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Grade: 12th pass
        
A ball of relative density 0.8 falls water from a height of 2m. The depth to which the ball will sink is ? (please do it by law of conservation of energy and if not possible please give valid reason)
3 years ago

Answers : (2)

Vikas TU
9205 Points
							
Buoyancy Force = (dV)g
m = d’V 
acceleration = net Force/m => dVg/d’V = > d*g/d’ = 10/0.8 = 12.5 m/s^2
net accln.  = 12.5 – 10 = 2.5 m/s^2.
 
depth it gets sinked down:
=>
(2g*2) = 0^2 + 2*g*s
s = 8 meter.
 
3 years ago
Dr Arvind Mittal
13 Points
							
This question has three parts, viz.,
  1. Speed of ball just before entering water
  2. Acceleration of ball in water due to buoyancy of water
  3. Depth attained by ball due to changed acceleration.
 
1.         Speed of ball just before entering water
                u = 0 m/sec
                v = ? m/sec
                s = 2 m
                Since v2-u2=2gs
                Therefore,
                v = 2√g m/sec
 
  1. Acceleration of ball in water due to buoyancy of water
Density(d) = Mass(m)/Volume(V)
Therefore Mass of ball(m) = 0.8V Kg
Density of water (at 4°C) = 1g/mL or 1000 Kg/m3
            Downward force exerted by ball
            F = mg
Or       Fb = 0.8Vg (Downwards direction)
            Since ball will displace equal volume of water
Therefore Force due to buoyancy of water
            Fw = 1.0Vg (Upward direction)
Therefore Net force
            Fb – Fw = 0.8Vg – 1.0Vg = -0.2Vg (upward direction)
However,
            F = ma or a = F/m
            Where m is mass of the ball
Therefore,
            a = -0.2Vg/0.8V = -g/4 m/sec2
 
  1. Depth attained by ball due to changed acceleration:
Initial velocity of ball (u) = 2√g m/sec
Final velocity of ball (v) = 0 m/sec
g = -g/4 m/sec2.
Hence
            (0)2-(2√g)2 = 2(-g/4) x s
Solving this gives us
            s = 8 m
one month ago
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