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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. |
Four similar charges each of charge +q are placed at four corners of a square of side a. Find the value of integral -int_(infty)^(a//sqrt(2)) vec(E).vec(d)r. (in volts) if value of integral -int_(a//sqrt(2))^(0) vec(E).vec(dr)=(sqrt(2)Kq)/(a)("where" K=(1)/(4 pi epsilon_(0))) |
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Answer» `(4Kq)/(a)` |
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| 2. |
The ceiling of a long hall is 20 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^(-1)can go without hitting the ceiling of the hall (g = 10 ms^(-2) ) ? |
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Answer» Solution :Here, H = 20 m, u = 40 `ms^(-1)`. Suppose the ball is thrown at an angle `theta` with the horizontal. Now, `H=(u^(2)sin^(2)theta)/(2G) rArr 20 = ((40)^(2) sin^(2)theta)/(2 xx 10)` or, `sin theta = 0.5` or, `theta = 30^(@)` Now, `R=(u^(2)sin 2THETA)/g =((40)^(2) xx sin 120^(@))/10` `=((40)^(3) xx 0.866)/10 = 138.56` cm |
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| 3. |
Magnetic lines of force |
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Answer» Always intersect |
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| 4. |
A circular conducting loop of radius R and resistance per unit length lambda is pulled out from the regio of uniform magnetic filed with constant velocity v. The situation shown in the figure corresponds to that is at t=0. Mark out the correct statement (s) |
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Answer» Just after `t=0 i.e.,` the MOTION starts, the induced CURRENT in the loop is `(sqrt(3)Bv)/(2pilambda)`  `PO=sqrt(R^(2)-((R)/(2)-vt)^(2))` Emf induced `e=Bvxx2PO` `i=(e)/(lambdaxx2piR)` As flux isdecrasing with time, induced emf will try to oppose the DECREASING flux and hence induced current would be in clockwise direction. |
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| 5. |
A ray of light travels from a medium of refractive, index muto air. Its angle of incidence in the medium is i, measured from the normal to the boundary, and its angle of deviation is delta. delta is plotted against i which of the following best represents the resulting curve |
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Answer»
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| 6. |
What are thesalient features of corpuscular theory of light? |
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Answer» Solution :(i) Accoding this theory, LIGHT is emitted as tiny, massiess (negligibly small mass) and perfectly elastic particles called corpuscles. (ii) As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it EMITS light for a long time. (iii) On account of high speed, they are unaffected by the force of gravity and their path is a straigh line in a medium of unifrom refractive index. (IV) The energy of light is the kinetic of these corpuscles. When these corpuscles impinge on the retain of the eye, the vision i produced. (v) The different size of the corpusles is the reason for different colours of light. (VI) Whenthe corpuscles approacha surface between two MEDIA. they are either attracted or repelied. (vii) The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium. |
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| 7. |
A charged particle is travelling in the figure. ( ## EXP_SPS_PHY_XII_C04_E04_026_Q01 .png" width="80%"> :Give the expression for the net force experienced by the charge in the region II. |
| Answer» SOLUTION :LORENTZ FORCE | |
| 8. |
A current loop consist of two straight segments (OA and OB), each of length , having an angle theta between them and a semicircle (ACB). The loop is placed on an incline plane making an angle theta with horizontal (see figure). The loop carries a current I. A uniform vertical magnetic field B is switched on. (a)Write the value of magnetic torque on the loop. (b)Tell whether the normal contact force between the incline and the loop increases or decreases when magnetic field is switched on. Assume that the loop remains stationary on the incline. |
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Answer» (d) No CHANGE |
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| 9. |
Write the expression for electric field at a point on the axis of a short electric dipole. |
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Answer» Solution :Figure Expression for `E_(1)` Expression foar `E_(2)` Substitution in `E=E_(2)+E_(1)` ARRIVING at final expression `E=(1)/(4piepsi_(0))*(2pr)/((r^(2)-a^(2))^(2))` Detailed Answer:  Consider an electric dipole consisting of +q and -q charges. 2l is the length of the dipole consider a point P at DISTANCE r from the centre .O. of the dipole on the axial LINE of the dipole. Let a UNIT positive charge be placed at point P. Now electric field intensity at P due to +q charge is given by `vecE_(+)=(1)/(4piepsi_(0))(q)/((r-l)^(2))hati` Electric field intensity at .P. due -q charge is given by `vecE_(-)=(-q)/(4piepsi_(0)(r+l)^(2))hati` According to the principle of superposition `vecE=(vecE_(+)+vecE_(-))` `=(q)/(4piepsi_(0))[(1)/((r-l)^(2))-(1)/((r+l)^(2))]hati` `=(q)/(4piepsi_(0))[(4rl)/((r^(2)-l^(2))^(2))]hati` `=(1)/(4piepsi_(0))xx(2r(qxx2l))/((r^(2)-l^(2))^(2))hati``[becausep=qxx2l]`. `vecE=(2rp)/(4piepsi_(0)(r^(2)-l^(2))^(2))hati`. |
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| 10. |
A refrigerator is the reverse of a heat engine. Explain. |
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Answer» Solution :A heat engine is a device that extracts energy from its surroundings in the form of heat and does useful work as it CONTINUOUSLY repeats a thermodynamic cycle. In ONE cycle, the working substance absorbs energy as heat from a hot reservoir at constant temperature `T_(h),` converts PART of it into useful mechinical work and rejects the remaining as heat to a cold reservoir at constant temperature `T_(c) lt T_(h).` A refrigerator is a device that uses work to transfer energy in the form of heat from a cold reservoir to a hot reservoir as it continuously repeats a thermodynamic cycle. The net result is absorption of some energy as heat from a reservoir at low temperature, a net amount of work done on the system and the rejection of a larger amount of energy as heat to a reservoir at a higher temperature. Thus, a refrigerator performs a thermodynamic cycle in a direction opposite to that of a heat engine. Hence, a refrigerator is said to be the reverse of a heat engine. |
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| 11. |
The mean lives of a radioactive substance are 1620 year and 405 year for alpha-emission and beta-emission respectively. Find the time during which three - forth of a sample will decay if it is decayed both by alpha -emission and beta-emission simultaneously. |
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Answer» Solution :The DECAY constant `lambda` is the reciprocal of the mean life, `tau` Thus, `lambda_alpha=1/1620` PER year and `lambda_beta=1/405` Per year `therefore` Total decay constant , `lambda=lambda_alpha+lambda_beta` `lambda=1/1620+1/405=1/324` year We know that `N=N_0 e^(-lambdat)` When `3/4`th part of the sample has disintegrated, `N=N_0//4 therefore N_0/4 = N_0e^(-lambdat)` or `e^(lambdat)`=4 Taking logarithm on both SIDES , we get `lambdat=log_e 4` `t=1/lambdalog_e 2^2 = 2/lambda log_e2` =2 x 324 x 0.693 = 449 years. |
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| 12. |
An alpha-particle of energy 1/2 mv^(2) bombards a heavy nuclear target of charge Ze. Then distance of closest approach for a nucleus is proportional to : |
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Answer» `1/(Ze)` `r_(0)=(1)/(4pi epsi_(0))(4Ze^(2))/(mv^(2))` `r_(0) ALPHA 1/m` |
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| 13. |
How many types of wavefronts are considered ? |
| Answer» SOLUTION :THREE, SPHERICAL, CYLINDRICAL and PLANE | |
| 14. |
The overall gain of a multistate amplifier is 100. When negative feedback is applied, the gain reduces to 10. Find the fraction of the output that is feedback to the input. |
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Answer» SOLUTION :We KNOW that `A_(f)=(A)/(1+Abeta)` Here, `A_(f)=10 and A=100` `therefore 10=(100)/(1+100beta) or 10+1000beta=100` `BETA=(100-10)/(1000)=(90)/(1000)=(9)/(100)=0.09` |
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| 15. |
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is |
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Answer» `3q//epsilon_(0)1` |
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| 16. |
A plane electromagnetic wave of frequency 20 MHz travels through a space along x-direction. If the electric field vector at a certain point in space is 6V m^(-1), what is the magnetic field vector at that point ? |
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Answer» `2 xx 10^(-3) T` `B=(6)/(3 xx 10^(8))=2 xx 10^(-8) T` |
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| 17. |
When cathode C of photocell is incident with photon of 5 eV energy ,maximum KE of electron emitted is 2 eV.For which value of stopping potential on anode (A) photo incident with energy of 6 eV,number of electron reaching to anode will become zero? |
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Answer» `-1 V` 2=5-`phi_(0)` `THEREFORE phi_(0)=3 EV` Again `eV_(0)=hf-phi_(0)` `therefore V_(0)=3eV` `therefore V_(C)-V_(A)=3V` `therefore 0-V_(A)=3V` `therefore V_(A)=-3V` |
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| 18. |
The vertical component of Earth's magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place ? |
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Answer» `30^(@)` |
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| 19. |
If the nucleons bound in a nucleus are separated apart form each other, the sum of their masses is greater than the mass of the nucleus. Where does this mass difference come form? Explain briefly. |
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Answer» Solution :When nucleons are BOUND in a nucleus, some of their mass is converted into energy that binds them together in the nucleus. The energy equal to binding energy is SPENT to separate these nucleons apart form one another. It is this energy which APPEARS in the form of INCREASED mass `[DELTA m=B.E.//c^2]`. |
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| 20. |
In Young's double slit experiment, light of wavelength 6000Å is used to get an interference pattern on a screen.The fringe width changes by 1.5 mm, when the screen is brought towards the double slit by 50 cm. The distance between the two slits is |
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Answer» `0.2` MM |
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| 21. |
If I_0 is the intensity of principle maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled |
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Answer» `I_0` |
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| 22. |
Solve Problem 25.1 assuming the capacitor to have been initially charged and subsequently disconnected before the mica sheet is inserted in it. |
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Answer» Solution : If the capacitor is disconnected from the SOURCE, the quantity remaining constant is the CHARGE on the plates, the potential and the FELD strength CHANGE by a FACTOR of 1/8 when the dielectric is inserted between the plates. Therefore `sigma_("pol")=((epsi-1) epsi_(0) varphi)/(epsid)` |
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| 23. |
(A): Repeaters are used to extend the range of communication (R): Repeater is a combination of a receiver and a transmitter |
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Answer» Assertion and reason are TRUE and reason is the CORRECT explanation of assertion |
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| 24. |
In the arrangement as shown in the figure, S_(1) and S_(3) are already closed for a long time. Now at t=0 the S_(2) is closed and S_(1), S_(3) are opened [ epsilon=1 volt, L=1 henry, R=1Omega & C=1 Faraday ] |
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Answer» After `S_(2)`s is CLOSED the maximum current through the INDICTOR is `2A` `I_("max")=sqrt(2)` amp Also, `(Q_(i)^(2))/(2c)+1/2Li_(i)^(2)=(Q_("max")^(2))/(2c)` |
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| 25. |
Define decay constant and half-life and derive relation between them and explain graphically. |
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Answer» Decay constant i.E. `-(dN)/(dt) prop N_(0)` Decay constant is defined as the TIME after which the number of radioactive atoms reduce to 1/e times the original number ofatoms. RELATION between half life T and `lambda`. Half-life period The half-life of a radioactive substance is defined as the time during which half of the atoms of radioactive substancewill DISINTEGRATE. Let us denote the half-time of a substance by T. Then, by definition, after time T, number of atoms left behind will be `(N_(0))/(2)`. ...(1) Setting this condition i.e. when `t=T, N=(N_(0))/(2)` in the given equation (1), we get `(N_(0))/(2)=N_(0)e^(-lambda T)` or `e^(-lambdaT)=1/2 ` or `e^(lambdaT)=2` or `lambda T=log_(e)2=2.303 log_(10)2` `=2.303 xx 0.3010=0.693` Therefore, `T=(0.693)/(lambda)` ...(2) Thus, half life of radioactive substance is inversely proportional to its decay constant and is a characteristic property of its nucleus and cannot be altered by any known method. |
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| 26. |
What is meant by excitation energy? |
| Answer» Solution :The energy required to excite an ELECTRON from lower energy state to any HIGHER energy state is KNOWN as EXCITATION energy. | |
| 27. |
For a transistor, the current amplificationfactor alpha=0.9. This transistor connected in common emitter configuration. When the base current changes by 0.4 mA, the change in collector current will be |
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Answer» 36 mA |
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| 28. |
A student measured the diameter of a wire using a screw gauge with the least count 0.001 em and listed the measurements. The measures value should be recorded as: |
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Answer» 5.320 CM |
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| 29. |
The de-Broglie wavelength of helium atom at a temperature of 89^(@)C, will be: |
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Answer» `0.47Å` |
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| 30. |
The half-lives of radioisotopes p^(32) and p^(33)are 14 days and 25 days respectively. These radioisotopes are mixed in the ratio of 4 : 1 of their atoms. If the initial activity of the mixed sample is 3.0m Ci, find the activity of the mixed isotopes after 60 year. |
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Answer» |
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| 31. |
All questions refer to the diagram given below. The denoted lines represent the two surfaces of a lens that may be plane or curved. R_(1) and R_(2) are the radii of curvature of the surfaces on the left and right respectively. The refractive index of the medium to the left of the lens is n_(1), that of the material of the lens itself is n_(2), and that of the medium to the lens is n_(3). The first and second focal lengths of the lens are f_(1) and f_(2) respectively. Column A contains certain conditions and column B contains some results which may follow from such conditions. |
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Answer» |
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| 32. |
A metallic ring connected to a rod which oscillates freely like a pendulum. If now magnetic field is applied in horizontal direction so that the pendulum now swings through the field, the pendulum will |
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Answer» Keep oscillating with the OLD TIME period |
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| 33. |
Referring to the previous illustration, what is the (a) acceleration of the particle? (b) velocity of the particle at the end of 3rd second? (c) displacement of the particio in 3 seconds? d) displacement of the particle in 3rd second? |
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Answer» Solution :(a) SOLVING equation (1) & (2) of Illustration 7, we get a = `10m//s^(2)` (b)v=`v_(0)+at` |
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| 35. |
Eddy currents are also known as ___________ currents. |
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Answer» alternating |
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| 36. |
If 50% of CO_(2) converts to CO at the following equilibrium : 1/2C(s) + 1/2 CO_(2)(g) hArr CO(g) and the equilibrium pressure is 12 atm Calculate K_(P). |
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Answer» `4` `{:("Initial mole",1/2,0),("Eq"^(n),1/2-1/2x,x):}` `n_("total") = 0.25 + 0.5 = 0.75` `K_(P) = (P_(CO))/(P_(CO_(2))^(1//2)) = ((0.5)/(0.75)xx12)/(SQRT((0.25/(0.075)) xx 12)` `K_(p)= (0.5 xx 12)/(0.75 xx 2) = 2/3 xx 6 = 4` |
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| 37. |
The number of silicon atoms per m^(3) is 5 xx 10^(28). This is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of Arsenic and 5 xx 10^(20) per m^(3) atoms of Indium. Calcualte the number of electrons and holes. Given that n_(i) = 1.5 xx 10^(16) m^(-3). Is the material n-type or p-type? |
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Answer» SOLUTION :`n_(C) = 4.95 xx 10^(22), n_(h) = 4.75 xx 10^(9)`, n-type since `n_(e) gt gt n_(h)` For CHARGE neutrality `N_(D)-N_(A) = n_(e).n_(h) = n_(i)^(2)` Solving these equations, `n_(e) = 1/2 [(N_(D)-N_(A)) + sqrt((N_(D)-N_(A))^(2) + 4n_(i)^(2))]` |
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| 38. |
Give outline of experimental study of photoelectric effect. |
Answer» Solution :As shown in figure a metal plate (cathode C) and another metal plate (anode A) are placed inside evacuated glass TUBE.  A quartz window is sealed on tube from which ultravioletradiation can pass through it and ultraviolet radiation can pass through it and irradiate photosensitive plate C. Monochromatic ultraviolet radiation having sufficiently low wavelength is incident on cathode (C )Thorugh quartz window W. Electrons are EMITTED from plate (Cathode C) Whic are attracted by collector (anode) plate A and due to this flow of electron current is formed. Potential difference between emitter (C ) and collector (A) can be measured by voltmeter. current produced is of the order of `muA` RANGE which can be measured by micro ammeter connected. By CHANGING photosensitive surfaceon plate ALSO can be changed and change in photocurrent can be studied. Intensity and frequency of radiation incident of collector plate (C )can be varied and change in photoelectric current can be studied. Filter or coloured glass of different colour can beplaced between plate (C )and path of incident radiation and light of different frequencies can be used. By changing distance between light source and plate (C ) can be varied. |
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| 39. |
f:CrarrR f(Z)=IZI is |
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Answer» ONE - one and into |
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| 40. |
When air is replaced by dielectric medium of constant K, the maximum capacitance of the capacitor |
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Answer» INCREASES K times |
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| 41. |
Consider the case of two bodies of masses m_(1) and m_(2) which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations: (i) when m_(1) gt m_(2). In this case a=((m_(1)-m_(2))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g (ii) when m_(2) gt m_(1). In this case a=((m_(2)-m_(1))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g (iii) When m_(1)=m_(2)=m. In this case a = 0, T = mg If the pulley is pulled upward with acceleration equal to the acceleration due to gravity, what will be the tension in the string? |
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Answer» 75N `T.-2m_(2)g=m_(2)a. ""...(2)` On SOLVING (1) and (2) `a.=(g)/(2)=5m//s^(2)` `T.=150N` |
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| 42. |
A current loop is placed in a uniform magnetic field in the following orientations (1) and (2). Calculate the magnetic moment in each case. |
| Answer» SOLUTION :(i) `-MB` (II) ZERO | |
| 43. |
Graph plotted between stopping potential V_(0) and frequency v, where v gt v_(0), for a photosensitive material is a_____graph whose slope is ____. |
| Answer» SOLUTION :LINEAR, `(H)/E` | |
| 44. |
When a clean metal surface incident with radiation of sufficiently high frequency ,electron are emitted from it.This method of emission of electron is called……. |
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Answer» THERMIONIC emission |
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| 45. |
Two monochromatic radiations, blue and vilet, of the same intensity, are incident on a photosensitive surface and cause photoelectric emission. Would (i) the number of electrons emitted per second and (ii) the maximum kinetic energy of the electrons, be equal in the two cases ? Justify your answer. |
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Answer» Solution :Here intensity I of both monochromatic radiations of BLUE and violet colour are EXACTLY the same, but frequency of violet radiations is more than that of blue radiations. Hence (i) The number of electrons emitted per second are equal in two cases because number of photoelectrons depends on the number of incident radiation PHOTONS i.e., the intensity of incident radiation. (II) The maximum kinetic energy of photoelectrons emitted by violet radiation is more because `K_(max)=hv-phi_(0)`, where `phi_(0)` is the work function of given photosensitive surface. |
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| 46. |
If m_(e) is mass of an electron, then mass of pion plus (pi^(+)) particle is |
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Answer» `207m_(E)` |
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| 47. |
A voltmeter having a resistance of 1800 Omega employed to measure the potential difference across a 200 Omega resistor which is connected to the terminals of a dc power supply having an emf of 50 V and an internal resistance of 20 Omega. What is the percentage decrease in the potential difference across the 200 Omega resistor as a result of connecting the voltmeter across it? |
| Answer» ANSWER :A | |
| 49. |
For a block with dlmensions1 cm xx 1 cm xx 100 cmhavng resistivity 3 xx 10^(-7) Omegam, find its resistance between two rectangular sides. |
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Answer» `3 xx 10^(-9) Omega` Area of RECTANGULAR cross section A = 1 `xx 100 cm^(2)` `= 100 cm^(2)` `= 10^(-2) m^(2)`  Resistance R` = (rho L)/(A) = (3 xx 10^(-7) xx 1)/(10^(-2))` = 3 ` xx 10^(-5) Omega` |
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