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Three blocks of masses $ {{\text{m}}_{1}},{{\text{m}}_{2}} $ and $ {{\text{m}}_{3}} $ are placed on a horizontal frictionless surface. A force of 40 N pulls the system. Then calculate the value of T if
 $ {{\text{m}}_{1}}=10\text{ kg, }{{\text{m}}_{2}}=6\text{ kg, }{{\text{m}}_{3}}=4\text{ kg} $
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(A) 40 N
(B) 20 N
(C) 10 N
(D) 5 N

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Last updated date: 13th Jun 2024
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Answer
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Hint: We know that tension T can be formed by using the formula
T=ma
Now, at the point A tension T is due to sum of masses $ {{\text{m}}_{2}} $ and $ {{\text{m}}_{3}} $
So,
 $ \text{T=}\left( {{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a} $
Using this formula, tension T can be found.

Complete step by step solution
We know that F=ma
Here we have 3 masses, $ {{\text{m}}_{1}}=10\text{ kg, }{{\text{m}}_{2}}=6\text{ kg, }{{\text{m}}_{3}}=4\text{ kg} $
So F= $ \left( {{\text{m}}_{1}}+{{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a} $
 $ =\left( 10+6+4 \right)\text{a} $
F=20 a
F=40 N…… (Given)
So, 40=20a
a=2
Now tension T is given as:
T= $ \left( {{\text{m}}_{2}}+{{\text{m}}_{3}} \right)\text{a} $
  $ \begin{align}
  & =\left( 6+4 \right)\times 2 \\
 & =10\times 2 \\
\end{align} $
T=20 N
So, the correct option is (B).

Note
Tension is described as the pulling force transmitted axially by means of a string, cable, chain or 1-D objects. Tension never applies on its own. It has to be put on a system and it is always a pulling force.