A block of steel of size \[5cm \times 5cm \times 5cm\] is weighed in water. If the relative density of steel is 7. Its apparent weight is:
A) $6 \times 5 \times 5 \times 6gf$
B) $4 \times 4 \times 4 \times 7gf$
C) $5 \times 5 \times 5 \times 7gf$
D) $4 \times 4 \times 4 \times 6gf$
Answer
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Hint: Whenever an object is immersed partly or completely in a fluid, the fluid exerts an upward force on it. The property of a fluid, due to which it exerts an upward force on any object immersed in the fluid, is called buoyancy. It is a property exhibited by all fluids, whether liquids or gases. The upward force on an immersed object is called the buoyant force or upthrust or force of buoyancy.
Complete step by step solution:
When an object is wholly or partially immersed in a fluid two forces act on objects that are the weight of the object due to gravitational force in downward direction and buoyant force in upward direction due to fluid displaced.
So, the net downward force acting on object is called apparent weight and is denoted by
$W’ = \text{true weight of object} - \text{buoyant force$(F_B)$} $............(1)
Now mass is not given so we can write mass \[ = {{ }}density{{ }}\left( \rho \right){{ }} \times {{ }}volume{{ }}\left( V \right)\]
Density of steel \[{\rho _S}{{ }} = {{ }}7\]and that of water is ${\rho _w} = 1$
Now${\rho _s}$ will be 7 times of ${\rho _w}$ so we can write \[{\rho _s} = 7{\rho _w}\]
So now equation 1 becomes
$W = {\rho _s}Vg - {\rho _w}Vg$ (where g is the acceleration due to gravity)
As ${\rho _s} = 7{\rho _w}$
$\Rightarrow W = 7{\rho _w}Vg - {\rho _w}Vg$
Now it becomes
\[\Rightarrow (7{\rho _w} - {\rho _w})Vg = 6{\rho _w}Vg\]
Now putting values in this equation
\[\Rightarrow 6 \times 5 \times 5 \times 5 \times gf\]
Now let us match this value with given options :-
Option A: This value exactly matches the calculated value , thus this option is correct.
Option B :This value does not match with the value we find . Thus, this option is not correct.
Option C: As dimensions are \[5{{ }} \times {{ }}5{{ }} \times {{ }}5\], this value does not match with the value we find. Thus, this option is not correct.
Option D: This value does not match with the value we find. Thus, this option is not correct.
Our required answer is A that is $6 \times 5 \times 5 \times 5g$.
Note: Whenever an object is immersed wholly or partially in a fluid, there is an apparent decrease in its weight and this apparent decrease in weight is equal to the weight of the fluid displaced by the object. The buoyant force is directly proportional to volume of immersed part of object in fluid and to the density of fluid.
Complete step by step solution:
When an object is wholly or partially immersed in a fluid two forces act on objects that are the weight of the object due to gravitational force in downward direction and buoyant force in upward direction due to fluid displaced.
So, the net downward force acting on object is called apparent weight and is denoted by
$W’ = \text{true weight of object} - \text{buoyant force$(F_B)$} $............(1)
Now mass is not given so we can write mass \[ = {{ }}density{{ }}\left( \rho \right){{ }} \times {{ }}volume{{ }}\left( V \right)\]
Density of steel \[{\rho _S}{{ }} = {{ }}7\]and that of water is ${\rho _w} = 1$
Now${\rho _s}$ will be 7 times of ${\rho _w}$ so we can write \[{\rho _s} = 7{\rho _w}\]
So now equation 1 becomes
$W = {\rho _s}Vg - {\rho _w}Vg$ (where g is the acceleration due to gravity)
As ${\rho _s} = 7{\rho _w}$
$\Rightarrow W = 7{\rho _w}Vg - {\rho _w}Vg$
Now it becomes
\[\Rightarrow (7{\rho _w} - {\rho _w})Vg = 6{\rho _w}Vg\]
Now putting values in this equation
\[\Rightarrow 6 \times 5 \times 5 \times 5 \times gf\]
Now let us match this value with given options :-
Option A: This value exactly matches the calculated value , thus this option is correct.
Option B :This value does not match with the value we find . Thus, this option is not correct.
Option C: As dimensions are \[5{{ }} \times {{ }}5{{ }} \times {{ }}5\], this value does not match with the value we find. Thus, this option is not correct.
Option D: This value does not match with the value we find. Thus, this option is not correct.
Our required answer is A that is $6 \times 5 \times 5 \times 5g$.
Note: Whenever an object is immersed wholly or partially in a fluid, there is an apparent decrease in its weight and this apparent decrease in weight is equal to the weight of the fluid displaced by the object. The buoyant force is directly proportional to volume of immersed part of object in fluid and to the density of fluid.
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