Consider three blocks of mass $m_{1}, m_{2}, m_{3}$ have been interconnected by strings, which are being pulled by a common force $F$ on a frictionless horizontal table as shown in the figure. The tension $T_{1}$ and $T_{2}$ are also indicated.
a) $T_{2} > T_{1}$ if $m_{2} > m_{1}$
b) $T_{2} = T_{1}$ if $m_{2} = m_{1}$
c) $T_{2} > T_{1}$ always
d) acceleration of the system $= \frac{F}{m_{1} + m_{2} + m_{3}}$
Which of the above are true?

a, b

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

b, d

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

a, d

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

c, d

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

Open in App

Solution

The correct option is D c, d
For boundary 1, $F = \sum m a = (m_{1} + m_{2} + m_{3}) a$ where a= net acceleration of the system.
$a = \frac{F}{m_{1} + m_{2} + m_{3}}$
Boundary 2, $T_{2} = (m_{1} + m_{2}) a$
Boundary 3, $T_{1} = m_{1} a$ .
Therefore $T_{2}$ is always greater than $T_{1}$ .

Suggest Corrections

Similar questions

Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be: