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Byju's Answer

Standard XII

Physics

Tension in a String

Three blocks ...

Question

Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. A force of $40 N$ pulls the system. Then calculate the value of $T$ (in $N$ ) if $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ .

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Solution

The correct option is C $8$
Common acceleration of the system is given by
$a = \frac{F_{n e t}}{m_{t o t a l}} = \frac{40}{20} = 2 m / s^{2}$

From the FBD of mass $m_{3}$ , we have

$T = m_{3} \times a$
$⟹ T = 4 \times 2 = 8 N$

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Similar questions

Q. Three blocks of masses

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m_{2}

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are connected by massless strings as shown in the figure and placed on a frictionless table. They are pulled with a force

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m_{1} = 10 k g, m_{2} = 6 k g

and

m_{3} = 4 k g

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Q. Three blocks of masses

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Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. A force of $40 N$ pulls the system. Then calculate the value of $T$ (in $N$ ) if $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ .

Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be:

Q. Three block of masses

m_{1}, m_{2}

and

m_{3}

are connected by massless string as shown on a frictionless table. They are pulled with a force

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Three blocks of masses m1, m2 and m3 are connected by massless strings as shown in the figure and placed on a frictionless table. A force of 40 N pulls the system. Then calculate the value of T (in N) if m1=10 kg,m2=6 kg and m3=4 kg.

The correct option is C 8Common acceleration of the system is given by a=Fnetmtotal=4020=2 m/s2 From the FBD of mass m3, we have T=m3×a ⟹T=4×2=8 N

Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. A force of $40 N$ pulls the system. Then calculate the value of $T$ (in $N$ ) if $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ .

The correct option is C $8$
Common acceleration of the system is given by
$a = \frac{F_{n e t}}{m_{t o t a l}} = \frac{40}{20} = 2 m / s^{2}$

From the FBD of mass $m_{3}$ , we have

$T = m_{3} \times a$
$⟹ T = 4 \times 2 = 8 N$