JEE Main 2015 — Physics Mechanics

Consider an infinitesimal small element of length $dx$dx and mars density $\lambda$λ

$${x}_{cm}=\frac{\int (dm)x}{\int dm}=\frac{\int (\lambda dx)x}{\int \lambda dx}$$xcm​=∫dm∫(dm)x​=∫λdx∫(λdx)x​

$$=\frac{{\int }_{0}^{L}(a+\frac{b}{L}x)xdx}{{\int }_{0}^{L}(a+\frac{b}{L}x)dx}$$=∫0L​(a+Lb​x)dx∫0L​(a+Lb​x)xdx​

$$=\frac{{\int }_{0}^{L}axdx+{\int }_{0}^{L}\frac{b}{L}{x}^{2}dx}{{\int }_{0}^{L}adx+{\int }_{0}^{L}\frac{b}{L}xdx}$$=∫0L​adx+∫0L​Lb​xdx∫0L​axdx+∫0L​Lb​x2dx​

$$=\frac{a{[\frac{{x}^{2}}{2}]}_{0}^{L}+\frac{b}{L} {[\frac{{x}^{3}}{3}]}_{0}^{L}}{a{[x]}_{0}^{L}+\frac{b}{L}{[\frac{{x}^{2}}{2}]}_{0}^{L}}$$=a[x]0L​+Lb​[2x2​]0L​a[2x2​]0L​+Lb​[3x3​]0L​​

$$=\frac{\frac{a}{2}({L}^{2}-{0}^{2})+\frac{b}{3L} ({L}^{3}-{0}^{3})}{a(L-0)+\frac{b}{2L}({L}^{2}-{0}^{2})}$$=a(L−0)+2Lb​(L2−02)2a​(L2−02)+3Lb​(L3−03)​

$$=\frac{\frac{a}{2}{L}^{2}+\frac{b{L}^{2}}{3}}{aL+\frac{bL}{2}}=\frac{\frac{a}{2}L+\frac{bL}{3}}{a+\frac{b}{2}}$$=aL+2bL​2a​L2+3bL2​​=a+2b​2a​L+3bL​​

$$\because {x}_{cm}=\frac{7L}{12}=\frac{(\frac{a}{2}+\frac{b}{3})L}{a+\frac{b}{2}}\Rightarrow \frac{7}{12}= \frac{\frac{3a+2b}{6}}{\frac{2a+b}{2}}$$∵xcm​=127L​=a+2b​(2a​+3b​)L​⇒127​=22a+b​63a+2b​​

$$\Rightarrow \frac{7}{12}=\frac{3a+2b}{3}\times \frac{1}{(2a+b)}\Rightarrow 7(2a+b)=4(3a+2b)$$⇒127​=33a+2b​×(2a+b)1​⇒7(2a+b)=4(3a+2b)

$$\Rightarrow 14a+7b=12a+8b\Rightarrow 2a=b$$⇒14a+7b=12a+8b⇒2a=b