The system shown in figure is released from rest. Calculate the tension in the string connecting $3$ kg and $1$ kg, assuming pulleys and strings are masseless

\frac{7 g}{6}

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\frac{5 g}{3}

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\frac{6 g}{6}

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\frac{3 g}{5}

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Solution

The correct option is A $\frac{7 g}{6}$ N
option A is correct
For $2 k g$ block
$2 g - T_{1} = 2 a - (1)$
For $1 k g$ block $T_{2} - g = a - (2)$
$(1) + (2)$ gives $\Rightarrow T_{2} - T_{1} + g = 3 a - (3)$
For $3 k g$ block $T_{1} - T_{2} = 3 a - (4)$
$(3) + (4)$ gives $\Rightarrow g = 6 a$
$(5) \Rightarrow a = \frac{g}{6}$
Putting value of a from (5) in (2)
$T_{2} - g = a$
$T_{2} = a + g = \frac{g}{6} + g = \frac{7 g}{6} N$
Hence option A is correct

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The system shown in figure is released from rest. Calculate the tension in the string connecting 3kg and 1kg, assuming pulleys and strings are masseless

The correct option is A 7g6Noption A is correctFor 2 kg block2g−T1=2a−(1)For 1 kg block T2−g=a−(2)(1)+(2) gives ⇒T2−T1+g=3a−(3)For 3 kg block T1−T2=3a−(4)(3)+(4) gives ⇒g=6a(5)⇒a=g6Putting value of a from (5) in (2)T2−g=aT2=a+g=g6+g=7g6NHence option A is correct

The system shown in figure is released from rest. Calculate the tension in the string connecting $3$ kg and $1$ kg, assuming pulleys and strings are masseless