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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. |
k के किस मान के लिए निम्न समीकरण युग्म के अनंत हल होंगे :2x+3y=5.4x+ky=10 |
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Answer» 1 |
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| 2. |
Two light rays 1 and 2 are incident on two faces AB and AC on an isosceles prism as shown in the figure. The rays emerge from the side BC. Then, |
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Answer» minimum DEVIATION of ray `1gt`minimum deviation of `ray 2`` `mu` and A for both rays are same. Hence, value of `delta_m` is also same for both rays. |
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| 3. |
A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet while it is passing through the ring is |
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Answer» EQUAL to that due to gravity |
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| 4. |
Given magnetic field equation is vec(B)=3xx10^(-8) sin[omega t-kx + phi] hat(j) then appropriate equation for electric field (vec(E )) will be ……. |
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Answer» `3xx10^(-9)sin (omega t-kx + phi)hat(K)` `THEREFORE (E_(0))/(B_(0))=c` `therefore E_(0)=B_(0)c` `therefore E_(0)=3xx10^(-8)xx3xx10^(8) ""` [`therefore` From given equation `B_(0)=3xx10^(-8)T`] `therefore E_(0)=9N//C` `therefore` Equation of electric field, `vec(E )=E_(0)sin[omega t-kx + phi)hat(k)` `therefore vec(E ) = 9 sin [omega t-kx+phi)hat(k)` Electric field will be in z - DIRECTION. |
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| 5. |
The ozone layer on top of stratosphere is crucial for human survival, why? |
| Answer» SOLUTION :If an em WAVE transfers a total energy U to a SURFACE , the total momentum delivered to the surface is . P = U/C. | |
| 6. |
Abeam of light consisting of two wavelengths , 6500 A^(0) and 5200 A^(0)is usedto obtain interference frnges in a Young.s double slit experiment ( 1 A^(0) = 10^(-10) m) .The distancebetween the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm . Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 A^(0) |
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Answer» `0.117` CM |
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| 7. |
Crystalline solid are |
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Answer» anisotropic |
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| 8. |
The dimensional representation of Planck's constant will be |
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Answer» `MLT^-2` |
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| 9. |
The electric field at distance .r. from infinte line of charge ("having linear charge density" lambda) is |
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Answer» `(lambda PI epsilon_(0)R)` |
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| 10. |
The alternating voltage induced in the secondary coil of a transformer is mainly due to |
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Answer» A varying electric field |
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| 11. |
A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one meter away. What would be the fringe width ? |
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Answer» `2.45 XX 10^(-4)m` |
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| 12. |
The r.m.s. value of potential difference V shown in Fig. is |
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Answer» `V_(0)` `:. V_(RMS) = sqrt((V_(0)^(2) xx (T)/(2) xx 0^(2) xx T//2)/(T)) = (V_(0))/(sqrt2)` |
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| 13. |
Light of wavelength 9xx10^(-9) m is incident on a metal plate . If the wavelength of metal is negligible , calculate the de Broglie wavelength of the photoelectrons emitted. |
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Answer» |
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| 14. |
The binding energy of a hydrogen molecule is 4.75 eV. Energy required to dissociate 0.05% of hydrogen gas at NTP occupying volume 5.6 litres is: |
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Answer» 20 J nearly = `B.E. = 4. 75 XX 1.6 xx 10^(-19) J` Volume to be dissociated =`(5.6xx0.05)/100` Since 224 litre contain `6.02xx10^(23)` molecules. `THEREFORE` Number of molecules to be dissociated `therefore` Total energy required `=(6.02xx10^(23))/(22.4)xx(5.6xx0.05)/100xx4.75xx1.6xx10^(-19)` `=57.2J` Thus the correct CHOICE is (d). |
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| 15. |
The work function of caesium is 2.14 eV.Find (a) the threshold frequency for caesium,and (b)The wavelength of the incident light if the photocurrent is brought to zero by stopping potential of 0.60 V. |
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Answer» Solution :Here,work function, `phi_(0)`=2.14 EV `h=6.63xx10^(-34)JS` `1eV=1.6xx10^(-19)J` `C=3.0xx10^(8)m//s` Stopping potential `V_(0)=0.60`V `therefore=V_(0)=?,lambda=?` (a)Work function, `phi=hv_(0)` `therefore V_(0)(phi_(0))/(h)=(3.424xx10^(-19))/(6.63xx10^(-34))` `therefore V_(0)=0.51644xx10^(15)` `therefore V_(0)~~5.16xx10^(14)Hz` (b) Photoelectric equation, `eV_(0)=hv-phi_(0)` `therefore eV_(0)=hv-phi_(0)` `therefore eV_(0)=(hc)/(lambda)-phi_(0)` `therefore eV_(0)+phi_(0)=(hc)/(lambda)` `therefore lambda=(hc)/(eV_(0)+phi_(0))` =`(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx0.60+3.424xx10^(-19))` =`(19.89xx10^(-26))/(0.96xx10^(-19)+3.424xx10^(-19))` `therefore ~~ 454xx10^(-9)m=454nm` |
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| 16. |
The dipole moment of a current loop is independent of |
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Answer» CURRENT in the loop |
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| 17. |
In which of the following situations involving a source of sound and a detector of the sound is it possible that there is NO perceived Dopper shifts? |
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Answer» The SOURCE travels TOWARDS the stationary DETECTOR |
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| 18. |
The wax melts up to the length 15 cm and 30 cm for two identical rods of different metals in Ingen Hausz apparatus. The ratio of thermal conductivities is : |
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Answer» `2:3` `rArr(k_(1))/(k_(2))=1/4` Thus correct CHOICE is (C ). |
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| 19. |
Find maximum voltage across AB in the circuit shown in Fig. Assume that diode is ideal. |
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Answer» Solution :As the DIODE is treated ideal, its forward RESISTANCE `R_(f)=` zero. It acts as short circuit. So, `10 kOmega` is in parallel with `15kOmega` and the effective resistance across AB is `R_(AB)=(10xx15)/(10+15)=(10xx15)/(25)=6kOmega` `6kOmega` in series with `5kOmega`. `therefore` total resistance `=R_(T)=6kOmega+5kOmega=11kOmega` `V=30V` Current drawn from the battery is `I=(V)/(R_(T))=(30V)/(11kOmega)=2.72mA` `V_(AB)=IR_(AB)=2.72mA xx6kOmega=16.32V` |
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| 20. |
An electron is displaced by 1.5 cm opposite to the direction of a field of intensity 2xx10^(4)N//C. Then a proton covers the same distance towards the direction of the previous field. Mass of a proton and an electron are -1.73xx10^(-27)kg and 9.1xx10^(-31)kg respectively. Determine the amount of time taken by the proton and the electron. |
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Answer» |
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| 21. |
A plane wavefront ABC is incident on a pIane ABC as shownin [fig.6.26]. veIocity ofIightis 3xx10^(8). s^(-1)(ii) During that time what wiII be the distance covered by the waveIet ftom A ? |
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Answer» |
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| 22. |
A plane wavefront ABC is incident on a pIane ABC as shownin [fig.6.26]. veIocity ofIightis 3xx10^(8). s^(-1)What wiII be the distance covered by waveIet from D at that moment?Draw the diagram of the refIected wavefront. Given that, AE = 12 m andAD = 6m |
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Answer» |
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| 23. |
If E is emissive power and 'a' is the coefficient of absorption of a bbody at any temperature, E_(b) is the emissive power of a perfectly black body at that temperature, then according to Kirchhoff's law |
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Answer» `E=E_(B)//a` |
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| 24. |
Atomic number of Iodine is 53 and its mass number is 125. Radius of iodine is approximately.(R_0 = 1.2 xx 10^(-15) m) |
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Answer» `6 XX 10^(-5)m` |
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| 25. |
Coulomb's Law, Superposition Law Coulomb's law supports ....... |
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Answer» LENZ's LAW |
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| 26. |
The tangent of polarizing angle is numerically equal to |
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Answer» DIVERSITY of the REFLECTING medium |
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| 27. |
When a piece of polythene is rubbed with wool, a charge of -2 xx 10^(-7)C is developed on polythene,what mass is transferred to polythene? |
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Answer» `5.69 XX 10^(-19)` KG |
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| 28. |
A card sheet divided into squares each of size 1 mm^2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens ? How much is the area of each square in the virtual image ? (b) What is the angular magnification (magnifying power) of the lens ? (c) Is the magnification in (a) equal to the magnifying power in (b) ? Explain. |
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Answer» Solution :Here, distance of card SHEET from magnifying glass `|u| = 9 cm`, focal length of magnifying glass f= + 9 cm and size of SQUARE `h= 1 mm^(2)`or a side of square `h = sqrt(1 mm^(2)) = 1MM` (a) As |u| = f= 9 cm,hence image formed by magnifying glass lies at infinite distance. `therefore` Magnification `=D/f = (25 cm)/(9 cm) = 2.8` `therefore` Side of each square in the image `h. = mh = 2.8 xx 1 mm = 2.8 mm` `therefore` Area of each square in the image `=(2.8)^(2) mm^(2) = 7.8 mm^(2)` (b) ANGULAR magnification (magnifying power) of the lens = `(1+D/f) = 1+ 25/9 = 3.8` (c) The magnification in (a) is not EQUAL to the magnifying power in (b). In fact the two values are equal only when the image is formed at the least distance of distinct vision. |
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| 29. |
Humen eye is sensitive to |
| Answer» Solution :Visible spectrum (WAVE lengths from 400 NM to 750 nm) | |
| 30. |
A vehicle has a driving mirror of focal length 30 cm. Another vehicle of dimension 2 xx 4 xx 1.75 m^3is 9 m away from the mirror of first vehicle. Position of the second vehicle as seen in the mirror of first vehicle is |
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Answer» 30cm |
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| 31. |
Draw a circuit diagram to study characteristics of n-p-n transistor in CE configuration. Draw the sketch of input and output characteristics of this configuration . Define current amplification factor ? |
Answer» Solution :Diagram for CE characteristics of n-p-n transistor  Output characterisitics  Current AMPLIFICATION FACTOR is the ratio of collector current to base current. i.e. `beta=(I_(c))/(I_(b))`. |
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| 32. |
If the feedback voltage is increased in a negative feedback amplifier , then |
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Answer» a. both GAIN and DISTORTION DECREASES |
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| 33. |
If vecB,vecB, vecC represents the three sides of an equilateral triangle taken in the same order then find the angle between (i) vecA and vecB (ii) vecB and vecC (iii) A and vecC. |
| Answer» Solution : From the DIAGRAM the ANGLE between the VECTORS `vecA and VECB` is `120^(@)` the angle between `vecB and vecC` is `120^(@)` the angle between `vecA and vecC` is `120^(@)` | |
| 34. |
In producing a pure specturm, the incident light is passed through a narrow slit placed i the focal plane of an chromatic lens because a narrow slit |
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Answer» a PRODUCES LESS diffraction |
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| 35. |
In photoelectric effect , if a week intensity radiation instead of strong intensity of suitablefrequency is used then |
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Answer» Photoelectric EFFECT will get delayed |
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| 36. |
Assertion: An electrostatic field line never form closed loop Reason:Electrostatic field is a conservative field. |
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Answer» Both Assertion and REASON are true and Reason is the CORRECT EXPLANATION of Assertion |
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| 37. |
Which are universal gates,Why? |
| Answer» SOLUTION :NAND GATE and NOR gate are called universal gate because all other gates can be CONSTRUCTED using these gates. | |
| 38. |
A person wear normal spectacles in which the distance of glasses and eyes is approximately 2 cm. then power required is -5D if the wears contact lens, then the requires power is |
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Answer» <P>`-5.2D`  For glasses `(1)/(V)-(1)/(u)=(1)/(f)implies(1)/(-x)-(1)/(oo)=(1)/(f)` `f=-x" "cm=-(x)/(100)` power`=(1)/(f)=-(100)/(x)=-5impliesx=20cm` if he used lens.  `v=-oo,u=-22cm` `(1)/(v)-(1)/(u)=(1)/(f)implies(1)/(-oo)=(1)/(f)` `f=-22,p'=-(100)/(22)D=-4.54D` |
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| 39. |
When a charged particle moves in a uniform magnetic field at right angles to the direction of the field, which of the following changes? Speed of the particle, Energy of the particle or Path of the particle. |
| Answer» SOLUTION :PATH of the PARTICLE. | |
| 40. |
A vessel contain oil (Density=0.8gm/cm^3) over mercury ( density=13.6gm/cm^3) . Ahomogenous sphere floats with half of it's volume immersed in mercury and the other half in oil . The density of the material of the sphere in (gm)/(cm)^3 is |
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Answer» a)3.3 |
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| 41. |
The base current of a transistor is l00muA and the collector current is 2 mA. Determine the values of beta,I_E and alpha. |
| Answer» SOLUTION :`I_B=100muA=100xx10^-3mA,I_c=2mAbeta=I_C/I_B=2/(100xx10^-3)=20I_B=I_C+I_B=2+100xx10^-3=2+0.1=2.1mAalpha=I_C/I_E=2/2.1=0.95` | |
| 42. |
if the velocity of electron is 25% of the velocity of photon, then (E_(e))/(E_(ph)): |
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Answer» `1:2` |
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| 43. |
Light is incident from glass (mu=1.50) to water (mu=1.33). Find the range of the angle of deviation for which there are two angles of incidence. |
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Answer» 0 to `COS^(-1) (8//9)` |
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| 44. |
The resistance of a 10m long wire is 10 Omega. Its length is increased by 25% by stretching. The resistance of the wire is now |
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Answer» `12.5 OMEGA` |
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| 45. |
If the planes of two identical concentric coils are perpendicular and the magnetic moment of each coil is M, then the resultant magnetic moment of the two coils wil be |
| Answer» ANSWER :D | |
| 46. |
The transfer characteristics of a base biased common in the figure. Which of the following statements are true? |
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Answer» At `V_(i) = 1 V,` it can be used as an amplifier. (i) When `V_(i)=1V`, which is in between 0.6 V to 2 V, the transistor circuit is in active state and is used as an amplifier. (ii) When `V_(i)=0.5V`, there is no collector CURRENT. The transistor is in CUT off state. The transistor circuit can be used as a switch turned off. (iii) When`V_(i)=2.5V`, the collector current becomes maximum and transistor is in saturation state and can be used as switch turned on. So option (a), (b) and (c) are correct. 
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| 47. |
A simple pendulum of length has a maximum angular displacement theta. The maximum kinetic energy of the bob of masss m will be : |
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Answer» `mg(1-cos theta)`  `:.""cos theta=(OC)/(OB)=(l-h)/(l)=1-(h)/(l)` `(h)/(l)=1-cos thetaimpliesh=(l-cos theta)` `:.` P.E. of bob at extreme position `="mgl"(1-cos theta)`. Correct choice is ( c ) |
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| 48. |
What will be the angular momentum in fourth orbit if L is the angular momentum of the electron in the second orbit of hydrogen atom? |
| Answer» ANSWER :C | |
| 49. |
n resistances, each of r Omega, when connected in parallel give an equivalent resistance of R Omega. If these resistances were connected in series, the combination would have a resistance in homs equal to |
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Answer» `n^(2)R` Resistance in SERIES combination, `R.=nr=n^(2)R` |
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| 50. |
Passage : In the circuit shown in figure : For what ratio (R_(2))/(R_(4))current through R_(3) will be zero ? |
| Answer» Answer :D | |