NEETPhysics

Mechanical Properties of Solids

5 NEET Physics previous year questions on Mechanical Properties of Solids — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.

Q1 FREE PREVIEW
PYQ

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are \(8\times {10}^{8}\mathrm{N}{\mathrm{m}}^{-2}\) and \(2\times {10}^{11}\mathrm{N}{\mathrm{m}}^{-2}\), is :

[NEET 2024]

a

4 mm

b

0.4 mm

c

40 mm

d

8 mm

✓ Correct answer: a)

4 mm

Explanation

Under the elastic limit, the maximum permissible stress = \(8\times {10}^{8}\text{ }\mathrm{N}\mathrm{/}{\mathrm{m}}^{2}\). Since

\(\text{strain}\text{  }=\text{  }\frac{\text{stress}}{Y},\ \Delta L\text{  }=\text{  }\text{(strain)}\times L,\)

we get

\({\mathrm{ΔL}}_{\max }=(\frac{8\times {10}^{8}}{2\times {10}^{11}})\times 1\text{ }\mathrm{m}=4\times {10}^{−3}\text{ }\mathrm{m}=4\text{ }\mathrm{m}\mathrm{m}\mathrm{.}\)

Q2
PYQ

Match List I with List II :

List I List II
A. Young's Modulus I. \(\frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)\)
B. Compressibility II. \(\frac{FL}{A(\Delta L)}\)
C. Bulk Modulus III. \(-\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)\)
D. Poisson's Ratio IV. \(-P\left(\frac{V}{\Delta V}\right)\)
a

A-I, B-IV, C-III, D-II

b

A-IV, B-I, C-II, D-III

c

A-III, B-II, C-I, D-IV

d

A-II, B-III, C-IV, D-I

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Q3
PYQ

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

a

\( \frac{\mathrm{Mg}\left(\mathrm{L}_{1}-\mathrm{L}\right)}{\mathrm{AL}} \)

b

\( \frac{\mathrm{MgL}}{\mathrm{AL}_{1}} \)

c

\( \frac{\mathrm{MgL}}{\mathrm{A}\left(\mathrm{L}_{1}-\mathrm{L}\right)} \)

d

\( \frac{\mathrm{MgL}_{1}}{\mathrm{AL}} \)

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Q4
PYQ

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

[NEET 2020]

a

\( \frac{\mathrm{Mg}\left(\mathrm{L}_{1}-\mathrm{L}\right)}{\mathrm{AL}} \)

b

\( \frac{\mathrm{MgL}}{\mathrm{AL}_{1}} \)

c

\( \frac{\mathrm{MgL}}{\mathrm{A}\left(\mathrm{L}_{1}-\mathrm{L}\right)} \)

d

\( \frac{\mathrm{MgL}_{1}}{\mathrm{AL}} \)

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Q5
PYQ

An aluminium rod with Young's modulus \(Y=7.0 \times 10^{10}\) \(\mathrm{N} / \mathrm{m}^{2}\) undergoes elastic strain of \(0.04 \%\). The energy per unit volume stored in the rod in SI unit is:

[JEE Main 2023, 8 Apr (Shift 1)]

a

5600

b

8400

c

2800

d

11200

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