
A beaker containing water is balanced on the pan of common balance. A solid of specific gravity 1 and mass of 5g is tied to the arm of the balance and is immersed in water contained in the beaker. The scale pan with the beaker
a) goes down
b) goes up
c) remains unchanged
d) none of these
Answer
574.8k+ views
Hint: The specific gravity of any substance is defined as the ratio of the density of that substance to the density of liquid. It is given that the density of the substance is 1. Hence we can imply that its density will be equal to the density of water. When a substance is immersed in liquid, the volume of the liquid that gets displaced is equal to the volume of the object immersed into the liquid. From this we can calculate if there is a decrease in mass or there is an increase in the mass contained in the beaker and accordingly will determine the change in position of the pan.
Complete answer:
In the above question it is given to us that the beaker containing the water is balanced on a pan. When the solid is immersed into the beaker, the water will spill off from the beaker. Let us say the mass contained in the beaker initially as X and the water that spilled off has a mass of Y. We know that the volume of the displaced water is equal to the volume of the solid and that the density of the two is equal. Density of a substance is given by,
$Density=\dfrac{mass}{volume}$. Since the density of water is equal to the density of the solid we can write,
$Density(water)=Density(solid)$. The given mass of the solid is 5g and the volume of the displaced water is equal to the volume of the solid i.e. V. Hence the mass of the water displaced is,
$\begin{align}
& Density(water)=Density(solid) \\
& \dfrac{M(\text{displaced water})}{Volume}=\dfrac{M(solid)}{Volume} \\
& \dfrac{M(\text{displaced water})}{V}=\dfrac{5}{V} \\
& M(\text{displaced water})=5g \\
\end{align}$
From this we can conclude that we removed 5 g from the beaker and inserted back 5g to the beaker. Hence the pan will not move from its equilibrium position.
So, the correct answer is “Option C”.
Note:
Whatever water gets spilled from the beaker is assumed to be falling on the ground and not back into the pan. In that case the weight on the pan will increase and consequently move down. We have only considered the changes that take place in the beaker.
Complete answer:
In the above question it is given to us that the beaker containing the water is balanced on a pan. When the solid is immersed into the beaker, the water will spill off from the beaker. Let us say the mass contained in the beaker initially as X and the water that spilled off has a mass of Y. We know that the volume of the displaced water is equal to the volume of the solid and that the density of the two is equal. Density of a substance is given by,
$Density=\dfrac{mass}{volume}$. Since the density of water is equal to the density of the solid we can write,
$Density(water)=Density(solid)$. The given mass of the solid is 5g and the volume of the displaced water is equal to the volume of the solid i.e. V. Hence the mass of the water displaced is,
$\begin{align}
& Density(water)=Density(solid) \\
& \dfrac{M(\text{displaced water})}{Volume}=\dfrac{M(solid)}{Volume} \\
& \dfrac{M(\text{displaced water})}{V}=\dfrac{5}{V} \\
& M(\text{displaced water})=5g \\
\end{align}$
From this we can conclude that we removed 5 g from the beaker and inserted back 5g to the beaker. Hence the pan will not move from its equilibrium position.
So, the correct answer is “Option C”.
Note:
Whatever water gets spilled from the beaker is assumed to be falling on the ground and not back into the pan. In that case the weight on the pan will increase and consequently move down. We have only considered the changes that take place in the beaker.
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