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 xa b (where \sigma_0 is a constant), would be \underline\hspace2cm .

Center of Mass PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q54 with Solution

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Correct answer: B

Set up the plate and density

Take the rectangular plate to occupy $0 \le x \le a, \qquad 0 \le y \le b.$

Given surface mass density: $\sigma(x)=\frac{\sigma_0 x}{ab}.$

So the density depends only on $x$ , not on $y$ .

Find the total mass

A small area element is $dA=dx\,dy,$ so $dm=\sigma(x)\,dA=\frac{\sigma_0 x}{ab}\,dx\,dy.$

Hence total mass, $M=\int_0^a\int_0^b \frac{\sigma_0 x}{ab}\,dy\,dx.$

First integrate over $y$ :

=\frac{\sigma_0}{a}\int_0^a x\,dx.$$ $$M=\frac{\sigma_0}{a}\left[\frac{x^2}{2}\right]_0^a =\frac{\sigma_0}{a}\cdot \frac{a^2}{2} =\frac{\sigma_0 a}{2}.$$ --- 3. **Find the $x$-coordinate of center of mass** $$x_{\text{cm}}=\frac{1}{M}\int x\,dm =\frac{1}{M}\int_0^a\int_0^b x\cdot \frac{\sigma_0 x}{ab}\,dy\,dx.$$ So, $$x_{\text{cm}}=\frac{1}{M}\int_0^a\int_0^b \frac{\sigma_0 x^2}{ab}\,dy\,dx.$$ Integrating over $y$: $$x_{\text{cm}}=\frac{1}{M}\int_0^a \frac{\sigma_0 x^2}{ab}(b)\,dx =\frac{1}{M}\frac{\sigma_0}{a}\int_0^a x^2\,dx.$$ $$x_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{a}\left[\frac{x^3}{3}\right]_0^a =\frac{1}{M}\frac{\sigma_0}{a}\cdot \frac{a^3}{3} =\frac{1}{M}\frac{\sigma_0 a^2}{3}.$$ Using $M=\frac{\sigma_0 a}{2}$, $$x_{\text{cm}}=\frac{\sigma_0 a^2/3}{\sigma_0 a/2} =\frac{2a}{3}.$$ --- 4. **Find the $y$-coordinate of center of mass** $$y_{\text{cm}}=\frac{1}{M}\int y\,dm =\frac{1}{M}\int_0^a\int_0^b y\cdot \frac{\sigma_0 x}{ab}\,dy\,dx.$$ Separate the integrals: $$y_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{ab}\left(\int_0^a x\,dx\right)\left(\int_0^b y\,dy\right).$$ Now, $$\int_0^a x\,dx=\frac{a^2}{2}, \qquad \int_0^b y\,dy=\frac{b^2}{2}.$$ Thus, $$y_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{ab}\cdot \frac{a^2}{2}\cdot \frac{b^2}{2} =\frac{1}{M}\cdot \frac{\sigma_0 ab}{4}.$$ Using $M=\frac{\sigma_0 a}{2}$, $$y_{\text{cm}}=\frac{\sigma_0 ab/4}{\sigma_0 a/2}=\frac{b}{2}.$$ This also makes sense physically because density varies only along $x$, so the distribution remains symmetric about the horizontal midline. --- 5. **Final answer** Therefore, the center of mass is $$\boxed{\left(\frac{2a}{3},\frac{b}{2}\right)}.$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Set up the plate and density

Take the rectangular plate to occupy $0 \le x \le a, \qquad 0 \le y \le b.$

Given surface mass density: $\sigma(x)=\frac{\sigma_0 x}{ab}.$

So the density depends only on $x$ , not on $y$ .

Find the total mass

A small area element is $dA=dx\,dy,$ so $dm=\sigma(x)\,dA=\frac{\sigma_0 x}{ab}\,dx\,dy.$

Hence total mass, $M=\int_0^a\int_0^b \frac{\sigma_0 x}{ab}\,dy\,dx.$

First integrate over $y$ :

=\frac{\sigma_0}{a}\int_0^a x\,dx.$$ $$M=\frac{\sigma_0}{a}\left[\frac{x^2}{2}\right]_0^a =\frac{\sigma_0}{a}\cdot \frac{a^2}{2} =\frac{\sigma_0 a}{2}.$$ --- 3. **Find the $x$-coordinate of center of mass** $$x_{\text{cm}}=\frac{1}{M}\int x\,dm =\frac{1}{M}\int_0^a\int_0^b x\cdot \frac{\sigma_0 x}{ab}\,dy\,dx.$$ So, $$x_{\text{cm}}=\frac{1}{M}\int_0^a\int_0^b \frac{\sigma_0 x^2}{ab}\,dy\,dx.$$ Integrating over $y$: $$x_{\text{cm}}=\frac{1}{M}\int_0^a \frac{\sigma_0 x^2}{ab}(b)\,dx =\frac{1}{M}\frac{\sigma_0}{a}\int_0^a x^2\,dx.$$ $$x_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{a}\left[\frac{x^3}{3}\right]_0^a =\frac{1}{M}\frac{\sigma_0}{a}\cdot \frac{a^3}{3} =\frac{1}{M}\frac{\sigma_0 a^2}{3}.$$ Using $M=\frac{\sigma_0 a}{2}$, $$x_{\text{cm}}=\frac{\sigma_0 a^2/3}{\sigma_0 a/2} =\frac{2a}{3}.$$ --- 4. **Find the $y$-coordinate of center of mass** $$y_{\text{cm}}=\frac{1}{M}\int y\,dm =\frac{1}{M}\int_0^a\int_0^b y\cdot \frac{\sigma_0 x}{ab}\,dy\,dx.$$ Separate the integrals: $$y_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{ab}\left(\int_0^a x\,dx\right)\left(\int_0^b y\,dy\right).$$ Now, $$\int_0^a x\,dx=\frac{a^2}{2}, \qquad \int_0^b y\,dy=\frac{b^2}{2}.$$ Thus, $$y_{\text{cm}}=\frac{1}{M}\frac{\sigma_0}{ab}\cdot \frac{a^2}{2}\cdot \frac{b^2}{2} =\frac{1}{M}\cdot \frac{\sigma_0 ab}{4}.$$ Using $M=\frac{\sigma_0 a}{2}$, $$y_{\text{cm}}=\frac{\sigma_0 ab/4}{\sigma_0 a/2}=\frac{b}{2}.$$ This also makes sense physically because density varies only along $x$, so the distribution remains symmetric about the horizontal midline. --- 5. **Final answer** Therefore, the center of mass is $$\boxed{\left(\frac{2a}{3},\frac{b}{2}\right)}.$$ So the correct option is **B**.

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