
What is the terminal speed of a $6.00\;kg$ spherical ball that has a radius of $3.00\;cm$ and a drag coefficient of $1.60\;$? The density of the air through which the ball falls is $1.20kg/m^3$
Answer
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Hint:We know that terminal velocity is the maximum possible speed an object falling through any liquid can attain. When the body has net zero force acting on it, then we can say that the acceleration of the body is also zero and thus the body is said to attain terminal velocity.
Formula used:
$v_t=\sqrt{\dfrac{2mg}{\rho AC_d}}$
Complete step-by-step solution:
We know that in fluid dynamics, when an object tries to move through the fluid at terminal velocity, then it also experiences a resistive force, thus the velocity with which it moves remains constant.
The resistive force here is called the drag force, which is expressed in terms of the drag coefficient. Since drag coefficient depends on the surface area of the object it is resisting, we can say that less the drag force i.e. the less the aerodynamic or hydrodynamic, the more streamlined is the object.
The terminal velocity is mathematically expressed as $v_t=\sqrt{\dfrac{2mg}{\rho AC_d}}$, where, $m$ is the mass of the object, $g$ is the acceleration due to gravity, $C_d$ is the drag coefficient due to the fluid with density $\rho$ and $A$ is the surface area exposed to the fluid.
Given that, $m=6kg$ is the mass of the object, $g=9.8m/s$ is the acceleration due to gravity, $C_d=1.6$ is the drag coefficient due to the fluid with density $\rho=1.2kg/m^3$ and assuming that the object is a ball or sphere, then the surface area exposed to the fluid is$A=\pi r^2=\pi (0.03)^2m^2$
Thus, substituting the given we get
$\implies v_t=\sqrt{\dfrac{2\times 6\times 9.8}{1.6\times 1.2 \times \pi \times{ 0.03}^2}}$
$\therefore v_t=147m/s$
Thus the terminal speed of the sphere with the given dimensions is$147m/s$
Note: The terminal velocity of any given object is a constant for that given object i.e. its mass, the area which is exposed to the fluid and it depends on the density of fluid, the drag force acting on the body and majorly. Also, the net force acting on the body when it is attaining the terminal velocity is zero.
Formula used:
$v_t=\sqrt{\dfrac{2mg}{\rho AC_d}}$
Complete step-by-step solution:
We know that in fluid dynamics, when an object tries to move through the fluid at terminal velocity, then it also experiences a resistive force, thus the velocity with which it moves remains constant.
The resistive force here is called the drag force, which is expressed in terms of the drag coefficient. Since drag coefficient depends on the surface area of the object it is resisting, we can say that less the drag force i.e. the less the aerodynamic or hydrodynamic, the more streamlined is the object.
The terminal velocity is mathematically expressed as $v_t=\sqrt{\dfrac{2mg}{\rho AC_d}}$, where, $m$ is the mass of the object, $g$ is the acceleration due to gravity, $C_d$ is the drag coefficient due to the fluid with density $\rho$ and $A$ is the surface area exposed to the fluid.
Given that, $m=6kg$ is the mass of the object, $g=9.8m/s$ is the acceleration due to gravity, $C_d=1.6$ is the drag coefficient due to the fluid with density $\rho=1.2kg/m^3$ and assuming that the object is a ball or sphere, then the surface area exposed to the fluid is$A=\pi r^2=\pi (0.03)^2m^2$
Thus, substituting the given we get
$\implies v_t=\sqrt{\dfrac{2\times 6\times 9.8}{1.6\times 1.2 \times \pi \times{ 0.03}^2}}$
$\therefore v_t=147m/s$
Thus the terminal speed of the sphere with the given dimensions is$147m/s$
Note: The terminal velocity of any given object is a constant for that given object i.e. its mass, the area which is exposed to the fluid and it depends on the density of fluid, the drag force acting on the body and majorly. Also, the net force acting on the body when it is attaining the terminal velocity is zero.
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