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Practice each question, then open the answer or worked solution.
To what depth must a rubber ball be taken in the deep sea so that its volume is decreased by 0.1%.
Bulk modulus of rubber = 9.8 × 108 N/m2
Density of sea water = 103 kg/m3
Percentage decrease in volume = 0.1%
Change in pressure, p = hρg
B = ∆P/(∆V/V)
∆P = hρg
h = 100 m
A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched?
Length of steel cable, l = 9.1 m
Radius, r = 5 mm
Tension on the cable = 800 N
Young’s modulus = Y = 2 × 1011 N/m2
Y = (F/A)/( ∆L/L)
∆L = 0.5 × 10-3 m
Two identical solid balls, one of ivory and the other of wet clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the flood and why?
Ivory ball will rise to a greater height after striking the flood because the ivory ball is more elastic than clay ball and the amount of energy and momentum transferred is more on the ivory ball.
Long AnswersConsider a long steel bar under tensile stress due to forces F acting at the edges along the length of the bar. Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?
a) for what angle is the tensile stress a maximum?
b) for what angle is the shearing stress a maximum?
Given,
F is applied along the horizontal
When F is perpendicular = F sin θ
When F is parallel = F cos θ
A is the cross-sectional area of the bar.
The plane aa’ is equilibrium.
Area of the plane aa’ is
Tensile stress on the plane aa’ is
Shear stress on the plane aa’ is
F sin 2θ/2A
a) When the tensile stress is maximum θ = 90o
b) When the shearing stress is maximum θ = 45o
a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg/m3.
b) If the yield strength of steel is 2.5 × 108 N/m2, what is the maximum weight that can be hung at the lower end of the wire?
Let dx be the small element
Let dm be the mass
Let L be the length of the wire
Let x be the distance from the end where it is hung
Let μ be the mass per unit length
Therefore,
dm = μ.dx
r = 0.1 cm
A = π(10)-3
M = 25 kg
L = 10 m
a) The downward force is used for calculating the mass, m = 0.25 kg
b) The maximum weight that can be hung at the lower end of the wire is calculated using the tension in the wire which is the maximum at x = L
Therefore, the mass is = 78.25 kg
A steel rod of length 2l, cross-sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod.
T(r) and T(r+dr) are the tensions that act as an external force to the rod at points A and B.
T(r+dr)-T(r) is the centrifugal force on the element due to the tension difference
dr = dm.rω2 is the centripetal force due to rotation on the element
When the below equation is integrated,
We get the extension in the rod from both sides as:
An equilateral triangle ABC is formed by two Cu rods AB and BC, and one Al rod. It is heated in such a way that the temperature of each rod increases by ∆T. Find the change in the angle ABC.
Trigonometry is used for determining the change in the angle ABC
When the above equation is differentiated, the change in the angle ABC is given as
In nature, the failure of structural members usually results from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling, and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis if the tree is given by
. Y is the Young’s modulus, r is the radius of the trunk, and R is the radius of curvature of the bent surface along the height of the trees containing the centre of gravity. Estimate the critical height of the trees for a given radius of the trunk.
Using Pythagoras theorem,
R2 = (R – d)2 + (h/2)2
Solving the equation, we get d as
d = h2/8R
W0 is the weight of the trunk per unit volume
The weight of the trunk is = (Volume)(W0)
Substituting the values for volume and weight, we can calculate the critical height, h as
h = [2Y/W0]1/3 r2/3
A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at point P. Initially, the stone is at the same level as point P. The stone is dropped vertically from point P.
a) find the distance y from the top when the mass comes to rest for an instant, for the first time
b) what is the maximum velocity attained by the stone in this drop?
c) what shall be the nature of the motion after the stone has reached its lowest point?
PE of stone = mechanical energy of string
mgy = ½ K(y-L)2
Where K is the spring constant
L is the length of the string
Solving the above equation, we get
Ky2 – 2(KL + mg)y + KL2 = 0
Using the above equation, we can determine the distance y from the top when the mass comes to rest for an instant
D = b2 – 4ac
Where a = K, b = -2(KL +mg) and c = KL2
Substituting the values we get
Therefore, y is solved as:
b) When the velocity is maximum, acceleration is zero, and therefore the force is also zero.
Let v be the maximum velocity of the stone.
The maximum velocity attained by the stone is determined with the help of the law of conservation of energy
PE lost by the stone from p to Q’ = KE of the stone + PE gained by the string
mg (L + x) = 1/2 mv2 + 1/2 Kx2
Solving the above equation gives maximum velocity v as
c) The nature of the motion after the stone has reached its lowest point is z0
z0 = [L + mg/K]