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A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter 2 mm breaks under a tension of
- a)500 N
- b)1000 N
- c)10000 N
- d)4000 N
Correct answer is option 'D'. Can you explain this answer?
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A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire...
Given:
Diameter of first wire = 1 mm
Tension at which first wire breaks = 1000 N
Diameter of second wire (made of same material as first wire) = 2 mm
To find: Tension at which second wire breaks
Formula used:
The stress (σ) on a wire is given by σ = F/A, where F is the force applied on the wire and A is the cross-sectional area of the wire.
The strain (ε) is the ratio of the change in length (∆L) to the original length (L) of the wire. So, ε = ∆L/L.
The Young's modulus (Y) is the ratio of the stress to the strain, i.e., Y = σ/ε.
Calculation:
Let's assume that the first wire has a cross-sectional area of A1 and the second wire has a cross-sectional area of A2.
We know that the two wires are made of the same material. Therefore, the Young's modulus (Y) of the two wires will be the same.
For the first wire:
Diameter = 1 mm
Radius (r1) = 0.5 mm = 0.0005 m
Area (A1) = πr1^2 = 0.000785 m^2
Tension (F1) = 1000 N
For the second wire:
Diameter = 2 mm
Radius (r2) = 1 mm = 0.001 m
Area (A2) = πr2^2 = 0.00314 m^2
Tension (F2) = ?
Using the formula Y = σ/ε, we can write:
Y = F1/(ε1L) = F2/(ε2L)
where ε1 and ε2 are the strains produced in the wires, and L is the original length of both wires.
Since both wires have the same material, the value of Y will be the same for both wires. Therefore:
F1/(ε1L) = F2/(ε2L)
We can rearrange this equation to get:
F2 = F1 (A2/A1) (ε2/ε1)
We know that the first wire broke under a tension of 1000 N. Therefore, the stress (σ1) produced in the first wire just before it broke is given by:
σ1 = F1/A1 = 1000/0.000785 = 1,273,885 Pa
We also know that the strain produced in the first wire just before it broke is given by:
ε1 = σ1/Y
We don't know the value of Y, but we can use the fact that the two wires are made of the same material. The value of Y for the material can be looked up in a table or found experimentally. Let's assume that Y = 2 x 10^11 Pa.
Then:
ε1 = σ1/Y = 1,273,885/2 x 10^11 = 6.37 x 10^-6
Now, we can use the formula for F2 derived earlier to find the tension at which the second wire will break:
F2 = F1 (A2/A1) (ε2/ε1)
We know that A2/A1
Diameter of first wire = 1 mm
Tension at which first wire breaks = 1000 N
Diameter of second wire (made of same material as first wire) = 2 mm
To find: Tension at which second wire breaks
Formula used:
The stress (σ) on a wire is given by σ = F/A, where F is the force applied on the wire and A is the cross-sectional area of the wire.
The strain (ε) is the ratio of the change in length (∆L) to the original length (L) of the wire. So, ε = ∆L/L.
The Young's modulus (Y) is the ratio of the stress to the strain, i.e., Y = σ/ε.
Calculation:
Let's assume that the first wire has a cross-sectional area of A1 and the second wire has a cross-sectional area of A2.
We know that the two wires are made of the same material. Therefore, the Young's modulus (Y) of the two wires will be the same.
For the first wire:
Diameter = 1 mm
Radius (r1) = 0.5 mm = 0.0005 m
Area (A1) = πr1^2 = 0.000785 m^2
Tension (F1) = 1000 N
For the second wire:
Diameter = 2 mm
Radius (r2) = 1 mm = 0.001 m
Area (A2) = πr2^2 = 0.00314 m^2
Tension (F2) = ?
Using the formula Y = σ/ε, we can write:
Y = F1/(ε1L) = F2/(ε2L)
where ε1 and ε2 are the strains produced in the wires, and L is the original length of both wires.
Since both wires have the same material, the value of Y will be the same for both wires. Therefore:
F1/(ε1L) = F2/(ε2L)
We can rearrange this equation to get:
F2 = F1 (A2/A1) (ε2/ε1)
We know that the first wire broke under a tension of 1000 N. Therefore, the stress (σ1) produced in the first wire just before it broke is given by:
σ1 = F1/A1 = 1000/0.000785 = 1,273,885 Pa
We also know that the strain produced in the first wire just before it broke is given by:
ε1 = σ1/Y
We don't know the value of Y, but we can use the fact that the two wires are made of the same material. The value of Y for the material can be looked up in a table or found experimentally. Let's assume that Y = 2 x 10^11 Pa.
Then:
ε1 = σ1/Y = 1,273,885/2 x 10^11 = 6.37 x 10^-6
Now, we can use the formula for F2 derived earlier to find the tension at which the second wire will break:
F2 = F1 (A2/A1) (ε2/ε1)
We know that A2/A1
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A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter 2 mm breaks under a tension ofa)500 Nb)1000 Nc)10000 Nd)4000 NCorrect answer is option 'D'. Can you explain this answer?
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