A stone of density `2.5g*cm^(-3)` is completely immersed in sea water and is allowed to sink from rest. Calculate the depth attained by the stone in 2s. Neglect the effect of friction. The specific gravity of sea water is 1.025, acceleration due to gravity = `980cm*s^(-2)`.
Text Solution
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Let the mass of the stone be mg. So, the volume of the stone = `m/2.5cm^(3)`. When the stone is allowed to sink in sea water, the resultant downward force = weight of the stone - upthrust or, `ma=mg-m/2.5xx1.025xxg` [a = downward acceleration of the stone] or, `a=980-1.025/2.5xx980-578.2cm*s^(-2)`. If the stone attains a depth h in 2 s, then `h=1/2at^(2)=1/2xx578.2xx(2)^(2)=1156.4cm=11.564m`.
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