Exam

JEE Main

Year

2015

Date

11 April

Mode

Online

Subject

Physics

Chapter

Center of Mass, Momentum and Collision

JEE Main 2015 · 11 April · Online · Physics · Center of Mass, Momentum and Collision

A uniform thin rod AB of length L has linear mass density μ(x) = a...

A uniform thin rod AB of length $L$ has linear mass density $\mu\left(x\right) = a + \dfrac{bx}{L}$, where $x$ is measured from A. If the CM of the rod lies at a distance of \left(\dfrac{7}{12}L\right)$from A, then$a and b$ are related as:

Step-by-step solution

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

${x}_{cm} = \dfrac{∫ \left(dm\right)x}{∫ dm} = \dfrac{∫ \left(\lambdadx\right)x}{∫ \lambdadx}$

$= \dfrac{{∫}_{0}^{L}\left(a + \dfrac{b}{L}x\right)xdx}{{∫}_{0}^{L}\left(a + \dfrac{b}{L}x\right)dx}$

$= \dfrac{{∫}_{0}^{L}axdx + {∫}_{0}^{L}\dfrac{b}{L}{x}^{2}dx}{{∫}_{0}^{L}adx + {∫}_{0}^{L}\dfrac{b}{L}xdx}$

$= \dfrac{a{\left[\dfrac{{x}^{2}}{2}\right]}_{0}^{L} + \dfrac{b}{L}{\left[\dfrac{{x}^{3}}{3}\right]}_{0}^{L}}{a{\left[x\right]}_{0}^{L} + \dfrac{b}{L}{\left[\dfrac{{x}^{2}}{2}\right]}_{0}^{L}}$

$= \dfrac{\dfrac{a}{2}\left({L}^{2} - {0}^{2}\right) + \dfrac{b}{3L}\left({L}^{3} - {0}^{3}\right)}{a\left(L - 0\right) + \dfrac{b}{2L}\left({L}^{2} - {0}^{2}\right)}$

$= \dfrac{\dfrac{a}{2}{L}^{2} + \dfrac{b{L}^{2}}{3}}{aL + \dfrac{bL}{2}} = \dfrac{\dfrac{a}{2}L + \dfrac{bL}{3}}{a + \dfrac{b}{2}}$

$∵ {x}_{cm} = \dfrac{7L}{12} = \dfrac{\left(\dfrac{a}{2} + \dfrac{b}{3}\right)L}{a + \dfrac{b}{2}} ⇒ \dfrac{7}{12} = \dfrac{\dfrac{3a + 2b}{6}}{\dfrac{2a + b}{2}}$

$⇒ \dfrac{7}{12} = \dfrac{3a + 2b}{3} \times \dfrac{1}{\left(2a + b\right)} ⇒ 7\left(2a + b\right) = 4\left(3a + 2b\right)$

$⇒ 14a + 7b = 12a + 8b ⇒ 2a = b$