The figure shows a pulley-mass system which is kept in an elevator moving up with acceleration g.Then, the tension T in the string is given by

\frac{m_{1} m_{2} g}{m_{1} + m_{2}}

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\frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}

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\frac{4 m_{1} m_{2} g}{m_{1} + m_{2}}

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\frac{(m_{1} m_{2}) g}{m_{1} + m_{2}}

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Solution

The correct option is C $\frac{4 m_{1} m_{2} g}{m_{1} + m_{2}}$
$\begin{matrix} \frac{a_{1} + a_{2}}{2} = g \Rightarrow a_{1} + a_{2} = 2 g . . . . . . . . (1) T - m_{1} g = m_{1} a_{1} . . . . . . . . . . . . . . (2) T - m_{2} g = m_{2} a_{2} . . . . . . . . . . . . . . . . . . (3) \Rightarrow T - m_{2} g - m_{2} [2 g - a_{1}] T - m_{2} g = 2 m_{2} g - m_{2} a_{1} \Rightarrow T = 3 m_{2} g - m_{2} [\frac{T - m_{1} g}{m_{1}}] \Rightarrow T = 3 m_{2} g - \frac{m_{2}}{m_{1}} . T + m_{2} g \Rightarrow (\frac{m_{1} + m_{2}}{m_{1}}) T = 4 m_{2} g ∴ T = \frac{4 m_{1} m_{2} g}{m_{1} + m_{2}} \end{matrix}$
$∴$ Option $C$ is correct.

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The figure shows a pulley-mass system which is kept in an elevator moving up with acceleration g.Then, the tension T in the string is given by

The correct option is C 4m1m2gm1+m2a1+a22=g⇒a1+a2=2g........(1)T−m1g=m1a1..............(2)T−m2g=m2a2..................(3)⇒T−m2g−m2[2g−a1]T−m2g=2m2g−m2a1⇒T=3m2g−m2[T−m1gm1]⇒T=3m2g−m2m1.T+m2g⇒(m1+m2m1)T=4m2g∴T=4m1m2gm1+m2∴ Option C is correct.