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Grade 12th passGeneral Physics

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)

Profile image of satyam sahu
9 Years agoGrade 12th pass
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3 Answers

Profile image of Vikas TU
9 Years ago
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.
 
Profile image of Dr Arvind Mittal
6 Years ago
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
Profile image of ankit singh
5 Years ago
    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

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