Question 418: Properties of Matter

Question

Two solid spheres A and B of equal volumes but of different densities d
A
and d
B
are connected by a string. They are fully immersed in a fluid of density d
F
. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2mkuwuo/bd1760f2-e07f-4ae7-9d10-bb1ad686e6ae/dd3ccd00-c8e8-11ec-8c2d-13f33da2fbe1/file-1l2mkuwup.png?format=png)

Accepted Answer

Let V be the volume of each sphere and let T be the tension in the string.
Buoyant force on sphere A is U
A
= d
F
Vg
Buoyant force on sphere B is U
B
= d
F
Vg
Weight of A is W
A
= d
A
Vg
Weight of B is W
B
= d
B
Vg
The free body diagrams of A and B are as follows. (see Fig. 11.34)
For equilibrium,
U
A
= T + W
A
and U
B
+ T = W
B
i.e. d
F
Vg = T + d
A
Vg ..... (i)
and d
F
Vg + T = d
B
Vg .... (ii)
From Eq. (i) ${d_F} = {T \over {Vg}} + {d_A}$. Hence, d
F
> d
A
. So choice (a) is correct.
From Eq. (ii) ${d_B} = {T \over {Vg}} + {d_F}$. Hence d
B
> d
F
. So choice (b) is also correct.
Eliminating T from Eqs. (i) and (ii) we get $2{d_F} = {d_A} + {d_B}$, which is choice (d).
So, the correct choices are (a), (b) and (d).
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2ml18kb/5df75cce-7990-49d2-91e0-d215ab13af3b/8d23c6b0-c8e9-11ec-8c2d-13f33da2fbe1/file-1l2ml18kc.png?format=png)

Answer

d
A
< d
F

Answer

d
B
> d
F

Answer

d
A
> d
F

Answer

d
A
+ d
B
= 2d
F

Properties of Matter

Two solid spheres A and B of equal volumes but of different densities d A and d B are connected by a string. They are fully immersed in a fluid of density d F . They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Choose the correct answer: