JEE Challenger
More from Mechanical Properties of Solids
PhyHardAdv 2018 • Paper 2

Vernier Scale Division Coincidence under Elastic Wire Elongation

A steel wire with diameter 0.5 mm0.5\text{ mm} and Young's modulus 2×1011 N m−22 \times 10^{11}\text{ N m}^{-2} carries a mass MM. The wire's length with the load is 1.0 m1.0\text{ m}. A vernier scale having 1010 divisions is attached at the wire's end. Next to this steel wire, a reference wire with a main scale of least count 1.0 mm1.0\text{ mm} is positioned. The 1010 divisions on the vernier scale match 99 divisions of the main scale. Initially, the zero mark of the vernier scale aligns with the zero mark of the main scale. When the load on the steel wire is increased by 1.2 kg1.2\text{ kg}, find the vernier scale division that coincides with a main scale division. (Take g=10 m s−2g = 10\text{ m s}^{-2} and π=3.2\pi = 3.2)

Official Numerical Answer3

Topics & Concepts

Mechanical Properties of SolidsElasticityMeasuring Instruments

Step-by-Step Solution

To find the vernier scale division that coincides with a main scale division, we first determine the elongation of the steel wire due to the additional load, and then use the properties of the Vernier scale to find the coincident division.

Step 1: Calculate the elongation of the steel wire (ΔL\Delta L)

The given parameters for the steel wire are:

  • Diameter of the wire, d=0.5 mm=0.5×10−3 md = 0.5\text{ mm} = 0.5 \times 10^{-3}\text{ m}
  • Radius of the wire, r=d2=0.25×10−3 mr = \frac{d}{2} = 0.25 \times 10^{-3}\text{ m}
  • Initial length of the wire with load, L=1.0 mL = 1.0\text{ m}
  • Young's modulus, Y=2×1011 N m−2Y = 2 \times 10^{11}\text{ N m}^{-2}
  • Increase in load, Δm=1.2 kg\Delta m = 1.2\text{ kg}
  • Acceleration due to gravity, g=10 m s−2g = 10\text{ m s}^{-2}
  • Value of π=3.2\pi = 3.2

The cross-sectional area AA of the wire is: A=πr2=3.2×(0.25×10−3 m)2=3.2×0.0625×10−6 m2=0.2×10−6 m2=2×10−7 m2A = \pi r^2 = 3.2 \times \left(0.25 \times 10^{-3}\text{ m}\right)^2 = 3.2 \times 0.0625 \times 10^{-6}\text{ m}^2 = 0.2 \times 10^{-6}\text{ m}^2 = 2 \times 10^{-7}\text{ m}^2

The additional force applied to the wire is: ΔF=Δm⋅g=1.2 kg×10 m s−2=12 N\Delta F = \Delta m \cdot g = 1.2\text{ kg} \times 10\text{ m s}^{-2} = 12\text{ N}

Using Hooke's Law, the elongation ΔL\Delta L produced in the wire is: ΔL=ΔF⋅LA⋅Y\Delta L = \frac{\Delta F \cdot L}{A \cdot Y}

Substitute the known values into the equation: ΔL=12×1.02×10−7×2×1011=124×104 m=3×10−4 m=0.3 mm\Delta L = \frac{12 \times 1.0}{2 \times 10^{-7} \times 2 \times 10^{11}} = \frac{12}{4 \times 10^4}\text{ m} = 3 \times 10^{-4}\text{ m} = 0.3\text{ mm}

Step 2: Determine the Least Count of the Vernier Scale

It is given that:

  • 1 Main Scale Division (MSD)=1.0 mm1\text{ Main Scale Division (MSD)} = 1.0\text{ mm}
  • 10 Vernier Scale Divisions (VSD)=9 Main Scale Divisions (MSD)10\text{ Vernier Scale Divisions (VSD)} = 9\text{ Main Scale Divisions (MSD)}

Therefore, the value of 1 VSD1\text{ VSD} is: 1 VSD=910 MSD=0.9 mm1\text{ VSD} = \frac{9}{10}\text{ MSD} = 0.9\text{ mm}

The Least Count (LC\text{LC}) of the vernier scale is: LC=1 MSD−1 VSD=1.0 mm−0.9 mm=0.1 mm\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1.0\text{ mm} - 0.9\text{ mm} = 0.1\text{ mm}

Step 3: Find the coincident Vernier Scale Division

When the zero mark of the vernier scale shifts by ΔL=0.3 mm\Delta L = 0.3\text{ mm} relative to the main scale zero mark, the total displacement measured by the scale is given by: Displacement (ΔL)=Main Scale Reading+(N×LC)\text{Displacement } (\Delta L) = \text{Main Scale Reading} + (N \times \text{LC})

Since 0.3 mm<1.0 mm0.3\text{ mm} < 1.0\text{ mm} (1 MSD1\text{ MSD}), the Main Scale Reading remains 0 mm0\text{ mm}. Thus: 0.3 mm=0 mm+N×0.1 mm0.3\text{ mm} = 0\text{ mm} + N \times 0.1\text{ mm} N=0.30.1=3N = \frac{0.3}{0.1} = 3

Hence, the 3rd3^{\text{rd}} vernier scale division coincides with a main scale division.

PreviousNext