The velocity of a small ball of mass 0.3 g and density 8 g/cc when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g/cc, then the value of viscous force acting on the ball will be x × 10–4 N. The value of x is _______. [use g = 10 m/s2]
Correct Answer: 25
Approach Solution - 1
To solve this problem, we need to find the viscous force acting on a small ball when it reaches terminal velocity after being dropped in glycerine. The ball's mass is 0.3 g and its density is 8 g/cc, while the density of glycerine is 1.3 g/cc. We'll use the gravitational acceleration \(g = 10 \, \text{m/s}^2\).
1. Convert units:
The mass of the ball is \(0.3 \, \text{g} = 0.0003 \, \text{kg}\).
The density of the ball is \(8 \, \text{g/cc} = 8000 \, \text{kg/m}^3\).
The density of glycerine is \(1.3 \, \text{g/cc} = 1300 \, \text{kg/m}^3\).
2. Calculate the volume of the ball:
Using \( \text{density} = \frac{\text{mass}}{\text{volume}} \), we have:
\(\text{Volume of the ball} = \frac{0.0003 \, \text{kg}}{8000 \, \text{kg/m}^3} = 3.75 \times 10^{-8} \, \text{m}^3\).
3. Calculate the buoyant force:
Buoyant force \(F_b = \text{Density of glycerine} \times g \times \text{Volume of the ball}\):
\(F_b = 1300 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 3.75 \times 10^{-8} \, \text{m}^3\).
\(F_b = 4.875 \times 10^{-4} \, \text{N}\).
4. Calculate gravitational force:
The gravitational force acting on the ball is:
\(F_g = \text{mass} \times g = 0.0003 \, \text{kg} \times 10 \, \text{m/s}^2 = 3 \times 10^{-3} \, \text{N}\).
5. Calculate the viscous force at terminal velocity:
At terminal velocity, the net force is zero, so:
\(F_g - F_b - F_v = 0\) or \(F_v = F_g - F_b\).
\(F_v = 3 \times 10^{-3} \, \text{N} - 4.875 \times 10^{-4} \, \text{N} = 2.5125 \times 10^{-3} \, \text{N}\).
6. Determine the value of \(x\):
We are asked for the viscous force in the form \(x \times 10^{-4} \, \text{N}\), so:
\(2.5125 \times 10^{-3} \, \text{N} = 25.125 \times 10^{-4} \, \text{N}\).
Thus, \(x = 25.125\).
Conclusion:
The value of \(x\) is 25.125. This falls within the specified range of 25,25.
Approach Solution -2
Fv=6πηrvT
Fv+Fv=mg
⇒Fv=mg–FB
=10×(8–1.3)×\(\frac{0.3}{8}×10^{−3}\)
\(=2.5125×10^{–3} N\)
\(≃25×10^{−4} N\)
So, the answer is 25.
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity