Two spheres P and Q for equal radii have densities \rho 1 and \rho 2, respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities \sigma 1 and \sigma 2 and viscosities \eta _1 and \eta _2 , respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure) If sphere P along in L2 has terminal velocity vP and Q alone in L1 ha terminal velocity vQ, then

|v_P| \over |v_Q| = \eta _1 \over \eta _2, vP . vQ < 0

|v_P| \over |v_Q| = \eta _1 \over \eta _2

|v_P| \over |v_Q| = \eta _2 \over \eta _1

Properties of Matter PYQ — JEE Advanced 2015 · Shift 2 · Q56 with Solution

View written solutionFree

Correct answer: A, D

Given setup

Two spheres $P$ and $Q$ have equal radii, hence equal volumes: $V=\frac{4}{3}\pi r^3$

Their densities are $\rho_1$ and $\rho_2$ respectively.

They are connected by a massless string and float in equilibrium at the interface, with:

sphere $P$ in liquid $L_1$ of density $\sigma_1$ , viscosity $\eta_1$
sphere $Q$ in liquid $L_2$ of density $\sigma_2$ , viscosity $\eta_2$

The string is taut.

We are also told:

$v_P$ = terminal velocity of sphere $P$ when placed alone in $L_2$
$v_Q$ = terminal velocity of sphere $Q$ when placed alone in $L_1$

We must determine the correct options.

Use equilibrium of the connected system

Since the two-sphere system is in equilibrium and the string is taut, tensions act internally. For the whole system, net external force is zero.

Weight of both spheres: $W=(\rho_1 Vg+\rho_2 Vg)=(\rho_1+\rho_2)Vg$

Buoyant force from the two liquids: $B=(\sigma_1 Vg+\sigma_2 Vg)=(\sigma_1+\sigma_2)Vg$

Equilibrium gives: $(\rho_1+\rho_2)Vg=(\sigma_1+\sigma_2)Vg$

So, $\rho_1+\rho_2=\sigma_1+\sigma_2$

Hence, $\rho_1-\sigma_2=\sigma_1-\rho_2$

or equivalently, $\rho_2-\sigma_1=\sigma_2-\rho_1=-(\rho_1-\sigma_2)$

This relation is the key.

Terminal velocity formula using Stokes' law

For a sphere of radius $r$ moving slowly through a viscous liquid, terminal velocity is obtained by balancing effective weight with viscous drag: $6\pi \eta r |v|=|\rho-\sigma|Vg$

Since $V=\frac{4}{3}\pi r^3$ we get $|v|=\frac{2r^2g}{9\eta}|\rho-\sigma|$

Direction depends on the sign of $(\rho-\sigma)$ :

if $\rho>\sigma$ , sphere moves downward
if $\rho<\sigma$ , sphere moves upward

Find $v_P$ when sphere $P$ is alone in $L_2$

Here density difference is: $\rho_1-\sigma_2$

Therefore, $|v_P|=\frac{2r^2g}{9\eta_2}|\rho_1-\sigma_2|$

Find $v_Q$ when sphere $Q$ is alone in $L_1$

Here density difference is: $\rho_2-\sigma_1$

Therefore, $|v_Q|=\frac{2r^2g}{9\eta_1}|\rho_2-\sigma_1|$

But from equilibrium relation above, $\rho_2-\sigma_1=-(\rho_1-\sigma_2)$

Hence, $|\rho_2-\sigma_1|=|\rho_1-\sigma_2|$

So,

\frac{\dfrac{2r^2g}{9\eta_2}|\rho_1-\sigma_2|}{\dfrac{2r^2g}{9\eta_1}|\rho_2-\sigma_1|} =\frac{\eta_1}{\eta_2}$$ Thus **Option A is correct** and **Option B is incorrect**. --- 6. **Determine the sign of $v_P\cdot v_Q$** From $$\rho_2-\sigma_1=-(\rho_1-\sigma_2)$$ we see that the two density differences have opposite signs. That means: - if $P$ in $L_2$ moves downward, then $Q$ in $L_1$ moves upward, - or vice versa. So $v_P$ and $v_Q$ are opposite in sign. Therefore, $$v_P\cdot v_Q<0$$ Thus **Option D is correct** and **Option C is incorrect**. --- 7. **Final answer** Correct options are: $$\boxed{A,\ D}$$ --- 8. **Comparison with stored answer** Stored correct answer: $A, D$ My derived answer matches the stored answer exactly.

More from Properties of Matter

View written solutionFree

Correct answer: A, D

Given setup

Two spheres $P$ and $Q$ have equal radii, hence equal volumes: $V=\frac{4}{3}\pi r^3$

Their densities are $\rho_1$ and $\rho_2$ respectively.

They are connected by a massless string and float in equilibrium at the interface, with:

sphere $P$ in liquid $L_1$ of density $\sigma_1$ , viscosity $\eta_1$
sphere $Q$ in liquid $L_2$ of density $\sigma_2$ , viscosity $\eta_2$

The string is taut.

We are also told:

$v_P$ = terminal velocity of sphere $P$ when placed alone in $L_2$
$v_Q$ = terminal velocity of sphere $Q$ when placed alone in $L_1$

We must determine the correct options.

Use equilibrium of the connected system

Since the two-sphere system is in equilibrium and the string is taut, tensions act internally. For the whole system, net external force is zero.

Weight of both spheres: $W=(\rho_1 Vg+\rho_2 Vg)=(\rho_1+\rho_2)Vg$

Buoyant force from the two liquids: $B=(\sigma_1 Vg+\sigma_2 Vg)=(\sigma_1+\sigma_2)Vg$

Equilibrium gives: $(\rho_1+\rho_2)Vg=(\sigma_1+\sigma_2)Vg$

So, $\rho_1+\rho_2=\sigma_1+\sigma_2$

Hence, $\rho_1-\sigma_2=\sigma_1-\rho_2$

or equivalently, $\rho_2-\sigma_1=\sigma_2-\rho_1=-(\rho_1-\sigma_2)$

This relation is the key.

Terminal velocity formula using Stokes' law

For a sphere of radius $r$ moving slowly through a viscous liquid, terminal velocity is obtained by balancing effective weight with viscous drag: $6\pi \eta r |v|=|\rho-\sigma|Vg$

Since $V=\frac{4}{3}\pi r^3$ we get $|v|=\frac{2r^2g}{9\eta}|\rho-\sigma|$

Direction depends on the sign of $(\rho-\sigma)$ :

if $\rho>\sigma$ , sphere moves downward
if $\rho<\sigma$ , sphere moves upward

Find $v_P$ when sphere $P$ is alone in $L_2$

Here density difference is: $\rho_1-\sigma_2$

Therefore, $|v_P|=\frac{2r^2g}{9\eta_2}|\rho_1-\sigma_2|$

Find $v_Q$ when sphere $Q$ is alone in $L_1$

Here density difference is: $\rho_2-\sigma_1$

Therefore, $|v_Q|=\frac{2r^2g}{9\eta_1}|\rho_2-\sigma_1|$

But from equilibrium relation above, $\rho_2-\sigma_1=-(\rho_1-\sigma_2)$

Hence, $|\rho_2-\sigma_1|=|\rho_1-\sigma_2|$

So,

\frac{\dfrac{2r^2g}{9\eta_2}|\rho_1-\sigma_2|}{\dfrac{2r^2g}{9\eta_1}|\rho_2-\sigma_1|} =\frac{\eta_1}{\eta_2}$$ Thus **Option A is correct** and **Option B is incorrect**. --- 6. **Determine the sign of $v_P\cdot v_Q$** From $$\rho_2-\sigma_1=-(\rho_1-\sigma_2)$$ we see that the two density differences have opposite signs. That means: - if $P$ in $L_2$ moves downward, then $Q$ in $L_1$ moves upward, - or vice versa. So $v_P$ and $v_Q$ are opposite in sign. Therefore, $$v_P\cdot v_Q<0$$ Thus **Option D is correct** and **Option C is incorrect**. --- 7. **Final answer** Correct options are: $$\boxed{A,\ D}$$ --- 8. **Comparison with stored answer** Stored correct answer: $A, D$ My derived answer matches the stored answer exactly.

Properties of Matter question

Properties of Matter question

More from Properties of Matter

Properties of Matter question

Properties of Matter question

More from Properties of Matter