A wire of cross-section \(4 \mathrm{~mm}^{2}\) is stretched by $0.1 \mathrm{~mm}$ by a certain weight. How far (length) will be wire of same material and length but of area \(8 \mathrm{~mm}^{2}\) stretched under the action of same force. (A) \(0.05 \mathrm{~mm}\) (B) \(0.10 \mathrm{~mm}\) (C) \(0.15 \mathrm{~mm}\) (D) \(0.20 \mathrm{~mm}\)

The given information is as follows: - Initial wire cross-section area: \( A_1 = 4 \,mm^2 \) - Elongation of the initial wire: \( \Delta l_1 = 0.1 \,mm \) - Second wire cross-section area: \( A_2 = 8 \,mm^2 \) - We need to find the elongation of the second wire: \( \Delta l_2 \) The formula for Young's modulus (\( Y \)) is as follows: \( Y = \frac{stress}{strain} \) Stress is defined as the force acting on the object divided by its area, and strain is the change in length divided by the original length: \( Y = \frac{F/A}{\Delta l/L} \) Now we can rearrange this formula to make the change in length, \(\Delta l\), the focus: \( \Delta l = \frac{FL}{YA} \) We know that the wires are made of the same material and are under the same force, so they have the same Young's modulus and the same value for the force.

Let's express the original elongation of the wire with area \(A_1\): \( \Delta l_1 = \frac{F_1L_1}{YA_1} \) Now let's express the elongation of the second wire with area \(A_2\): \( \Delta l_2 = \frac{F_2L_2}{YA_2} \) Since both wires are under the same force ( \(F_1 = F_2\) ) and assuming they have the same length ( \(L_1 = L_2\) ), we can set up a proportion between these two equations and solve for \( \Delta l_2 \): \( \frac{\Delta l_1}{\Delta l_2} = \frac{F L_1}{Y A_1} \cdot \frac{YA_2}{F L_2} \)

After rearranging the equation from Step 2, we only need to plug in the given values to find \( \Delta l_2 \): \( \Delta l_2 = \Delta l_1 \cdot \frac{A_1}{A_2} \) Now substitute the given values: \( \Delta l_2 = 0.1 \,mm \cdot \frac{4 \,mm^2}{8 \,mm^2} \) Calculate the result: \( \Delta l_2 = 0.1 \,mm \cdot \frac{1}{2} = 0.05 \,mm \) The elongation of the second wire under the same force is \( 0.05 \,mm \). So the correct answer is (A).