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Tension in a String

In two pulley...

Question

In two pulley - particle figures (i) and (ii), the acceleration and force imparted by the string on the pulley and tension in the strings are, $(a_{1}, a_{2}), (N_{1}, N_{2})$ , and $(T_{1}, T_{2})$ respectively. Ignoring friction in all contacting surfaces, which of the following relations is true?

A

\frac{a_{1}}{a_{2}} = 1

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B

\frac{T_{1}}{T_{2}} < 1

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C

\frac{N_{1}}{N_{2}} > 1

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D

\frac{a_{1}}{a_{2}} < 1

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Solution

The correct option is D $\frac{a_{1}}{a_{2}} < 1$
$a_{1} = \frac{(m_{2} - m_{1}) g}{(m_{1} + m_{2})}$ and $a_{2} = \frac{(m_{2} - m_{1}) g}{m_{1}}$
Hence $\frac{a_{1}}{a_{2}} = \frac{m_{1}}{(m_{1} + m_{2})} = \frac{1}{(1 + \frac{m_{2}}{m_{1}})}$
$\frac{a_{1}}{a_{2}} < 1$
As $T_{1} = \frac{2 m_{1} m_{2} g}{(m_{1} + m_{2})}$ and $T_{2} = m_{2} g$
Hence $\frac{T_{1}}{T_{2}} = \frac{2 m_{1}}{(m_{1} + m_{2})} = \frac{2}{1 + \frac{m_{2}}{m_{1}}}$
Hence, $\frac{T_{1}}{T_{2}}$ will depend upon the values of $m_{1}$ and $m_{2}$
$∴ N_{2} = 2 T_{1}, N_{2} = 2 T_{2}$
So the relation of $\frac{N_{1}}{N_{2}}$ will be same as $\frac{T_{1}}{T_{2}} .$

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