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MHT CET202313 May 2023Evening ShiftPhysicsMotion in One DimensionActual

A small steel ball is dropped from a height of 1.5 ~m into a glycerine jar. The ball reaches the bottom of the jar 1.5 second after it was dropped. If the retardation is 2.66 ~m / s ^2 , the height of the glycerine in the jar is about (acceleration due to gravity g=9.8 ~m / s ^2 )

Options

  1. A7.0 ~m
  2. B7.5 ~m
  3. C5.5 ~m
  4. D3.2 ~m

Correct answer

C. 5.5 ~m

Step-by-step solution

The velocity of the ball after it has been dropped from height till it reaches the glycerine surface is v _ i ^2=0+2 gh . ( v ^2- u ^2=2 gh )v_i^2=2 9.8 1.5v_i^2=29.4 The velocity of the ball after it enters glycerine is v_f^2=v_i^2-2 g h0=29.4-(2 2.66 h) h = 29.4 2 2.66 =5.5 ~m

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