Three blocks of masses 2 kg- 4 kg and 6 kg are connected by string and resting on a frictionless inclined of 53- as shown- A force of 120 N is applied upward along the inclined to the 6 kg block- If the strings are ideal- the ratio T1-T2 will be g- 10m s-2-

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Byju's Answer

Standard XII

Physics

Pulley

Three blocks ...

Question

Three blocks of masses 2 kg, 4 kg and 6 kg are connected by string and resting on a frictionless inclined of $53^{\circ}$ as shown. A force of 120 N is applied upward along the inclined to the 6 kg block. If the strings are ideal, the ratio $T_{1} / T_{2}$ will be (g= 10m $s^{- 2}$ ):

1:1

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1:3

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Solution

The correct option is C 1:3
The correct option is C

Let all the three blocks as a system.

Then $f = m a$

$a = \frac{f}{M}$ Where $M + m_{1} + m_{2} + m_{3}$

$a = \frac{120}{2 + 4 + 6} = 10 m / s^{2}$

So, This is the common acceleration of all three blocks.

Since,

We have,

$f - t_{2} = 6 \times 10$

$t_{2} = 120 - 60 = 60 N$

$t_{2} - t_{1} = 4 \times 10$

$t - 1 = 60 - 40 = 20 N$

Thus the ratio of $t_{2}$ with $t_{1}$ is

$\frac{t_{1}}{t_{2}} = \frac{20}{60}$

$1 : 3$

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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: