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A light strin...

Question

A light string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ ( $m_{1}$ > $m_{2}$ ). The masses are moving vertically. If the acceleration of the system is g/8, then ___________

\frac{m_{1}}{m_{2}} = \frac{9}{7}

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\frac{m_{1}}{m_{2}} = \frac{4}{3}

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T e n s i o n i n s t r i n g = \frac{3}{4} (m_{1} g)

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T e n s i o n i n s t r i n g = \frac{9}{8} (m_{2} g)

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Solution

The correct options are
A $\frac{m_{1}}{m_{2}} = \frac{9}{7}$
D $T e n s i o n i n s t r i n g = \frac{9}{8} (m_{2} g)$
Given : $a = \frac{g}{8}$
For block $m_{1}$ : $m_{1} a = m_{1} g - T$ ..........(1)
For block $m_{2}$ : $m_{2} a = T - m_{2} g$ ..........(2)
Adding (1) and (2), $(m_{1} + m_{2}) a = (m_{1} - m_{2}) g$
$∴$ $(m_{1} + m_{2}) \frac{g}{8} = (m_{1} - m_{2}) g$ $⟹ \frac{m_{1}}{m_{2}} = \frac{9}{7}$

Also from (2), $T = m_{2} (a + g) = m_{2} (\frac{g}{8} + g) = \frac{9}{8} m_{2} g$

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A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 ( m1 > m2). The masses are moving vertically. If the acceleration of the system is g/8, then ___________

The correct options are A m1m2=97 D Tension in string=98(m2g)Given : a=g8For block m1 : m1a=m1g−T ..........(1)For block m2 : m2a=T−m2g ..........(2)Adding (1) and (2), (m1+m2)a=(m1−m2)g∴ (m1+m2)g8=(m1−m2)g ⟹m1m2=97Also from (2), T=m2(a+g)=m2(g8+g)=98m2g

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