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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Laws of Motion

BITSAT 2011 · Physics · Laws of Motion

Three blocks of masses m_ 1 , m_ 2 and m_ 3 are connected by...

Three blocks of masses m1,m2\mathrm{m}_{1}, \mathrm{ m}_{2} and m3\mathrm{m}_{3} are connected by massless strings, as shown, on a frictionless table. They are pulled with a force T3=40N\mathrm{T}_{3}=40 \mathrm{ N}. If m1=10kg,m2=6kgm_{1}=10 \mathrm{ kg}, \mathrm{ m}_{2}=6 \mathrm{ kg} and m3=4kg,\mathrm{m}_{3}=4 \mathrm{ kg}, the tension T2\mathrm{T}_{2} will be Three blocks of masses m_ 1 , m_ 2 and m_ 3 are connected by massless strings, as shown, on a frictionless table. They

Options

  1. A20N20 \mathrm{ N}
  2. B40N40 \mathrm{ N}
  3. C10N10 \mathrm{ N}
  4. D32N32 \mathrm{ N}

Correct answer

D. 32N32 \mathrm{ N}

Step-by-step solution

For equilibrium of all 3 masses, a=T3m1+m2+m3\frac{\mathrm{T}_{3}}{\mathrm{ m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}} For equilibrium of m1  and m2T2=(m1+m2)⋅a\mathrm{m}_{1} \ \text{ and } \mathrm{ m}_{2}\mathrm{T}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \cdot \mathrm{a} or, T2=(m1+m2)T3m1+m2+m3\mathrm{T}_{2}=\frac{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{T}_{3}}{\mathrm{ m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}} Given m1=10kg,m2=6kg,m3=4kgT3=40N∴T2=(10+6)⋅4010+6+4=32Nm_{1}=10 \mathrm{ kg}, \mathrm{ m}_{2}=6 \mathrm{ kg}, \mathrm{ m}_{3}=4 \mathrm{ kg}\mathrm{T}_{3}=40 \mathrm{ N}\therefore \mathrm{T}_{2}=\frac{(10+6) \cdot 40}{10+6+4}=32 \mathrm{ N}
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