
A boy has $60kg$ weight. He wants to swim in a river with the help of a wooden log. If the relative density of wood is $0.6$ . What is the minimum volume of wooden log? (Density of river water is $1000kg{m^{ - 3}}$ )
A. $0.66{m^3}$
B. $150{m^3}$
C. $\dfrac{3}{1}{m^3}$
D. $\dfrac{3}{{20}}{m^3}$
Answer
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Hint: In this question, we need to determine the minimum volume of wooden log such that a boy of weight 60 kilograms wants to swim in the river on the wooden log. For this, the weight of the man and the log together must be balanced by the buoyancy force experienced by the log so, we will equate the two expressions and find out the value of the volume.
Complete step by step answer:
Let $V$ be the volume of the log.
Given that the density of water is $1000kg{m^{ - 3}}$
And the relative density of the log is $0.6$
Thus the absolute density of the log is $0.6 \times 1000kg{m^{ - 3}} = 600kg{m^{ - 3}}$
The total weight of the log and the boy must be balanced by buoyancy force experienced by thatlog
Buoyancy force is the upward force exerted on a body by the liquid wholly or partially immersed in a liquid. It is mathematical the product of volume of the object, density of liquid and acceleration due to gravity. For a partially immersed body, only the portion inside the liquid is taken into consideration for buoyancy force.
Hence equating both the sides we get that
$
{W_{\log }} + {W_{boy}} = {F_{buoyancy}} \\
\left( {600 \times V + 60} \right)g = 1000 \times V \times g \\
\\
$
Now we put cancel the acceleration due to gravity and put the data given in the question
$
600 \times V + 60 = 1000 \times V \\
\implies 400V = 60 \\
\implies V = \dfrac{{60}}{{400}} = \dfrac{3}{{20}}{m^3} \\
$
Thus the minimum volume of log required is $\dfrac{3}{{20}}{m^3}$
So, the correct answer is “Option D”.
Note:
Force of buoyancy is the upward thrust exerted by the displaced liquid on the body in the liquid.
It is worth noting down here that the students should not forget to include the weight of the log as well in the equation.
The concept of absolute and relative density should also be known.
Complete step by step answer:
Let $V$ be the volume of the log.
Given that the density of water is $1000kg{m^{ - 3}}$
And the relative density of the log is $0.6$
Thus the absolute density of the log is $0.6 \times 1000kg{m^{ - 3}} = 600kg{m^{ - 3}}$
The total weight of the log and the boy must be balanced by buoyancy force experienced by thatlog
Buoyancy force is the upward force exerted on a body by the liquid wholly or partially immersed in a liquid. It is mathematical the product of volume of the object, density of liquid and acceleration due to gravity. For a partially immersed body, only the portion inside the liquid is taken into consideration for buoyancy force.
Hence equating both the sides we get that
$
{W_{\log }} + {W_{boy}} = {F_{buoyancy}} \\
\left( {600 \times V + 60} \right)g = 1000 \times V \times g \\
\\
$
Now we put cancel the acceleration due to gravity and put the data given in the question
$
600 \times V + 60 = 1000 \times V \\
\implies 400V = 60 \\
\implies V = \dfrac{{60}}{{400}} = \dfrac{3}{{20}}{m^3} \\
$
Thus the minimum volume of log required is $\dfrac{3}{{20}}{m^3}$
So, the correct answer is “Option D”.
Note:
Force of buoyancy is the upward thrust exerted by the displaced liquid on the body in the liquid.
It is worth noting down here that the students should not forget to include the weight of the log as well in the equation.
The concept of absolute and relative density should also be known.
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