Question

Two wires A and B are of same materials. Their lengths are in the ratio 1:2 and diameters are in the ratio 2:1. When stretched by forces \(F_A\) and \(F_B\) respectively, they get equal increase in their lengths. Then the ratio \(F_A : F_B\) should be

• A. 1:2
• B. 1:1
• C. 2:1
• D. 8:1

Hint:

Force required is proportional to area and inversely proportional to length.

Answer 1

The Correct Option is D

Let's solve the problem using the principles of physics related to the stretching of wires. The problem deals with two wires, A and B, that are made of the same material, and we need to find the ratio of forces applied to them.

We are given:

• The ratio of their lengths (\(L_A : L_B\)) is 1:2.
• The ratio of their diameters (\(d_A : d_B\)) is 2:1.
• When stretched by forces \(F_A\) and \(F_B\), they experience equal increase in lengths, implying equal strain.

According to Young's modulus formula:

\(Y = \frac{F \cdot L}{A \cdot \Delta L}\)

where \(F\) is the force applied, \(L\) is the original length, \(A\) is the cross-sectional area, and \(\Delta L\) is the change in length.

Since both wires are made of the same material, they have the same Young's modulus, \(Y\), and for equal extension, \(\Delta L\), we can equate the expression for both wires:

\(\frac{F_A \cdot L_A}{A_A} = \frac{F_B \cdot L_B}{A_B}\)

The cross-sectional area \(A\) of a wire is given by \(\pi \left(\frac{d}{2}\right)^2\), thus:

\(A_A = \pi \left(\frac{d_A}{2}\right)^2\) and \(A_B = \pi \left(\frac{d_B}{2}\right)^2\)

Given \(d_A : d_B = 2:1\), the area ratio becomes:

\(\frac{A_A}{A_B} = \left(\frac{2}{1}\right)^2 = 4\)

Substituting these ratios:

\(\frac{F_A \cdot 1}{4} = \frac{F_B \cdot 2}{1}\)

Simplifying gives:

\(F_A \cdot 1 = F_B \cdot 8\)

Therefore, the ratio \(F_A : F_B\) is \(8:1\).

Hence, the correct answer is 8:1.