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Young’s modulus of a metallic wire: NEET MCQs with answers

48 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In an experiment to determine Young's modulus, which of the following precautions is specifically taken to eliminate backlash error in the spherometer or micrometer screw?

  1. Keeping a very large load suspended permanently from the reference wire
  2. Taking measurements of the wire diameter at three mutually perpendicular directions
  3. Rotating the screw in only one direction while taking readings during loading or unloading
  4. Using a vernier calliper instead of a screw gauge to measure the diameter
Show answer and explanation

Answer: (C) Rotating the screw in only one direction while taking readings during loading or unloading

Backlash error occurs due to looseness or wear in the screw threads. It is avoided by continuously rotating the micrometer screw in the same direction when setting the spirit level bubble to the center.

medium

2. In a Searle's apparatus experiment, the screw attached to the spirit level has a pitch of 0.5 mm and 100 divisions on its circular scale. When a load of 4.0 kg is added, the circular scale must be rotated by 1 complete revolution and 20 additional divisions to restore the bubble to the center. If g = 9.8 m/s², the elongation produced is:

  1. 0.60 mm
  2. 0.50 mm
  3. 0.70 mm
  4. 1.20 mm
Show answer and explanation

Answer: (A) 0.60 mm

The least count of the screw = Pitch / Number of circular divisions = 0.5 mm / 100 = 0.005 mm. The elongation corresponding to 1 revolution (0.5 mm) and 20 circular divisions is: l = 0.5 mm + (20 × 0.005 mm) = 0.5 mm + 0.10 mm = 0.60 mm.

hard

3. A student performs the Young's modulus experiment and plots tensile stress (σ) along the y-axis versus longitudinal strain (ε) along the x-axis for two wires X and Y of the same initial length and cross-section. The line for wire X makes an angle of 60° with the strain axis, while for wire Y it makes an angle of 30° with the strain axis. The ratio of Young's modulus of wire X to that of wire Y (Y_X / Y_Y) is:

  1. 3 : 1
  2. 1 : 3
  3. √3 : 1
  4. 1 : √3
Show answer and explanation

Answer: (A) 3 : 1

Here, Young's modulus is the slope of the stress versus strain graph, Y = σ/ε = tan θ, where θ is the angle with the strain axis. Thus, Y_X / Y_Y = tan(60°) / tan(30°) = (√3) / (1/√3) = 3 : 1.

medium

4. In an experiment to measure Young's modulus Y of a wire, a student plots a graph between the elongation ΔL (on the y-axis, in mm) and the suspended mass M (on the x-axis, in kg). The slope of the resulting straight line is S. If the original length of the wire is L, its radius is r, and acceleration due to gravity is g, what is the expression for Young's modulus?

  1. Y = (g · L) / (π · r² · S)
  2. Y = (S · g · L) / (π · r²)
  3. Y = (π · r² · S) / (g · L)
  4. Y = (g · L · S) / (2 · π · r)
Show answer and explanation

Answer: (A) Y = (g · L) / (π · r² · S)

From Hooke's law, ΔL = (M · g · L) / (π · r² · Y). A plot of ΔL versus M gives a straight line with slope S = ΔL / M = (g · L) / (π · r² · Y). Rearranging for Y yields Y = (g · L) / (π · r² · S).

easy

5. The diameter of the wire in the Young's modulus experiment is measured with a:

  1. Screw gauge
  2. Metre scale
  3. Vernier calipers only
  4. Spherometer
Show answer and explanation

Answer: (A) Screw gauge

The wire is thin, so its diameter is measured accurately with a screw gauge at several places and the mean is taken.

NCERT Class 11

hard

6. In Searle's method, the spirit level has a length of 10 cm between its supporting points. One end is supported by the frame of the reference wire and the other by the frame of the experimental wire via a micrometer screw of pitch 0.5 mm and 100 divisions on the circular scale. If the tilt of the spirit level can be detected to an angular precision of 10⁻⁴ radian, the minimum elongation that can theoretically be detected by the spirit level tilt is:

  1. 5.0 × 10⁻⁶ m
  2. 1.0 × 10⁻⁵ m
  3. 1.0 × 10⁻⁴ m
  4. 2.5 × 10⁻⁵ m
Show answer and explanation

Answer: (B) 1.0 × 10⁻⁵ m

The angle of tilt θ is related to the difference in height (elongation ΔL) and the distance between supports d = 10 cm = 0.1 m by θ = ΔL / d. Thus, ΔL_min = d · θ_min = 0.1 m × 10⁻⁴ rad = 1.0 × 10⁻⁵ m = 0.01 mm.

medium

7. During the measurement of the diameter of an experimental wire using a screw gauge, a student notes that with no object between the studs, the 5th circular division lies above the reference line. The pitch is 0.5 mm and there are 50 divisions on the circular scale. While measuring the wire, the main scale reads 1.0 mm and the 24th circular division coincides with the reference line. The correct diameter of the wire is:

  1. 1.19 mm
  2. 1.24 mm
  3. 1.29 mm
  4. 1.14 mm
Show answer and explanation

Answer: (A) 1.19 mm

Least count = 0.5 mm / 50 = 0.01 mm. Since the 5th division lies above the reference line when closed, the gauge has a positive zero error of + (5 × 0.01 mm) = +0.05 mm. The observed reading = 1.0 mm + 24 × 0.01 mm = 1.24 mm. Correct reading = Observed reading - Zero error = 1.24 mm - 0.05 mm = 1.19 mm.

medium

8. A student performs the experiment to determine the Young's modulus of a metallic wire. A graph is plotted with the square of the diameter (d²) on the x-axis and the elongation (ΔL) on the y-axis for wires of the same material, equal lengths, and subjected to the same stretching force. The shape of the graph obtained is:

  1. A straight line passing through the origin
  2. A rectangular hyperbola
  3. A parabola opening towards the y-axis
  4. An exponential curve approaching zero
Show answer and explanation

Answer: (B) A rectangular hyperbola

From the expression for Young's modulus, Y = (4FL)/(πd²ΔL), which gives ΔL = (4FL / πY) × (1/d²). Therefore, ΔL × d² = constant (for constant F, L, and Y). A plot of ΔL versus d² yields a rectangular hyperbola (xy = constant).

medium

9. In a Young's modulus experiment, a wire of length L and uniform cross-sectional area A has density ρ. Due to its own weight suspended vertically from a rigid support, the elongation produced in the wire is (Y is Young's modulus, g is acceleration due to gravity):

  1. ρgL² / Y
  2. 2ρgL² / Y
  3. ρgL² / (2Y)
  4. ρgL² / (4Y)
Show answer and explanation

Answer: (C) ρgL² / (2Y)

The tension at a distance x from the lower free end is T(x) = ρ·A·x·g. The total elongation is the integral from 0 to L of [T(x) dx / (A·Y)] = (ρ·g/Y) ∫ x dx = ρgL² / (2Y).

medium

10. Why must both the experimental wire and the auxiliary (reference) wire in Searle's apparatus be suspended from the exact same rigid ceiling support?

  1. To ensure that both wires carry equal tension at all times
  2. To make the effective radius of cross-section identical for both wires
  3. To eliminate errors caused by the yielding or bending of the common support
  4. To reduce air resistance acting on the hanging weights
Show answer and explanation

Answer: (C) To eliminate errors caused by the yielding or bending of the common support

If the main support flexes or yields slightly when heavy weights are placed on the experimental wire, both wires descend by the exact same amount because they share the same rigid support. This common-mode displacement cancels out, ensuring that only the true elastic stretch of the experimental wire tilts the spirit level.

medium

11. In an experiment to determine Young's modulus of a wire, a student measures the initial length L using a meter scale (least count 1 mm), diameter d using a screw gauge (least count 0.001 cm), and elongation ΔL using a spherometer screw (least count 0.001 mm). Which measurement contributes the maximum fractional error if L = 2 m, d = 0.5 mm, and ΔL = 0.25 mm?

  1. Measurement of length L
  2. Both diameter d and elongation ΔL equally
  3. Measurement of elongation ΔL
  4. Measurement of diameter d
Show answer and explanation

Answer: (D) Measurement of diameter d

The fractional error in Y = 4MgL / (πd²ΔL) gives contributions: ΔL/L = 10⁻³ / 2 = 0.0005; 2(Δd/d) = 2 × (0.01 mm / 0.5 mm) = 0.04; Δ(ΔL)/ΔL = 0.001 mm / 0.25 mm = 0.004. The term 2(Δd/d) contributes 4%, which is the largest error contribution.

medium

12. In an experiment to determine Young's modulus of a wire, a student measures the following parameters: Length of the wire, L = (2.000 ± 0.002) m Diameter of the wire, d = (0.500 ± 0.005) mm Applied load, M = (4.000 ± 0.004) kg Elongation produced, ΔL = (0.400 ± 0.004) mm Assuming the acceleration due to gravity g is constant and exact, what is the maximum percentage fractional error in the calculated Young's modulus?

  1. 2.5%
  2. 1.8%
  3. 4.1%
  4. 3.2%
Show answer and explanation

Answer: (D) 3.2%

Young's modulus is given by Y = (4MgL) / (πd²ΔL). The fractional error is ΔY/Y = ΔM/M + ΔL/L + 2(Δd/d) + Δ(ΔL)/(ΔL). Substituting the values: ΔM/M = 0.004/4.000 = 0.001 (0.1%), ΔL/L = 0.002/2.000 = 0.001 (0.1%), 2(Δd/d) = 2 × (0.005/0.500) = 0.020 (2.0%), and Δ(ΔL)/(ΔL) = 0.004/0.400 = 0.010 (1.0%). Total maximum percentage error = (0.1 + 0.1 + 2.0 + 1.0)% = 3.2%.

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