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A block of material of specific gravity ...

A block of material of specific gravity 0.4 is held submerged at a depth of 1m in a vessel filled with water. The vessel is accelerated upwards with acceleration of `a_(o) = g//5`. If the block is released at t = 0, neglecting viscous effects, it will reach the water surface at t equal to `(g = 10 m/s^(2))` :

A

0.6 sec

B

0.4 sec

C

1.2 sec

D

1 sec

Text Solution

Verified by Experts

The correct Answer is:
B
Buoyant force =`Vrho(g+a)` (a is acceleratuion of vessel=`(g)/(2)`)
Weight `mg=(Vrhog)/(2)` (density of material is `rho//2`)
Applying second law, `B-mg=ma_(1)` (`a_(1)` is acceleration of block) `" "Rightarrowa_(1)=2g`
Therefore, acceleration with respect to vessel `a_("relative")=a_(1)-a=(3g)/(2)`
`S_(relative)=(1)/(2)a_(relative)t^(2)`, Solving t=0.4 sec
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