QJEEM25Ph061f: JEE Main 2025 Physics - Class 11 JEEM25Ph061f Solution: The center of mass of a thin rectangular plate (fig - x ) with…

$\sigma = {{{\sigma _0}x} \over {ab}}$

Let's consider a strip of infinitesimal width dx at x distance away from O.

As $\sigma$ depends only on x i.e. it's constant for a specific value of x so centre of mass of the strip will be $\left( {x,{b \over 2}} \right)$ .

Now, we know,

$COM = \left( {{{\int {xdm} } \over {\int {dm} }},{{\int {ydm} } \over {\int {dm} }}} \right)$

$= \left( {{{\int {x\sigma dA} } \over {\int {\sigma dA} }},{{\int {{b \over 2}\sigma dA} } \over {\int {\sigma dA} }}} \right)$

$= \left( {{{\int {xbdx{{{\sigma _0}x} \over {ab}}} } \over {\int {{{{\sigma _0}x} \over {ab}}bdx} }},{{{b \over 2}\int {\sigma dA} } \over {\int {\sigma dA} }}} \right)$

For whole plate, x varies from o to a.

So, $= \overleftarrow C OM = \left( {{{{{{\sigma _0}} \over a}\int_0^a {{x^2}dx} } \over {{{{\sigma _0}} \over a}\int_0^a {xdx} }},{b \over 2}} \right)$

$= \left( {{{{1 \over 3}\left[ {{x^3}} \right]_0^a} \over {{1 \over 2}\left[ {{x^2}} \right]_0^a}},{b \over 2}} \right)$

$= \left( {{2 \over 3}{{{a^3}} \over {{a^2}}},{b \over 2}} \right)$

$\Rightarrow COM = \left( {{2 \over 3}a,{b \over 2}} \right)$