StudentAidUnit 5: Solids and Fluid Dynamics — Numericals
11th Class Physics · Unit 5: Solids and Fluid Dynamics
5.1.A steel wire of length 2 metres and cross-sectional area of 2 x 10^-6 m^2 is stretched by a force of 400 N. If the Young's modulus of steel is 2 x 10^11 N m^-2, calculate the extension of the wire.
Given
Length
L = 2 m
Cross-sectional area
A = 2 × 10-6 m2
Force
F = 400 N
Young's modulus
Y = 2 × 1011 N m-2
Formula
Young's modulus definition
Y = F × LA × ΔL
Rearranging for extension
ΔL = F × LA × Y
Working
Substituting values
ΔL = frac{400 × 2}{2 × 10-6 × 2 × 1011
= 8004 × 105
= 200 × 10-5 m
Result
Extension
ΔL = 0.002 m
5.2.A spring with a spring constant 200 N m^-1 is stretched by 0.5 m. Find the elastic P.E. stored in the spring.
Given
Spring constant
K = 200 N m-1
Extension
X = 0.5 m
Formula
Elastic potential energy
P.E = 12kx2
Working
Substituting values
P.E = 12 × 200 × (0.5)2
= 100 × 0.25
Result
Elastic potential energy
P.E = 25 J
5.3.A copper wire of length 3 metres and cross-sectional area of 1 x 10^-6 m^2 is subjected to a force of 500 N. Calculate the stress and strain produced in the wire.
Given
Length
L = 3 m
Cross-sectional area
A = 1 × 10-6 m2
Force
F = 500 N
Formula
Stress
sigma = FA
Working
Calculating stress
sigma = frac{500}{1 × 10-6
Result
Stress
sigma = 5 × 108 N m-2
Formula
Young's modulus relation
Y = frac{stress{strain = sigmavarepsilon
Strain from Young's modulus
varepsilon = sigmaY
Given
Young's modulus for copper
Y = 1.1 × 1011 N m-2
Working
Calculating strain
varepsilon = frac{5 × 108}{1.1 × 1011
Result
Strain
varepsilon = 4.54 × 10-3
5.4.A block of wood of mass 10 kg and density of 600 kg m^-3 is floating in water. Calculate the buoyant force acting on the block. (Density of water = 1000 kg m^-3.)
Given
Mass
m = 10 kg
Density of wood
rho = 600 kg m-3
Density of water
rhow = 1000 kg m-3
Formula
Archimedes' principle for floating object
Buoyant force = Weight of object
Weight calculation
Buoyant force = mg
Working
Substituting values
Fb = 10 × 9.8
Result
Buoyant force
Fb = 98 N
5.5.Water flows through a pipe with a diameter of 0.05 m at a velocity of 2 m s^-1. If the pipe narrows to a diameter of 0.03 m, calculate the velocity of water at narrow section.
Given
Initial diameter
d1 = 0.05 m
Initial velocity
v1 = 2 m s-1
Diameter of narrow section
d2 = 0.03 m
Formula
Equation of continuity
A1 v1 = A2 v2
Cross-sectional area of circle
A = pi r2 = pi d24, quad r = d2
Areas in terms of diameter
A1 = pi d124 quad and quad A2 = pi d224
Working
Substituting into continuity equation
pi d124 v1 = pi d224 v2
Simplifying
d12 v1 = d22 v2
Solving for v₂
v2 = d12 v1d22
Substituting values
v2 = (0.05)2 × 2(0.03)2
= 0.0025 × 20.0009
= 0.0050.0009
Result
Velocity at narrow section
v2 = 5.56 m s-1
5.6.Water flows through a horizontal pipe with a velocity of 3 m s^-1 and pressure of 200,000 Pa at point 1. At the nozzle (point 2), the pressure decreases to atmospheric pressure 101,300 Pa and the velocity increases to 14 m s^-1. Calculate the velocity of the water exiting the nozzle.
Given
Initial velocity
v1 = 3 m s-1
Initial pressure
P1 = 200,000 Pa
Final pressure
P2 = 101,300 Pa
Formula
Bernoulli's equation for horizontal pipe
P1 + 12rho v12 = P2 + 12rho v22
Given
Density of water
rho = 1000 kg m-3
Working
Substituting values
200000 + 12 × 1000 × (3)2 = 101300 + 12(1000)v22
200000 + 4500 = 101300 + 500 v22
204500 = 101300 + 500 v22
204500 - 101300 = 500 v22
103200 = 500 v22
v22 = 206.4
Taking square root
v2 = sqrt{206.4} = 14.37 m s-1
v2 = 14.37 m s-1
5.7.A tank filled with water has a hole at a depth of 5 m from the water surface. Calculate the velocity of water flowing out of the hole.
Given
Depth of hole below water surface
h = h1 - h2 = 5 m
Acceleration due to gravity
g = 9.8 m s-2
Formula
Torricelli's theorem
v = sqrt{2g(h1 - h2)} = sqrt{2gh}
Working
Substituting values
v = sqrt{2 × 9.8 × 5}
= sqrt{98}
Result
Velocity of water flowing out
v = 9.9 m s-1
5.8.Calculate the terminal velocity of a spherical raindrop with a radius 0.5 mm falling through the air. (η for air = 19×10^-6 kg m^-1 s^-1 and ρ = 1000 kg m^-3 for water)
Given
Radius of raindrop
r = 0.5 mm = 0.5 × 10-3 m
Coefficient of viscosity of air
eta = 19 × 10-6 kg m-1s-1
Density of water
rho = 1000 kg m-3
Acceleration due to gravity
g = 9.8 m s-2
Formula
Terminal velocity formula
vt = 2r2 rho g9eta
Working
Substituting values
vt = 2 × (0.5 × 10-3)2 × 1000 × 9.89 × (19 × 10-6)
Result
Terminal velocity
vt = 28.65 m s-1