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Three equal weights of$3kg$ each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses$II$ and$III$ will be (Take$g = 10m{s^{ - 2}}$)

(A) $5N$
(B) $6N$
(C) $10N$
(D) $20N$

Last updated date: 20th Jun 2024
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Hint We are given the situation of three blocks of equal masses put up in a certain configuration and are asked to find the tension of the string connecting two of the three blocks. Thus, we will just apply the basics of force and mass configuration.

Complete Step By Step Solution
For the tension and gravitational force on the block$I$, we get
$T - mg = ma$
Putting the values of mass, we get
$T - 3g = 3a$
Thus, we get
$T = 3a + 3g \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
For the blocks$II$ and$III$, we get
$2mg - T = 2ma$
Putting in the values of mass, we get
$6g - T = 6a$
Further, using equation$(1)$, we get
$6g - 3a - 3g = 6a$
After further evaluation, we get
$3g = 9a$
Then, we get
$a = \dfrac{g}{3}$
The $T$we considered till now is the net tension of the string.
We will take $T'$ as the tension between the blocks$II$ and$III$.
Thus, the equation will be
$mg - T' = ma$
Further, putting in the values, we get
$3g - T' = 3a$
Further, we get
$T' = 3g - 3a$
Thus, putting the evaluated value of$a$, we get
$T' = 3g - 3\left( {\dfrac{g}{3}} \right)$
Then we get
$T' = 2g$
Putting the value of$g$, we get
$T' = 2 \times 10 = 20N$

Hence, the correct option is (D).

Additional Information The diagram we are considering is called the free body diagram which shows us the bodies and the forces acting on it. The equations we are forming are on the basis of the fundamental idea that all the external forces are in balance for an isolated body.

Note The tension of the string will hold the blocks in equilibrium and will determine the motion of the blocks as the string moves on the pulley. But the tension of the string between the two blocks will make sure that while the motion of the block, the block remains in equilibrium.