KCET · Physics · Mechanical Properties of Solids
There are two wires of same material and same length while the diameter of second wire is two times the diameter of the first wire. Then the ratio of extensions produced in the wires by applying same load will be
- A \(1:1\)
- B \(1:2\)
- C \(2:1\)
- D \(4:1\)
Answer & Solution
Correct Answer
(D) \(4:1\)
Step-by-step Solution
Detailed explanation
The extension produced in a wire is given by the formula:
\(\Delta L = \dfrac{FL}{AY}\)
Since the material and length are the same, Young's modulus \(Y\) and length \(L\) are constant. The same load \(F\) is applied to both wires. Therefore, the extension is inversely proportional to the cross-sectional area \(A\).
\(\Delta L \propto \dfrac{1}{A}\)
The cross-sectional area \(A\) is proportional to the square of the diameter \(d\) (\(A = \dfrac{\pi d^2}{4}\)).
\(\Delta L \propto \dfrac{1}{d^2}\)
Given that the diameter of the second wire is twice the diameter of the first wire (\(d_2 = 2d_1\)), the ratio of their extensions is:
\(\dfrac{\Delta L_1}{\Delta L_2} = \left(\dfrac{d_2}{d_1}\right)^2\)
\(\dfrac{\Delta L_1}{\Delta L_2} = \left(\dfrac{2d_1}{d_1}\right)^2 = 4\)
The ratio of extensions produced in the wires is \(4:1\).
Answer: \(4:1\)
\(\Delta L = \dfrac{FL}{AY}\)
Since the material and length are the same, Young's modulus \(Y\) and length \(L\) are constant. The same load \(F\) is applied to both wires. Therefore, the extension is inversely proportional to the cross-sectional area \(A\).
\(\Delta L \propto \dfrac{1}{A}\)
The cross-sectional area \(A\) is proportional to the square of the diameter \(d\) (\(A = \dfrac{\pi d^2}{4}\)).
\(\Delta L \propto \dfrac{1}{d^2}\)
Given that the diameter of the second wire is twice the diameter of the first wire (\(d_2 = 2d_1\)), the ratio of their extensions is:
\(\dfrac{\Delta L_1}{\Delta L_2} = \left(\dfrac{d_2}{d_1}\right)^2\)
\(\dfrac{\Delta L_1}{\Delta L_2} = \left(\dfrac{2d_1}{d_1}\right)^2 = 4\)
The ratio of extensions produced in the wires is \(4:1\).
Answer: \(4:1\)
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