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 (Yal^{2}/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 W_{1}/W_{2} =_{ }l_{1}/l_{2 }where l_{1} and l_{2} are the increases in length of the wires. But l_{1} = FL_{1}/A_{1}Y and l_{2} = FL_{2}/A_{2}Y so that

W_{1}/W_{2} = l_{1}/l_{2} = (L_{1}/L_{2})×(A_{2}/A_{1}) = (3/1)×(1/4) = 3/4 [Option (b)].

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

**same material and length but of area 8 mm ^{2} 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} m^{2}/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} m^{3} = 0.75 ml.

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

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