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A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23?
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A copper ring has a diameter of exactly 25mm at its temperature is zer...
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Option (c)
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A copper ring has a diameter of exactly 25mm at its temperature is zer...
Given Information:
- Diameter of copper ring at 0°C: 25mm
- Diameter of aluminium sphere at 100°C: 25.05mm
- Coefficient of linear expansion (α) for copper: 17 × 10^-6
- Coefficient of linear expansion (α) for aluminium: 2.3 × 10^-5
- Specific heat of copper (Cu): 0.0923 cal/g°C
- Specific heat of aluminium (Al): 0.215 cal/g°C
Approach:
1. Calculate the change in diameter of the copper ring and the aluminium sphere due to temperature change.
2. Determine the change in volume of the copper ring and the aluminium sphere.
3. Find the ratio of the mass of the aluminium sphere to the mass of the copper ring.
Calculation:
1. Change in Diameter:
Change in diameter of the copper ring (ΔDcopper) = 0 - 25mm = -25mm
Change in diameter of the aluminium sphere (ΔDaluminium) = 25.05mm - 25mm = 0.05mm
2. Change in Volume:
Change in volume of the copper ring (ΔVcopper) = π * (Dcopper + ΔDcopper)^2 * h - π * Dcopper^2 * h
= π * (25 + (-25))^2 * h - π * 25^2 * h
= π * 0^2 * h - π * 25^2 * h
= 0 - π * 625 * h
= -625πh
Change in volume of the aluminium sphere (ΔValuminium) = 4/3 * π * ((Daluminium/2)^3 - (Dcopper/2)^3)
= 4/3 * π * ((25.05/2)^3 - (25/2)^3)
= 4/3 * π * (12.525^3 - 12.5^3)
= 4/3 * π * (19597.265625 - 19531.25)
= 4/3 * π * 66.015625
= 88.021
3. Ratio of Mass:
Ratio of mass (m) = (ΔValuminium * density of aluminium) / (ΔVcopper * density of copper)
= (88.021 * density of aluminium) / (-625πh * density of copper)
Density of copper (ρcopper) = mass of copper / volume of copper
= (m * ΔVcopper * density of copper) / ΔVcopper
= m * density of copper
Similarly, density of aluminium (ρaluminium) = m * density of aluminium
Ratio of mass (m) = (88.021 * density of aluminium) / (-625πh * density of copper)
= (88.021 * ρaluminium) / (-625πh * ρcopper)
= (88.021 * ρaluminium) / (625πh * ρcopper) ... (1)
Volume Expansion:
The volume expansion
- Diameter of copper ring at 0°C: 25mm
- Diameter of aluminium sphere at 100°C: 25.05mm
- Coefficient of linear expansion (α) for copper: 17 × 10^-6
- Coefficient of linear expansion (α) for aluminium: 2.3 × 10^-5
- Specific heat of copper (Cu): 0.0923 cal/g°C
- Specific heat of aluminium (Al): 0.215 cal/g°C
Approach:
1. Calculate the change in diameter of the copper ring and the aluminium sphere due to temperature change.
2. Determine the change in volume of the copper ring and the aluminium sphere.
3. Find the ratio of the mass of the aluminium sphere to the mass of the copper ring.
Calculation:
1. Change in Diameter:
Change in diameter of the copper ring (ΔDcopper) = 0 - 25mm = -25mm
Change in diameter of the aluminium sphere (ΔDaluminium) = 25.05mm - 25mm = 0.05mm
2. Change in Volume:
Change in volume of the copper ring (ΔVcopper) = π * (Dcopper + ΔDcopper)^2 * h - π * Dcopper^2 * h
= π * (25 + (-25))^2 * h - π * 25^2 * h
= π * 0^2 * h - π * 25^2 * h
= 0 - π * 625 * h
= -625πh
Change in volume of the aluminium sphere (ΔValuminium) = 4/3 * π * ((Daluminium/2)^3 - (Dcopper/2)^3)
= 4/3 * π * ((25.05/2)^3 - (25/2)^3)
= 4/3 * π * (12.525^3 - 12.5^3)
= 4/3 * π * (19597.265625 - 19531.25)
= 4/3 * π * 66.015625
= 88.021
3. Ratio of Mass:
Ratio of mass (m) = (ΔValuminium * density of aluminium) / (ΔVcopper * density of copper)
= (88.021 * density of aluminium) / (-625πh * density of copper)
Density of copper (ρcopper) = mass of copper / volume of copper
= (m * ΔVcopper * density of copper) / ΔVcopper
= m * density of copper
Similarly, density of aluminium (ρaluminium) = m * density of aluminium
Ratio of mass (m) = (88.021 * density of aluminium) / (-625πh * density of copper)
= (88.021 * ρaluminium) / (-625πh * ρcopper)
= (88.021 * ρaluminium) / (625πh * ρcopper) ... (1)
Volume Expansion:
The volume expansion
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A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23?
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A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23?.
A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23?.
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A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23?, a detailed solution for A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? has been provided alongside types of A copper ring has a diameter of exactly 25mm at its temperature is zero degree Celsius. An aluminium sphere has a diameter 25.05mm and its temperature is 100 degree Celsius .The sphere is placed on top of the ring are the two are allowed to come to thermal equilibrium. The ratio of the mass of the sphere and the ring is Given alpha of copper =17×10^-6 alpha of aluminium =2.3×10^-5 Specific heat of cu=0.0923 cal/g C for al =.215cal/g c 1) 1/5 2)23/108 3)23/54 4)216/23? theory, EduRev gives you an
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