JEE Main 2025 Jan 28 Shift 1, Physics Q19: Centre of Mass

JEE Main2025Jan 28, Shift 1Physics

The centre of mass of a thin rectangular plate with sides of length $a$ and $b$ , whose mass per unit area ( $σ$ ) varies as $σ = \frac{σ _{0} x}{ab}$ (where $σ_{0}$ is a constant), would be ______

A
$(\frac{2}{3} a, \frac{b}{2})$
B
$(\frac{2}{3} a, \frac{2}{3} b)$
C
$(\frac{a}{2}, \frac{b}{2})$
D
$(\frac{1}{3} a, \frac{b}{2})$

Solution

The surface density $σ = \frac{σ _{0} x}{ab}$ depends only on $x$ , not on $y$ . So by symmetry along $y$ , $y_{cm} = \frac{b}{2}$ .

For $x_{cm}$ , use a vertical strip of width $d x$ at position $x$ , area $b d x$ , mass $d m = σ \cdot b d x = \frac{σ _{0} x}{ab} \cdot b d x = \frac{σ _{0} x}{a} d x$ .

$x_{cm} = \frac{\int _{0}^{a} x d m}{\int _{0}^{a} d m} = \frac{\int _{0}^{a} x ^{2} d x}{\int _{0}^{a} x d x} = \frac{a ^{3} /3}{a ^{2} /2} = \frac{2 a}{3}$ .

$∴$ Centre of mass at $(\frac{2 a}{3}, \frac{b}{2})$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q19: Centre of Mass

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