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6.System of Particles and Rotational Motion
hard
A uniform rod $AB$ is suspended from a point $X$, at a variable distance from $x$ 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 variable that give a straight line, when plotted, are
A
$m,\frac{1}{x}$
B
$m,\frac{1}{x^2}$
C
$m,x$
D
$m,x^2$
(JEE MAIN-2018)
Solution
Balancing torque w.r.t point of suspension
$\begin{array}{l}
mg\,x\, = Mg\left( {\frac{{.\ell }}{2} – x} \right)\\
\Rightarrow mx = M\frac{\ell }{2} – Mx\\
m = \left( {M\frac{\ell }{2}} \right)\frac{1}{x} – M\\
y = \alpha \frac{1}{x} – C,\,\,\,\,\,\,\,\,\, straight\,\, line\,\, equation.
\end{array}$
Standard 11
Physics
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