Question

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

• A. $w_{A} = w_{B} = w_{C} = w_{D}$
• B. $w_{B} = w_{C} > w_{D} > w_{A}$
• C. $w_{B} = w_{C} > w_{A} > w_{D}$
• D. $w_{B} > w_{C} >w_{D} > w_{A}$

Answer 1

Answer: (B)

CaseA Here only force acting on weighing pan is weight of water

So, $w_{A}=m g$

In this case, downward forces are weight and reaction of buoyant force.
$\therefore \quad w_{B}=m g+F_{B}$

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.
$\therefore w_{C}=m g+F_{B}$

In this case, forces acting on bottom of beaker are
$m' g$ : weight of ball
$mg$ : weight of water
$F_B$ : reaction of buoyant force
$T$ : tension in string
Also, $T=F_B$
So, net downward force on bottom of beaker is $mg + m' g$.
$\therefore w_{D}=mg +m'g$
So, $w_{B} = w_{C} > w_{D} > w_{A}$