Four identical beakers contain the same amoun | vedclass

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(B) Case $A$: Here,only the force acting on the weighing pan is the weight of the water. So,$w_{A} = mg$.Case $B$: In this case,the downward forces on

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Case $A$: Here,only the force acting on the weighing pan is the weight of the water. So,$w_{A} = mg$.Case $B$: In this case,the downward forces on the pan are the weight of the water and the reaction force equal to the buoyant force $(F_{B})$ exerted by the water on the ball. So,$w_{B} = mg + F_{B}$.Case $C$: In this case,the downward acting forces are the weight of the water and the reaction of the buoyant force,which is the same as in case $B$ because the balls are of the same size. So,$w_{C} = mg + F_{B}$.Case $D$: In this case,the forces acting on the bottom of the beaker are the weight of the water $(mg)$,the weight of the ball $(m'g)$,and the tension $(T)$ in the string pulling upwards. The buoyant force $(F_{B})$ acts upwards on the ball. For the ball to be in equilibrium,$T + F_{B} = m'g$,so $T = m'g - F_{B}$. The total force on the pan is the weight of the water plus the weight of the ball minus the upward tension: $w_{D} = mg + m'g - T = mg + m'g - (m'g - F_{B}) = mg + F_{B}$.Wait,let's re-evaluate Case $D$: The total downward force on the pan is the weight of the water $(mg)$ plus the weight of the ball $(m'g)$ minus the tension $(T)$ in the string. Since the ball is submerged,$F_{B} + T = m'g$,so $T = m'g - F_{B}$. Thus,$w_{D} = mg + m'g - (m'g - F_{B}) = mg + F_{B}$.Therefore,$w_{B} = w_{C} = w_{D} > w_{A}$. However,if the $TT$ ball is very light $(m' \approx 0)$,then $w_{D} \approx mg + F_{B}$. If the lead ball is heavy,$w_{B} = mg + F_{B}$. The correct relationship is $w_{B} = w_{C} = w_{D} > w_{A}$.

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