Question:
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 :
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 :
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Take torques about the suspension point X: the hanging mass balances the weight of the rod acting at its centre. Rearrange to the form \(m=\) constant \(\times\frac1x+\) constant.
Updated On: Nov 11, 2026
- \(m,\ x\)
- \(m,\ \frac{1}{x}\)
- \(m,\ \frac{1}{x^2}\)
- \(m,\ x^2\)
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The Correct Option is B
Solution and Explanation
Let the rod have mass \(M\) and length \(L\); its weight acts at the centre, at distance \(\frac L2-x\) from X on the side of B.
Torque balance about X: \(mg\,x=Mg\left(\frac L2-x\right)\).
\(mx=\frac{ML}{2}-Mx\), so \(m=\frac{ML}{2}\cdot\frac1x-M\).
This is of the form \(y=c\,X-M\) with \(y=m\) and \(X=\frac1x\), a straight line.
Answer: (2) \(m,\ \frac1x\)
Torque balance about X: \(mg\,x=Mg\left(\frac L2-x\right)\).
\(mx=\frac{ML}{2}-Mx\), so \(m=\frac{ML}{2}\cdot\frac1x-M\).
This is of the form \(y=c\,X-M\) with \(y=m\) and \(X=\frac1x\), a straight line.
Answer: (2) \(m,\ \frac1x\)
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