# Important Questions for Class 11 Chapter 9 - Mechanical Properties of Solids

**Important questions based on NCERT syllabus for Chapter 9 - Mechanical Properties of Solids:**

*Question-1*: Calculate the pressure on a litre of water if it is to be compressed by 0.15%.

*Solution*:

Given, volume of water,V= 1 L

And water needs to be compressed by 0.15%.

∴ Fractional change, ∆V / V = 0.15 / (100 × 1) = 1.5 x 10^(-3)

We know,

Bulk modulus, B = ρ / (∆V/V)

=> ρ = B × (∆V/V)

We know, bulk modulus of water, B = 2.2 × 10^9 Nm^-2

=> ρ = 2.2 × 10^9 × 1.5 x 10^-3 = 3.3 × 10^6 Nm^-2

Thus, a pressure of 3.3 ×10^6 Nm–2 should be applied on the water.

*Question-2*: Two metal bars are riveted together at their ends by four rivets, each having a diameter of 5 mm. Calculate the maximum tension the riveted bars can bear if the maximum shearing stress a rivet can take is 6.9 x 107 Pa. Consider that each rivet carries ¼ of the total load.

*Solution*:

Given,

Diameter of the metal bar, d = 5.0 mm = 5.0 × 10^–3 m

Radius, r = d/2 = 2.5 × 10^-3 m

Maximum shearing stress = 6.9 × 10^7 Pa

We know,

Maximum stress = Maximum force or tension / Area

=> Maximum force = Maximum stress × Area

= 6.9 × 10^7 × π × (2.5)^2

= 6.9 × 10^7 × π × (2.5 ×10^–3)^2

= 1354.125 N

Since each rivet carries ¼ of the load.

∴ Maximum tension on each rivet = 4 × 1354.125 = 5416.5 N.

*Question-3*: The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

*Solution*: (a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus =(Stress /Strain) is greater for A than that of B.

(b) Strength of a material is determined by the amount of stress required to cause fracture. This stress corresponds to the point of fracture. The stress corresponding to the point of fracture in A is more than for B. So, material A is stronger than material B.