DPP
DAILY PRACTICE PROBLEMS
CH - 9 MECHANICAL PROPERTIES OF FLUIDS
SOLUTIONS
1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
πR2V = nπr2(v) ⇒ v =
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure = upthrust
Access pressure in air bubble =
⇒ ⇒
3. (d) From the figure it is clear that liquid 1 floats on liquid 2. The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ρ1 < ρ2
Also ρ3 < ρ2 otherwise the ball would have sink to the bottom of the jar.
Also ρ3 > ρ1 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
ρ1 < ρ3 < ρ2.
4. (c) Upthrust = weight of 40 cm3 of water
= 40 g = down thrust on water
5. (d) T1 + T cos (π – θ) = T2
∴ cos (π – θ) =
6. (a) The condition for terminal speed (vt) is
Weight = Buoyant force + Viscous force
7. (a) As A1v1 = A2v2 (Principle of continuity)
or, (Efflux velocity = )
∴ or
8. (b) Pressure inside tube =
∴ (since r2 > r1)
Hence pressure on side 1 will be greater than side 2. So air from end 1
flows towards end 2
9. (c) Velocity of ball when it strikes the water surface
...(i)
Terminal velocity of ball inside the water
...(ii)
Equation (i) and (ii) we get
10. (a) Fluid resistance is given by R =
When two capillary tubes of same size are joined in parallel, then
equivalent fluid resistance is
= =
Rate of flow = = X
11. (c) Vertical distance covered by water before striking ground = (H –
h). Time taken is, : Horizontal velocity of water
coming out of hole at P,
∴ Horizontal range =
=
12. (a) Because film tries to cover minimum surface area.
13. (c) Pressure at interface A must be same from both the sides to be in
equilibrium.
∴ (R cos α + R sin α)d2g = (R cos α – R sin α)d1g
14. (c) Angle of contact θ
when water is on a waxy or oily surface
TSA < TSL cos θ is negative i.e., 90° < θ < 180°
i.e., angle of contact θ increases
And for θ > 90º liquid level in capillary tube fall. i.e., h decreases
15. (c) Sum of volumes of 2 smaller drops
= Volume of the bigger drop
Surface energy = T.4πR2
= T4π22/3r2 = T.28/3 πr2
16. (a) When a body falls through a viscous liquid, its velocity increases
due to gravity but after some time its velocity becomes uniform
because of viscous force becoming equal to the gravitational force.
Viscous force itself is a variable force which increases as velocity
increases, so curve (a) represents the correct alternative.
17. (c)
New mass m2 =
18. (d) The excess pressure inside a bubble of soap is given by . It
means excess pressure inside the bubbles A and C is more than
inside B. So, air will go towards B from A and C. So, A and C will
start collapsing with the increasing volume of B.
19. (c)
20. (a)
21. (8) Inside pressure must be greater than outside pressure in bubble.
This excess pressure is provided by charge on bubble.
22. (0.1) Terminal velocity,
23. (10.67) Let Vi be the volume of the iceberg inside sea water and V is
the total volume of iceberg.
Total weight of iceberg
= weight of water displaced by iceberg.
Thus the fraction of total volume of iceberg above the sea level
24. (650) Volume of ball =
Downthrust on water = 50 g.
Therefore reading is 650 g.
25. (0.025) At equilibrium, weight of the given block is balanced by force
due to surface tension, i.e.,
2L. S = W
or
