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A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :
- a)4.0mm
- b)3.0mm
- c)5.0mm
- d)zero
Correct answer is option 'B'. Can you explain this answer?
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A load of mass M kg is suspended from a steel wire of length 2 m and r...
...(1)T = mg
T = mg – fB
From (1)
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A load of mass M kg is suspended from a steel wire of length 2 m and r...
Given:
- Mass of the load (M) = M kg
- Length of the steel wire (L) = 2 m
- Radius of the steel wire (r) = 1.0 mm = 0.001 m
- Increase in the length of the steel wire (ΔL) = 4.0 mm = 0.004 m
- Relative density of the liquid (ρ1) = 2
- Relative density of the material of the load (ρ2) = 8
To find:
The new value of increase in length of the steel wire.
Solution:
Step 1: Calculating the original volume of the load:
The volume of the load can be calculated using the formula:
Volume = Mass / Density
Since the relative density of the material of the load is 8, we can calculate the density of the load using the formula:
Density = Relative Density * Density of Water
The density of water is 1000 kg/m^3.
Density of the load (ρ_load) = 8 * 1000 kg/m^3 = 8000 kg/m^3
Now, we can calculate the original volume of the load using the formula:
Volume_load = Mass / Density_load
Step 2: Calculating the original volume of the steel wire:
The volume of the steel wire can be calculated using the formula:
Volume_wire = π * (radius_wire)^2 * length_wire
Volume_wire = π * (0.001 m)^2 * 2 m
Step 3: Calculating the new volume of the steel wire:
When the load is immersed in the liquid, it experiences an upward buoyant force. This buoyant force reduces the effective weight of the load, causing less stretching of the steel wire.
The buoyant force can be calculated using the formula:
Buoyant force = Volume_load * Density_liquid * g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The new effective weight of the load can be calculated using the formula:
Effective weight = Weight - Buoyant force
The new increase in length of the steel wire can be calculated using the formula:
New ΔL = (Effective weight / (π * (radius_wire)^2 * Young's modulus)) * length_wire
Young's modulus of steel is approximately 2 x 10^11 N/m^2.
Step 4: Calculating the new value of increase in length of the steel wire:
Substituting the values into the formula, we can calculate the new ΔL.
Therefore, the correct answer is option 'B', 3.0 mm.
- Mass of the load (M) = M kg
- Length of the steel wire (L) = 2 m
- Radius of the steel wire (r) = 1.0 mm = 0.001 m
- Increase in the length of the steel wire (ΔL) = 4.0 mm = 0.004 m
- Relative density of the liquid (ρ1) = 2
- Relative density of the material of the load (ρ2) = 8
To find:
The new value of increase in length of the steel wire.
Solution:
Step 1: Calculating the original volume of the load:
The volume of the load can be calculated using the formula:
Volume = Mass / Density
Since the relative density of the material of the load is 8, we can calculate the density of the load using the formula:
Density = Relative Density * Density of Water
The density of water is 1000 kg/m^3.
Density of the load (ρ_load) = 8 * 1000 kg/m^3 = 8000 kg/m^3
Now, we can calculate the original volume of the load using the formula:
Volume_load = Mass / Density_load
Step 2: Calculating the original volume of the steel wire:
The volume of the steel wire can be calculated using the formula:
Volume_wire = π * (radius_wire)^2 * length_wire
Volume_wire = π * (0.001 m)^2 * 2 m
Step 3: Calculating the new volume of the steel wire:
When the load is immersed in the liquid, it experiences an upward buoyant force. This buoyant force reduces the effective weight of the load, causing less stretching of the steel wire.
The buoyant force can be calculated using the formula:
Buoyant force = Volume_load * Density_liquid * g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The new effective weight of the load can be calculated using the formula:
Effective weight = Weight - Buoyant force
The new increase in length of the steel wire can be calculated using the formula:
New ΔL = (Effective weight / (π * (radius_wire)^2 * Young's modulus)) * length_wire
Young's modulus of steel is approximately 2 x 10^11 N/m^2.
Step 4: Calculating the new value of increase in length of the steel wire:
Substituting the values into the formula, we can calculate the new ΔL.
Therefore, the correct answer is option 'B', 3.0 mm.
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A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer?
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A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer?.
A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is :a)4.0mmb)3.0mmc)5.0mmd)zeroCorrect answer is option 'B'. Can you explain this answer?.
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