In each case $m_{1} = 4 k g$ and $m_{2} = 3 k g$ . If $a_{1}, a_{2}$ and $a_{3}$ are the respective accelerations of the block $m_{1}$ in given situations, then:

a_{1} > a_{2} > a_{3}

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a_{1} > a_{2} > a_{3}

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a_{1} = a_{2} = a_{3}

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a_{1} > a_{3} > a_{2}

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Solution

The correct option is B $a_{1} > a_{2} > a_{3}$
We write equations for all three cases. $T$ is tension in string.
Case 1:
$m_{2} \times g - T = m_{2} \times a_{1}$
$T = m_{1} \times a_{1}$
Thus, $a_{1} = \frac{m_{2} \times g}{(m_{1} + m_{2})} = \frac{3 \times g}{7}$
Case 2:
$T - m_{2} \times g = m_{2} \times a_{2}$
$m_{1} \times g - T = m_{1} \times a_{2}$
Thus, $a_{2} = \frac{(m_{1} - m_{2}) \times g}{(m_{1} + m_{2})} = \frac{g}{7}$
Case 3:
$m 2 \times g - T = m_{2} \times a_{3}$
$T - m_{1} \times g \times s i n 30 = m_{1} \times a_{3}$
Thus, $a_{3} = \frac{(m_{2} - m_{1} \times s i n 30) \times g}{(m_{1} + m_{2})} = \frac{g}{14}$

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In each case m1=4kg and m2=3kg. If a1,a2 and a3 are the respective accelerations of the block m1 in given situations, then:

In each case $m_{1} = 4 k g$ and $m_{2} = 3 k g$ . If $a_{1}, a_{2}$ and $a_{3}$ are the respective accelerations of the block $m_{1}$ in given situations, then: