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A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific gravity of the liquid will be
- a)0.65
- b)0.66
- c)0.67
- d)0.68
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A beaker is filled with a liquid up to the mark of one litre and weigh...
Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).
Thus, S = 0.66.
Thus, S = 0.66.
Most Upvoted Answer
A beaker is filled with a liquid up to the mark of one litre and weigh...
To find the specific gravity of the liquid, we need to understand the concept of specific gravity first.
Specific gravity is the ratio of the density of a substance to the density of a reference substance. In this case, the reference substance is usually water at a specified temperature. It is a dimensionless quantity and does not have any units.
The formula to calculate specific gravity is:
Specific Gravity = Density of the Substance / Density of the Reference Substance
Given information:
Weight of the liquid = 6.5 N
Volume of the liquid = 1 litre
Density is defined as mass per unit volume. Therefore, we can find the density of the liquid using the given information.
Density = Weight / Volume
Substituting the values, we get:
Density = 6.5 N / 1 L
Now, we need to convert the volume from liters to cubic meters because the SI unit of density is kg/m^3.
1 L = 0.001 m^3
Density = 6.5 N / 0.001 m^3
Density = 6500 kg/m^3
Now, we can calculate the specific gravity of the liquid using the formula mentioned above.
Specific Gravity = Density of the Substance / Density of the Reference Substance
Since the reference substance is water, we need to know the density of water at a specified temperature. At room temperature (approximately 25°C), the density of water is 1000 kg/m^3.
Specific Gravity = 6500 kg/m^3 / 1000 kg/m^3
Specific Gravity = 6.5
However, the options provided are in decimal form, so we need to convert the specific gravity to decimal form.
Specific Gravity = 0.65 (after dividing by 10)
Therefore, the specific gravity of the liquid is 0.65, which corresponds to option B.
Specific gravity is the ratio of the density of a substance to the density of a reference substance. In this case, the reference substance is usually water at a specified temperature. It is a dimensionless quantity and does not have any units.
The formula to calculate specific gravity is:
Specific Gravity = Density of the Substance / Density of the Reference Substance
Given information:
Weight of the liquid = 6.5 N
Volume of the liquid = 1 litre
Density is defined as mass per unit volume. Therefore, we can find the density of the liquid using the given information.
Density = Weight / Volume
Substituting the values, we get:
Density = 6.5 N / 1 L
Now, we need to convert the volume from liters to cubic meters because the SI unit of density is kg/m^3.
1 L = 0.001 m^3
Density = 6.5 N / 0.001 m^3
Density = 6500 kg/m^3
Now, we can calculate the specific gravity of the liquid using the formula mentioned above.
Specific Gravity = Density of the Substance / Density of the Reference Substance
Since the reference substance is water, we need to know the density of water at a specified temperature. At room temperature (approximately 25°C), the density of water is 1000 kg/m^3.
Specific Gravity = 6500 kg/m^3 / 1000 kg/m^3
Specific Gravity = 6.5
However, the options provided are in decimal form, so we need to convert the specific gravity to decimal form.
Specific Gravity = 0.65 (after dividing by 10)
Therefore, the specific gravity of the liquid is 0.65, which corresponds to option B.
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A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific gravity of the liquid will bea)0.65b)0.66c)0.67d)0.68Correct answer is option 'B'. Can you explain this answer?
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