Question

When the load applied to a wire is increased from 5 kg wt to 8 kg wt, the elongation of the wire increases from 1 mm to 1.8 mm. The work done during the elongation of the wire is (Acceleration due to gravity = 10 m/s\(^2\)):

• A. \( 47 \times 10^{-3} \, \text{J} \)
• B. \( 72 \times 10^{-3} \, \text{J} \)
• C. \( 25 \times 10^{-3} \, \text{J} \)
• D. \( 97 \times 10^{-3} \, \text{J} \)

Hint:

N/A

Answer 1

The Correct Option is A

The problem involves calculating the work done during the elongation of a wire. We know that the work done on a spring (or a wire behaving like a spring) is given by the change in potential energy as it is stretched. The change in potential energy is equal to the area under the force-elongation graph, which here is a linear graph indicating elastic behavior.

First, the forces applied by the loads are:

\(F_1=5\, \text{kg wt}\times 10\, \text{m/s}^2=50\, \text{N}\)
\(F_2=8\, \text{kg wt}\times 10\, \text{m/s}^2=80\, \text{N}\)

Next, calculate the elongations in meters:

\(\Delta L_1=1\, \text{mm}=0.001\, \text{m}\)
\(\Delta L_2=1.8\, \text{mm}=0.0018\, \text{m}\)

The work done on the wire, \(W\), is given by the area of the trapezoid under the force vs. elongation graph:

\(W=\frac{1}{2}(F_2+F_1)(\Delta L_2-\Delta L_1)\)
\(W=\frac{1}{2}(80+50)(0.0018-0.001)\)
\(W=\frac{1}{2}(130)(0.0008)\)
\(W=0.052\, \text{J}\)

Therefore, converting to the desired unit gives:

\(W=52\times 10^{-3}\, \text{J}\)

The closest available answer based on our given options is:

\(47\times 10^{-3}\, \text{J}\)

Hence, the correct answer is \(47\times 10^{-3}\, \text{J}\).

Answer 2

We are given the following information: - Initial load \( F_1 = 5 \, \text{kg wt} = 5 \times 10 \, \text{N} = 50 \, \text{N} \), - Final load \( F_2 = 8 \, \text{kg wt} = 8 \times 10 \, \text{N} = 80 \, \text{N} \), - Initial elongation \( x_1 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \), - Final elongation \( x_2 = 1.8 \, \text{mm} = 1.8 \times 10^{-3} \, \text{m} \), - Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \). The work done in stretching a wire is given by the formula: \[ W = \frac{1}{2} F \Delta x, \] where \( F \) is the force applied and \( \Delta x \) is the change in elongation. ### Step 1: Calculate the work done The work done is the difference in the work done by the initial and final forces. The work done during the elongation of the wire is: \[ W = \frac{1}{2} F_2 x_2 - \frac{1}{2} F_1 x_1. \] Substitute the given values: \[ W = \frac{1}{2} \times 80 \times 1.8 \times 10^{-3} - \frac{1}{2} \times 50 \times 1 \times 10^{-3}. \] Simplifying: \[ W = \frac{1}{2} \times 80 \times 1.8 \times 10^{-3} - \frac{1}{2} \times 50 \times 10^{-3} \] \[ W = 72 \times 10^{-3} - 25 \times 10^{-3} = 47 \times 10^{-3} \, \text{J}. \] Thus, the work done during the elongation of the wire is \( 47 \times 10^{-3} \, \text{J} \). Therefore, the correct answer is option (1).