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A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)
- a)0.85 cc
- b)0.46 cc
- c)0.153 cc
- d)0.05 cc
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A glass flask is filled up to a mark with 50 cc of mercury at 18°C. I...
Due to volume expansion of both mercury and flask, die change in volume of mercury relative to flask is given by
= 50 [180 x 10-6- 3 x 9 x 10-6] (38 -18)
=0.153 cc
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Community Answer
A glass flask is filled up to a mark with 50 cc of mercury at 18°C. I...
Given data:
- Initial volume of mercury = 50 cc
- Temperature at initial state (T1) = 18°C
- Final temperature (T2) = 38°C
- Coefficient of volume expansion for glass (α) = 9 × 10^−6/°C
- Coefficient of real expansion of mercury (β) = 180 × 10^−6/°C
Calculation:
1. Calculate the volume of mercury at the final temperature using the formula:
V2 = V1 * (1 + β * (T2 - T1))
V2 = 50 * (1 + 180 × 10^−6 * (38 - 18))
V2 = 50 * 1.009
V2 = 50.45 cc
2. Calculate the change in volume of mercury:
ΔV = V2 - V1
ΔV = 50.45 - 50
ΔV = 0.45 cc
3. Calculate the change in volume of glass using the formula:
ΔV_glass = V1 * α * (T2 - T1)
ΔV_glass = 50 * 9 × 10^−6 * (38 - 18)
ΔV_glass = 50 * 9 × 10^−6 * 20
ΔV_glass = 0.009 cc
4. Calculate the total change in volume:
Total change in volume = ΔV + ΔV_glass
Total change in volume = 0.45 + 0.009
Total change in volume = 0.459 cc
5. The amount of mercury that will be above the mark:
Amount above the mark = Total change in volume - ΔV_glass
Amount above the mark = 0.459 - 0.009
Amount above the mark = 0.45 cc
Therefore, the amount of mercury above the mark when the flask and contents are heated to 38°C is 0.45 cc, which is closest to option (c) 0.153 cc.
- Initial volume of mercury = 50 cc
- Temperature at initial state (T1) = 18°C
- Final temperature (T2) = 38°C
- Coefficient of volume expansion for glass (α) = 9 × 10^−6/°C
- Coefficient of real expansion of mercury (β) = 180 × 10^−6/°C
Calculation:
1. Calculate the volume of mercury at the final temperature using the formula:
V2 = V1 * (1 + β * (T2 - T1))
V2 = 50 * (1 + 180 × 10^−6 * (38 - 18))
V2 = 50 * 1.009
V2 = 50.45 cc
2. Calculate the change in volume of mercury:
ΔV = V2 - V1
ΔV = 50.45 - 50
ΔV = 0.45 cc
3. Calculate the change in volume of glass using the formula:
ΔV_glass = V1 * α * (T2 - T1)
ΔV_glass = 50 * 9 × 10^−6 * (38 - 18)
ΔV_glass = 50 * 9 × 10^−6 * 20
ΔV_glass = 0.009 cc
4. Calculate the total change in volume:
Total change in volume = ΔV + ΔV_glass
Total change in volume = 0.45 + 0.009
Total change in volume = 0.459 cc
5. The amount of mercury that will be above the mark:
Amount above the mark = Total change in volume - ΔV_glass
Amount above the mark = 0.459 - 0.009
Amount above the mark = 0.45 cc
Therefore, the amount of mercury above the mark when the flask and contents are heated to 38°C is 0.45 cc, which is closest to option (c) 0.153 cc.
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A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer?
Question Description
A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer?.
A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark? (a for glass is 9 × 10 –6/ºC and coefficient of real expansion of mercury is 180 × 10–6/ºC)a)0.85 ccb)0.46 ccc)0.153 ccd)0.05 ccCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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