Here is a question involving Young’s modulus. It’s a simple question, but there is a good chance of committing a mistake if you are not careful enough.

The expression for the depression (sag) ‘δ’ of a centrally loaded beam is,

δ = (WL

In the above problem, no separate load is suspended from the centre of the beam, but the weight of the beam acts through its centre to depress it. When a beam of twice the thickness is used, its weight appearing in the numerator is doubled and the value of d

Let us consider another question which also may baffle you slightly:

l = FL/YA. Here F = ALρg, which is the weight of the rope. You have to substitute the length L as it is, in the expression for F. But the weight of the rope acts through its centre of gravity and hence you have to substitute L/2 instead of L in the expression for the increase in length. The answer therefore is (ρgL

**A wooden plank, used as a bridge over a canal sags by 10mm due to its own weight. If the thickness of the wooden plank is doubled, the sag will be**

(a) 5mm (b) 2.5mm (c) 1.25mm (d) 10mm (e) 20mm(a) 5mm (b) 2.5mm (c) 1.25mm (d) 10mm (e) 20mm

The expression for the depression (sag) ‘δ’ of a centrally loaded beam is,

δ = (WL

^{3})/4Ybd^{3}where W is the load (W=mg), L is the length of the beam, Y is the Young’s modulus, ‘b’ is the breadth of the beam and ‘d’ is its thickness(depth).In the above problem, no separate load is suspended from the centre of the beam, but the weight of the beam acts through its centre to depress it. When a beam of twice the thickness is used, its weight appearing in the numerator is doubled and the value of d

^{3}appearing in the denominator becomes 8 times. Therefore, the depression is one-fourth, equal to 2.5 mm [option (b)].Let us consider another question which also may baffle you slightly:

**A thick rope of length L is hanging from the ceiling of a room. If Y and ρ are respectively the Young’s modulus and density of its material, its increase in length due to its own weight is**

(a) (ρgL

Since the Young’s modulus is given by the usual expression,Y = FL/Al where L is the original length, ‘l’ is the increase in length, F is the elongating force and A is the area of cross section, we have,(a) (ρgL

^{2})/2Y (b) (ρgL^{2})/Y (c) (ρgL)/Y (d) Y/ρgL^{2}(e) data insufficient.l = FL/YA. Here F = ALρg, which is the weight of the rope. You have to substitute the length L as it is, in the expression for F. But the weight of the rope acts through its centre of gravity and hence you have to substitute L/2 instead of L in the expression for the increase in length. The answer therefore is (ρgL

^{2})/2Y [option (a)].
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