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| 1. |
A ball of material of specific gravity d_(1) falls from a height 'h' on the surface of a liquid of relative density d_(2) such that d_(2) > d_(1). The time for which the body will be falling into the liquid is : |
| Answer» <html><body><p>`(d_(1))/(d_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))sqrt((<a href="https://interviewquestions.tuteehub.com/tag/2h-300377" style="font-weight:bold;" target="_blank" title="Click to know more about 2H">2H</a>)/<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)`<br/>`(d_(2))/(d_(1))sqrt((2h)/g)`<br/>`(d_(1))/(d_(2)-d_(1))sqrt((2h)/g)`<br/>`(d_(2)-d_(1))/(d_(2))sqrt((2h)/g)`</p>Solution :Velocity of <a href="https://interviewquestions.tuteehub.com/tag/fall-983217" style="font-weight:bold;" target="_blank" title="Click to know more about FALL">FALL</a> of the body just before striking <br/> `v=sqrt(2gh)` <br/> Retarding force F = V `(d_(2)- d_(1)) g` <br/> `therefore` Retardation a =`F/m=(V(d_(2)-d_(1))/(Vd_(1))g` <br/> `=(d_(2)-d_(1))/(d_(1))g` <br/> Time of fall `t=v/a=(sqrt(2gh))/((d_(2)-d_(1))/(d_(1))g)` <br/> `=(d_(1))/(d_(2)-d_(1))xxsqrt((2h)/g)` <br/> `therefore` Correct <a href="https://interviewquestions.tuteehub.com/tag/choice-915996" style="font-weight:bold;" target="_blank" title="Click to know more about CHOICE">CHOICE</a> is (c).</body></html> | |