The distance of centre of mass from end A of a one dimensional rod (AB) having mass density \rho=\rho_0(1-\fracx^2L^2) kg / m and length L (in meter) is \frac3 L\alpha m . The value of \alpha is \underline\hspace2cm . (where x is the distance from end A)

Center of Mass PYQ — JEE Main 2022 · 28 Jul · Shift 2 · Q58 with Solution

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Correct answer: 8

Given mass density

The rod extends from end $A$ at $x=0$ to end $B$ at $x=L$ .

Its linear mass density is

ho(x)=\rho_0\left(1-\frac{x^2}{L^2}\right).$$ We need the position of the centre of mass from end $A$. --- 2. **Formula for centre of mass of a rod with variable density** The coordinate of centre of mass is $$x_{\text{cm}}=\frac{\int_0^L x\,dm}{\int_0^L dm}.$$ Since $$dm=\rho(x)\,dx,$$ we get $$x_{\text{cm}}=\frac{\int_0^L x\rho(x)\,dx}{\int_0^L \rho(x)\,dx}.$$ --- 3. **Calculate total mass** $$M=\int_0^L \rho_0\left(1-\frac{x^2}{L^2}\right)dx$$ $$M=\rho_0\int_0^L \left(1-\frac{x^2}{L^2}\right)dx$$ $$M=\rho_0\left[ x-\frac{x^3}{3L^2} \right]_0^L$$ $$M=\rho_0\left(L-\frac{L^3}{3L^2}\right)$$ $$M=\rho_0\left(L-\frac{L}{3}\right)=\rho_0\cdot \frac{2L}{3}.$$ --- 4. **Calculate numerator for centre of mass** $$\int_0^L x\rho(x)\,dx=\rho_0\int_0^L x\left(1-\frac{x^2}{L^2}\right)dx$$ $$=\rho_0\int_0^L \left(x-\frac{x^3}{L^2}\right)dx$$ $$=\rho_0\left[\frac{x^2}{2}-\frac{x^4}{4L^2}\right]_0^L$$ $$=\rho_0\left(\frac{L^2}{2}-\frac{L^4}{4L^2}\right)$$ $$=\rho_0\left(\frac{L^2}{2}-\frac{L^2}{4}\right)=\rho_0\cdot \frac{L^2}{4}.$$ --- 5. **Compute $x_{\text{cm}}$** $$x_{\text{cm}}=\frac{\rho_0\frac{L^2}{4}}{\rho_0\frac{2L}{3}}$$ Cancel $\rho_0$: $$x_{\text{cm}}=\frac{L^2}{4}\cdot \frac{3}{2L}$$ $$x_{\text{cm}}=\frac{3L}{8}.$$ --- 6. **Compare with given form** Given, $$x_{\text{cm}}=\frac{3L}{\alpha}.$$ So, $$\frac{3L}{\alpha}=\frac{3L}{8} \implies \alpha=8.$$ --- 7. **Final answer** $$\boxed{\alpha=8}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: 8

Given mass density

The rod extends from end $A$ at $x=0$ to end $B$ at $x=L$ .

Its linear mass density is

ho(x)=\rho_0\left(1-\frac{x^2}{L^2}\right).$$ We need the position of the centre of mass from end $A$. --- 2. **Formula for centre of mass of a rod with variable density** The coordinate of centre of mass is $$x_{\text{cm}}=\frac{\int_0^L x\,dm}{\int_0^L dm}.$$ Since $$dm=\rho(x)\,dx,$$ we get $$x_{\text{cm}}=\frac{\int_0^L x\rho(x)\,dx}{\int_0^L \rho(x)\,dx}.$$ --- 3. **Calculate total mass** $$M=\int_0^L \rho_0\left(1-\frac{x^2}{L^2}\right)dx$$ $$M=\rho_0\int_0^L \left(1-\frac{x^2}{L^2}\right)dx$$ $$M=\rho_0\left[ x-\frac{x^3}{3L^2} \right]_0^L$$ $$M=\rho_0\left(L-\frac{L^3}{3L^2}\right)$$ $$M=\rho_0\left(L-\frac{L}{3}\right)=\rho_0\cdot \frac{2L}{3}.$$ --- 4. **Calculate numerator for centre of mass** $$\int_0^L x\rho(x)\,dx=\rho_0\int_0^L x\left(1-\frac{x^2}{L^2}\right)dx$$ $$=\rho_0\int_0^L \left(x-\frac{x^3}{L^2}\right)dx$$ $$=\rho_0\left[\frac{x^2}{2}-\frac{x^4}{4L^2}\right]_0^L$$ $$=\rho_0\left(\frac{L^2}{2}-\frac{L^4}{4L^2}\right)$$ $$=\rho_0\left(\frac{L^2}{2}-\frac{L^2}{4}\right)=\rho_0\cdot \frac{L^2}{4}.$$ --- 5. **Compute $x_{\text{cm}}$** $$x_{\text{cm}}=\frac{\rho_0\frac{L^2}{4}}{\rho_0\frac{2L}{3}}$$ Cancel $\rho_0$: $$x_{\text{cm}}=\frac{L^2}{4}\cdot \frac{3}{2L}$$ $$x_{\text{cm}}=\frac{3L}{8}.$$ --- 6. **Compare with given form** Given, $$x_{\text{cm}}=\frac{3L}{\alpha}.$$ So, $$\frac{3L}{\alpha}=\frac{3L}{8} \implies \alpha=8.$$ --- 7. **Final answer** $$\boxed{\alpha=8}$$ This matches the stored correct answer.

Center of Mass question

Center of Mass question

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Center of Mass question

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