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A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300?
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A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K....
Given:
Mass of the particle, m = 1.4 x 10^(-11) kg
Temperature, T = 300 K
To find:
The root mean square (rms) speed of the particle due to random collisions with air molecules.
Solution:
1. Calculation of Boltzmann constant:
The rms speed of a particle can be calculated using the formula:
v = √(3kT/m)
where,
v = rms speed of the particle
k = Boltzmann constant
T = Temperature in Kelvin
m = mass of the particle
To calculate the value of k, we can rearrange the equation as:
k = (v^2 * m) / (3T)
2. Conversion of mass to kg:
Given mass of the particle is in grams, we need to convert it to kilograms.
1 gram = 10^(-3) kg
So, the mass of the particle in kilograms is:
m = 1.4 x 10^(-11) kg
3. Conversion of temperature to Kelvin:
The given temperature is already in Kelvin, so no conversion is required.
T = 300 K
4. Calculation of rms speed:
Using the above values, we can calculate the Boltzmann constant:
k = (v^2 * m) / (3T)
5. Substituting the values:
Substituting the given values:
k = (v^2 * 1.4 x 10^(-11)) / (3 x 300)
6. Calculation:
Solving the equation, we get:
v^2 = (3 x 300 x k) / (1.4 x 10^(-11))
v^2 = (900 x k) / (1.4 x 10^(-11))
v^2 = (6.43 x 10^11) x k
Taking the square root of both sides, we get:
v = √((6.43 x 10^11) x k)
7. Calculation of rms speed:
Finally, substituting the value of Boltzmann constant, k = 1.38 x 10^(-23) J/K, we get:
v = √((6.43 x 10^11) x 1.38 x 10^(-23))
v = √(8.87 x 10^(-12))
v = 2.98 x 10^(-6) m/s
8. Conversion to micrometers per second:
1 meter = 10^6 micrometers
So, the rms speed of the particle in micrometers per second is:
v = 2.98 x 10^(-6) x 10^6
v = 2.98 m/s
Therefore, the rms speed of the particle due to random collisions with air molecules is closest to 2.98 μm/s.
Answer:
The correct option is (b) 3.
Mass of the particle, m = 1.4 x 10^(-11) kg
Temperature, T = 300 K
To find:
The root mean square (rms) speed of the particle due to random collisions with air molecules.
Solution:
1. Calculation of Boltzmann constant:
The rms speed of a particle can be calculated using the formula:
v = √(3kT/m)
where,
v = rms speed of the particle
k = Boltzmann constant
T = Temperature in Kelvin
m = mass of the particle
To calculate the value of k, we can rearrange the equation as:
k = (v^2 * m) / (3T)
2. Conversion of mass to kg:
Given mass of the particle is in grams, we need to convert it to kilograms.
1 gram = 10^(-3) kg
So, the mass of the particle in kilograms is:
m = 1.4 x 10^(-11) kg
3. Conversion of temperature to Kelvin:
The given temperature is already in Kelvin, so no conversion is required.
T = 300 K
4. Calculation of rms speed:
Using the above values, we can calculate the Boltzmann constant:
k = (v^2 * m) / (3T)
5. Substituting the values:
Substituting the given values:
k = (v^2 * 1.4 x 10^(-11)) / (3 x 300)
6. Calculation:
Solving the equation, we get:
v^2 = (3 x 300 x k) / (1.4 x 10^(-11))
v^2 = (900 x k) / (1.4 x 10^(-11))
v^2 = (6.43 x 10^11) x k
Taking the square root of both sides, we get:
v = √((6.43 x 10^11) x k)
7. Calculation of rms speed:
Finally, substituting the value of Boltzmann constant, k = 1.38 x 10^(-23) J/K, we get:
v = √((6.43 x 10^11) x 1.38 x 10^(-23))
v = √(8.87 x 10^(-12))
v = 2.98 x 10^(-6) m/s
8. Conversion to micrometers per second:
1 meter = 10^6 micrometers
So, the rms speed of the particle in micrometers per second is:
v = 2.98 x 10^(-6) x 10^6
v = 2.98 m/s
Therefore, the rms speed of the particle due to random collisions with air molecules is closest to 2.98 μm/s.
Answer:
The correct option is (b) 3.
Community Answer
A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K....
Vrms=√3KT/m
= √(3×1.4×^-23×300)/1.4×^-11
=30
option: C
= √(3×1.4×^-23×300)/1.4×^-11
=30
option: C
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A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300?
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A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300?.
A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tiny particle of mass 1.4times10^(-11)kg is floating in air at 300K.Ignoring gravity its rm.s.speed (in mu m/s ) due to random collisions with air molecules will be closest to JAM 2012 (a) 0.3 (b) 3 (c) 30 (d) 300?.
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