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.

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

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

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?

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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