A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light stringpassing over a pulley. At the end of 5 s, the string breaks . How much high from now the 2 kg masswill go? (Take, 9 = 9.8 ms-2)
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a
4.9 m
b
9.8 m
c
19.6 m
d
2.45 m
answer is A.
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Detailed Solution
Acceleration of system before breaking the string,a= Net pulling force Total mass =3g-2g5=g5After 5 s velocity of system, v=at=g5×5=g ms-1Now, h=v22g=g22g=g2=49 m
A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light stringpassing over a pulley. At the end of 5 s, the string breaks . How much high from now the 2 kg masswill go? (Take, 9 = 9.8 ms-2)