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| 37301. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: Charge stored in P, when a dielectric slab is inserted in between the plates |
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Answer» increases |
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| 37302. |
Two polaroids P_(1)" and "P_(2) are placed with their axis perpendicular to each other. Unpolarised light I_(0) is incident of P_1. A third polaroid P_3 is kept in between P_(1)" and "P_(2) such that its axis makes an angle 45^(@) with that of P_(i). The intensity of transmitted light through P_(2) is : |
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Answer» `I_(0)"/"16` |
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| 37303. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: If a dielectric slab of dielectric constant 5 is inserted in between the plates of P so that the entire space is occupied by the dielectric. If E is theelectric field intensity between the plates before the insertion of dielectric then electric field intensity after the dielectric slab is introduced is |
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Answer» E |
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| 37304. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: If dielectric slab of dielectric constant 5 is inserted in between the plates of Q so that entire space is occupied by the dielectric. If E is the electric field intensity between the plates before the insertion of dielectric then electric field intensity after the dielectric slab is introduced is |
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Answer» E |
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| 37305. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: If the separation between the plates of P is increased then electric field intensity in between the plates of capacitor |
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Answer» increases |
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| 37306. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: If the separation between the plates of Q is increased then electric field intensity in between the plates of capacitor |
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Answer» increases |
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| 37307. |
PassageII We have two parallel plate capacitors P and Q having same capacitance C_(0). P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: Charge stored in Q, when a dielectric slab is inserted in between the plates |
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Answer» INCREASES |
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| 37308. |
Keeping the grid voltage constant a change in the plate potential of 50 V, changes the plate current by 10 mA. And keeping the plate potential constant a change in the grid potential of 2V changes the plate current by 10 ma again , the amplification factor of the triode will be : |
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Answer» 100 |
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| 37309. |
A lamp is hanging 1 m above the centre of a circular table of diameter 1 m . The ratio of illuminaces at the centre and the edge is |
| Answer» Answer :b | |
| 37310. |
A copper wire of 10^(-6)m^(2) are of cross section, carries a current of 2A. If the number of electrons per cubic meter is 8xx10^(28), calculate the current density and average drift velocity. |
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Answer» Solution : Cross-sections area of copper wire, A -=`10^(-6)m^(2)` 1 = 2A Number of electron, n = `8xx10^(28)` Current density, J= `(1)/(A)=(2)/(10^(-6))` J=`2xx10^(6)Àm^(-2)` AVERAGE drift velocity, `V_(d)=(1)/(n EA)` e is the charge of electron = `1.6xx10^(-9)C` `V_(d)=(2)/(8xx10^(28)xx1.6xx10^(-19)xx10^(-6))=(1)/(64xx10^(3))` `V_(d)=0.15625xx10^(-3)`,`V_(d)=15.6xx10^(-5)MS^(-1)` |
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| 37311. |
A thin spherical shell radius of r has a charge Q uniformly distributed on it. At the centre of the shell, a negative point charge -q is placed. If the shell is cut into two identical hemispheres, still equilibrium is maintained. Then find the relation between Q and q? |
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Answer» Solution :Here the outward electric pressure at every point on the shell due to its own charge is `P_(1) = (sigma^(2))/(2in_(0)) =(1)/(2 in_(0)) ((Q)/(4pi r^(2)))^(2), P_(1) = (Q^(2))/(32pi^(2) in_(0)r^(4))` Due to -q, the electric field on the surface of the shell is `E= (1)/(4pi in_(0)) (q)/(r^(2))` This electric field pulls every point of the shell in INWARD direction. The inward pressure on the surface of the shell due to the negative charge is `P_(2)= sigma E` `= ((Q)/(4pi r^(2))) ((1)/(4pi in_(0))(q)/(r^(2)))= (Qq)/(16pi^(2) in_(0)r^(4))` For equilibrium of the HEMISPHERICAL shells `P_(2) ge P_(1)` or `(Qq)/(16pi^(2) in_(0)r^(4)) ge (Q^(2))/(32pi^(2) in_(0)r^(4)) q ge (Q)/(2)` |
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| 37312. |
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity – v. At this instant, |
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Answer» the electric forces on both the particles cause identical accelerations. 2. Both particle gain and LOST equal amount of rate of energy because force on both particles are equal in magnitude and OPPOSITE in directions. 3. C.M. of system does not change so we can find it by only `vecB`. So, (B, C and D) OPTIONS are TRUE. |
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| 37313. |
In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be |
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Answer» `1.944 R` |
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| 37314. |
You are given two graphs. What conclusion do you draw from the graphs? |
| Answer» Solution :The FIRST figure SHOWS the variation of FIELD intensity of a point CHARGE and the SECOND figure shows the potential of a point charge. | |
| 37315. |
The graph of x = -2 is a line parallel to the |
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Answer» X- axis |
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| 37316. |
What inference do you draw from above result ? |
| Answer» Solution :The length of DIPOLE antenna decreases as FREQUENCY of TRANSMISSION is INCREASED. | |
| 37317. |
In an ac circuit , the flow of current is opposed by inductors and capacitors.The is called reactance.(a) Fill in the blanks If omega is the angular frequency of AC, then the reactance offered by L and C are respectively X_I and X_C (##ANE_RJB_PHY_XII_C07_E03_003_Q01##) |
| Answer» SOLUTION :`X_L=L(OMEGA) ,X_C=1/C(omega)` | |
| 37318. |
Two fixed, identical conductingplates (alpha and beta), each of surface area S are charged to -Q and q, respectively, where Q gt q gt 0. A thirdindentical plate (gamma), free to move is locatedon the other side of the plate with charge q at a distanced, fig. The third plate is released andcollidies with theplate beta. Assumethe collsion is electricand the time of collision is sufficient to redistributechargeamongst beta and gamma. (a) Find the electricfieldacting on the plate gamma before collision. (b) Find the charge on beta and gamma after the collision. (c) Find the velocity of the plate gammaafter the collision and at a distance d from the plate beta. |
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Answer» Solution :The given set up is shown in FIG. (a) The electric field at plate `gamma` due to plate `alpha` is `E_(1) = - (Q)/(S (2in_(0))` , to the left The electricfield at plate `gamma` to plate `beta` is `E_(2) = (q)/(S(2 in_(0))` , to the right. HENCE, the net electric field at plate `lamma` beforecollision is `E = E_(1) + E_(2) = (q-Q)/(S (2in_(0))` , to the left if `Q gt q` (b) During collision, plates`beta and gamma` are together. Their potentials become same. Suppose chargeon plane`beta is q_(1)` and chargeon plate `gamma is q_(2)`. At any point 0, inbetween the two plates, Fig, the electric field must be zero. Electric field at 0 due to plate `alpha = ((-) Q)/(S (2in_(0)))` , to the left Electric field at 0 due to plate `beta = (q_(1))/(S(2in_(0)))` , to the right Electric field at 0 due to plate `gamma = (q_(2))/(S(2in_(0)))` , to the left As the electric field at 0 is zero, therefore , `(Q+q_(2))/(S(2in_(0))) = (q_(1))/(S(2 in_(0)))` `:.Q + q_(2) = q_(1) or Q = q_(1) - q_(2)`...(i) As there is no loss of charge on collision, `Q + q = q_(1) + q_(2)`...(ii) On solving (i) and (ii),we get`q_(1) = (Q + q//2)` = charge on plate `gamma`. (c) After collision,at a distance d from plate `beta`, let the velocity of plate`gamma` be v. After the collision, electric field at plate `gamma` is`E_(2) = (-Q)/(2 in_(0) S) + ((Q + q//2))/(2 in_(0) S) = (q//2)/(2 in_(0) S)` to the right. Just before collision, electric field at plate `gamma` is `E_(1) = (Q-q)/(2 in_(0) S)` If `F_(1)` is force on plate`gamma` before collision, then`F_(1) = E_(1) Q = ((Q-q)Q)/(2 in_(0) S)` Similarly, force `F_(2)` on plate `gamma` after collision,`F_(2) = E_(2) (q)/(2) = ((q//2)^(2))/(2in_(0) S)` Total work DONE by the electricfield in round trip movement of plate `gamma` `W = (F_(1) + F_(2))d = ([(Q-q) Q+ (q//2)^(2)] d)/(2in_(0) S) = ((Q-q//2)^(2) d)/(2in_(0) S)` If m is MASS of plate `gamma`, the KE gainedby plate `gamma = (1)/(2) mv^(2)` ACCORDINGTO work energy principle, `(1)/(2) mv^(2) = W= ((Q-q//2)^(2) d)/(2in_(0) S)` `v = (Q - q//2) ((d)/(m in_(0) s))^(1//2)` 
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| 37319. |
A very long straight wire carries a current of 5 A. An electron moves with a velocity of 10^(6)ms^(-1) remaining parallel to the wire at a distance of 10 cm from wire in a direction opposite to that of electric current. Find the force on this electron. (Here the mass of electron is taken as constant) (e=-1.6xx10^(-19)C,mu_(0)=4pixx10^(-7)SI) |
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Answer» Solution :1. I = 5 A, V = `10^(6)` m/s y = 0.1 m e = `1.6xx10^(-19)C` `mu_(0)=4pixx10^(-7)SI` Magnetic field at a DISTANCE y from the conducting wire, `B=(mu_(0)I)/(2piy)` = `(4pixx10^(-7)xx5)/(2pixx0.1)` `B=10^(-5)T` 2. This magnetic field is perpendicular to PAGE. Electron moves in the direction opposite to current so v and B are perpendicular to each other. `thereforetheta=pi/2""thereforesintheta="sin"pi/2=1` Now, `vecF=e(vecvxxvecB)=BeVsintheta=BeV` = `10^(-5)xx1.6xx10^(-19)xx10^(6)` = `16xx10^(-19)N` |
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| 37320. |
What is ratio of Bohr magneton and nuclear magneton ? |
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Answer» `(m_(p))/(m_(e))` `mu_(B)=(EH)/(2m_(e))` `m_(e)`= electronic mass, e = electronic charge In nuclear physics magnetic moments associated with the spins of protons and neutrons are expressed in nuclear magnetons `(mu_(N))` `mu_(N)=(eh)/(2m_(p))` where `m_(p)` = mass of proton `rArr (mu_(B))/(mu_(N))=(m_(p))/(m_(e))` |
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| 37321. |
Match the situations in column I to the accelerations of blocks in the column II (acceleration due to gravity is g and F is an additional force applied to one of the blocks ? |
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Answer» |
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| 37322. |
The value indicated by fermi-energy level in an intrinsic semiconductor is |
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Answer» The AVERAGE ENERGY of electrons and holes |
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| 37323. |
Elinstein got Nobal prize in physics in 1921 for hisw explanation of photo electric effect. Write down Einstein's photoelecric equation |
| Answer» SOLUTION :`hV=phi+KE_max orhV=hV+1/2mv^2` | |
| 37324. |
Define temperature coefficient or resistance. |
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Answer» SOLUTION :It is defined as the ratio of increase in RESISTIVITY per degree rise in temperature to its resistivity at `T_(0)`. `THEREFORE ALPHA=(rho_(T)-rho_(o))/(rho_(o)(T-T_(o)))=(Deltarho)/(rho_(o)DeltaT)` |
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| 37325. |
Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. |
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Answer» Solution :At resonance, `|Z|=` MINIMUM `= R = 3 Omega` Power dissipated at resonance is `P = I_(rms) ^(2) R` `= { ( V_(rms))/( |Z| ) }^(2) R` `= { ( V_(rms))/( R ) }^(2) R` (At resonance, `|Z| = R )` `= ( V_(rms)^(2))/( R)` `= (((V_(m))/(sqrt(2)))^(2))/(R )` `= ( V_(m)^(2))/( 2R)` `:. P = (( 283)^(2))/( 2 xx 3 )` `:. P = 13348 W` (Watt ) Above is the maximum power fetched by the CIRCUIT from the source and the same is wasted in the form of heat in the resistance ( at the time of resonance ). |
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| 37326. |
The energy equivalent to a substance of mass 1 g is |
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Answer» `18xx10^(13)J` |
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| 37327. |
Match the following {:(,,"Table-1",,"Table-2"),(,(A),L,(P),[M^(0)L^(0)T^(-2)]),(,(B),"Magnetic Flux",(Q),[ML^(2)T^(-2)A^(-1)]),(,(C),LC,(R),[ML^(2)T^(-2)A^(-2)]),(,(D),CR^(2),(S),"None"):} |
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Answer» |
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| 37328. |
If the electron gains the energy for the jump from energy level E_(1) to energy level E_(3) by absorbing light, what light wavelength is required ? |
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Answer» Solution :KEY IDEAS (1) If light is to transfer energy to the electron, the transfer must be by photon absorption. (2) The photon.s energy must equal the energy difference `DeltaE` between the initial energy level of the electron and a higher level, according to Eq. 38 - 6 `(hf=DeltaE)`. Otherwise, a photon can not be absorbed. Wavelength : Substituting `c//lambda` for f, we can rewrite Eq. 36-6 as `lambda=(HC)/(DeltaE)`. For the energy difference `DeltaE_(31)` we found in (B), this EQUATION gives us `lambda=(hc)/(DeltaE_(31))` `=((6.63xx10^(-34)J.s)(2.998xx10^(8)m//s))/(4.83xx10^(-17)J)` `=4.12xx10^(-9)m.` |
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| 37329. |
Calculate the mass defect, binding energy and building energy per nucleon of an alpha particle (An alpha -particle is nothing but helium nucleus. Hence its symbol is ""_(2)He^4 . It contains 2 protons , 2 neutrons with a mass number 4. Mass hydrogen atom m_H = 1.007825u : Mass of neutron m_(n) = 1.008665u : Atomic number of helium Z = 2 , Mass number of helium A = 4 , Mass of helium atom m_(a) = 4.00260u ) |
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Answer» Solution :Mass defect, `Deltam = Zm_(H) +(A-Z) m_n-m_(a)` `[(2) (1.007825) +(4-2)(1.008665)-4.00260]U` `= (2xx1.007825+2xx1.008665-4.00260)u` Mass defect, `Deltam = 0.03038` u `:.` Binding energy of the nucleus `=(Deltam)C^2` `= (0.03038) u xxC^2` `= 0.030 38 xx 931.5 MeV ( :. 1u xxC^(2) = 931.5 MeV)` = 28 . 3 MeV Binding Energy per nucleon `= (28.3)/4` MeV Biding energy per nucleon = 7.075 MeV |
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| 37330. |
On looking downward, an electron appears moving anti-clockwise in a horizontal circle under a magnetic field. What is the direaction of the field ? |
| Answer» SOLUTION :VERTICALLY UPWARD | |
| 37331. |
A65 kg horizontal force is just sufficient to draw 1300 kg block at level table surface at uniform speed. Then, the coefficient of friction is: |
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Answer» 0.5 Now `F_(f)=mu=(F)/(R)=(65xx9.8)/(1300xx9.8)=0.05` Hence correct CHOICE is (d) |
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| 37332. |
Drift velocity of holes is ...... |
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Answer» in the DIRECTION of CURRENT DENSITY. |
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| 37333. |
Ultraviolet radiation of different frequencies v_(1) and v_(2) are incident on two photosensitive materials having work functions phi_(1) and phi_(2)(phi_(1) gtphi_(2)) respectively. The kinetic energy of the emitted electrons is same in both the cases. which one of the two radiations will be of higher frequency ? |
| Answer» Solution :As PER relation `hv=phi+(K.E.)_(max)` we find that `v_(1) gt v_(2)`. | |
| 37334. |
1 picometre is equal to. |
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Answer» `10^-14` m |
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| 37335. |
If for an ideal step-up transformer current in primary is I_P and current in secondary is I_S, their respective voltage are V_P and V_S, then…… |
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Answer» `I_S V_S = I_P V_P` `therefore`INSTANTANEOUS output power = instantaneous input power `therefore I_SV_S=I_PV_P` [ `because` power P=VI] |
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| 37336. |
In figure, the e.m.f. of the cell is 120V and internal resistance is negligible. The resistance of the voltmeteris 80 ohm. The % error in reading of voltmeter will be |
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Answer» `20%` `=(120xx80)/(100)=96"Volt"` Voltmeter will measure (V) `=(120xx40)/(60)` = 80 volt `THEREFORE%"error"=(96-80)/(96)xx100` `=(16)/(96)xx100=(50)/(3)=16.7%` |
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| 37337. |
A square wire of side 1 cm is placed perpendicular to the principal axis of a concave mirror of focal length 15 cm at a distance of 20 cm. The area enclosed by the image of the wire is : |
| Answer» Answer :D | |
| 37338. |
A metallic wire has variable cross sectional area. Cross sectional area at cd is twice the area at ab. The wire is connected to a cell as shown. find the ratio of heat dissipated per unit volume at section cd to that at section ab. |
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Answer» |
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| 37339. |
यदि समुच्चय {1,2,3} में R={(1,2)} द्वारा परिभाषित एक संबंध R हे तो R |
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Answer» स्वतुल्य है। |
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| 37340. |
If the sound level in a room is increased from 50dB to 60 dB, by water factor is the pressure amplitude increased ? |
| Answer» ANSWER :B | |
| 37341. |
If the particle is moving along a straight line given by the relation x=2-3t+4t^(3) where s is in cms.,and t in sec.Its average velocity during the third sec is |
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Answer» 73 cm/s |
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| 37342. |
Electric charge q is uniformly distributed in a spherical space from which a cavity of diameter equal to the radius Rof thespherical space has been removed. Calculate the potential and field at a point P lying on the diameter of the cavity at a distance rgtr//2 but lt R from the centre of the sphere. |
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Answer» |
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| 37343. |
Which one is unstable among neutron, proton,electron and alpha-particle? |
| Answer» Solution :NEUTRON is unstable out of the FOUR GIVEN PARTICLES. | |
| 37344. |
In which of the following mixtures, the London dispersion force acts as major intermolecular force of attraction ? |
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Answer» Sodium chloride and water |
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| 37345. |
Tow short electric dipoles are placed as shown The energy or electric interaction between these dipoles will be. |
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Answer» ` (2 K P_1p_2 COS theta)/r^2 ` ` =- P_2 E_1 cos theta ` ` =- P_2 (2 KP_1 COSQ)/r^3` ` U=- (2 KP_1OP_2)/r^3 cos theta` ` (##NG_PHY_C01_E01_047_S01.png" width="80%">. |
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| 37346. |
Distinguish between avalanche and zenerbreakdown. |
Answer» SOLUTION :
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| 37347. |
A source of soud is moving along a circular orhit of dins 3 m with an angular velocity of 10 rad/s. sound detector located far away from the source is executing linear simple harmonic motion along the line BD with amplitude BC = CD = 6m. The frequency oscillation of the detector is (5//pi) rev/sec. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 H, find the maximum and the minimum frequencies recorded by the detector velocity of sound :: 330 m/s) |
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Answer» Solution :Time period of circular motion `T=(2pi// OMEGA)=(2pi//10)` is same as that of SHM. i.e., `T=(1//f)=(pi//5)`, so both will COMPLETE one periodic motion in same time. Furthermore source is moving on a circle, its speed `V_(s)=r omega=3xx10=30m//s` and as detector is executing SHM. `V_(D)=omegasqrt(A^(2)-y^(2))=10 sqrt(6^(2)-y^(2))` i.e, `(V_(D))_(max)=60m//s` when y=0 So `f_(AP)` WIL be maximum when both move towards each other. `f_(max)=f|(V+V_(D))/(V-V_(S))|` with `V_(D)`= max i.e., the source is at M and detector at C and moving towards B, so `f_(max)=340[(330+60)/(330-30)]=442Hz` Similarly `f_(AP)` will be minimum when both are moving away from each other, i.e `f_(max)=f[(V-V_(D))/(V+V_(s))]` with `V_(D)` = max i.,e the source is at Nadn detector at C but moving towards D, `so f_(MIN) =340[(330+60)/(330+30)]=225Hz` |
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| 37348. |
Two batteries each of emf E and internal resistance r are connected in series and in parallel, and are used to find current in an external resistance R. If the current in series is equal to that in parallel, the internal resistance of each battery is |
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Answer» <P> Solution : `I_(s)=(2E)/(2r+R) and I_(P)=(E)/((r)/(2)+R)=(2E)/(r+2R)` If `I_(S)+I_(P)`, (2E)/(2R+R)=(2E)/(r+2R), 2r+R=r+2R RARR r=R` |
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| 37349. |
Light of wavelength 0.6mumfrom a sodium lamp fallson a photocell and causes the emission of Photoelectronsfor which the stopping potential is 0.5 V. With light of wavelength 0.4 mum from a sodium lamp, stopping potential is 1.5 V . With this data, the value of h/e is |
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Answer» ` 4 XX 10^(-19) V s ` ` eV= (hc)/(lamda) = phi_(0)` Where , VStopping POTENTIAL ` lamda ` = Incidentwavelength ` phi_(0)` = Workfunction ` V = ( h/e)c/lamda - (phi_(0))/e` ` V_(1) = ( h/e) c/lamda_(1) = (phi_(0))/e ` ` V_(2) = (h/e) c/lamda_(2) - (phi_(0))/e ` Solvingthese two EQUATIONS ,we get ` h/e= (lamda_(1)lamda_(2)) (V_(1)-V_(2))/( c(lamda_(2)-lamda_(1))` ` ((0.6 xx 0.4 xx 10^(-12)) (1.0))/ ( 3xx 10^(8)) (0.2 xx 10^(-6)) = 4 xx 10^(-15) V s ` |
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| 37350. |
Explain the Young's double slit experimental setup and obtain the equation for path difference. |
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Answer» Solution :(i) An opaque screen with two small openings called double slit `S_1` and `S_2` is kept equidistance from a SOURCE S. (ii) The width of each slit is about 0.03 mm and they are separated by a distance of about 0.3 mm. (iii) As `S_1` and `S_2` are equidistant from S, the light Superposition PRINCIPLE waves from S reach `S_1` and `S_2` in-phase. So, `S_1` and `S_2` act as COHERENT sources which are the requirement of obtaining interference pattern. Experimental setup: (i) WAVEFRONTS from `S_1` and `S_2` spread out and overlapping takes place to the right side of double slit. (ii) When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. (iii) These are called interference fringes or bands. Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from `S_1` and `S_2` travel equal distances and arrive in-phase as shown in Figure.  (iv) These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.(v) The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference. Equation for path DIFFERENCE: (i) Let d be the distance between the double slits `S_1` and `S_2` which act as coherent sources of wavelength `lamda`. A screen is placed parallel to the double slit at a distance D from it. (ii) The mid-point of `S_1` and `S_2` is C and the i mid-point of the screen O is equidistant from `S_1` and `S_2` P is any point at a distance y from O.(iii) The waves from `S_1` and `S_2` meet at P either n-phase or out-of-phase depending uponthe path difference between the two waves. (iv) The path difference `delta` between the light waves from `S_1` and `S_2` to the point P is, ` delta= S_2P - S_1 P` (v) A perpendicular is dropped from the point `S_1` to the line `S_2P ` at M to find the path difference more precisely. ` delta = S_2 P - MP= S_2M` The angular position of the point P from C is `theta. angle OCP = theta` From the geometry, the angles `angleOCP` and `angleS_2S_1M` are equal. ` angle OCP = angle S_2 S_1 M= theta` In right angle triangle `Delta S_1 S_2 M` , the path difference, ` S_2 M = d sin theta` ` delta = d sin theta` If the angle `theta` is small, `sin theta ~~ tan theta ~~ theta` .From the right angle triangle `DeltaOCP, tan theta = y/D` The path difference,` delta= (dy)/(D)` Based on the condition on the path difference, the point P may have a bright or dark fringe. |
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