Question

PLS ANSWER MY QUESTION ,"who will answer correct I'll mark them as brainlist ","FOLLOW THEM ​

Answer 1

hope it helps you

my friend harini

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Balloon of Mass = M.
• Mass released = m.
• Acceleration = $\large{ \alpha}$

$\huge{\bold{\underline{Explanation:-}}}$

Let the balloon experiences a upward thrust of "F" .

Applying Newton's second law,

As the balloon is coming down with acceleration α.

Therefore,

$\large{Mg - F = M \alpha \: ----(1)}$

Now, A mass of "m" is removed so that acceleration becomes upward,

$\large{F - (M - m)g = (M - m) \alpha}$

$\large{F - Mg + mg = M \alpha - m \alpha \: ----(2)}$

Add equation (1) and (2)

$\large{Mg - F + F - Mg + mg = M \alpha + M \alpha - m \alpha}$

$\large{ \cancel{Mg} - \cancel{F} + \cancel{F} - \cancel{Mg} + mg = M \alpha + M \alpha - m \alpha}$

Now,

$\large{mg = 2M \alpha - m \alpha}$

$\large{mg + m \alpha = 2M \alpha}$

$\large{m( g + \alpha) = 2M \alpha}$

$\huge{\boxed{\boxed{m = \dfrac{2M \alpha}{g + \alpha}}}}$

So, the option is (2) [m = 2Mα/g + α]