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?

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

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

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

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\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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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, (a1,a2),(N1,N2), and (T1,T2) respectively. Ignoring friction in all contacting surfaces, which of the following relations is true?

The correct option is D a1a2<1a1=(m2−m1)g(m1+m2) and a2=(m2−m1)gm1 Hence a1a2=m1(m1+m2)=1(1+m2m1) a1a2<1 As T1=2m1m2g(m1+m2) and T2=m2g Hence T1T2=2m1(m1+m2)=21+m2m1 Hence, T1T2 will depend upon the values of m1 and m2 ∴N2=2T1, N2=2T2 So the relation of N1N2 will be same as T1T2.