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Mechanical Properties of Solids

Mechanical Properties of SolidsNEET Physics · Class 11 · NCERT Chapter 8

29 NEET previous-year questions on Mechanical Properties of Solids, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

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16
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24
Stress & strain
1
1
1
Young's modulus
1
2
2
1
1
2
Bulk modulus
1
1
2
Shear modulus
1
1
1
Poisson's ratio
1
1
Elastic PE
1
1
1
1
1
Wires (series/parallel)
2
1

Darker = more questions in our PYQ bank for that topic and year.

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All (29)
Stress & strain (3)
Young's modulus (9)
Bulk modulus (4)
Shear modulus (3)
Poisson's ratio (2)
Elastic PE (5)
Wires (series/parallel) (3)

A

B

C

D

Solution

.

A

B

C

D

Solution

(same , ).

.

A

stress × strain

B

C

stress / strain

D

Solution

Within the elastic limit, .

A

B

C

D

Solution

.

A

B

C

D

Solution

. To double length, :

.

A

B

C

D

Solution

Force varies linearly from 0 to . Average force = . Work = average force × distance = .

A

Force

B

Energy

C

Pressure

D

Power

Solution

Stress = Force/Area, dimensions , same as pressure.

A

to

B

to

C

to

D

to

Solution

Theoretical thermodynamic bounds: . Most real materials have .

A

B

C

D

Solution

.

A

Stress to strain

B

Pressure change to volume strain

C

Force to area

D

Strain to stress

Solution

— bulk stress (pressure change) divided by volumetric strain.

A

B

C

D

Solution

Shear stress = (force per top-face area).

Shear strain = stress/G = .

A

B

C

D

Solution

Same force, same material → strain .

.

A

B

C

D

Solution

where .

.

A

Returns to original shape on removing the force

B

Shows permanent deformation

C

Develops cracks immediately

D

Has zero stress

Solution

Beyond the elastic limit, deformation becomes plastic — permanent. The body does not return fully to its original shape.

A

Material

B

Temperature

C

Length of the wire

D

Type of strain (within elastic limit)

Solution

Y is a property of the material itself. It does NOT depend on the dimensions (length, area) of the specimen — those determine the elongation under a given force, but Y stays the same.

A

Greater than its bulk modulus

B

Equal to its bulk modulus

C

Zero

D

Infinite

Solution

Fluids cannot sustain shear stress at rest — they flow under any shear. So the shear modulus is zero.

A

B

C

D

Solution

with :

.

A

Strain

B

Compressibility

C

Young's modulus

D

Modulus of rigidity

Solution

Compressibility . Higher compressibility means easier to squeeze.

A

B

C

D

Solution

In series, both have same , so .

.

A

B

C

D

Solution

Volume conservation requires .

So , giving .

Rubber has for this reason.

A

Smaller stress

B

Greater stress

C

Same stress

D

Cannot be determined

Solution

Stress = F/A. With the same force , the smaller area produces larger stress. So the thinner wire experiences greater stress (and greater strain since is the same).

A

B

C

D

Solution

. Same form as elastic PE in a stretched wire.

A

Same

B

Half

C

One quarter

D

Twice

Solution

. Twice diameter → 4× area → becomes 1/4 of original.

A

Twice

B

Three times

C

Four times

D

Same

Solution

. Doubling multiplies by 4.

A

Elastic limit

B

Young's modulus

C

Yield stress

D

Ultimate strength

Solution

In the linear region, , so the slope of stress vs strain is — Young's modulus.

A

B

C

D

Solution

.

.

A

Higher bulk modulus

B

Lower bulk modulus

C

Equal bulk modulus

D

Zero bulk modulus

Solution

Gases are highly compressible — small pressure produces large fractional volume change — so their bulk modulus is much smaller than liquids.

A

B

C

D

Solution

For most common solids, . Exact relation involves Poisson's ratio: .

A

B

C

D

Solution

Energy stored: .

Volume: .

Density: .

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