
A test tube of thin walls has some lead shots in it at its bottom and the system floats vertically in water, sinking by a length \[{{l}_{0}}=10cm\]. A liquid of density less than that of water, is poured into the tube till the levels inside and outside the tube are even. If the tube now sinks to a length \[l=40cm\], the specific gravity of the liquid is \[\dfrac{(70+x)}{100}\]. Find \[x\].
Answer
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Hint: Archimedes principle has been used in this question. The formula for the amount of liquid displaced has been used. Upward force on the test tube and liquid has been calculated. Calculate the weight of water displaced and liquid poured in the system.
Formula used:
\[W=Al\rho g\]
Complete answer:
According to the question, the test tube has some lead shots in its bottom and floats vertically in water. Therefore, using the concept of classical mechanics we know upward thrust will be equal to weight of lead shots and the test tube.
Let us consider \[A\] to be the area of cross section, \[l\] be the length of tube and \[\rho \] be the density of liquid.
First of all we will calculate the weight of water displaced. Here \[{{W}_{1}}\] is the sum of weight of lead shots and weight of liquid poured. This can be written as
\[\begin{align}
& {{W}_{1}}=Al{{\rho }_{w}}g \\
& {{W}_{1}}={{W}_{lead}}+{{W}_{liquid}} \\
\end{align}\]
\[={{W}_{lead}}+Al{{\rho }_{1}}g\]
\[{{W}_{lead}}={{W}_{1}}-Al{{\rho }_{1}}g\]
\[=Al{{\rho }_{w}}g-Al{{\rho }_{1}}g\]
From above equation
\[Al{{\rho }_{w}}g-Al{{\rho }_{1}}g=A{{l}_{0}}{{\rho }_{w}}g\]
\[l{{\rho }_{w}}-l{{\rho }_{1}}={{l}_{0}}{{\rho }_{w}}\]
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{w}}}=1-\dfrac{{{l}_{0}}}{l}=1-\dfrac{10}{40}=\dfrac{3}{4}=0.75\]
So the specific gravity of liquid is found to be \[0.75\].
Additional information:
When an object is immersed in a liquid it experiences buoyant force which is equal to the force acting on the displaced fluid. It has various applications and can be used for calculating volume of the object.
Note:
Archimedes Principle basically tells us about the behavior of any object immersed in any liquid. Formula plays a significant role and the upward thrust should be calculated carefully. Density of different liquids should be named properly to avoid confusion.
Formula used:
\[W=Al\rho g\]
Complete answer:
According to the question, the test tube has some lead shots in its bottom and floats vertically in water. Therefore, using the concept of classical mechanics we know upward thrust will be equal to weight of lead shots and the test tube.
Let us consider \[A\] to be the area of cross section, \[l\] be the length of tube and \[\rho \] be the density of liquid.
First of all we will calculate the weight of water displaced. Here \[{{W}_{1}}\] is the sum of weight of lead shots and weight of liquid poured. This can be written as
\[\begin{align}
& {{W}_{1}}=Al{{\rho }_{w}}g \\
& {{W}_{1}}={{W}_{lead}}+{{W}_{liquid}} \\
\end{align}\]
\[={{W}_{lead}}+Al{{\rho }_{1}}g\]
\[{{W}_{lead}}={{W}_{1}}-Al{{\rho }_{1}}g\]
\[=Al{{\rho }_{w}}g-Al{{\rho }_{1}}g\]
From above equation
\[Al{{\rho }_{w}}g-Al{{\rho }_{1}}g=A{{l}_{0}}{{\rho }_{w}}g\]
\[l{{\rho }_{w}}-l{{\rho }_{1}}={{l}_{0}}{{\rho }_{w}}\]
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{w}}}=1-\dfrac{{{l}_{0}}}{l}=1-\dfrac{10}{40}=\dfrac{3}{4}=0.75\]
So the specific gravity of liquid is found to be \[0.75\].
Additional information:
When an object is immersed in a liquid it experiences buoyant force which is equal to the force acting on the displaced fluid. It has various applications and can be used for calculating volume of the object.
Note:
Archimedes Principle basically tells us about the behavior of any object immersed in any liquid. Formula plays a significant role and the upward thrust should be calculated carefully. Density of different liquids should be named properly to avoid confusion.
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