You visited us 1 times! Enjoying our articles? Unlock Full Access!

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?

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $\frac{T_{1}}{T_{2}} .$

Suggest Corrections

Similar questions

(a_{1}, a_{2}), (N_{1}, N_{2})

(T_{1}, T_{2})

(a_{1}, a_{2}), (N_{1}, N_{2})

(T_{1}, T_{2})

m

T_{1}

T_{2}

T_{1} : T_{2}

^{2}

m

T_{1}

T_{2}

T_{1} : T_{2}

Join BYJU'S Learning Program