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♪⚝Sanjiv❀Jaiswal❁⩸

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Elasticity

Calculate the longest length of a steel wire that can hang vertically without breaking. Breaking stress for steel = 7.982 × 108 N/m2 and density for steel = 8.1 × 103 kg/m3 . (g = 9.8 m/s2 ).

Solution:
The breaking stress = 7.982 10⁸N/m²;
the density of steel,ρ= 8.1 10³kg/m³
Suppose A is the area of cross-section of the steel wire and L is the longest length of the steel wire.
We have,
the weight of the wire = mg = (volume density)g
or F = LAρg

A uniform steel wire of density 7800 kg/m3 weighs 16 g and is 250 cm long. It lengthens by 1.2 mm when stretched by a force of 80 N. Calculate (i) the value of Young's modulus for steel, (ii) the e...

A cylindrical of copper wire and a cylindrical steel wire, each of length 1.5 m and diameter 2 mm, are joined at one end to form a composite wire 3 m long. The wire is loaded until its length becom...

Solution:
The length of a copper wire, LC = 1.5 m;
the length of a steel wire, LS = 1.5 m;
the diameter of the wires, d = 2 mm = 2 10^–3m;
the composite length, L = 3 m;
the extended composite length, L' = 3.003 m;
the total extension, l = L' – L = 3.003 – 3 = 0.003 m;
the extension of the copper wire, l_C= ?;
the extension of the steel wire, l_S= ?

l_C+ l_S= 0.003 … (1)

For the copper wire, Young modulus,

Find the extension produced in a copper wire of length 2 m and diameter of 3 mm when a force of 30 N is applied. (Young's modulus for copper = 1.1 × 1011 N/m2 ).

Solution:
The extension, l = ?;
the force applied, F = 30 N;
the length of a copper wire, L = 2 m;
the diameter of the wire, d = 3 mm = 3 10^–3m
The cross-section area of the wire,

An aluminum cube of each side 4 cm is subjected to a tangential force. The top face of the cube is sheared 0.012 cm w.r.t. the bottom. Find (i) shearing strain (ii) shear stress and (iii) shearing ...

Solution:
The side of a cube, L = 4 cm = 0.04 m;
the shear length, x = 0.012 cm = 1.2 10^–4m

A solid sphere of radius 10 cm is subjected to a uniform pressure equal to 5 × 108 N/m2 . The bulk modulus of the material of the sphere is 3.14 × 1011 N/m2 . Find the change in its volume.

Solution:
The radius of a sphere, r = 10 cm = 0.10 m;
the increase in pressure, p = 5 10⁸N/m²;
the bulk modulus, B = 3.14 10¹¹N/m²;
the change in volume, V = ?

What stress would cause a wire to increase in length by one-tenth of one percent if the Young modulus for the wire is 12 × 1010 N/m2 ? What force would produce this stress if the diameter of the wi...

Solution:
The Young modulus of a wire, Y = 12 1010 N/m2 ;
the diameter of the wire, d = 0.56 mm = 0.56 10^–3 mm
The cross-section area of the wire,

The strain = one-tenth of 1 percent

Explain which one is more elastic–rubber or steel.

Suppose two wires of steel and rubber of equal length L and cross-section A are stretched by the same force F. The Young modulus for steel is steel steel and similarly that for rubber iswhere l_steelis the extension of the steel wire and l_rubberis that of the rubber wire. Thus taking the ratio, we have

As l_rubber> l_steel, Y_steel> Y_rubber. The more the value of the elastic constant, the more the object is elastic because a very large force or stress is needed to make it break. This means...

The material in human bones and elephant bones essentially the same, but an elephant has thicker legs. Explain why, in terms of breaking stress.

Breaking stress may occur when the material is excessively stretched or compressed. To withstand the heavy weight of an elephant, it has thicker legs so that the ratio of the weight to the cross-section area of the legs does not exceed the breaking stress of the bone material. To withstand the weight of a man, he can have thin legs and the ratio of his weight to the cross-section area of the leg will not exceed the breaking stress for the bone material. This is why an elephant has thicker...