- The weight of the water displaced by the stone is equal to the weight of the stone itself.
4. Balance reading after immersing the stone:
- The balance reading will be the sum of the weight of the beaker with water and the weight of the displaced water (which is equal to the weight of the stone).
- Balance reading = Weight of beaker with water + Weight of stone
- Balance reading = 1.5 kg + 0.5 kg = 2 kg
Therefore, the balance reading after immersing the stone will be 2 kg. Hence, the correct answer is option B.
- Mass of beaker with water = 1.5 kg
- Mass of stone = 0.5 kg
- Density of stone = 500 kg/m³
To find:
- Balance reading after the stone is immersed in water
Solution:
1. Weight of the beaker with water:
- The balance reading of 1.5 kg indicates the weight of the beaker with water.
2. Weight of the stone:
- Weight = Mass x Acceleration due to gravity
- Mass of the stone = 0.5 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Weight of the stone = 0.5 kg x 9.8 m/s² = 4.9 N
3. Weight of the water displaced by the stone:
- When the stone is immersed in water, it displaces some water. The weight of the displaced water can be calculated using Archimedes' principle.
- According to Archimedes' principle, the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
- The weight of the water displaced by the stone is equal to the weight of the stone itself.
4. Balance reading after immersing the stone:
- The balance reading will be the sum of the weight of the beaker with water and the weight of the displaced water (which is equal to the weight of the stone).
- Balance reading = Weight of beaker with water + Weight of stone
- Balance reading = 1.5 kg + 0.5 kg = 2 kg
Therefore, the balance reading after immersing the stone will be 2 kg. Hence, the correct answer is option B.
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A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3is immersed in water without touching the walls of beaker. What will be the balance reading now ?a)2 Kgb)2.5 Kgc)1 KGd)3 KgCorrect answer is option 'B'. Can you explain this answer?
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