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Four identical beakers contain same amount of water as shown below. Beaker $A$ contains only water. A lead ball is held submerged in the beaker $B$ by string from above. A same sized plastic ball, say a table tennis $(TT)$ ball, is held submerged in beaker $C$ by a string attached to a stand from outside. Beaker $D$ contains same sized $TT$ ball which is held submerged from a string attached to the bottom of the beaker. These beakers (without stand) are placed on weighing pans and register readings $w_{A}, w_{B}, w_{C}$ and $w_{D}$ for $A, B, C$ and $D$, respectively. Effects of the mass and volume of the stand and string are to be neglected.
$w_{A}=w_{B}=w_{C}=w_{D}$
$w_{B}=w_{C} > w_{D} > w_{A}$
$w_{B}=w_{C} > w_{A} > w_{D}$
$w_{B} > w_{C} > w_{D} > w_{A}$
Solution
$(b)$ $Case$ $A$ Here only force acting on weighing pan is weight of water.
So, $w_{A}=m g$
$Case$ $B$ In this case, downward forces are weight and reaction of buoyant force.
$\therefore \quad w_{B}=m g+F_{B}$
$Case$ $C$ In this case, downward acting forces are weight and reaction of buoyant force which is same is as that of case $B$ as balls are of same size.
$\therefore \quad w_{C}=m g+F_{B}$
$Case$ $D$ ln this case, forces acting on bottom of beaker are
$m^{\prime} g$ : weight of ball
$m g$ : weight of water
$F_{B}$ : reaction of buoyant force
$T$ : tension in string
Also, $\quad T=F_{B}$
So, net downward force on bottom of beaker is $m g+m^{\prime} g$.
$\therefore \quad w_{D}=m g+m^{\prime} g$
So, $w_{B}=w_{C} > w_{D} > w_{A}$
