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A piece of Ice floats in a vessel with water above which a layer of lighter oil is poured. When ice melts 1. The level of oil water interface falls , 2. The level of oil water interface rises 3. The thickness of oil layer decreases ,4. The thickness of oil layer remain same 5. The thickness of oil layer increases , 6. The level of oil-air interface falls 7. The level of oil-air interface remains same , 8. The level of oil-air interface rises Select the correct alternatives :

Answer»

Only 4 & 7 are CORRECT
2,3 & are correct
1,5,7 are correct
only is correct

Solution :From fig. (a) `h_(2)A`=VOLUME of oil +some volume of ice From fig. (b) `h_(2)` A =volume of oil
`rArr (h_(2)-H'_(2))` A =Some volume of ice gt0
`rArr h_(2) gt h'_(1)`
`:.` Statement3 correct
PRESSURE at bottom in fig. (a) is given by
`rArr P_(0)+rho_(oil) h_(2)g +rho_("water")h_(1)g`
`:. (P_(0)+rho_(oil)h_(2)g+rho_("water")h_(1)g)A=P_(0)A+W_(oil) +W_("water")+W_("Ice")(i)`
similarly from fig. (b)
`(P_(0)+rho_(oil)h'_(2)g+rho_("water")h'_(2)g+rho_("water") h'_(1)g)A=P_(0)A+W_(oil) +W_("water")+W_(Ice) (ii)`
`rho_(oil)(h_(2)-h'_(2))+rho_("water") h'_(1)=rho_("oil")h_(2)+rho_("water")h_(1)`
`rArr rho_("oil") (h_(2)-h'_(2))=rho_("water")(h'_(1)-h_(1))`
`rArr h'_(1)-h_(1)=(rho_(oil))/(rho_("water"))(h_(2)-h'_(2)) gt 0`
`:.` Statement 2 is correct.
Now fall in level `=|h_(2)-h'_(2)|`
and rise in level `=|h'_(1)-h_(1)|`
`=(rho_(oil))/(rho_("water"))(h_(2)-h'_(2)) LT h_(2)-h'_(2)`
`rArr ` Fall is more
Statement b is correct


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