In the arrangement, shown in figure, pulleys are massless and frictionless and threads are inextensible. Blocks of mass m1 will remain at rest if

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m_{1} = m_{2} + m_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}$
If $m_{1}$ remains at rest

$2 T = m_{1} g \dots (i)$

$T = \frac{2 m_{2} m_{3} g}{m_{3} + m_{2}} \dots (i i)$

From (i) and (ii)

$\frac{4 m_{2} m_{2} g}{m_{2} + m_{3}} = m_{1} g$

$\frac{1}{m_{1}} = \frac{m_{1} + m_{2}}{4 m_{2} m_{3}}, \frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}$

Suggest Corrections

Similar questions

m_{1}

m_{1}

m_{2}

m_{1} = 1 k g

m_{2} = \frac{1}{3} k g

m_{3}

m_{3}

M 1, M 2, M 3

Join BYJU'S Learning Program