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You are given marbles of diameter 10mm. ...

You are given marbles of diameter `10mm`. They are to be placed such that their centres are laying in a square bound by four lines each of length `40mm`. What will be the arrangements of marbles in a plane so that maximum number of marbles can be placed inside the area? Sketch the diagram and derive expression for the number of molecules per unit area.

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To solve the problem of arranging marbles of diameter 10 mm in a square area of 40 mm x 40 mm, we will follow these steps: ### Step 1: Understand the dimensions of the square and marbles - The square has a side length of 40 mm. - The diameter of each marble is 10 mm, which means the radius is 5 mm. ### Step 2: Determine the arrangement of the marbles To maximize the number of marbles, we can arrange them in a close-packed manner. The most efficient arrangement for spheres in a two-dimensional plane is a hexagonal close packing (HCP). However, for simplicity, we can also consider a square packing arrangement. ...
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The root mean square speed of an ideal gas is given by : u_("rms") = sqrt((3RT)/M) Thus we conclude that u_("rms") speed of the ideal gas molecules is proportional to square root of the temperature and inversely proportional to the square root of the molar mass. The translational kinetic energy per mole can also be given as 1/2Mu_(rms)^(2) . The mean free path ( lambda ) is the average of distances travelled by molecules in between two successive collisions whereas collision frequency (C.F.) is expressed as number of collisions taking place in unit time. The two terms lambda and C.F. are related by : C.F = (u_("rms")/lambda) If n represents number of moles, n0 is number of molecules per unit volume, k is Boltzmann constant, R is molar gas constant, T is absolute temperature and NA is Avogadro's number then which of the following relations is wrong ?

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) For a given gas the mean free path at a particular pressure is :

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) Three ideal gas samples in separate equal volume containers are taken and following data is given : {:(" ","Pressure","Temperature","Mean free paths","Mol.wt."),("Gas A",1atm,1600K,0.16nm,20),("Gas B",2atm,200K,0.16nm,40),("Gas C",4atm,400K,0.04nm,80):} Calculate ratio of collision frequencies (Z_(11)) (A:B:C) of following for the three gases.

In a tyre of "Ferrari" car of Mr. Obama, a tube having a volume of 12.3 litres is filled with air at a pressure of 4 atm at 300 K.Due to travelling, the temperature of the tube and air inside it raised to 360 K and pressure reduced to 3.6 atm in 20 minutes. If the porosity (number of pores per unit area) of the tube material is 5xx10^3 pores/ cm^2 and each pore can transfer air from inside to outside of the tube with the rate of 6.023xx10^8 molecules per minute.Calculate the total surface area (m^2) of the tube.(R=0.082 Lt-atm/mole-K) Give your answer divide by 100.

Two metallic plate A and B , each of area 5 xx 10^(-4)m^(2) , are placed parallel to each at a separation of 1 cm plate B carries a positive charge of 33.7 xx 10^(-12) C A monocharonatic beam of light , with photoes of energy 5 eV each , starts falling on plate A at t = 0 so that 10^(16) photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10^(6) incident photons fall on it per square meter per second. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remain constant at the value 2 eV Determine (a) the number of photoelectrons emitted up to i = 10s, (b) the magnitude of the electron field between the plate A and B at i = 10 s, and (c ) the kinetic energy of the most energotic photoelectrons emitted at i = 10 s whenit reaches plate B Negilect the time taken by the photoelectrons to reach plate B Take epsilon_(0) = 8.85 xx 10^(-12)C^(2)N- m^(2)

VMC MODULES ENGLISH-THE SOLID STATE-JEE Advance ( Archive )
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  2. A substance A(x)B(y) crystallizes in a face-centred cubic lattice in ...

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  3. You are given marbles of diameter 10mm. They are to be placed such tha...

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