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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 stringbreaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. `9.8m`B. `19.6m`C. `2.45m`D. `4.9m` |
Answer» Correct Answer - d Acceleration of the system before the string breaks `a = ("net pulling force")/("total mass") = (3 g -2g)/((3 + 2)) = (g)/(5)` After 5s, velocity of the system, `upsilon = "at" = (g)/(5) xx 5 = gm//s` Required height `h = (upsilon^(2))/(2 g) = (g xx g)/(2 g) = (g)/(2)` ` = (9.8)/(2) = 4.9m` . |
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