Question

Statement 1: The length of an elastic string of initial length L is ‘a’ meter when the tension is 4N and ‘b’ meter when the tension is 5N. The length of the string (in meter) when tension is ‘9N’ is 4b-5a because:
Statement 2: The extension of an elastic string is proportional to the initial length of the string, if the deforming force is constant (neglect lateral strain).
A. Both assertion and reason are correct and reason is the correct explanation for assertion.
B. Both assertion and reason are correct but reason is not the correct explanation for assertion.
C. Assertion is correct but reason is incorrect.
D. Both assertion and reason are incorrect.

Answer 1

Hint: Practically no material is rigid (a body in between any two particles of which the distances can’t be changed). Every body gives some deformation when subjected to a force. Hooke’s law gives the relation between the deformation of body and force applied.
Formula used:
$Y =\dfrac{stress}{strain} = \dfrac{ F/A}{ {\Delta l}/{l}} = \dfrac{Fl}{A\Delta l}$

Complete answer:
Hooke’s law states that if some stress is applied to a material of young modulus ‘Y’, a strain =$\dfrac{stress}{Y}$is produced in it.
Stress: Tension acting per unit area of the cross section of the wire is called stress.
Strain: The ratio of change in dimension to the original dimension is called strain.
Now, let the young’s modulus of the wire be ‘Y’ and a constant stress ‘$\sigma$’ be applied on the body. Since stress is constant hence force is also constant.
Hence $strain = \dfrac{Y}{\sigma}=\dfrac{\Delta l}{l}$
Since both ‘Y’ and ‘$\sigma$’ both are constant, thus strain is also constant.
Hence, $\Delta l \propto l$, statement 2 is correct.
Now, when tension is 4N, length is ‘a’:
$Y = \dfrac{Fl}{A\Delta l}$
$Y = \dfrac{4L}{A(a-L)}$
Or $\dfrac{L}{AY} = \dfrac{(a-L)}{4}$ . . . . ①
Also, when tension is 5N, length is ‘b’:
$Y = \dfrac{5L}{A(b-L)}$
 $\dfrac{L}{AY} = \dfrac{(b-L)}{5}$ . . . . ②
Equating ① and ②, we get:
$\dfrac{L-b}{5} = \dfrac{L-a}{4}$
$4L - 4b = 5L -5a$
Or, $L = 5a-4b$
Also, when tension is 9N, let length is ‘x’:
$Y = \dfrac{9L}{A(x-L)}$
$\dfrac{YA}{L} = \dfrac{9}{x-L}$
$\dfrac{x-L}{9} = \dfrac{a-L}{4}$ [From ①]
Or $4x = 9a-9L+4L = 9a-5L$
Putting the value of $L = 5a-4b$, we get;
$4x = 9a-5(5a-4b) = 20b-16a$
Or $x =5b-4a$
Hence statement 1 is also correct.
Since we’ve used information given in statement 2 to get statement 1 correct, thus statement 2 is the correct explanation of statement 1 as well.

So, the correct answer is “Option A”.

Note:
Any material behaves like elastic material up to a certain limit called the elastic limit. After this limit, the material goes on permanent deformation and this behavior of material is called plastic behavior. This suggests that after complete removal of load, the length of string won’t come back to its original length.