
A fisherman hooks an old log of wood of weight 12N and volume$1000c{m^3}$. He pulls the log half way out of water. The tension in the tension in the string at this instant is:
A) 12N.
B) 8N.
C) 10N.
D) 7N.
Answer
580.2k+ views
Hint:Buoyancy force is defined as the upthrust force that acts on the body if a body is submerged into any liquid. The tension always acts on the opposite side of the attached object and for a string attached with the object the string develops tension which is equal to the weight of the body.
Formula used: The buoyancy force on the submerged body is equal to,
${F_c} = {\rho _{liq.}}Vg$
Where buoyancy force is${F_c}$, the density of the liquid is${\rho _{liq.}}$, the volume of liquid is displaced by the object and acceleration due to gravity is g.
Complete step by step answer:
It is given in the problem that a fisherman hooks an old log of wood of weight 12N and volume $1000c{m^3}$ the fisherman pulls the log half way out of water and we need to find the tension in the string.
The buoyancy force on the submerged body is equal to,
${F_c} = {\rho _{liq.}}Vg$
Where buoyancy force is ${F_c}$, the density of the liquid is ${\rho _{liq.}}$, the volume of liquid is displaced by the object and acceleration due to gravity is g.
As the density of the liquid is $1000\dfrac{{kg}}{{{m^3}}}$, the volume displaced for this case is equal to $1000c{m^3}$ which can also be represented as $0 \cdot 001{m^3}$ the acceleration due to gravity is
$g = 10m{s^{ - 2}}$. Therefore the buoyancy on the object is equal to,
$ \Rightarrow {F_c} = {\rho _{liq.}}Vg$
$ \Rightarrow {F_c} = \dfrac{{\left( {{{10}^3} \times 0 \cdot 001 \times 10} \right)}}{2}$
$ \Rightarrow {F_c} = 5N$………eq. (1)
The weight of the wooden block is equal to$W = 12N$.
The tension in the string is equal to,
$ \Rightarrow T + {F_c} = W$………eq. (2)
Replace the value of weight and buoyancy from equation (1) in the equation (2)
$ \Rightarrow T + {F_c} = W$
$ \Rightarrow T + 5 = 12$
$ \Rightarrow T = 12 - 5$
$ \Rightarrow T = 7N$.
The tension in the string is equal to$T = 7N$.
The correct answer for this problem is option D.
Note: It is advisable for students to understand and remember the formula of the buoyancy as it can help students to solve problems of these kinds. The volume in the formula of the buoyancy force is equal to the volume of the liquid displaced by the object.
Formula used: The buoyancy force on the submerged body is equal to,
${F_c} = {\rho _{liq.}}Vg$
Where buoyancy force is${F_c}$, the density of the liquid is${\rho _{liq.}}$, the volume of liquid is displaced by the object and acceleration due to gravity is g.
Complete step by step answer:
It is given in the problem that a fisherman hooks an old log of wood of weight 12N and volume $1000c{m^3}$ the fisherman pulls the log half way out of water and we need to find the tension in the string.
The buoyancy force on the submerged body is equal to,
${F_c} = {\rho _{liq.}}Vg$
Where buoyancy force is ${F_c}$, the density of the liquid is ${\rho _{liq.}}$, the volume of liquid is displaced by the object and acceleration due to gravity is g.
As the density of the liquid is $1000\dfrac{{kg}}{{{m^3}}}$, the volume displaced for this case is equal to $1000c{m^3}$ which can also be represented as $0 \cdot 001{m^3}$ the acceleration due to gravity is
$g = 10m{s^{ - 2}}$. Therefore the buoyancy on the object is equal to,
$ \Rightarrow {F_c} = {\rho _{liq.}}Vg$
$ \Rightarrow {F_c} = \dfrac{{\left( {{{10}^3} \times 0 \cdot 001 \times 10} \right)}}{2}$
$ \Rightarrow {F_c} = 5N$………eq. (1)
The weight of the wooden block is equal to$W = 12N$.
The tension in the string is equal to,
$ \Rightarrow T + {F_c} = W$………eq. (2)
Replace the value of weight and buoyancy from equation (1) in the equation (2)
$ \Rightarrow T + {F_c} = W$
$ \Rightarrow T + 5 = 12$
$ \Rightarrow T = 12 - 5$
$ \Rightarrow T = 7N$.
The tension in the string is equal to$T = 7N$.
The correct answer for this problem is option D.
Note: It is advisable for students to understand and remember the formula of the buoyancy as it can help students to solve problems of these kinds. The volume in the formula of the buoyancy force is equal to the volume of the liquid displaced by the object.
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