FSC 1st Year Physics Chapter 6 Exercise & Numericals Notes PDF

In this post, I am sharing FSC 1st Year Physics Chapter 6 Solved Exercise & Numericals Notes PDF for the students of FSC Part 1. This is the 6th chapter from the 11th Physics Textbook and its name is Fluid Dynamics. Students can download the 11th Class Physics Chapter 6 Fluid Dynamics Solved Exercise & Numericals Notes PDF on their laptop or mobile to read it offline. These Physics Notes for class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Here are the complete FSC 1st Year Physics Notes PDF chapter-wise.

FSC Part 1 Physics Chapter 6 Fluid Dynamics Solved Exercise & Numericals Notes PDF Download

Certain globular protein particle has a density of 1246kgm^-3. It falls through pure water with a terminal speed of 3.0 cmh^-1. Find the radius of the particle.

solution:
data:
            the density of protein particle p = 1246kgm-3
            coefficient of viscosity = 8.0 x 10^-4 Nms^-3
                                    = 3.0 x 10²/ 3600 ms^-1

                                    V_i = 8.3 x 10 ^-6 ms^-1

To find:            radius r = ?
calculation :
            as the expression of the terminal viscosity of the particle is
                        vi = 2gr2p / 9n
                        r2 = vi x 9n/2gp
substituting the value in the equation we get

 r² 0.02447 x 10^-10 = 2.447 x 10^-6
r = 1.6 x 10 ^-6 m

what flow through a hose, whose internal diameter is 1cm at a speed of 1ms^-1. What should be the diameter of the nozzle if the water is to emerge at 21 ms^-1 ?

Solution:
Data:
                        internal diameter of hose = d₁ = 1cm = 10²m
                        speed of water in the hose v₁ = 1m/s
                        speed of emerging water v₂ = 21m/s
                        diameter of nozzle = d₂ = ?
to find:
            as the equation of continuity is:
                        A₁v₁ = A₂v₂
                        V₁(p₁²) – V₂(p₂²)
                        = V₂ (d₂/2)²
                        v₁d₁² = v₂d₂²
                        d₂² = 1/21 x (10^-2)² = 0.0476 x 10^-4
                                d₂  = 0.22 sm ans

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