Question

pls explain clearly​

Answer 1

pls refer the attachment

Answer 2

Answer:

$\boxed{(2) \: \frac{(3 \alpha + 2 \beta L)L}{3(2 \alpha + \beta L)}}$

Given:

$Linear \: mass \: density \: (\lambda) = \alpha + \beta x$

Explanation:

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Consider a small element of length dx having small mass dm.

$dm = \lambda dx \\ \: \: \: \: \: \: \: \: \: = (\alpha + \beta x)dx$

$Position \: of \: center \: of \: mass, \\ x_{cm} = \frac{ \int\ {x} \, dm }{ \int\ \,dm } \\ \\ \: \: \: \: \: \: \: \: \: = \frac{\int\limits^L_0 {x( \alpha + \beta x)} \, dx }{\int\limits^L_0 {( \alpha + \beta x)} \, dx } \\ \\ \: \: \: \: \: \: \: \: \: = \frac{\int\limits^L_0 { \alpha x } \, dx + \int\limits^L_0 { \beta {x}^{2} } \, dx }{\int\limits^L_0 { \alpha } \, dx + \int\limits^L_0 { \beta x } \, dx } \\ \\ \: \: \: \: \: \: \: \: \: = \frac{( \frac{ \alpha {L}^{2} }{2} + \frac{ \beta {L}^{3} }{3} )}{(\alpha L + \frac{ \beta {L}^{2} }{2} )} \\ \\ \: \: \: \: \: \: \: \: \: = \frac{{L}^{ \cancel{2}} ( \frac{ \alpha}{2} + \frac{ \beta L}{3}) }{ \cancel{L}( \alpha + \frac{ \beta L}{2} )} \\ \\ \: \: \: \: \: \: \: \: \: = \frac{L ( \frac{ \alpha}{2} + \frac{ \beta L}{3}) }{ ( \alpha + \frac{ \beta L}{2} )} \\ \\ \: \: \: \: \: \: \: \: \: \: = \frac{L (\frac{3 \alpha + 2 \beta L}{6} )}{ (\frac{2 \alpha + \beta L}{2} )} \\ \\ x_{cm} = \frac{L(3 \alpha + 2 \beta L)}{3(2 \alpha + \beta L)}$