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
- A7.0 ~m
- B7.5 ~m
- C5.5 ~m
- 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