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In two pulley-particle systems (i) and (ii), the acceleration, 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, study the following statements.
i) $\frac{a_{1}}{a_{2}} = 1$ ii) $\frac{T_{1}}{T_{2}} < 1$ iii) $\frac{N_{1}}{N_{2}} > 1$ iv) $\frac{a_{1}}{a_{2}} < 1$
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Only relation (iv) always follows

Relations (ii) and (iv) always follow

Relations (ii) and (iii) always follow

Only relation (i) always follows

answer is D.

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Detailed Solution

$a_{1} = \frac{(m_{2} {-m}_{1}) g}{(m_{1} {+m}_{2})}$

${and a}_{2} \frac{(m_{2} {-m}_{1}) g}{m_{1}}$

$Hence \frac{a_{1}}{a_{2}} = \frac{m_{1}}{(m_{1} + m_{2})} = \frac{1}{(1 + \frac{m_{2}}{m_{1}})} \Rightarrow \frac{a_{1}}{a_{2}} < 1$

${As T}_{1} = \frac{{2m}_{1} m_{2} g}{(m_{1} {+m}_{2})} {and T}_{2} {=m}_{2} g$

$Hence , \frac{T_{1}}{T_{2}} = \frac{{2m}_{1}}{(m_{1} {+m}_{2})} = \frac{2}{(1+ \frac{m_{2}}{m_{1}})}$

Therefore , $T_{1} {/T}_{2}$ will depend upon the values of m₁ and m_{2 .}

$N_{1} {=2T}_{1} {,N}_{2} {=2T}_{2}$

So the relation of N₁/N₂ will be same as $T_{1} {/T}_{2}$

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