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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) 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)
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.
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
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
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