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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Arrange the order of power dissipated in the given circuits, if the same current is passing through all circuits and each resistor is 'r' |
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Answer» `P_(2)gtP_(3)gtP_(4)gtP_(1)` `P_(2)=I^(2)xx3r...(2)` `P_(3)=I^(2)xx(3r)/2.....(3)` `P_(4)=I^(2)xx(2r)/3....(4)` comparing the equations `P_(2)gtP_(3)gtP_(4)gtP_(1)` |
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| 2. |
Solve for x and y :0.4x - 1.5y = 6.5,0.3x + 0.2x = 0.9 |
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Answer» X = -5 and y = -3 |
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| 3. |
A ring of radius R carries a non - uniform charge of linear density lambda = lambda_(0)costheta sec in the figure) Magnitude of the net dipolement of the ring is : |
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Answer» `PIR^(2)lambda_(0)` |
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| 4. |
What was Hertz's experiment ? |
| Answer» Solution :In 1888, HERTZ understood the existence of em. Wave. By using oscillting a.c. CIRCUITS he not ONY produced and DETECTED em. | |
| 5. |
Define electrical resistivity of a material of a conductor. |
| Answer» Solution :It is the resistance of the conductor PER UNIT length and unit AREA of cross-section. | |
| 6. |
A short bar magnet has a magnetic moment of 0.5 J T^(-1). Calculate magnitude and direction of the magnetic field produced by the bar magnet which is kept at a distance of 0.1 m from the centre of the bar magnet (a) axial line of the bar magnet and (b) normal bisector of the bar magnet. |
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Answer» Solution :GIVEN magnetic 0.5 J `T^(-1)`and distance r = 0.1 m (a) When the point lies on the axial line of the bar magnet, the magnetic field for SHORT magnet is given by . `vec(B)_("axial") = (mu_(0))/(4 pi) ( (2p_(m))/(r^(3)) ) hat(i)` `vec(B)_("axial") = 10^(-7) xx ( (2xx 0.5)/((0.1)^(3)) ) hat(i) = 1 xx 10^(-4) T hat(i)` hence, the magnitude of the magnetic field along axial is `B_("equatorial ") = 0.5 xx 10^(-4) `T and direction is towards North to South. (b)When the piont lies on the normal bisector (equatorial ) line of the bar magnet, the magnetic field for short magnet is given by `vec(B)_("axial") = (mu_(0))/(4 pi) (p_(m))/(r^(3)) hat(i)` `vec(B)_("axial") = 10^(-7) xx ( ( 0.5)/((0.1)^(3)) ) hat(i) = -0.5 xx 10^(-4) T hat(i)` Note that magnitude of `B_("axial")` is twice that of magnitude of `B_("equatorial")` and the direction of `B_("axial") and B_("equatorial") ` are opposite. |
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| 7. |
The Brewster's angle for a transparent medium is 60^@ The angle of incidence is: |
| Answer» ANSWER :D | |
| 8. |
Green light of wavelength 5460 Å is incident on an air and glass interface. If refractive index of glass is 1.5, then wavelength of light in glass would be ..... |
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Answer» 5460 Å `lambda`=WAVELENGTH in AIR = `5460^@`, `lambda.`=wavelength in MEDIUM n=1.5 `thereforelambda.=(lambda)/(n)=(5460)/(1.5)=3640` Å |
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| 9. |
See the figure here. We have two hollow thin spherical shells A and B of radii R, and R, respectively. Initially as shown in Fig.(i) shell B has a charge + Q which is uniformly distributed over its outer surface but shell B has no charge. At a particular instant the two spherical shells are connected by a thin copper wire as shown in Fig.(ii). After a couple of minutes two spherical shells A and B are disconnected again as shown in Fig.(iii). Now answer the following questions : Find the ratio of surface charge densities of shell A and shell B. |
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Answer» Solution :LET finally CHARGES on two shells be `q_A and q_B` respectively, then `q_A/q_B = (C_AV)/(C_BV) = C_A/C_B = R_1/R_2` As SURFACE density of CHARGE `sigma = q/(4pi R^2)` `rArr sigma_A/sigma_B = q_A/q_Bxx(R_2/R_1)^2 = R_1/R_2 xx (R_2/R_1)^2 = R_2/R_1` |
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| 10. |
Magnetic susceptibility of paramagnetic substance is ...... . |
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Answer» zero |
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| 11. |
Select the correct alternative |
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Answer» The charge gained by the uncharged body from a charged body DUE to conduction is equal to half of the total charge initially present. |
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| 12. |
See the figure here. We have two hollow thin spherical shells A and B of radii R, and R, respectively. Initially as shown in Fig.(i) shell B has a charge + Q which is uniformly distributed over its outer surface but shell B has no charge. At a particular instant the two spherical shells are connected by a thin copper wire as shown in Fig.(ii). After a couple of minutes two spherical shells A and B are disconnected again as shown in Fig.(iii). Now answer the following questions : Find the ratio of electric field on the surfaces of shell A and B. |
| Answer» SOLUTION :SINCE `E = sigma/epsi_0`, HENCE `E_A/E_B = sigma_A/sigma_B = R_2/R_1` | |
| 13. |
A shell is fired a cannon with a velocity V at an angle theta with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is |
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Answer» `3V cos THETA` |
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| 14. |
See the figure here. We have two hollow thin spherical shells A and B of radii R, and R, respectively. Initially as shown in Fig.(i) shell B has a charge + Q which is uniformly distributed over its outer surface but shell B has no charge. At a particular instant the two spherical shells are connected by a thin copper wire as shown in Fig.(ii). After a couple of minutes two spherical shells A and B are disconnected again as shown in Fig.(iii). Now answer the following questions : What is the ratio of final electric potentials of shell A and shell B ? |
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Answer» Solution :When two spherical shells are CONNECTED, ELECTRIC charge flows from shell A toshell B TILL the two acquires a COMMON potential V. So, we have `V_A/V_B = V/V =1/1` |
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| 15. |
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in figure (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in figure (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in figure (c). What is the expected repulsion of A on the basis of Coulomb's law ? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres. |
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Answer» Solution :Coulombian force between A and B initially, `F = k(q_(A)q_(B))/r^(2)`…….(1) When A and C are brought in contact and then SEPARATED charge on each of them will be, `q_(A)^(.) = q_( C)^(.) =(q_(A) + 0)/2 =q_(B)/2`……..(2) Similarly when B and D are brought in contact and then separated charge on each of them will be, `q_(B)^(.) = q_(D)^(.) = (q_(B) + 0)/2 = q_(B)/2`.......(3) Now, when A and B are kept at `r. = r/2` if new Coulombian force between them is F. then, `F. = k(q_(A)^(.).q_(B)^(.))/r^(2) = k(q_(A//2))(q_(B//2))/(r/2)^(2) = k(q_(A).q_(B))/r^(2)`........(4) From equation (1) and (4), `F.=F` |
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| 16. |
If a wire is stretched to make it 0.1% longer, its resistance will |
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Answer» DECREASE by 0.2% |
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| 17. |
Draw the symbol of a p-n-p transistor. Show the biasing of a p-n-p transistor and explain the transistor action. |
Answer» Solution :Symbol for p-n-p transistor  Action of p-n-p Transistor. In a p-n-p transistor, when the emitter is forward biased, the holes in the emitter and the electrons in the base begin the move towards the junction, holes in the emitter and the electrons in the base begin to move towards the junction, holes being attracted by the negative terminal and the electrons by the positive terminal of the battery. On reaching the base emitter junction a small fraction of total number of holes with electrons to GET neutralized . As the base layer is very thin the collector is kept at high negative potential , almost all the holes are attracted by the collector, producinga hole-current between the emitter and collector (See figure)  The emitter current `I_(E)` is the SUM of base current and the collector current. i.e. `I_(e)=I_(b)+I_(e)` So `I_(c) lt I_(e)` Working of p-n-p In working of `p-n-p, as each hole reaches the collector an electron is EMITTED from the negative terminal of battery and the hole is neutralized. Simultaenously a covalent bond is broken near th emitter electron the electron so produced enter the positive terminal of the forward bias battery and the hole starts its journey towards the emitter base junction. Holes are the current CARRIERS in p-n-p transistor however the current in the outer circuit is always due to flow of electrons. |
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| 18. |
See the figure here. We have two hollow thin spherical shells A and B of radii R, and R, respectively. Initially as shown in Fig.(i) shell B has a charge + Q which is uniformly distributed over its outer surface but shell B has no charge. At a particular instant the two spherical shells are connected by a thin copper wire as shown in Fig.(ii). After a couple of minutes two spherical shells A and B are disconnected again as shown in Fig.(iii). Now answer the following questions : What is the ratio of capacitances of shell A and shell B ? |
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Answer» Solution :As capacitance of an ISOLATED spherical SHELL `C = 4PI epsi_0 R` `rArr C_A/C_B = R_1/R_2` |
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| 19. |
A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle theta with the horizontal. A horizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take g = 10 m/s^(2)) |
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Answer» `theta=45^(@)` `therforemg=0.1 XX 10= 1 N` The various forces acting on the block are as shown in the figure  If `theta` = 45°, then `costheta = sintheta` So frictional force acts on the block towards Q. If plane is rough and `theta < 45°`, then `costheta > sintheta` So frictional force acts on the block towards P. |
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| 20. |
What is a toroid ? |
| Answer» Solution :An ANCHOR ring, around which a LARGE number of turns of .METALLIC wires WOUND. is called a toroid. | |
| 21. |
Define distance of closest approach and impact parameter. |
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Answer» Impact parameter of `alpha`-particle is defined as the perpendicular distance of the velocity VECTOR of `alpha`-particle from the centre of the nucleus, when it is far away from the atoms. |
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| 22. |
Assertion : ._(1)^(3)Hisotope does not undergo fusion of the type._(1)^(3)H+._(1)^(2)Hrarras it is rarely found in nature. Reason : ._(1)^(3)H has half life of~~12 years. |
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Answer» If both assertion and reason are true and reason is the CORRECT EXPLANATION of assertion. |
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| 23. |
n alpha particles per second are emitted from N atoms of a radioactive element. Find the half-life of radioactive element? |
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Answer» SOLUTION :`(DN)/(dt) = - lamdaN IMPLIES n = - lamdaN implies LAMDA=(-n)/N` Half life `=(0.693)/(lamda) = (0.693N)/n` |
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| 24. |
A hydrogen atom absorbs photon of energy 12.1 eV in ground state, then the change in angular momentum will be ..... Js. |
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Answer» `210xx10^(-34)` `E_(n)=E_(1)=12.1impliesE_(n)=12.1-13.6=-1.51eV` but `E_(n)=(E_(1))/(n^(2))impliesn^(2)=(E_(1))/(E_(n))=(-13.6)/(-1.51)-=9` `:.n=3` `:.` CHANGE in ANGULAR momentum, `=(3h)/(2pi)-(h)/(2pi)=(2h)/(2pi)` `=(6.62xx10^(-34))/(3.14)` `=2.1xx10^(-34)Js` |
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| 25. |
A person cannot see an object lying beyond 80 cm, where as a normal person can easily see the object distant 160 cm. The focal length and nature of the lens used to rectify this defect will be |
| Answer» Solution :`1/f=1/v-1/u` | |
| 26. |
The velocity-time graph of a body moving in one dimension is as shown below. The displacement of body in 10 second is : |
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Answer» 40 m |
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| 27. |
Find rms value in the following cases (a) I = 5 +3 sin omegat (b)I = a sin omegat + b cos omegat (c ) I =i_(1) sin omegat + i_(2) cos omegat + i_(3) sin 2 omegat . |
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Answer» Solution :`bari^(2)= (int_(0)^(T) i^(2)dt)/(int_(0)^(T) dt) = (1)/(T) int_(0)^(T) (5 + 3sin omegat)^(2)dt` `= (1)/(T) int_(0)^(T) (25 + 30sin omegat + 9sin^(2) omegat)dt` `= (1)/(T) [25 T + 30 int_(0)^(T) sin omegat dt + 9 int_(0)^(T) sin^(2) omega dt]` `(1)/(T) [25T + 0 + 9 .(T)/(2)] =25+ (9)/(2) =5^(2) + (3^(2))/(2)` `i_(rms) = sqrt(bari^(2)) = sqrt(5^(2) + (3^(2))/(2) ) =sqrt59/(2)` `BAR(i^(2))=(a^(2))/(2) + (b^(2))/(2)` `i _(rms) = sqrt((a^(2) +b^(2))/(2))` `bar(i^(2))=(i_(1)^(2))/(2) + (i_(2)^(2))/(2) + (i_(3)^(2))/(2)` `i_(mas) = sqrt((i_(1)^(2) + i_(2)^(2) +i_(3)^(2))/(2))` . |
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| 28. |
What is the angular magnification(of magnifying power) of an optical instrument? |
| Answer» SOLUTION :they are TWO (I) SIMPLE, (II) COMPOUND. | |
| 29. |
See the figure here. We have two hollow thin spherical shells A and B of radii R, and R, respectively. Initially as shown in Fig.(i) shell B has a charge + Q which is uniformly distributed over its outer surface but shell B has no charge. At a particular instant the two spherical shells are connected by a thin copper wire as shown in Fig.(ii). After a couple of minutes two spherical shells A and B are disconnected again as shown in Fig.(iii). Now answer the following questions : Compare the final electrostatic energy stored in shell A with that in shell B. |
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Answer» SOLUTION :As ELECTROSTATIC energy STORED `u = 1/2 CV^2` `RARR u_A/u_B = (C_1V^2)/(C_2V^2) = C_1/C_2 =R_1/R_2` |
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| 30. |
Frequency of photon having 66 eV energy will be……(h=6.6xx10^(-34)Js) |
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Answer» `8XX10^(-15)Hz` `therefore F=(E )/(H)=(66.1.6xx10^(-19))/(6.6xx10^(-34))=16xx10^(-15)Hz` |
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| 31. |
A current of 70 mA is passed through a tangent galvanometer of 50 turns having a coil connected to a battery of 6 V having internal resistance of 15Omega. Find the galvanometer resistance. |
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Answer» |
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| 32. |
In Young.s double slit experiment the source is white light. One slit is covered with red filter and the other with blue filter. There shall be |
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Answer» ALTERNATE RED & BLUE fringes |
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| 33. |
If the anglularmomentum of an electronis J then themagnitude of themagnetic moment will be |
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Answer» `(EJ)/(m)` |
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| 34. |
When air is replaced by a dielectricmedium of dielectricconstant K, the force at attractionbetweentwo charges q_1 and q_2separated by a finite distance 'd' |
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Answer» DECREASES K TIMES |
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| 35. |
A potentiometer wire of length 100 cm has a resistance 5 ohm. It is connected in series with a resistance and a cell of emf 2v and of negligible internal resistance. A source of emf 5my balanced by 10cm length of potentiometer wire. The value of external resistance is |
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Answer» `540 OMEGA ` |
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| 36. |
The dimension of magnetic Induction B is____and IT is_____maxwell. |
| Answer» SOLUTION :`[M^1L^@T^(-2)A^(-1), 10^4]` | |
| 37. |
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly : |
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Answer» 1.2 nm |
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| 38. |
At netutral points the resultant magnetic field due to the magnet and earth is .............. . |
| Answer» SOLUTION :MAGNETIC FIELD of the EARTH | |
| 39. |
A magnet suspended so as to swing horizontally makes 50 vibrations per minutes at a place where dip is 30^(@),and 40 vibrations where sip is 45^(@) . Compare the earth's total fields at the two places. |
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Answer» |
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| 40. |
Figure shows, in cross section , two solid spheres with uniformly distributed charge throughout their volumes. Each has radius R. Point O lies on a line connecting the centers of the spheres, at radial distance R//2 from the center of sphere 1. If the net electric field at point P is zero, and Q_1 is 64muC and Q_2 = 8xmuC, what is the value of x. |
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Answer» Solution :`E_(P)=(kQ_(2))/(((3R)/(2))^(2))-(kQ_(1)(R)/(2))/(R^(3))=0` or `(4)/(9)Q_(2)=(Q_(1))/(2)` or `(Q_(2))/(Q_(1))=(9)/(8)` `thereforeQ_(2)=(9xx64)/(8)=72muC=8xx9muC` |
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| 41. |
The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch S_(1) while keeping switch S_(2) open.The capacitor can be connected in series with an series with an inductor L by closing switchS_(2) and opening S_(1). After the capacitor gets fully charged S_(1) is opened and S_(2) is closed so that the inductor is connected in series with the capacitor.Then |
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Answer» at `t=0`,energy stored in the circuit is purely in the form of MAGNETIC energy `i=-(dq)/(dt)=Q_(0) omega sin omegat` `RARR i_(max)=Comega V=Vsqrt(C/L)` |
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| 42. |
In Fig. 27-44, epsi_(1)=8.00 V, epsi_(2)=12.0V, R_(1)=100 Omega, R_(2)=200Omega, R_(3)=300 Omega. One point of the circuit is grounded (V =0). What are the (a) size and (b) direction (up or down) of the current through resistance 1, the (c) size and (d) direction (left or right) of the current through resistance 2, and the (e) size and (f) direction of the current through resistance 3? (g) What is the electric potential at point A? |
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Answer» |
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| 43. |
The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch S_(1) while keeping switch S_(2) open.The capacitor can be connected in series with an series with an inductor L by closing switchS_(2) and opening S_(1). Initially the capacitor was uncharged.Now switch S_(1) is closed and S_(2) is kept open.If time constant of this circuit is t, then |
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Answer» after TIME INTERVAL `t`, charge on the CAPACITOR `Q=CV(1-e^(-t//tau))` after time interval `2TAU`. |
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| 44. |
If the equation of state of a gas isexpressed as (P+(a)/(V^(2)))(V-b)=RT where Pis the pressure V is the volume and T the absolute temperature and a,b,R are cosntants then find the dimensions of a and b . |
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Answer» Solution :By principle of homogeneity of dimensions P can added to P only. ITMEANS `(a)/(V^(2))` also gives pressure. Dimension formulae for pressure (P) =` M^(1)L^(-1)T^(-2)` and Volume (V) =` M^(0)L^(3)T^(0)` . Since `(a)/(V^(2)) ` = pressure `:.(a)/((M^(0)L^(3)T^(0))^(2))=M^(1)L^(-1)T^(-2)IMPLIES (a)/(M^(0)L^(6)T^(0))=M^(1)L^(-1)T^(-2)` `:. a = M^(1)L^(5)T^(-2)` Similarly b will have same dimension as volume V-b = volume `:. b = M^(0)L^(3)T^(0)` |
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| 45. |
(a) The energy released in the explosion of 1.00 mol of TNT is 3.40 MJ. The molar mass of TNT is 0.227 kg/mol. What weight of TNT is needed for an explosive release of 1.80xx10^(14)J? (b) Can you carry that weight in a back-pack, or is a truck or train required? (c ) Suppose that in an explosion of a fission bomb, 0.080% of the fissionable mass is converted to released energy. What weight of fissionable material is needed for an explosive release of 1.80xx10^(14)J? (d) Can you carry that weight in a backpack, or is a truck or train required? |
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| 46. |
When light gets polarised by reflection, the reflected beam is __ to the refracted beam . |
| Answer» SOLUTION :PERPENDICULAR | |
| 47. |
In a full-wave rectifier, the load current flows for |
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Answer» The complete CYCLE of the INPUT signal |
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| 48. |
In the above question, the angular momentum of the stone is : |
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Answer» `64pikgxxm^(2)//s` |
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| 49. |
In the given circuit if point C is connected to the earth and a potential of + 2000 V is given to point A, the potential at B is :- |
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