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Standard XII

Physics

Law of Conservation of Mechanical Energy

A meter stick...

Question

A meter stick is hung from two spring balances $A$ and $B$ of equal lengths that are located at $20 c m$ and $70 c m$ mark of the meter stick. Weights of $2 N$ are placed at $10 c m$ and $40 c m$ marks, while a weight of $1.0 N$ is placed at the $90 c m$ mark. The weight of the uniform meter stick is $1.5 N$ . Determine the scale readings of the two balances $A$ and $B$ .

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Solution

From Newton's laws
$T_{1} + T_{2} = (1.5 + 1) N T_{1} + T_{2} = 2.5 N (1)$
Balancing the torque about point A
$30 (1.5) + 70 (1) = 50 T_{1} \Rightarrow T_{1} = \frac{4.5 + 7}{5} = \frac{11.5}{5} = 2.3 N$
$T_{2} = 2.5 - 2.3 = 0.2 N$
Here, $T_{1} & T_{2}$ are corresponding readings of spring balances on the right and left.

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From Newton's lawsT1+T2=(1.5+1)NT1+T2=2.5N(1)Balancing the torque about point A30(1.5)+70(1)=50T1⇒T1=4.5+75=11.55=2.3NT2=2.5−2.3=0.2NHere, T1&T2 are corresponding readings of spring balances on the right and left.