13 NEET previous-year questions on Mechanical Properties of Solids, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
PYQ frequency · topic × year
Darker = more questions in our PYQ bank for that topic and year.
Want to time yourself?
Take a free 10-question chapter mock test on Mechanical Properties of Solids — no login needed for your first attempt.
∆ l versus 1/ l
∆ l versus l 2
∆ l versus 1/ l 2
∆ l versus l
Solution
The extension in a wire is given by , where is the cross-sectional area and is Young's modulus. Since for a fixed volume , substituting gives . This shows that is directly proportional to , so the graph of versus is a straight line, making option (b) correct.
1 : 2
2 : 1
4 : 1
1 : 1
Solution
www.vedantu.com 44 Y= W A ∙ℓ Δℓ So Δℓ= Wℓ AY Δe1=Δe2 w1ℓ AY1 = w2ℓ AY2 w1 w2 = Y1 Y2 =2
p B
3 B p
3p B
3 p B
Solution
pB V V = Δ⎛⎞ ⎜⎟⎝⎠ Vp VB Δ = 3 rp rB Δ = 3 rp rB Δ =
4 F
6 F
9 F
F
Solution
The force required to stretch a wire is given by , where is Young's modulus, is the cross-sectional area, is the change in length, and is the original length. Since the volumes are the same, . For the same , the force required for the second wire is , so option (c) is correct.
A + B
𝐴0 + √𝐴2 + 𝐵2
√𝐴2 + 𝐵2
√𝐴0 2 +(𝐴 + 𝐵)2
Solution
x = 𝐴0 + A sin wt + B cos wt x – 𝐴0 = A sin wt + B cos wt ⸫ Amplitude = √𝐴2+ 𝐵2 + 2𝐴𝐵 𝑐𝑜𝑠90 = √𝐴2+ 𝐵2
1 MgL AL
1 Mg(L L) AL −
1 MgL AL
1 MgL A(L L) −
Solution
Young's modulus . Substituting, , so option (b) is correct.
0.01 mm
0.25 mm
0.5 mm
1.0 mm
Solution
The least count of a screw gauge is given by . Given and 50 divisions, . Therefore, option (c) is correct.
Both (A) and (R) are true and (R) is the correct explanation of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
(A) is false but (R) is true
Solution
It is true that stretching of spring is determined by shear modulus of the spring as when coil spring is stretched neither its length nor its volume changes, there is only change in its shape. Tensile strength of steel is more than that of copper. Hence Assertion is true and reason is false.
Zero
2W/A
W/A
W/2A
Solution
Longitudinal stress is defined as the force per unit area, given by . Here, the force is the weight applied at the free end, so the stress is . Therefore, option (c) is correct.
19.8 mN
198 N
1.98 mN
99 N
Solution
Excess force = T × 2πR 7 4.52 3.14100 100= × × × = 197.82 × 10–4 = 19.8 × 10–3 N = 19.8 mN
4 mm
0.4 mm
40 mm
8 mm
Solution
In the case for maximum elongation, Stress = Elastic limit 8 –3elastic max 11 8 10 1 4 10Young's modulus 2 10 σ× ××δ = = = × × L = 4 mm i.e. maximum elongation is 4 mm
5 × 103 N
50 × 103 N
100 × 103 N
2 × 103 N
Solution
Thermal strain = Longitudinal strain = α Δ T ⇒ Longitudinal strain, δ = 10–5 × 102 = 10–3 ⇒ Compressive stress = δ × Young’s Modulus = 10–3 × 0.5 × 1011 = 0.5 × 108 ⇒ Compressive force = 0.5 × 108 × 10–3 = 0.5 × 105 = 4 105 10 10×× = 50 × 103 N - 21 - NEET (UG)-2024 (Code-Q1)
Solution
Sol.
By hit and trial (using option (1))
Put
Same NCERT order — keep the PYQ practice rolling chapter by chapter.
Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.
Free 14-day trial · No credit card required