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Question: In the following diagram find value of \[{T_2}\]. ![](https://www.vedantu.com/question-sets/b9334d...

In the following diagram find value of T2{T_2}.

A. 12N12N
B. 6N6N
C. 4N4N
D. 1N1N

Explanation

Solution

The tension acting on T2{T_2} andT3{T_3} can be considered to be almost the same. This is because they are acted upon by the same mass pulling the strings downwards. Also, the motion of masses on the surface will almost be the same due to the hanging mass.
Tension on the strings can be derived by considering the equation of motion of masses.

Formula Used:
The tension in the string is given by the formula: T2=(m1+m2)m3(m1+m2)+m3g{T_2} = \dfrac{{({m_1} + {m_2}){m_3}}}{{({m_1} + {m_2}) + {m_3}}}g

Complete step by step answer:
The force exerted through a thread, rope, chain or any stretched string is called tension. The direction of tension is so as to pull the body. As seen in the given figure, three bodies of masses 1kg, 1kg and 3kg are given. Consider the names for the masses are m1,m2{m_1},{m_2} and m3{m_3} respectively such that m3{m_3} is the mass of 3kg. Tension T1{T_1} ,T2{T_2} and T3{T_3} are the tension on the string that connects the masses.
Then, equation of motion of mass m1{m_1} and m2{m_2} will be as follows
T = \left( {{m_1} + {m_2}} \right)a$$$$ \to (1)
where, TT is the tension produced in the stringT1{T_1} and T2{T_2}andaa is the acceleration produced due to this tension.
On the other hand, equation of motion of mass m3{m_3} will be as follows:
m3gT=m3a{m_3}g - T = {m_3}a
Here, T{T_{}} is the tension in T3{T_3} part of the string.
The acceleration can be written as
a=m3g(m1+m2)+m3a = \dfrac{{{m_3}g}}{{({m_1} + {m_2}) + {m_3}}}
And, thus tension T2{T_2} will be
T2=(m1+m2)m3(m1+m2)+m3g{T_2} = \dfrac{{({m_1} + {m_2}){m_3}}}{{({m_1} + {m_2}) + {m_3}}}g
T2=(1+1)×3(1+1)+3×10=605=12N\therefore {T_2} = \dfrac{{(1 + 1) \times 3}}{{(1 + 1) + 3}} \times 10 = \dfrac{{60}}{5} = 12N

Hence, option (A) is the correct answer.

Note: Here gg is approximated to 10 meters per seconds squared. Students must also note that the two blocks of masses are resting on a smooth surface. Therefore, there will be no frictional force between the two blocks and the surface.