Elasticity and Elastic Moduli: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
In a Searle-type experiment, a steel wire of length 2.00m (from the clamp to the reference mark) and diameter 0.500mm hangs vertically. A load of 1.0kg is kept on it throughout to remove kinks, and the reading of the measuring scale is noted. When the total load is raised to 5.0kg, the reading increases by 2.00mm. A reference wire of the same material hangs beside it, so room-temperature drift needs no correction. Take g=10m s−2 and neglect the change in the wire's cross-section. The Young's modulus of the steel is closest to ______. Exactly one option is correct.
A5.09×1010Pa
B2.55×1011Pa
C2.04×1011Pa
D8.15×1011Pa
Show answer and solution
Answer: C
Only the change in load produces the measured change in length, so ΔF=(5.0−1.0)×10=40N.
Area: A=πr2=π(0.250×10−3)2=1.963×10−7m2.
Y=AΔlΔFL=1.963×10−7×2.00×10−340×2.00=2.04×1011Pa.
Distractors: using the total 50 N gives 2.55×1011; using the diameter as the radius gives 5.09×1010; using radius d/4 gives 8.15×1011.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A uniform rod of mass M, natural length L, cross-sectional area A and Young's modulus Y hangs vertically from a rigid ceiling. A small block of mass m is attached to its lower end, and the system is in equilibrium. Assume the stress is uniform over each horizontal cross-section, g is uniform, and the deformation is small and within the linear elastic range. The total elastic potential energy stored in the rod is ______. Exactly one option is correct.
A6AYg2L(3m2+3mM+M2)
B2AYg2L(m2+mM+M2)
C2AYg2L(m+M/2)2
D3AYg2L(3m2+3mM+M2)
Show answer and solution
Answer: A
Let x be the distance measured up from the lower end. The tension there supports the block and the rod below: T(x)=mg+LMgx.
Energy density is 2Yσ2, so for a slice of volume Adx: dU=2AYT2dx.
U=2AYg2∫0L(m+LMx)2dx=2AYg2L(m2+mM+3M2)=6AYg2L(3m2+3mM+M2).
Check: M→0 gives 2AYm2g2L (a light wire); m→0 gives 6AYM2g2L.
Trap: using the average tension (m+M/2)g in F2L/(2AY) misses M2/12 because energy depends on the mean of T2, not the square of the mean.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A rigid slab of mass 400kg rests on four identical rubber pads that are glued to the slab above and to the floor below. Each pad is a square block of face area 0.010m2 (the face glued to the slab) and thickness 0.020m, with modulus of rigidity 1.0×106Pa. The slab is given a small horizontal displacement and released. Assume each pad deforms in pure shear (its top and bottom faces stay parallel), the pads' mass and any vertical motion are negligible, there is no damping, and the floor does not move. The period of the slab's horizontal oscillations is closest to ______. Exactly one option is correct.
A0.044s
B0.178s
C0.126s
D0.089s
Show answer and solution
Answer: D
For one pad, a horizontal force F gives shear stress F/A and shear strain θ=x/h, so F=hηAx. Each pad acts like a spring of stiffness k1=ηA/h=0.0201.0×106×0.010=5.0×105N/m.
All four pads have the same displacement, so they act in parallel: k=4k1=2.0×106N/m.
T=2πM/k=2π400/(2.0×106)=0.0889s≈0.089s.
Distractors: one pad's stiffness gives 0.178 s, two pads 0.126 s, and four times the correct stiffness gives 0.044 s.
Difficulty: Very hard
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