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Byju's Answer

Standard XII

Physics

Motion Under Gravity

A mass of 3...

Question

A mass of $3 k g$ descending vertically downward support a mass of $2 k g$ by means of a light string passing over a pulley. At the end of $5 s$ the string breaks. How much high from now the $2 k g$ mass will go ? ( $g = 9.8 m / s^{2}$ )

4.9 m

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9.8 m

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16.9 m

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2.45 m

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Solution

The correct option is A $4.9 m$
Given that,

Mass $m_{1} = 3 k g$

Mass $m_{2} = 2 k g$

Now, the initial acceleration a of the system

$a = \frac{3 g - 2 g}{3 + 2}$

$a = \frac{g}{5}$

At $5 s e c$

Now, from equation of motion

$v = u + a t$

$v = 0 + \frac{g}{5} \times 5$

$v = 9.8 m / s$

When the string breaks at $t = 5 s$ ,

The mass $2 k g$ more under gravity

So, height is

$h = \frac{v^{2}}{2 g}$

$h = \frac{{(9.8)}^{2}}{2 \times 9.8}$

$h = 4.9 m$

Hence, the height is $4.9 m$

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A mass of 3 kg descending vertically downward support a mass of 2 kg by means of a light string passing over a pulley. At the end of 5 s the string breaks. How much high from now the 2 kg mass will go ? (g=9.8 m/s2)

A mass of $3 k g$ descending vertically downward support a mass of $2 k g$ by means of a light string passing over a pulley. At the end of $5 s$ the string breaks. How much high from now the $2 k g$ mass will go ? ( $g = 9.8 m / s^{2}$ )