Physics
Mechanics
Easy

Question

A space s t o n e space o f space m a s s space o f space 16 space k g space i s space a t t a c h e d space t o space a space s t r i n g space 144 space m space l o n g space a n d space i s space w h i r l e d space i n space a space h o r i z o n t a l space c i r c l e. space
T h e space m a x i m u m space t e n s i o n space t h e space s t r i n g space c a n space w i t h s tan d space i s space 16 space N e w t o n. space T h e space m a x i m u m space v e l o c i t y space o f space r e v o l u t i o n
t h a t space c a n space b e space g i v e n space t o space t h e space s t o n e space w i t h o u t space b r e a k i n g space i t comma space w i l l space b e space
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton. The maximum velocity of revolution that can be given to the stone without breaking it, will be 

  1. 20 theta subscript 2 over theta subscript 1 equals 3
  2. 16 m g equals 1 cross times 10 equals 10 N
  3. 14 space fraction numerator m v squared over denominator r end fraction equals fraction numerator 1 cross times left parenthesis 4 right parenthesis squared over denominator 1 end fraction equals 16
  4. 12 fraction numerator m v squared over denominator r end fraction minus m g equals 6 N

The correct answer is: 12 fraction numerator m v squared over denominator r end fraction minus m g equals 6 N


    table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell text  Maximum tension  end text equals fraction numerator m v squared over denominator r end fraction plus m g equals 26 N end cell row cell not stretchy rightwards double arrow omega equals square root of g ⇄→ divided by t with not stretchy rightwards arrow on top end root R not stretchy rightwards double arrow v equals 12 straight m divided by straight s end cell end table

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