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A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are :

  • Option 1)

    m, x

     

     

     

  • Option 2)

    m,\frac{1}{x}

  • Option 3)

    m,\frac{1}{x^{2}}

  • Option 4)

    m,{x^{2}}

 

Answers (2)

As we learned

 

Torque -

\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}   

 

- wherein

This can be calculated by using either  \tau=r_{1}F\; or\; \tau=r\cdot F_{1}

r_{1} = perpendicular distance from origin to the line of force.

F_{1} = component of force perpendicular to line joining force.

 

 

 

   Balancing Torque w.r.t point of suspension

mgx = Mg\left ( \frac{l}{2}-x \right )

\Rightarrow mx = M\frac{l}{2}-Mx \Rightarrow m = \frac{Ml}{2}\cdot \frac{1}{x}-M

This represents a straight line


Option 1)

m, x

 

 

 

Option 2)

m,\frac{1}{x}

Option 3)

m,\frac{1}{x^{2}}

Option 4)

m,{x^{2}}

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