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Byju's Answer

Standard XII

Physics

Second Law of Motion

Two masses ...

Question

Two masses $M_{1}$ and $M_{2}$ are attached to the ends of a string which passes over a attached to the of a double inclined plane of angles of inclination $α$ and $β$ . If $M_{2} > M_{1}$ the acceleration a of the system can be given by (neglect friction)

\frac{M_{2} g (sin β)}{M_{1} + M_{2}}

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\frac{M_{1} s sin α)}{M_{g} + M_{2}}

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(\frac{M_{2} sin β - M_{1} sin α}{M_{1} + M_{2}}) g

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Solution

The correct option is C $(\frac{M_{2} sin β - M_{1} sin α}{M_{1} + M_{2}}) g$
$\begin{matrix} a c c = \frac{F}{(m_{1} + m_{2})} a c c = (\frac{m_{2} sin β - m_{1} sin α}{m_{1} + m_{2}}) g \end{matrix}$
Hence,
option $C$ is correct answer.

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Two masses M1 and M2 are attached to the ends of a string which passes over a attached to the of a double inclined plane of angles of inclination α and β. If M2>M1 the acceleration a of the system can be given by (neglect friction)

The correct option is C (M2sinβ−M1sinαM1+M2)gacc=F(m1+m2)acc=(m2sinβ−m1sinαm1+m2)gHence,option C is correct answer.

Two masses $M_{1}$ and $M_{2}$ are attached to the ends of a string which passes over a attached to the of a double inclined plane of angles of inclination $α$ and $β$ . If $M_{2} > M_{1}$ the acceleration a of the system can be given by (neglect friction)

The correct option is C $(\frac{M_{2} sin β - M_{1} sin α}{M_{1} + M_{2}}) g$
$\begin{matrix} a c c = \frac{F}{(m_{1} + m_{2})} a c c = (\frac{m_{2} sin β - m_{1} sin α}{m_{1} + m_{2}}) g \end{matrix}$
Hence,
option $C$ is correct answer.