SECTION ‘A’ (MULTIPLE CHOICE QUESTIONS)
1. Choose the most appropriate answerfor each from the given options:
i) The maximum work done is possible in this process:
(ii) If the area of the plates of a parallel plates capacitor is doubled, the capacitance:
* remains unchanged
* is half
* is double
* is increased four times
(iii) The electric intensity between two uniformly oppositely charged parallel plates is:
(iv) Resisters of 2 ohms, 3 ohms, 4 ohms and 5 ohms are connected in series. If the current flowing through 2 ohms resistor is one ampere, the current throllgh the other resistors
* 4 ampere
* 14 ampere
* 1 ampere
* 0.1 ampere
(v) According to Einstein’s special theory of relativity, the mass of a particle moving with the speed of light will become:
* ten times
(vi) After alpha decay, the nucleus has its:
* charge number decreased by four
* charge number increased by four
* mass number increased by four
* mass number decreased by four
(vii) laser produces:
* an electron beam
* a neutron beam
* a coherent beam of light
* all of these
(viii) Absolute zero corresponds to this temperature on Fahrenheit scale:
(ix) Joule per coulomb is called:
(x) One kilowatt hour energy is equal to:
* 3.6 x 105 J
* 36 x 105 J
* 746 watt
* 6.3 x 105J
(xi) With the increase of temperature, the resistance of a semi conductor:
* remains constant
* becomes zero
(xii) The mathematical expression A.nax x T = constant is called:
* Stefan’s Law
* Wein’s Displacement law
* Rayleigh-Jean’s Law
* Planck’s Law
(xiii) This narrow beam from cobalt-60 is used in treating localized cancerous tumour:
* β rays
* All of these
(xiv) According to uncertainty principle:.
(xv) The practical application of the phenomenon of mutual inductance is:
* A.C. generator
(xvi) An electric current on passing through a conductor, produces around it:
* an electric field
* a magnetic field
* both electric and magnetic fields
* first a magnetic field then an electric field
(xvii) AVO meter is used to measure:
* electric current
* all of these
SECTION ‘B’ (SHORT – ANSWER QUESTIONS
2.i) You are given three resistors each of 2 ohms. How would you arrange these to obtain equivalent resistance of:
a) 1.33 ohms,
b) 3 ohms and
c) 6 ohms? Verify the results mathematically.
R4 are so adjusted that on closing the Keys K, and K2 no current passes through galvanometer, the bridge is said to be balanced. In this condition galvanometer gives null deflection, and relation between four resistances is given by:
Explanation: Under balanced condition when no current passes through galvanometer, the potnential at B & D must be same and the potenetial difference across galvanometer must be zero.
Potential difference across R, = Potential difference across R3
VAS = VAD
I,R, = bRJ –(1)
Potential difference across R2= Potential difference across R;
VBC = Voc
I,R2 = 12R; – – (2)
Dividing equation (1) by (2)
This is known as wheat stone bridge principle.
vi) In what way is the motional emf, produced in a conductor placed in a magnetic field, translated into the relevant mathematical relation?
Ans: See on Page # 266.
vii) Explain why the photograph of the path of Alpha particles is thick and made up of continuous lines whereas that of Beta particles in thin and made up of broken lines in the Wilson
In Wilson Cloud Chamber: A chamber contains super cooled vapours in a dust free atmosphere. When radians pass through the chamber, ions are formed, which serve an nucleation centres. The super saturated vapours get condensed on them. So the path shows up a as a fog trail. In this way, the instrument is used to visualize the tracks of charged particles.
Different Particles Produce Different Types of Tracks:
(i) Alpha Particles Produce thick, broad (dense) and continuous tracks
viii) Do α, β and y rays come from the same element?’ Why are all these three found in many radioactive substances?
A radioactive element either emits a-partices or ~-particles, but not both. Gamma radiations generally accompany p-emissions and in some cases with a-emission.
A ratioactive element (or sample) is a mixture of various nuclides of different relative abundances and wit~ different modes of distintegration.
Hence we can find all the three in a radioactive sample at the same
ix) What is semi-conductor diode? How is it used for half-wave rectification?
Semi-Conductor Diode: A block of a p-type semi-conductor is put adjacent to a block of n-tYpe semi-conductor the common plane is termed as p – n junction diode. This junction is formed by melting p and n semi-condutors by a special manufacturing process (called “thermo compression proess”). The p-n-junction is extremely thin and is of the order of 10-3mm.
Half Wave Rectifier: When a diode is connected to a transformer the
diode is made positive and it acts as a closed switch. The current flows in the circuit as a pulse of half-sine “wave-form”
During the next half cycle of the transformer voltage, anods of the diode is negative (reverse biased). And it acts as = 1 an open circuit. No current exists in the load (resistance = R) In this way, current flows in one direction. The output across the load is “pulsating d.e.” These pulsations in the output are further Sft’IOothed with the help of “filter circuit” .
x) A Carnot engine whose low temperature reservoir is 200K has an efficiency of 50%. It is desired to increase this to 75% By how m_any degrees must the temperature of low temperature reservoir be decreased if the temperature of the higher temperature reservoir remains constant.
SECTION ‘C’ (DETAILED – ANSWER QUESTIONS)
3.a) Derive a relation for the pressure on an ideal gas in terms of its density and mean square velocity.
Interpretation of Pressure of a Gas:
5.a) State the postulates of Bohr’s Atomic theory and derive the
expression for the radius of nth orbit of a hydrogen atom.
Ans: See on Page # 245.
b) State Gauss’s Law. Apply the law to find electric intensity due to a uniformly charged metallic sphere at a point Plying outside the charged sphere.
Gauss’s Law: The relation between electric flux and net charge enclosed by a surface was enunciated by E.E. Gauss (1775 – 1857). a German physicist and one of the greatest mathematicians of all time.
Statement: “The total electric flux through any closed surface is 1/E 0 times the total charge enclosed in it.”
Proof: Consider a closed surface S of some arbitary shape. Suppose it encloses point charge qt, qz, q3•…. qn.
Total charge Ø = Øt + Ø2 + Ø3 + ….. + Øn
To determine flux over the closed surface, it is divided into very small but equal pars, each part being so small that it is almost a plane.
The direction of E and M is thus the same. ,
The total flux will be the algebraic sum of the flux through all the patches.
Ø = Øt + Ø2 + Ø3 + ….. + Øn
Consider flux due to Charge q-. Imagine a sphere with ql as center. The flux through the sphere is the same as through surface S.
1. Electric flux through a closed surface is independent of the number of charges.
2. The flux through a closed surface is independent of the way in which the charge to distributed.
3. Flux is independent 0 the geometry (size and shape) of the closed surface (Gaussian surface), i.e., the choice of the Gaussian surface is arbitary.
4. Flux depends on the medium. •
5. The imaginary closed surface (Gaussian surfacej.shoutd be chosen such that it has the same symmetry as that of the charge
Limitation: “This law is applicable only to closed surface containing source charges.
Application of Gauss’s Law: Gauss’s Law can be used to evaluate electric intensity by the proper choice of a Gaussian surface.
Case a (i): Electric Intensity of a Charged Sphere at a Point Ou(side the Sphere: Consider an insulating sphere of radius ‘a’. Let charge q be uniformly distributed on its surface.
Let P be a point outside the sphere at a distance r from the centre O. Consider a closed sphere as a Gaussian surface.
Due to spherically symmetric charge distribution, the field has the san magnitude with respect to every point on the closed surface.