Question
Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by mass less strings as shown in figure, on a frictionless table. They are Pulled with a force $\mathrm{T}_{3}=40 \mathrm{~N}$. If $\mathrm{m}_{1}=10 \mathrm{~kg}$, $\mathrm{m}_{2}=6 \mathrm{~kg}$ and $\mathrm{m}_{3}=4 \mathrm{~kg}$ the tension $\mathrm{T}_{2}$ will be $=$$\mathrm{N}$(A) 20(B) 40(C) 10(D) 32
Step 1
The total mass of the system is the sum of the masses of all three blocks. So, the total mass $m_{total}$ is given by: \[m_{total} = m_{1} + m_{2} + m_{3} = 10 \, kg + 6 \, kg + 4 \, kg = 20 \, kg\] Show more…
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