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71 Questions around this concept.
A jar is filled with two non - mixing liquids 1 and 2 having densities $\rho_1$ and $\rho_2$ respectively. A solid ball, made of a material of density $\rho_3$ is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $\rho_1, \rho_2$ and $\rho_3$ ?
A ball is made of a material of density $\rho$ where $ \rho_{\text {oil }}<\rho<\rho_{\text {water }}$ with $\rho_{\text {oil }}$ and $\rho_{\text {water }}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?
When system has zero acceleration, the tension in the string connected to a submerged body
Archimedes principle states that when a body is immersed partly or wholly in a fluid, then the liquid exerts an upward force/upthrust/buoyant force on the body which is equal to the weight of the fluid displaced by the body.
Buoyant force-
The buoyant force is given as
$
F_B=\rho V g
$
Where $\mathrm{F}_{\mathrm{B}}=$ Buoyant force
$\rho=$ density of the fluid
$\mathrm{V}=$ Volume of the solid body immersed in the liquid or Volume of the fluid displaced
$
\text { As } \rho=\frac{m}{V}
$
So we can write $F_B=m g_{\text {=weight of the fluid displaced }}$
Where $\mathrm{m}=$ mass of the fluid displaced
- The buoyant force acts vertically upwards (opposite to the weight of the body)
- The buoyant force is independent of mass, size, the density of the body inside the fluid.
- The buoyant force depends upon the nature/density of the displaced fluid.
Apparent weight-
- Apparent weight=(Actual weight)-(Buoyant force)
- The apparent weight of the body of density $(\rho)$ when immersed in a liquid of density $(\sigma)$ is given by
$
\begin{aligned}
& \quad W_{a p p}=W-F_B=V \rho g-V \sigma g=V \rho g\left(1-\frac{\sigma}{\rho}\right) \\
& W_{a p p}=W\left(1-\frac{\sigma}{\rho}\right)
\end{aligned}
$
Where W= Actual weight of the body
V= volume of the body immersed in a liquid
From this, we can say that
If a body of volume V is immersed in a liquid of density () Then its weight reduces
And Loss in weight is given by
$
W_{l o s s}=W-W_{a p p}=V \sigma g
$
- The relative density of a body
$
\text { R.D }=\frac{\text { density of body }}{\text { density of water }}
$
- Floatation-
When a body of density $\rho$ is immersed in a liquid of density $\sigma$,
Then the body will float if the buoyant force on the body which is equal to the weight of the fluid displaced by the body. Means body is in equilibrium.|
1. If the density of the body is equal to that of liquid i.e $\rho=\sigma$
Then the Weight of the body will be equal to upthrust.
And the body will float but the body will fully be submerged in liquid.
2. If the density of the body is less than that of liquid $\rho<\sigma$
Then the Weight of the body will be less than upthrust.
And the body will float but the body will partially be immersed in liquid.
- If the density of the body is greater than that of liquid $\rho>\sigma$
Then the Weight of the body will be greater than upthrust So body will sink.
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