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Standard XII

Physics

Problem Solving

Fig shows an ...

Question

Fig shows an arrangment of masses hanging from a ceiling. In equilibrium, each rod is horizontal, has negligible mass and extends three times as far to the right of the wire supporting it as to the left. If mass $m_{4}$ is 48 kg then mass of $m_{1}$ is equal to

1 k g

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2 k g

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3 k g

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4 k g

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Solution

The correct option is A $1 k g$
$m_{4} l = (m_{1} + m_{2} + m_{3}) \times 3 l \Rightarrow m_{4} = 3 (m_{1} + m_{2} + m_{3}) s i m i l a r l y m_{3} = 3 (m_{1} + m_{2}) a n d m_{2} = 3 (m_{1}) \Rightarrow m_{3} = 3 ({4 m}_{1}) = 12 m_{1} a n d m_{4} = 3 \times (m_{1} + {3 m}_{1} + {12 m}_{1}) = 48 t h u s, m_{1} = 1 k g$

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Fig shows an arrangment of masses hanging from a ceiling. In equilibrium, each rod is horizontal, has negligible mass and extends three times as far to the right of the wire supporting it as to the left. If mass m4 is 48 kg then mass of m1 is equal to

The correct option is A 1kgm4l=(m1+m2+m3)×3l⇒m4=3(m1+m2+m3)similarlym3=3(m1+m2)andm2=3(m1)⇒m3=3(4m1)=12m1andm4=3×(m1+3m1+12m1)=48thus,m1=1kg

Fig shows an arrangment of masses hanging from a ceiling. In equilibrium, each rod is horizontal, has negligible mass and extends three times as far to the right of the wire supporting it as to the left. If mass $m_{4}$ is 48 kg then mass of $m_{1}$ is equal to