Multiple Choice Questions on Elasticity

Here are three questions (MCQ) on elasticity which appeared in Kerala Engineering and Medical Entrance 2004 Examination question papers:

(1) Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, the ratio of energy stored in A to that in B is

(a) 2:3 (b) 3:4 (c) 3:2 (d) 6:1 (e) 12:1

The work (W) done in increasing the length of a wire or rod by ‘l’ by applying a force ‘F’ is given by

W = ½ Fl

[ Here is the proof for the above: The work dW done for increasing the length by dl is

F×dl. The total work done for increasing the length by ‘l’ is ∫F×dl where the limits of

integration are zero and ‘l’. Since the Young’s modulus, Y = (F/A)(L/l) where A is the

area of cross section and L is the length of the wire, we have F = YAl/L. The total work

done is therefore ₀ ∫^l (Yal/L)dl = ½ (Yal²/L) = ½ (YAl/L)×l = ½ Fl ]

The energy stored (which is equal to the work done) in a wire is therefore directly proportional to the increase in length. The ratio of energy stored is therefore W₁/W₂ = l₁/l₂ where l₁ and l₂ are the increases in length of the wires. But l₁ = FL₁/A₁Y and l₂ = FL₂/A₂Y so that

W₁/W₂ = l₁/l₂ = (L₁/L₂)×(A₂/A₁) = (3/1)×(1/4) = 3/4 [Option (b)].

(2) A wire of cross section 4 mm² is stretched by 0.1 mm. How far will a wire of the

same material and length but of area 8 mm² stretch under the action of the same

force?

(a) 0.05 mm (b) 0.01 mm (c) 0.15 mm

(d) 0.2 mm (e) 0.25 mm

This question as well as the previous one appeared in Kerala Medical Entrance 2004 question paper.

Since the increase in length is inversely proportional to the area of cross section of the wire, the correct option is 0.05 mm.

(3) Compressibility of water is 5×10^–10 m²/N. The change in volume of 100 ml of water subjected to 15×10^–6 Pa pressure will be

(a) no change (b) increase by 0.75 ml (c) increase by 1.5 ml

(d) decrease by 1.5 ml (e) decrease by 0.75 ml

This question appeared in Kerala Engineering Entrance 2004 question paper. You must definitely remember the expression for bulk modulus ‘B’ as

B = –P/(dV/V) where P is the pressure which produces a change in volume dV in a volume V. The negative sign indicates that an increase in pressure will produce a decrease in volume.

Compressibility is the reciprocal of bulk modulus. Therefore we have

1/(5×10^–10) = (15×10^–10 × 100×10^–6)/dV, from which

dV = 0.75×10^–6 m³ = 0.75 ml.

Since an increase in pressure produces a decrease in volume, the correct option is (e).

Questions on Elasticity

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.
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
The expression for the depression (sag) ‘δ’ of a centrally loaded beam is,
δ = (WL³)/4Ybd³ 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 d³ 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) (ρgL²)/2Y (b) (ρgL²)/Y (c) (ρgL)/Y (d) Y/ρgL² (e) data insufficient.
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,
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 (ρgL²)/2Y [option (a)].