Test: Introduction & Properties of Fluid Level- 2 - Mechanical Engineering MCQ

The viscosity of fluids (liquids and gases) generally behaves differently with changes in temperature. Here's how temperature affects viscosity:

- **Liquids**: In general, as temperature decreases, the viscosity of liquids tends to increase. This is because lower temperatures reduce the kinetic energy of the liquid molecules, causing them to move more slowly and thus increasing viscosity.

- **Gases**: For gases, as temperature decreases, viscosity tends to decrease. This is because lower temperatures reduce the frequency of collisions between gas molecules, decreasing internal friction and thus reducing viscosity.

Given these behaviors, the most accurate statement regarding the effect of temperature on viscosity is:

**Option 3: an increase in the viscosity of liquids and a decrease in that of gases**

This option correctly reflects the general trend observed in viscosity changes with temperature for both liquids and gases.

∴ dm = d(ρV)

0 = ρdV + Vdρ

−dV / V = dp / ρ

As per definition of Bulk Modulus

K = dp / (−dV / V)

Therefore

K = dp / dρ/ρ

K = ρ dp / dρ

Where

1 → Dilatants like quicksand, oobleck

2 → Newtonian fluids like water, air

3 → Pseudo plastics like blood, milk

4 → Bingham substances like mayonnaise, toothpaste

Bingham substances are those which undergo static deformation upto a limiting value (τ) of shear stress and behave as solids but beyond this limiting value they deform continuously and behave as fluids.

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ΔV / V× 100 = −0.2

⇒ ΔV / V= −2 × 10^-3

K = Δp / −Δ? / ?

= (15 − 7.5) × 10⁶ / −(−2 × 10−3)

= 3.75 × 10⁹

Compressibility = 1 / K = 2.666 × 10^-10

≃ 0.267 × 10−9 m²/N

For the raindrop to be in equilibrium,

σπd = Δp × π / 4 d²

4σ d = Δp

Substituting values,

Δp = 4 × 0.073 / 1 × 10^-3

Δp = 292 N/m2

⇒ p × (Ld) = σ × 2L

⇒ p = 2σ / d

For a fluid at rest, shear stress at any plane passing through a point is zero hence all the planes are subjected to equal normal stress (pressure). Thus Mohr’s circle is a point on the −ve σ axis. And the distance of the point from origin is equal to pressure.

d(pVn) = 0

⇒ pnVn−1 dV + dp Vn = 0

⇒ dp = − 1 / npdV − np dV / V

⇒dp / (−dV / V) = np

⇒ K = np

Thus Bulk modulus of an ideal gas is directly dependent upon pressure i.e., if p ↑ K ↑

Knudsen number is the ratio of the molecular mean free path length to a representative physical length scale.

For any liquid the density of the liquid decreases with increasing temperature. But water is an exception and shows maximum density at 4.4°C.