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A mass m of volume V=10cm^(2) is tied wi...


A mass m of volume `V=10cm^(2)` is tied with a string to the base of a container filled with a liquid of density `(P_(l)=10^(3)kg//m^(3))` as shown in the figure. The container is then accelerated with acceleration `(a_(x)=9m//s^(2))` and `(a_(y)=2m//s^(2))` a shown in the figure. [`g=10((m)/(s^(2)))` acceleration due to gravity] the net buoyancy force acting on the mass is

A

10 N

B

12 N

C

15 N

D

9 N

Text Solution

Verified by Experts

The correct Answer is:
C
`F_(B)=sqrt(F_(B//x)^(2)+F_(B//y)^(2))`
`F_(B)sqrt([Vrho_(l)(g+a_(y))]^(2)+[Vrho_(l)a_(x)]^(2))`
`=10xx10^(-4)xx10^(3)sqrt((12)^(2)+(9)^(2))=15N`
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