Physics Paper 1 - Deformation topic - multiple choice question solution?
A wire is stretched by 8 mm when a load of 60N is applied.
What will be the extension of a wire of the same material having four times the cross-sectional area and twice the original length when the same load is applied?
A. 2 mm
B. 4 mm
C. 8 mm
D. 16 mm
Kindly help me with any resources or material for this topic and to solve this question.
What will be the extension of a wire of the same material having four times the cross-sectional area and twice the original length when the same load is applied?
A. 2 mm
B. 4 mm
C. 8 mm
D. 16 mm
Kindly help me with any resources or material for this topic and to solve this question.
Reply 1
Original post by Qani_reads
A wire is stretched by 8 mm when a load of 60N is applied.
What will be the extension of a wire of the same material having four times the cross-sectional area and twice the original length when the same load is applied?
A. 2 mm
B. 4 mm
C. 8 mm
D. 16 mm
Kindly help me with any resources or material for this topic and to solve this question.
What will be the extension of a wire of the same material having four times the cross-sectional area and twice the original length when the same load is applied?
A. 2 mm
B. 4 mm
C. 8 mm
D. 16 mm
Kindly help me with any resources or material for this topic and to solve this question.
You should recall E = σ/ε
Where E is the Young’s modulus, σ is the stress and ε is the strain.
σ is essentially the force exerted on the string divided by its cross sectional area (F/A) and ε is the ratio of the extension to the length of the string (x/l)
So E = (F/A)/(x/l) = Fl/Ax
Try plugging in the variables for each string and see what expressions you get for E (let the unknown extension of the second string be x).
After getting the two different expressions, set them equal and solve for x.
Spoiler
Reply 2
9
Original post by TypicalNerd
You should recall E = σ/ε
Where E is the Young’s modulus, σ is the stress and ε is the strain.
σ is essentially the force exerted on the string divided by its cross sectional area (F/A) and ε is the ratio of the extension to the length of the string (x/l)
So E = (F/A)/(x/l) = Fl/Ax
Try plugging in the variables for each string and see what expressions you get for E (let the unknown extension of the second string be x).
After getting the two different expressions, set them equal and solve for x.
Where E is the Young’s modulus, σ is the stress and ε is the strain.
σ is essentially the force exerted on the string divided by its cross sectional area (F/A) and ε is the ratio of the extension to the length of the string (x/l)
So E = (F/A)/(x/l) = Fl/Ax
Try plugging in the variables for each string and see what expressions you get for E (let the unknown extension of the second string be x).
After getting the two different expressions, set them equal and solve for x.
Spoiler
Reply 3
Original post by Qani_reads
still unable to solve
Okay, let’s go for the easier option in the spoiler (where I’ve just said let’s say all variables needed to calculate E that we don’t know are 1).
String 1 has A = 1 m^2 (that’s a fat string now that I think about it), l = 1 m, x = 8 mm (0.008 m) and F = 60 N
Using these to calculate a value of E, we get
E = (60 N * 1 m)/(0.008 m * 1 m^2) = 7500 Pa
We have a second string of the same material (i.e E is still 7500 Pa) with an area four times as large (i.e 4 m^2), twice as long (i.e 2 m), an unknown extension (x mm, or x/1000 m) and we are told the load is the same (i.e 60 N)
Plugging these into E = Fl/Ax:
7500 Pa = (60 N * 2 m)/(4 m^2 * x/1000 m)
Simplifying the right-hand side:
7500 Pa = (120 Nm)/(0.004x m^3)
So 7500 Pa = 30000/x Pa
How would you go about solving the equation for x?
Reply 4
9
okay i got it. do you have any suggestions other than to practice more word problems? i find myself stuck at specifically word problems similar to these
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