1.In the arrangement shown, the pulleys are massless and frictionless and the strings are inextensible. The block of mass m1m_1m1 will remain at rest ifa.1m1=1m2+1m3\dfrac{1}{m_1} = \dfrac{1}{m_2} + \dfrac{1}{m_3}m11=m21+m31b.m1=m2+m3m_1 = m_2 + m_3m1=m2+m3c.4m1=1m2+1m3\dfrac{4}{m_1} = \dfrac{1}{m_2} + \dfrac{1}{m_3}m14=m21+m31d.1m3=1m2+1m1\dfrac{1}{m_3} = \dfrac{1}{m_2} + \dfrac{1}{m_1}m31=m21+m11Login to continueOnly logged in users canattempt or see the solution.