Laws of Motion

JEE Main level, single correct, medium

5 of 210

Question 5 of 210

Three masses m1=4m_1 = 4 kg, m2=4m_2 = 4 kg and m3=6m_3 = 6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T1/T2T_1/T_2 is _____.
(take g=10g = 10 m/s2^2)

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Answer: (A) 5/35/3

Let the acceleration of the system be aa. The total mass on the right side is m2+m3=4+6=10m_2 + m_3 = 4 + 6 = 10 kg, and the mass on the left side is m1=4m_1 = 4 kg. Since 1010 kg >4\gt 4 kg, the right side accelerates downwards and the left side accelerates upwards.

The common acceleration aa of the system is given by:
a=(m2+m3)−m1m1+m2+m3ga = \dfrac{(m_2 + m_3) - m_1}{m_1 + m_2 + m_3} g
a=10−44+4+6g=614g=37ga = \dfrac{10 - 4}{4 + 4 + 6} g = \dfrac{6}{14} g = \dfrac{3}{7} g

For mass m1m_1, the equation of motion is:
T1−m1g=m1aT_1 - m_1 g = m_1 a
T1=m1(g+a)=4(g+37g)=4(107g)=407gT_1 = m_1(g + a) = 4 \left(g + \dfrac{3}{7}g\right) = 4 \left(\dfrac{10}{7}g\right) = \dfrac{40}{7}g

For mass m3m_3, the equation of motion is:
m3g−T2=m3am_3 g - T_2 = m_3 a
T2=m3(g−a)=6(g−37g)=6(47g)=247gT_2 = m_3(g - a) = 6 \left(g - \dfrac{3}{7}g\right) = 6 \left(\dfrac{4}{7}g\right) = \dfrac{24}{7}g

The ratio of the tensions is:
T1T2=407g247g=4024=53\dfrac{T_1}{T_2} = \dfrac{\dfrac{40}{7}g}{\dfrac{24}{7}g} = \dfrac{40}{24} = \dfrac{5}{3}

Answer: 5/35/3

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