Two masses of 1, kg and 5, kg are attached to | vedclass

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(B) Let $m_1 = 1\, kg$ and $m_2 = 5\, kg$. The tension $T$ in the string is given by the formula:$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 	imes 1

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Less than $6\, kg$

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(B) Let $m_1 = 1\, kg$ and $m_2 = 5\, kg$. The tension $T$ in the string is given by the formula:$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 	imes 1 	imes 5}{1 + 5} g = \frac{10}{6} g = \frac{5}{3} g$The spring balance measures the total downward force exerted by the pulley system,which is equal to the sum of the tensions in the two segments of the string pulling downwards on the pulley. Thus,the reading $R$ is:$R = 2T = 2 	imes \left( \frac{5}{3} g ight) = \frac{10}{3} g$Since the spring balance is calibrated to read in $kg$ (where $1\, kg$ corresponds to a force of $g$),the reading in $kg$ is:$R_{kg} = \frac{10}{3} \approx 3.33\, kg$Since $3.33\, kg < 6\, kg$,the reading of the spring balance will be less than $6\, kg$.

Two masses of $1\, kg$ and $5\, kg$ are attached to the ends of a massless string passing over a pulley of negligible weight. The pulley itself is attached to a light spring balance as shown in the figure. When the masses are released and start moving,the reading of the spring balance will be:

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