Solution
We are given that the tension forces in the threads remain the same in magnitude whether the vessel is empty (air) or filled with water.
Case 1 (In Air):
The only downward force is gravity ($mg$). The vertical components of the tension $T$ support the weight.
$$ F_{net, vertical} = 2T_y – mg = 0 \implies 2T_y = mg $$
Case 2 (In Water):
There is now an upward buoyancy force $B$.
For the tension magnitude to remain the same in a symmetric setup, the net external vertical force acting on the ball must have the same magnitude as before, even if the direction might reverse.
In water, the forces are Gravity ($mg$, down) and Buoyancy ($B$, up).
If the ball were heavier than water ($mg > B$), the tension would decrease ($2T_y = mg – B$).
If the ball is lighter than water ($B > mg$), the ball tries to float up, and the threads must pull down to keep it in place. The equation becomes:
$$ 2T_y = B – mg $$
Since we are given that Tension is the same, the required restoring force must be the same: $$ (2T_y)_{air} = (2T_y)_{water} $$ $$ mg = B – mg $$ $$ 2mg = B $$
Substituting $m = \rho_{ball} V$ and $B = \rho_{water} V g$: $$ 2 (\rho_{ball} V g) = (\rho_{water} V g) $$ $$ 2 \rho_{ball} = \rho_{water} $$ $$ \rho_{ball} = \frac{1000}{2} = 500 \text{ kg/m}^3 $$
This result holds irrespective of the thread lengths but it depends on point locations (They should be at the same level)
Answer: (d)