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Mechanical Properties of Solids: NEET previous year questions

Every Mechanical Properties of Solids question from the official NEET papers on this site, with the answer and a worked solution.

The questions come from NEET 2026 and NEET 2020. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Mechanical Properties of Solids guide.

Practise on the CBT screen2 questions, 2 minutes, marked +4 / −1Download PDF2 pages, 0.7 MB, with answers and solutions

NEET 2026 · the full paper

  1. NEET 2026 · Q2

    Match List I with List II and choose the correct answer from the options given below:

    List I

    List II

    P) Young's Modulus

    I) ΔdΔL(Ld)\frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)

    Q) Compressibility

    II) FLA(ΔL)\frac{FL}{A(\Delta L)}

    R) Bulk Modulus

    III) −1ΔP(ΔVV)-\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)

    S) Poisson's Ratio

    IV) −P(VΔV)-P\left(\frac{V}{\Delta V}\right)

    1. (1)

      P-IV, Q-I, R-II, S-III

    2. (2)

      P-III, Q-II, R-I, S-IV

    3. (3)

      P-I, Q-IV, R-III, S-II

    4. (4)

      P-II, Q-III, R-IV, S-I

    Answer and solution

    Answer: (4)

    Young's modulus is stress/strain =FLA ΔL= \frac{FL}{A\,\Delta L} (II); compressibility is the reciprocal of bulk modulus =−1ΔPΔVV= -\frac{1}{\Delta P}\frac{\Delta V}{V} (III); bulk modulus =−PVΔV= -P\frac{V}{\Delta V} (IV); and Poisson's ratio is lateral strain over longitudinal strain =ΔdΔLLd= \frac{\Delta d}{\Delta L}\frac{L}{d} (I). Hence P-II, Q-III, R-IV, S-I, which is option (4).

NEET 2020 · the full paper

  1. NEET 2020 · Q137

    A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L1_{1} when mass M is suspended from its free end. The expression for Young's modulus is

    1. (1)

      Mg(L1_{1}-L)/(AL)

    2. (2)

      MgL/(AL1_{1})

    3. (3)

      MgL/[A(L1_{1}-L)]

    4. (4)

      MgL1_{1}/(AL)

    Answer and solution

    Answer: (3)

    Young's modulus Y=StressStrain=F/AΔL/LY = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L}. Substituting, Y=MgA×LL1−L=MgLA(L1−L)Y = \frac{Mg}{A} \times \frac{L}{L_1 - L} = \frac{MgL}{A(L_1 - L)}. Chapter: Mechanical Properties of Solids.

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Questions

Are these the official NEET questions on Mechanical Properties of Solids?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

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