NCERT solution Class 11 mechanical properties of solids

NCERT Solutions

Class 11 Physics

Mechanical Properties of Solids

NCERT Questions

Q. 8.1
Q. 8.2
Q. 8.3
Q. 8.4
Q. 8.5
Q. 8.6
Q. 8.7
Q. 8.8
Q. 8.9
Q. 8.10
Q. 8.11
Q. 8.12
Q. 8.13
Q. 8.14
Q. 8.15
Q. 8.16

Q. 8.13

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10⁵ Pa

Density of water at the surface, ρ₁ = 1.03 × 10³ kg m^–3

Let ρ₂ be the density of water at the depth h.

Let V₁ be the volume of water of mass m at the surface.

Let V₂ be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = (V₁ - V₂)

=m ((1/ ρ₁) – (1/ ρ₂))

Therefore, Volumetric strain = (ΔV/ V₁)

=m ((1/ ρ₁) – (1/ ρ₂)) x (ρ₁/m)

Therefore, ΔV/ V₁ = (1- ρ₁/ ρ₂) … (i)

Bulk modulus, B = (p V₁/ ΔV)

(ΔV/ V₁)= (p/B)

Compressibility of water = (1/B) =45.8x10^-11 Pa^-1

Therefore, (ΔV/ V₁) = (80x1.013x10⁵x45.8x10^-11) = 3.71x10^-3 … (ii)

For equations (i) and (ii), we get:

(1- ρ₁/ ρ₂) = 3.71 x10^-3

ρ₂ = 1.03x10³/(1-(3.71x10^-3)

=1.034x10³kgm^-3

Therefore, the density of water at the given depth (h) is 1.034 × 10³ kg m^–3.

Video solution:

Mechanical Properties of Solids Class 11 Physics NCERT Chapter 8 NCERT Solutions

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