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Question: A fisherman hooks an old log of wood of weight 12N and volume\(1000c{m^3}\). He pulls the log half w...

A fisherman hooks an old log of wood of weight 12N and volume1000cm31000c{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.

Explanation

Solution

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,
Fc=ρliq.Vg{F_c} = {\rho _{liq.}}Vg
Where buoyancy force isFc{F_c}, the density of the liquid isρliq.{\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 1000cm31000c{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,
Fc=ρliq.Vg{F_c} = {\rho _{liq.}}Vg
Where buoyancy force is Fc{F_c}, the density of the liquid is ρliq.{\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 1000kgm31000\dfrac{{kg}}{{{m^3}}}, the volume displaced for this case is equal to 1000cm31000c{m^3} which can also be represented as 0001m30 \cdot 001{m^3} the acceleration due to gravity is
g=10ms2g = 10m{s^{ - 2}}. Therefore the buoyancy on the object is equal to,
Fc=ρliq.Vg\Rightarrow {F_c} = {\rho _{liq.}}Vg
Fc=(103×0001×10)2\Rightarrow {F_c} = \dfrac{{\left( {{{10}^3} \times 0 \cdot 001 \times 10} \right)}}{2}
Fc=5N\Rightarrow {F_c} = 5N………eq. (1)
The weight of the wooden block is equal toW=12NW = 12N.
The tension in the string is equal to,
T+Fc=W\Rightarrow T + {F_c} = W………eq. (2)
Replace the value of weight and buoyancy from equation (1) in the equation (2)
T+Fc=W\Rightarrow T + {F_c} = W
T+5=12\Rightarrow T + 5 = 12
T=125\Rightarrow T = 12 - 5
T=7N\Rightarrow T = 7N.
The tension in the string is equal toT=7NT = 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.