Answer:
Abhinav be seated at "2.57 cm" far. The further explanation is given below.
Explanation:
The given values are:
Deepta's weight,
= 450 N
Abhinav's weight,
= 350 N
Deepta's distance from fulcrum,
= 2cm
i.e.,
= 0.02 m
Let Abhinav's distance be "x".
As we know,
⇒ [tex]Work \ done \ by \ Deepta = Work \ done \ by \ Abhinav[/tex]
⇒ [tex]Deepta's \ weight\times Deepta's \ distance = Abhinav's \\ weight\times Abhinav's \ distance[/tex]
⇒ [tex]450\times 0.02=350\times x[/tex]
⇒ [tex]9=350\times x[/tex]
⇒ [tex]\frac{9}{350} = x[/tex]
⇒ [tex]x=0.025 \ m[/tex]
i.e.,
⇒ [tex]2.57 \ cm[/tex]