Question

In two pulley-particle systems (i) and (ii), the acceleration and force imparted by the string on the pulley and tension in the strings are $\left(a_{1}, a_{2}\right),\left(N_{1}, N_{2}\right)$, and $\left(T_{1}, T_{2}\right)$, respectively. Ignoring friction in all contacting surfaces, study the following statements:

• A. $\frac{a_{1}}{a_{2}}=1$
• B. $\frac{T_{1}}{T_{2}} < 1$
• C. $\frac{N_{1}}{N_{2}} > 1$
• D. $\frac{a_{1}}{a_{2}} < 1$

Answer 1

Answer: (D)

$a_{1}=\frac{\left(m_{2}-m_{1}\right) g}{\left(m_{1}+m_{2}\right)}$ and $a_{2}=\frac{\left(m_{2}-m_{1}\right) g}{m_{1}}$
Hence $\frac{a_{1}}{a_{2}}=\frac{m_{1}}{\left(m_{1}+m_{2}\right)}=\frac{1}{\left(1+\frac{m_{2}}{m_{1}}\right)}$
$\Rightarrow \frac{a_{1}}{a_{2}}<1$
As $T_{1}=\frac{2 m_{1} m_{2} g}{\left(m_{1}+m_{2}\right)}$ and $T_{2}=m_{2} g$
Hence, $\frac{T_{1}}{T_{2}}=\frac{2 m_{1}}{\left(m_{1}+m_{2}\right)}=\frac{2}{\left(1+\frac{m_{2}}{m_{1}}\right)}$
Therefore, $T_{1} / T_{2}$ will depend upon the values of $m_{1}$ and $m_{2}$.
$N_{1}=2 T_{1}, N_{2}=2 T_{2}$
So the relation of $N_{1} / N_{2}$ will be same as $T_{1} / T_{2}$.