TS EAMCET · Physics · Rotational Motion
A thin uniform square lamina of side \(a\) is placed in the \(x y\)-plane with its sides parallel to \(X\) and \(Y\)-axes and with its centre coinciding with origin. Its moment of inertia about an axis passing through a point on the \(Y\)-axis at a distance \(Y=2 a\) and parallel to \(X\)-axis is equal to its moment of inertia about an axis passing through a point on the \(X\)-axis at a distance \(x=d\) and perpendicular to \(x y\)-plane. Then value of \(d\) is
- A \(\frac{7}{3} a\)
- B \(\frac{\sqrt{47}}{12} a\)
- C \(\frac{9}{5} a\)
- D \(\frac{\sqrt{51}}{12} a\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{47}}{12} a\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{rlrl}\frac{m a^2}{12}+m\left(4 a^2\right) & =\frac{m a^2}{6}+m d^2 \\ & \text { or } \quad d^2 & =\frac{47 a^2}{12} \\ \therefore & d & =\sqrt{\frac{47}{12}} a\end{array}\)
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