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Question: Which of the following are correct? This question has multiple correct options A. The product of...

Which of the following are correct?
This question has multiple correct options
A. The product of bulk modulus of elasticity and compressibility is 11
B. A rope 1cm1\,cm in diameter breaks if the tension in it exceeds 500N500\,N. The maximum tension that may be given to a similar rope of diameter 2cm2\,cm is 2000N2000\,N
C. According to Hooke’s law, the ratio of the stress and strain remains constant
D. None of the above.

Explanation

Solution

In order to solve this problem,we are going to apply the concept of modulus of elasticity and also the tension produced in a body.
Formula used:
Bulk modulus=1compressibility = \dfrac{1}{{compressibility}} and
Stressstrain  =Modulus of elasticity\dfrac{{Stress}}{{strain\;}} = Modulus{\text{ }}of{\text{ }}elasticity= constant (Hooke’s law)

Complete step by step answer:
let us take the first statement (A), it says that the product of bulk modulus and compressibility is 11.
We know that compressibility is reciprocal of bulk modulus
So Bulk modulus=1compressibility = \dfrac{1}{{compressibility}}
Which means Bulk modulus ×\timesCompressibility=1 = 1.
So statement (A) is correct.
Let us take second statement
If we take T as the tension
Young's modulus of the rope, Y=StressStrain=(TA)(ΔII)Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{\left( {\dfrac{T}{A}} \right)}}{{\left( {\dfrac{{\Delta I}}{I}} \right)}}
T=(Y×ΔII)×πd24T = \left( {Y \times \dfrac{{\Delta I}}{I}} \right) \times \dfrac{{\pi {d^2}}}{4}
where, d is the diameter of the rope
Tension   d2 \propto \;{d^2}
​ So we get
Tcase2Tcase1=d22d12\dfrac{{{T_{case2}}}}{{{T_{case1}}}}\, = \,\,\dfrac{{{d_2}^2}}{{{d_1}^2}}
In first case tension is 500 N500{\text{ }}N
  Tcase1=500  N  \;{T_{case1}} = 500\;N\;
d1=1  cm{d_1} = 1\;cm and d2=2  cm{d_{2}} = 2\;cm
Tcase2=  500 x 22=2000 N{T_{case2}} = \;500{\text{ }}x{\text{ }}{2^2} = 2000{\text{ }}N
So statement (B) is correct
Let us take the third statement which says that according to Hooke's law, the ratio of stress to strain remains constant. This is true according to Hooke’s law.
StressStrain=constant=\dfrac{{Stress}}{{Strain}} = cons\tan t = Modulus of elasticity or coefficient of elasticity.
There are three moduli of elasticity corresponding to three types of strain: Young’s modulus, Bulk Modulus and modulus of rigidity.
So, statement (C) is also correct.
Hence statements (A), (B) and (C) are correct.

Note: Hooke’s law holds good up to a certain limit called elastic limit. In such questions we need to find out the ratio of proportionality, we just need to find out the relation of Tension with diameter.