Consider two rods that are made of the same material i.e

Consider two rods that are made of the same material i.e. have the same young modulus in our of the same length. The radius of one rod is three times the radius of the second rod.

The correct answer and explanation is:

Sure! Let’s analyze the problem step-by-step.

Problem statement:
Two rods have the same length and are made of the same material (thus same Young’s modulus, EE). The radius of one rod is 3 times the radius of the second rod.

What is being asked?

Although you didn’t specify the exact quantity to find, the typical question in such problems is to compare their stiffness, stress, or elongation under the same applied force.

I will assume you want to compare the stiffness or elongation of the rods under the same tensile load, as that’s a common use of Young’s modulus.

Key concepts:

• Young’s modulus EE:
  E=StressStrain=F/AΔL/LE = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L}
• Stress σ=FA\sigma = \frac{F}{A}, where FF is force, AA is cross-sectional area.
• Strain ϵ=ΔLL\epsilon = \frac{\Delta L}{L}, relative change in length.

Step 1: Calculate cross-sectional areas

The rods are cylindrical, so their cross-sectional area is:
A=πr2A = \pi r^2

If the smaller rod has radius rr, then the larger rod has radius 3r3r.

• Smaller rod area:
  A1=πr2A_1 = \pi r^2
• Larger rod area:
  A2=π(3r)2=π9r2=9A1A_2 = \pi (3r)^2 = \pi 9r^2 = 9 A_1

Step 2: Understand elongation under the same load FF

The elongation ΔL\Delta L of a rod under tensile force FF is:
ΔL=FLAE\Delta L = \frac{F L}{A E}

Since LL and EE are the same for both rods:

• For smaller rod:
  ΔL1=FLA1E\Delta L_1 = \frac{F L}{A_1 E}
• For larger rod:
  ΔL2=FLA2E=FL9A1E=ΔL19\Delta L_2 = \frac{F L}{A_2 E} = \frac{F L}{9 A_1 E} = \frac{\Delta L_1}{9}

Step 3: Interpretation

• The larger rod (radius 3 times) stretches 9 times less than the smaller rod under the same force.
• This is because the larger rod has 9 times the cross-sectional area, making it stiffer and less prone to elongation.

Summary:

• Radius of larger rod = 3×3 \times radius of smaller rod
• Cross-sectional area of larger rod = 9×9 \times that of smaller rod
• Elongation under same force for larger rod = 19\frac{1}{9} elongation of smaller rod

Additional notes:

• If the force applied is the same, the larger rod deforms much less.
• If the rods are stretched by the same amount, the force needed for the larger rod would be 9 times greater.
• This is a direct result of how cross-sectional area influences the rod’s ability to resist deformation.