Laws of Motion

JEE Main level, single correct, medium

26 of 210

Question 26 of 210

Two masses M1{M}_{1} and M2{M}_{2} are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2{M}_{2} is twice that of M1{M}_{1}. the acceleration of the system is a1{a}_{1}. When the mass M2{M}_{2} is thrice that of M1{M}_{1}. The acceleration of The system is a2{a}_{2}. The ratio a1a2\dfrac{{a}_{1}}{{a}_{2}} will be

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Answer: (B) 23\dfrac{2}{3}

In both cases m1>m2{m}_{1}\gt {m}_{2}. Therefore, m1{m}_{1} will accelerate downward and m2{m}_{2} will accelerate upward.
For m1{m}_{1},
m1g−T=m1a{m}_{1}g-T={m}_{1}a
For m2{m}_{2},
T−m2g=m2aT-{m}_{2}g={m}_{2}a
Adding the above equations we get,
a=m1−m2m1+m2ga=\dfrac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}g
⇒a1=m1−m2m1+m2g=2m2−m22m2+m2g=g3\Rightarrow {a}_{1}=\dfrac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}g =\dfrac{2{m}_{2}-{m}_{2}}{2{m}_{2}+{m}_{2}}g =\dfrac{g}{3}
and
⇒a2=m1−m2m1+m2g=3m2−m23m2+m2g=g2\Rightarrow {a}_{2}=\dfrac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}g =\dfrac{3{m}_{2}-{m}_{2}}{3{m}_{2}+{m}_{2}}g =\dfrac{g}{2}
⇒a1a2=23\Rightarrow \dfrac{{a}_{1}}{{a}_{2}}=\dfrac{2}{3}

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