31 Year NEET Previous Year Questions: Mechanical Properties of Solids - NEET MCQ

We know that Young's modulus

Since Y, F are same for both the wires, we have,

So,    ℓ₁ : ℓ₂ = 1 : 8

Energy stored per unit volume

stress (stress / Young' s modulus)

 (stress)² /( Young ' s modulus)

So, extension,  

[∵ F and Y are constant]

The ratio of   is maximum for case (d).
Hence, option (d) is correct.

From formula, Increase in length

The key observation is that for material X, the ultimate strength point and the fracture point are close to each other, which typically indicates a material that doesn't undergo significant plastic deformation before breaking. This suggests that material X is brittle.

For material Y, the ultimate strength point and the fracture point are far apart, which usually means that the material undergoes significant plastic deformation before fracturing. This suggests that material Y is ductile.

Therefore, the correct answer is:

(b) brittle and ductile.

Given : initial length = L, area of cross section = A
New length after mass M is suspended on the wire = L
∴ Change in length, ΔL = L₁ – L.
Now Young’s modulus, Y = Stress/Strain = F/A x L/ΔL
= 

Stress = F/A = Mg/A,
Strain = 
Energy stored in the wire is,
U = 1/2 × Stress × Strain × Volume
= 
= 1/2Mgl

Young’s modulus, 
Since initial volume of wires are same and their areas of cross sections are A and 3A so lengths are 3l and l respectively. For wire 1,

For wire 1,

For wire 2, let F ′ force is applied

From eqns (i) and (ii),

⇒ F' = 9F

Bulk modulus B is given as

The volume of a spherical object of radius r is given as

∴ 
or

Put this value in eqn. (i), we get 

Fractional decrease in radius is  

Let L and A be length and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights W_s and W_b are added to steel and brass wires respectively. Then, by definition of Young’s modulus, the elongation produced in the steel wire is

and that in the brass wire is 

But ΔL_s = ΔL_b (given)

∴ 

As, 

Depth of ocean, d = 2700 m
Density of water, ρ = 10³ kg m^–3
Compressibility of water, K = 45.4 × 10^–11 Pa^–1
ΔV/V = ?
Excess pressure at the bottom, ΔP = ρgd
= 10³ × 10 × 2700 = 27 × 10⁶ Pa

We know,

= 45.4 × 10^–11 × 27 × 10⁶

= 1.2 × 10^–2

As V = Al ... (i)
where A is the area of cross-section of the wire.
Young’s modulus, 

  (Using (i))
Δl ∝ l²
Hence, the graph between Δl and l² is a straight line.

Young’s modulus,

where F is the force applied, L is the length, D is the diameter and ΔL is the extension of the wire respectively. As each wire is made up of same material therefore their Young’s modulus is same for each wire.
For all the four wires, Y, F (= tension) are the same.
∴ 
In (a)

In (b)

In (c)

In (d)

Hence, ΔL is maximum in (c).

As 

∴ 

where subscripts S and C refer to copper and steel respectively.
Here, F_S = (5m + 2m)g = 7mg
F_C = 5mg

∴ 

Using Hooke's law, , where 

⇒ 

As Y and ΔL are constant, thus F∝D²/L

⇒ 

Given: D_A : D_B = 2:1 and L_A : L_B = 1:2

∴