An object of mass m and volume v, is totally immersed in a liquid of density p. The tension of a string holding it is
mg
(m + pv)g
(m - pv)g
pvg
[M/m-pv]g
Explanation
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The correct answer is C. (m - pv)g. Here’s why:
When an object is totally immersed in a liquid, it experiences a buoyant force equal to the weight of the liquid displaced by the object. The buoyant force (Fb) is given by the formula:
Fb=pvg
where:
p is the density of the liquid,
v is the volume of the object, and
g is the acceleration due to gravity.
The weight of the object (W) is given by:
W=mg
where:
m is the mass of the object, and
g is the acceleration due to gravity.
The tension (T) in the string holding the object is the difference between the weight of the object and the buoyant force. So, the tension is given by:
T=W−Fb=mg−pvg=(m−pv)g
So, the correct answer is C. (m - pv)g1.

