A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. 4.9 mB. 9.8mC. 10.6 mD. 2.45 m

A mass of `3kg` descending vertically downwards supports a mass of `2k

1.	A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. 4.9 mB. 9.8mC. 10.6 mD. 2.45 m
Answer» Correct Answer - A (a) Acceleration of system before breaking the string was, `a=("Net pulling force")/("Total mass")=(3g-2g)/(5)=(g)/(5)` After 5s velocity of system,`v=at=(g)/(5)xx5=g ms^(-2)` Now, `h=(v^(2))/(2g)=(g^(2))/(2g)=(g)/(2)=4.6m`

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A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. 4.9 mB. 9.8mC. 10.6 mD. 2.45 m

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