Two blocks of masses 1,kg and 5,kg are attach | vedclass

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

Vedclass · Accepted Answer

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 when the blocks are moving is given by the formula:$T = \frac{2 m_1 m_2}{m_1 + m_2} g$Substituting the values:$T = \frac{2 	imes 1 	imes 5}{1 + 5} g = \frac{10}{6} g = \frac{5}{3} g$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 attached to the pulley.Reading of spring balance $= 2T = 2 	imes \frac{5}{3} g = \frac{10}{3} g$Since the reading in $kg$ is equivalent to the force divided by $g$,the reading is $\frac{10}{3} \approx 3.33\,kg$.Since $3.33\,kg < 6\,kg$,the reading of the spring balance will be less than $6\,kg$.

Two blocks of masses $1\,kg$ 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 spring balance as shown in the figure. The blocks start moving; during this interval,the reading of the spring balance will be:

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