Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area \(\mathrm{A}\) and wire 2 has cross-sectional Area \(3 \mathrm{~A}\). If the length of wire 1 increases by on applying force \(\mathrm{F}\). How much force is needed to stretch wire 2 by same amount. (A) \(\mathrm{F}\) (B) \(4 \mathrm{~F}\) (C) \(6 \mathrm{~F}\) (D) \(9 \mathrm{~F}\)

Since the wires have the same volume, the product of the cross-sectional area and the length for wire 1 must equal the product of the cross-sectional area and the length for wire 2, i.e.: \( A \cdot L_1 = 3A \cdot L_2 \) where \(A\) is the cross-sectional area of wire 1, and \(L_1\) and \(L_2\) are the lengths of the wires 1 and 2, respectively. Now, we express the length of wire 2 in terms of wire 1: \( L_2 = \frac{L_1}{3} \)

According to Hooke's Law, the force acting on an object is proportional to its elongation, and it can be expressed as: \( F = Y \cdot \frac{ΔL}{L} \cdot A \) Where \(F\) is the force applied, \(Y\) is the Young's modulus (a constant specific to the material), \(ΔL\) is the change in length, \(L\) is the original length, and \(A\) is the cross-sectional area of the wire. Since both wires are made of the same material, their Young's modulus is the same. Therefore, we have the following setup for forces acting on wire 1 (denoted as \(F_1\)) and wire 2 (denoted as \(F_2\)): \( F_1 = Y \cdot \frac{ΔL_1}{L_1} \cdot A \) \( F_2 = Y \cdot \frac{ΔL_2}{L_2} \cdot 3A \) The problem states that the length increase of both wires is the same: \( ΔL_1 = ΔL_2 \)

Now, we need to find an expression for \(F_2\), the force on wire 2, in terms of \(F_1\). From the expressions for the forces above: \( F_1 = Y \cdot \frac{ΔL_1}{L_1} \cdot A \) \( F_2 = Y \cdot \frac{ΔL_2}{\frac{L_1}{3}} \cdot 3A \) Since \(ΔL_1 = ΔL_2\), we can rewrite the equation for \(F_2\) as: \( F_2 = Y \cdot \frac{ΔL_1}{\frac{L_1}{3}} \cdot 3A \) Now, divide the expression for \(F_2\) by the expression for \(F_1\): \( \frac{F_2}{F_1} = \frac{Y \cdot \frac{ΔL_1}{\frac{L_1}{3}} \cdot 3A }{Y \cdot \frac{ΔL_1}{L_1} \cdot A } \)

Now, we can simplify the expression and solve for \(F_2\): \( \frac{F_2}{F_1} = \frac{3 \cdot \frac{ΔL_1}{L_1} \cdot 3A}{\frac{ΔL_1}{L_1} \cdot A} \) Cancel out terms, and we get: \( \frac{F_2}{F_1} = 3 \cdot 3 = 9 \) Thus: \(F_2 = 9F_1\) The force needed to stretch wire 2 by the same amount as wire 1 is 9 times the force on wire 1. Therefore, the correct answer is: (D) \(9 \mathrm{~F}\)