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Physics

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CBSE Class 12

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CBSE Physics 2013 Exam Questions

Long Answer Type

31.

Using Gauss’ laws deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell.

Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance from the centre of the shell). 


i) Consider a uniformly charged thin spherical shell of radius R carrying charge Q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius (r >R), concentric with given shell. If E is electric field outside the shell, then by symmetry electric field strength has same magnitude E0 on the Gaussian surface and is directed radially outward.

So, electric flux through Gaussian surface is given by, 

contour integral subscript s space equals space stack E subscript o with rightwards harpoon with barb upwards on top space. space d S with rightwards harpoon with barb upwards on top space

Therefore, 

contour integral space equals space E subscript o space d s space c o s space 0 space equals space E subscript o.4 pi r squared space 

Charge enclosed by the Gaussian surface is Q.

Therefore, using gauss’s theorem, we have

contour integral subscript s space equals space E with rightwards harpoon with barb upwards on top subscript o space d S space equals space 1 over epsilon subscript o space x space c h a r g e space e n c l o s e d

rightwards double arrow space E subscript o space 4 pi r squared space equals space 1 over epsilon subscript o x space Q

rightwards double arrow space space E subscript o italic space italic equals italic space fraction numerator italic 1 over denominator italic 4 pi epsilon subscript o end fraction Q over r to the power of italic 2 

Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre.

ii) Electric field inside the shell:

 

The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let’s consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward. 

Electric flux through the Gaussian surface is given by, 

equals space integral subscript S space E with rightwards harpoon with barb upwards on top subscript i. space d S with rightwards harpoon with barb upwards on top
 
integral E subscript i space d S space c o s space 0 space equals space E subscript i space. space 4 pi r squared
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.

Therefore, using Gauss’s theorem, we have

space space space space space space integral subscript straight S space E with rightwards harpoon with barb upwards on top subscript i. space d S with rightwards harpoon with barb upwards on top space equals space 1 over epsilon subscript o x space c h a r g e space e n c l o s e d

rightwards double arrow space E subscript i.4 pi r squared space equals space 1 over epsilon subscript o space x space 0
rightwards double arrow space E subscript i space equals space 0 space 

Thus, electric field at each point inside a charged thin spherical shell is zero.

 

The graph above shows the variation of electric field as a function of R.

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32.

Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni ) to the lower state, (nf ).

When electron in hydrogen atom jumps from energy state ni =4 to nf =3, 2, 1, identify the spectral series to which the emission lines belong.


According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus.

Therefore, 

fraction numerator m v squared over denominator r end fraction space equals space fraction numerator 1 space left parenthesis Z e right parenthesis space left parenthesis e right parenthesis over denominator 4 pi epsilon subscript o r squared end fraction                                       ... (1) 

rightwards double arrow m v squared space equals fraction numerator 1 space over denominator 4 pi epsilon subscript o end fraction fraction numerator Z e squared over denominator r end fraction
So, Kinetic Energy, K.E = 1 half m v squared
K. E equals fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction fraction numerator Z e squared over denominator r end fraction
Potential energy is given by, P.E = fraction numerator 1 over denominator 4 pi epsilon subscript o end fraction fraction numerator left parenthesis Z e right parenthesis space left parenthesis negative e right parenthesis over denominator r end fraction space equals space minus fraction numerator 1 space over denominator 4 pi epsilon subscript o end fraction fraction numerator Z e squared over denominator r end fraction

Therefore, total energy is given by, E = K.E + P.E = Error converting from MathML to accessible text.
E =  negative fraction numerator 1 space over denominator 4 pi epsilon subscript o end fraction fraction numerator Z e squared over denominator 2 r end fraction, is the total energy. 

For nth orbit, E can be written as En,

straight E subscript straight n space equals negative space fraction numerator 1 over denominator 4 πε subscript straight o end fraction fraction numerator Ze squared over denominator 2 straight r subscript straight n end fraction italic space                           ... (2) 
Now, using Bohr's postulate for quantization of angular momentum, we have

mvr space equals space fraction numerator nh over denominator 2 straight pi end fraction

rightwards double arrow straight v space equals space fraction numerator nh over denominator 2 πmr end fraction 

Putting this value of v in equation (1), we get

Error converting from MathML to accessible text.

rightwards double arrow space straight r space equals space fraction numerator straight epsilon subscript straight o straight h squared straight n squared over denominator πmZe squared end fraction

rightwards double arrow space straight r space equals space fraction numerator straight epsilon subscript straight o straight h squared straight n squared over denominator πmZe squared end fraction 

Now, putting value of rn in equation (2), we get

Error converting from MathML to accessible text. 
R is the rydberg constant. 

For hydrogen atom Z =1,

straight E subscript straight n space equals space fraction numerator negative Rch over denominator straight n squared end fraction
If ni and nf are the quantum numbers of initial and final states and Ei & Ef are energies of electron in H-atom in initial and final state, we have 


straight E subscript straight i space equals negative space Rhc over straight n subscript straight i squared space and space straight E subscript straight f space equals space fraction numerator negative Rhc over denominator straight n subscript straight f squared end fraction 
If comma space straight upsilon space is space the space frequency space of space emitted space radiation comma space we space get space

straight nu space equals space fraction numerator straight E subscript straight i space minus space straight E subscript space straight f end subscript over denominator straight h end fraction

straight nu space equals space fraction numerator negative Rc over denominator straight n subscript straight i squared end fraction minus open parentheses fraction numerator negative Rc over denominator straight n subscript straight f squared end fraction close parentheses space equals space Rc open square brackets 1 over straight n subscript straight f squared space minus space 1 over straight n subscript straight i squared close square brackets

That is, when electron jumps from ni = 4 to nf = 3.21 .

Radiation belongs to Paschen, Balmer and Lyman series.

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33.

(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two important conclusions that can be drawn regarding the nature of nuclear force.

(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.

(c) Write the basic nuclear process of neutron undergoing –decay. Why is the detection of neutrinos found very difficult?


Graphical representation of (BE/A) for nucleons with mass number A.

The variation of binding energy per nucleon VS. mass number is shown in the figure:

Characteristics of Nuclear force:

(i) Nuclear forces non-central and short ranged force.

(ii) Nuclear forces between proton-neutron and neutron-neutron are strong and attractive in nature.

b) When a heavy nucleus (A > 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.

When two very light nuclei (A £10) join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

c) During the decay process of neutron, we have

n presubscript 0 presuperscript 1 space rightwards arrow space p presubscript 1 presuperscript 1 space plus space beta presubscript negative 1 end presubscript presuperscript 0 space plus space nu with bar on top space 

Neutrinos show weak interaction with other particles. Hence, its detection is very different.

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34.

(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.


Electric dipole moment is the product of either charges or the distance between two equal and opposite charges.

It is a vector quantity.

Electric dipole moment at a point on the equatorial plane:

Consider a point P on broad side on the position of dipole formed of charges + q and - q at separation 2l. The distance of point P from mid-point O of electric dipole is r. 

 

Let E1 and E2 be the electric field strength due to charges +q and –q of electric dipole.

From the fig. we have

Error converting from MathML to accessible text. 
Now, inorder to find the resultant electric field, we resolve the components along and perpendicular to AB.

The components perpendicular to AB are sin components and they being equal and opposite to each other cancel each other.

Therefore,

Resultant electric field is given by, 

E1 = E1 cos space theta space plus space E subscript 2 space c o s space theta 

But, 

Error converting from MathML to accessible text. 

From the fig. we can see that,

Error converting from MathML to accessible text.  

If dipole is infinitesimal and point P is far away, then l2 can be neglected as compared to r2.

Therefore, 

straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight p over straight r cubed comma space parallel space to space BA with rightwards harpoon with barb upwards on top 

b) Equipotential surfaces due to an electric dipole is given by,  


                                 

Electric potential is zero at all points in the plane passing through the dipole equator.

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