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`        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

```							Buoyancy Force = (dV)gm = d’V acceleration = net Force/m => dVg/d’V = > d*g/d’ = 10/0.8 = 12.5 m/s^2net accln.  = 12.5 – 10 = 2.5 m/s^2. depth it gets sinked down:=>(2g*2) = 0^2 + 2*g*ss = 8 meter.
```
3 years ago
```							This question has three parts, viz.,	Speed of ball just before entering water	Acceleration of ball in water due to buoyancy of water	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 	Acceleration of ball in water due to buoyancy of waterDensity(d) = Mass(m)/Volume(V)Therefore Mass of ball(m) = 0.8V KgDensity of water (at 4°C) = 1g/mL or 1000 Kg/m3            Downward force exerted by ball            F = mgOr       Fb = 0.8Vg (Downwards direction)            Since ball will displace equal volume of waterTherefore 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 ballTherefore,            a = -0.2Vg/0.8V = -g/4 m/sec2 	Depth attained by ball due to changed acceleration:Initial velocity of ball (u) = 2√g m/secFinal velocity of ball (v) = 0 m/secg = -g/4 m/sec2.Hence            (0)2-(2√g)2 = 2(-g/4) x sSolving this gives us            s = 8 m
```
one month ago
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