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Byju's Answer

Standard XII

Physics

Lenz's Law

Two masses ...

Question

Two masses $M_{1}$ and $M_{2}$ are attached to the ends of a light string, which passes over a massless pulley, attached to the top of a double inclined smooth plane of angle s, of inclination $α$ and $β$ . If $M_{2}$ > $M_{1}$ and $β$ > $a l p h a$ , then the acceleration of block $M_{2}$ down the inclined will be ?

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Solution

$M_{2} > M_{1}$

$T - m_{1} g sin α = M_{1} a$
$M_{2} g sin β - T = M_{2} a$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ g (M_{2} sin β - M_{1} sin α) = a (M_{1} + M_{2})$

$\Rightarrow a = \frac{(M_{2} sin β - M_{1} sin α) g}{M_{1} + M_{2}}$

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Two masses M1 and M2 are attached to the ends of a light string, which passes over a massless pulley, attached to the top of a double inclined smooth plane of angle s, of inclination α and β. If M2>M1 and β>alpha, then the acceleration of block M2 down the inclined will be ?

M2>M1T−m1gsinα=M1aM2gsinβ−T=M2a¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯g(M2sinβ−M1sinα)=a(M1+M2)⇒a=(M2sinβ−M1sinα)gM1+M2