Question

A wire ‘AB' is suspended about the end A.
Marks C and D on it are 40 cm apart. When
a load is suspended from B, then the marks
C and D are displaced by 0.3 mm and 0.5 mm
respectively. The original length of AD is
1) 60 cm
2) 100 cm
3) 80 cm
4) 200/3 cm​

Answer 1

Answer:

100cm

Explanation:

Firstly, THANKS for the GREAT question!!!

The tension in a rope at distance x from a rigid support for a suspended wire is given by:-

$\boxed{T=mg(\frac{l-x}{l})}---(1)$

For reference lookup : https://brainly.in/question/3081197

Note: x is the distance from top of wire (or point of suspension)

Let the length of wire be $l$ and D be a point at distance $a$ from the top, therefore C is at a distance of $(a+40)$ from the point of suspension

Here let the load suspended be given by Mg,

So now the equation (1) can be written as:

$T=(Mg+mg)(\frac{l-x}{l} )$

Tension at D is given by:

$T=(Mg+mg)(\frac{l-a}{l} )$

Tension at C is given by:

$T'=(Mg+mg)(\frac{l-(a+40)}{l} )\\\\T'=(Mg+mg)(\frac{l-40-a}{l} )$

Now using Hooke's law that states that

Stress α Strain

i.e  $\frac{F}{A} =k(\frac{del(l)}{l} )$

Here length of point c and d is taken to be dl

Applying the above result for point D and C;

$T=kA\frac{0.05}{dl}$

$\boxed{(Mg+mg)(\frac{l-a}{l} )=kA\frac{0.05}{dl}} ---(A)$

$T'=kA\frac{0.03}{dl}$

$\boxed{(Mg+mg)(\frac{l-40-a}{l} )=kA\frac{0.03}{dl}} ---(B)$

Now finally, dividing $\frac{A}{B}$ :-

$\frac{l-a}{l-40-a}=\frac{5}{3}$

Put (l-a)=y

$\frac{y}{y-40}=\frac{5}{3}$

On solving this, you get

y=100cm

Hope this answer helped you