The length of wire, when M1 is hung from it, is l1 and is l2 with both M1 and M2 hanging. The natural length of wire is

Question

The length of wire, when M1 is hung from it, is l1 and is l2 with both M1 and M2 hanging. The natural length of wire is (A) (M1/M2)(l1-l2)+l1 (B) (M2 l1-M1 l2/M1+M2) (C) (l1+l2/2) (D) √l1 l2

Tardigrade Admin · Accepted Answer

Let the natural length of wire be =l When only M1 hanging Using Δ l=(F L/A Y) (l1-l)=(M1 g ⋅ l/A Y) ... (1) When both M1, M2 hanging (l2-l)=((M1+M2) g ⋅ l/A Y) ldots (2) Dividing (1) by (2) (l1-l/l2-l)=(M1/M1+M2) Solving this we get l=(M1/M2)(l1-l2)+l1

Q. The length of wire, when $M_{1}$ is hung from it, is $l_{1}$ and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is

$\frac{M _{1}}{M _{2}} (l_{1} - l_{2}) + l_{1}$

$\frac{M _{2} l _{1} - M _{1} l _{2}}{M _{1} + M _{2}}$

$\frac{l _{1} + l _{2}}{2}$

$l_{1} l_{2}$

Q. The length of wire, when M1​ is hung from it, is l1​ and is l2​ with both M1​ and M2​ hanging. The natural length of wire is

M2​M1​​(l1​−l2​)+l1​

M1​+M2​M2​l1​−M1​l2​​

2l1​+l2​​

l1​l2​​

Q. The length of wire, when $M_{1}$ is hung from it, is $l_{1}$ and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is

$\frac{M _{1}}{M _{2}} (l_{1} - l_{2}) + l_{1}$

$\frac{M _{2} l _{1} - M _{1} l _{2}}{M _{1} + M _{2}}$

$\frac{l _{1} + l _{2}}{2}$

$l_{1} l_{2}$