JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The center of mass of a thin rectangular plate (fig - x ) with sides of length \(a\) and \(b\), whose mass per unit area \((\sigma)\) varies as \(\sigma=\frac{\sigma_0 x}{a b}\) (where \(\sigma_0\) is a constant), would be \(\qquad\)
- A \(\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)\)
- B \(\left(\frac{a}{2}, \frac{\mathrm{~b}}{2}\right)\)
- C \(\left(\frac{1}{3} a, \frac{\mathrm{~b}}{2}\right)\)
- D \(\left(\frac{2}{3} a, \frac{2}{3} b\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & d m=\sigma d A \\ & \begin{aligned} d & =\sigma(d x)(d y)=\frac{\sigma_0 x}{a b}(d x)(d y) \\ x_{c o m} & =\frac{\int x d m}{\int d m}=\frac{\int x \frac{\left(\sigma_0 x\right)}{a b}(d x)(d y)}{\int_0 \frac{\sigma_0}{a b}(d x)(d y)} \\ & =\frac{\int_0^a x^2 d…
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