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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. |
Draw a graph showing variation of resistivity with temperature of nichrome. Which property of nichrome is used to make standard resistance coils ? |
Answer» SOLUTION :At LOW temperature (or ZERO) the RESISTIVITY increase as a higher poer of temperature. Then linearly increase with increase in temperature. 
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| 2. |
Diffusion current in a p-n junction is greater than the drift current in magnitude |
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Answer» if the JUNCTION is FORWARD - BIASED |
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| 4. |
In Young's double slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe if lambda = 2000 Å and d = 7000 Å? |
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Answer» 12 or `sin theta = (n lambda)/(d) = (n(2000))/(7000) = (n)/(3.5)` As , `sin theta le 1` `therefore n = 0, 1, 2, 3, `only. THUS only seven max.can be seen on both sides if the SCREEN. |
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| 5. |
Rainbow is due to a combination of |
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Answer» DISPERSION and TOTAL INTERNAL reflection |
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| 6. |
What is meant by activity or decay rate? Give its unit. |
| Answer» Solution :Theactivity (R) or decay rate is defined as the NUMBER of NUCLEI decayed per SECOND and it isdenoted as `R = |(dN)/(dt)|`. The SI UNIT of activity R is Becquerel. | |
| 7. |
What is transformer ? Write its principle and write its construction. |
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Answer» Solution :Transformer is a device which can incrase or DECREASE the AC VOLTAGE. The transformer whichincrease the voltage are called step-up transformers, while transformers which decreases the voltage are called step-down transformers. Principle `:` Transformer works on the principle of the electromagnetic induction. Construction `:` A transformer consists of two sets of COILS insulated from each other. As shown in figure ( a+b) , the coils are wound on a sof -iron core either ONE on top of the other or on separate limbs of the core. One of the coils called the primary coil has `N_(p)` turns . The other coil is called the secondary coil has `N_(s)` turns. The primary coils is the input coil the secondary coil is OUTPUT coil of the transformer. The coils are wounded by two ways `:` (1)Core type`:` Coils are wound on a soft iron core on separate limbs of core as in figure . (2) Shell type `:` Coils are wound on a soft ironcore, either one on top of the other as in figure. 
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| 8. |
If a copper wire is stretched to increase its length by 0.1%. Then percentage of increase in its resistance will be |
| Answer» ANSWER :A | |
| 9. |
A proton with a kinetic energy T- 1.5MeV us captured by a deuteron H^(2). Find the excitation energy of the formed nucleus. |
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Answer» Solution :The REACTION is `p+drarrHe^(3)` Excitation energy of `He^(3)` is just the energy available in centre of mass. The velocity of the centre of mass is `(SQRT(2m_(p)T_(p)))/(m_(p)+m_(d))~~(1)/(3)sqrt((2T_(p))/(m_(p)))` In the `CM` frame, the kinetic enrgy available is `(m_(d)~~2m_(p))` `(1)/(2)m_(p)((2)/(3)sqrt((2T_(p))/(m_(p))))^(2)+(1)/(2)2m_(p)((1)/(3)sqrt((2T)/(m_(p))))^(2)=(2T)/(3)` The total energy available is then `Q=(2T)/(3)` where `Q=c^(2)(Delta_(n)+DELTA(d)-Delta_(He)^(3))` `=c^(2)xx(0.00783+0.01410-0.01603)am u` `= 5.49MeV` Finally `E=6.49MeV` |
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| 10. |
In an organ pipe whose one end is at x = 0, the pressure is expressed by p =p_(0)cos(3pix)/2 where x is in meter and t in sec. The organ pipe can be |
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Answer» closed at ONE end, open at ANOTHER with length = 0.5m |
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| 11. |
Sketch the electric field lines for a uniformaly charged hollow cylinder shown in figure. |
Answer» SOLUTION :Electric FIELD, LINES are shown in fig  `)
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| 12. |
The light emitted in a LED is due to |
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Answer» RECOMBINATION of charge carriers |
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| 13. |
Find the component of vector A+B along i.X-axis, ii. C. Given, A=hat(i)-2hat(j)+3hat(k)andC+hat(i)+hat(j). |
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Answer» `3,(1)/(sqrt(2))` i. Component of A+B along X -axis is 3. ii. Component of A+B=R (say) along C is `:.R*C=RC cos theta` `Rcostheta+(R*C)/C=((31hati-2hat(j)+3hat(k))*(hati+hat(j)))/sqrt((1)^(2)+(1)^(2))` `=(3-2)/(sqrt(2))=(1)/(sqrt(2))` |
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| 14. |
Consider one design of electron gun, which has a barrel of length 1 m made of an insulating material. At one end of the barrel, a filament is electrically heated so that it emits electrons. Assume these electrons are emitted with a the sides negligible speed. What electric field should be applied parallel to the barrel so that electrons may achieve speed of light before leaving the barrel? Ignore the variation of mass of electron due to its speed. |
| Answer» SOLUTION :`2.56xx10^5` N/C | |
| 15. |
From the dimensional consideration which of the following equation is correct ? |
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Answer» `T=2pisqrt((R^(3))/(GM))` |
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| 16. |
Sketch the wavefronts corresponding to Diverging rays |
Answer» SOLUTION :
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| 17. |
Interestellar space has an extremely weak magnetic field of the order of 10^(-12)T. Can such a weak field be of any significant consequence ? Explain. |
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Answer» Solution :When a charged particle moves in a magnetic field, it is deflected along a circular path such that `BeV = (mV^(2))/(R):. r = (mV)/(Be)` When B is low, r high i.e., radius of curvature of path is very large. Therefore, over the GIGANTIC inter stellar distance, the deflection of charged particles becomes LESS noticeable. |
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| 18. |
Write down the applications of solar cell. |
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Answer» Solution :Solar cells are widely used in CALCULATORS WATCHES toys, PORTABLE POWER supplies, portable power supplies, etc. • Solar cells are used in satellites and space applications • Solar panels are used to GENERATE electricity |
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| 19. |
Shape of interference fringes, in general is____. |
| Answer» SOLUTION :HYPERBOLIC | |
| 20. |
A source of sound is moving with a velocity of 50ms^(-1) towards a stationary observer . The Observer measure the frequency of sound as 500Hz. The appartment frequency of sound as heard by the observer when source is moving away from hom with the same speed is (Speed of sound at room temperature 350ms^(-1) |
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Answer» 400 Hz `f_(app)^(1)=?` (source away from the listener) `v=v_("sound")=350ms^(-1)` General FORMULA: `f_(app)=f_("true")((v+v_(L))/(v+v_(s)))` Convention L = listener (observer) s = source Case (i) : Source towards the listener `f_(app)=f_("true")((v+0)/(v+v_(s)))""....(1)` Case (ii) : Source away from the listener `f_(app)=f_("true")((V+1)/(V+V_(S)))""....(2)` `((1))/((2))implies(f_(app))/(f._(app))=(v+v_(s))/(v-v_(s))` `(500)/(f._(app))=(350+50)/(350-50)=(400)/(300)=(4)/(3)` `f._(app)=500xx(3)/(4)=375Hz` |
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| 21. |
Define ac signal current amplification factor (beta). |
| Answer» Solution :The CURRENT amplification factor `beta_(dc)`. is DEFINED as ratio of COLLECTOR current `(I_E)` to the emitter current` (I_E)`. i.e.,` beta _(dc) = (I_C)/(I_E)` | |
| 22. |
For which combination of working temperatures of source and sink the efficiency of Carnot's heat engine is maximum? |
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Answer» 600 K, 400 K For given combinations `eta_(1)=(1)/(3), eta_(2)=(1)/(2), eta_(3)=(2)/(5), eta_(4)=(2)/(3)` So, it is maximum for `T_(1)=300 K and T_(2)=100 K` So, correct CHOICE is (d). |
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| 23. |
If a size of particle is a and wavelength of light is lamda for a lt lt lamda scattering is directly proportional to…. |
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Answer» `1/lamda^4` |
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| 24. |
In the question 3, the ratio of conduction current and the displacement current is |
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Answer» `((api)/(lamda))^(2)` `I^(d)=intJ_(d)sdsd theta` `=(2pi)/(LAMBDA^(2))I_(0)int_(theta=0)^(2pi)int_(s=0)^(a)ln.((a)/(s))SDS SIN(2piupsilont)d theta` `=((2pi)/(lambda))^(2)I_(0)int_(s=0)^(a)(1)/(2)DS^(2)ln((a)/(s))sin(2piupsilont)` `=(a^(2))/(4)((2pi)/(lambda))^(2)I_(0)int_(s=0)^(a)d((s)/(a))^(2)ln.((a)/(s))^(2)sin(2pi upsilont)` `=(a^(2))/(4)((2pi)/(lambda))^(2)I_(0)sin(2piupsilont)` `I^(d)=((api)/(lambda))^(2)I_(0)sin2pi upsilont=I_(0)^(d)sin2pi upsilontrArr(I_(0)^(d))/(I_(0))=((api)/(lambda))^(2)` |
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| 25. |
The line A A' is on charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge of magnitude q. B is connected by string from a point on the line A A'. The tangent of angle (theta) formed between the line A A' and the string is |
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Answer» `(Q SIGMA)/(2epsi_(0)mg)` |
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| 26. |
20.96 rounded off to 3 significant figures is |
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Answer» `21.0` |
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| 27. |
Assertion: Direction of electric field is tangent to the line of force Reason:A charge will always travel along the lines of force. |
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Answer» Both ASSERTION and Reason are TRUE and Reason is the correct EXPLANATION of Assertion |
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| 28. |
The maximum pressure variation that the human ear can tolerate in loud sound is about 30 N/m2. Find the corresponding maximum displacement for a sound wave in air having a frequency of 10^3 Hz : (take velocity of sound in air as 300 m/s and density of air 1.5kg//m^3) |
| Answer» SOLUTION :NEARLY `10^(-5) m` | |
| 29. |
In a common emitter transistor amplifier , the output resistance is 500 k Omega and the current gain beta = 49 . If the power gain of the amplifier is 5 xx 10^(6) , the input resistance is |
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Answer» a) `325 OMEGA` |
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| 30. |
The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown , has been 'removed out') at a very far off point, P , located as shown, would be (nearly) : (##DSH_NTA_JEE_MN_PHY_C07_E03_008_Q01.png" width="80%"> |
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Answer» `(5)/(6) (GM)/(X^(2))` Radius = `(R)/(2)` (from figure) `(M)/((4)/(3) pi R^(3)) = (m)/((4)/(3) pi ((R)/(2))^(3)) rArr m = (M)/(8)` Mass of the left over part of the sphere `M. = M - (M)/(8) = (7)/(8) ` M therefore gravitational field due to the left over part of the sphere `= (GM.)/(x^(2)) - (7)/(8) (GM)/(x^(2))` |
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| 31. |
This question has Statement - 1 and Statement - 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency vgtv_(0) (the threshold frequency). The maximum kinetic energy and the stopping potential are K_(max) and V_(0) respectively. If the frequency incident on the surface doubled, both the K_(max) and V_(0) are also doubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. |
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Answer» Statement-1 is true, Statement-2 is true, Statement 2 is the correct explanation of Statement-1. `h_(v)-hv_(0)=exx DeltaV` `V_(0)=(hv)/(e)-(hv_(0))/(e)` .V. is doubled `(KE_("max"))_(2)=hv-hv_(0)` `V_(0).=(DeltaV).=(2hv)/(e)-(hv_(0))/(e)` `((K_("max"))_(1))/((KE_("max"))_(2))` = may not be EQUAL to 2. `rArr (V_(0))/(V_(0))` my not equal to2. `KE_("max")=hv-hv_(0)` `V=(hv)/(e)-(hv_(0))/(e)` so K.E. and v depend upon v only. |
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| 32. |
What is the meaning of 'breeding'? |
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Answer» to have babies |
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| 33. |
The potentiometer wire AB shown in figure. Is 50 cm long. When AD=30cm, no deflection occurs in the galvanometer. Find R |
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Answer» |
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| 34. |
This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two statements. Statement 1. Davisson-Germer experiment established the wave nature of electrons. Statement 2. If electrons have wave nature, they can interfere and show diffraction. |
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Answer» Statement 1 is true, Statement 2 is true, Statement 2 is not the CORRECT explanation of Statement 1 |
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| 35. |
O-Crsol & benzyl alcohol are |
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Answer» Fuctional ISOMERS 
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| 36. |
Two bulbs 25W, 100 V (upper bulb in figure) and 100 W, 200 V (lower bulb in figure) are connected in the circuit as shown in figure. Choose the correct answer. |
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Answer» Heat LOST PER second in the CIRCUIT will be 80 J. Resistance of upper bulb, `R_1 = 100^2//25 = 400 Omega` Resistance of lower bulb, `R_2 = 200^2//100 = 400 Omega` Equivalent resistance of the circuit, `R_(eq) = 500 Omega` Heat lost per second in the circuit is `e^2//R_(eq) = 200^2//500 = 80Js^(-1)` Potential difference across BRANCHES AB and CD will be same. Hence, ratio of heat generated in them is `Heat_(AB)/Heat_(CD) = R_(CD)/R_(AB) = 1000/500 = 2:1` Hence, (c ) is incorrect. Since potential difference across branches AB and CD will be the same, but their resistances are different, so current in them will be different. As the resistances of the bulbs are same, so heat generated in them will be different. Hence, (b) is incorrect. Current drawn from the cell is `e//R_(eq) = 200//500 = 0.4A`. |
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| 37. |
Isogonic lines representthe lines of |
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Answer» ZERO DECLINATION |
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| 38. |
A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is A and its time period is T. At any instant, its speed is half that of the maximum speed. What is the displacement of the particle at the that point? |
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Answer» `(2A)/sqrt(3)` ACCORDING to QUESTION, `v_("max")/2 = (Aomega)/2 = omega sqrt(A^(2) -y^(2))` `A^(2)/4 = A^(2) -y^(2) rArr y^(2) = A^(2) -A^(2)/4 rArr y=(sqrt(3)A)/2` |
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| 39. |
Which energy state of triply ionized beryllium (Be^(+3)) has the same orbital radius as that of state of hydrogen atom |
| Answer» Solution :`v_n alpha n^2//Z` | |
| 40. |
For the double star system, the two stars having masses m_(1) and m_(2) are separated b a distance r. They are revolgin about their common centre of mass in radii r_(1) and r_(2) respectively. Which of the following is/are incorrect.? |
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Answer» The whole system can be studied can be studied by REPLACING it by a single MASS `(m_(1)m_(2))/(m_(1)+m_(2))` revolgin about the center of mass. |
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| 41. |
A plane wavefront AB Is incident on a concave mirroras shown .Then,the wave front just after reflection is - |
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Answer»
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| 42. |
Figure 16-48 shows the transverse velocity u versus time t of the point on a string at x=0, as a wave passes through it. The scale on the vertical axis is set by u_(s)=12m//s. The wave has the generic form y(x,t)= y_(m) sin (kx-omegat+phi). What them is phi? |
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Answer» |
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| 43. |
The core of a transformer is laminated to reduce |
| Answer» SOLUTION :To reduce loss DUE to EDDY currents. | |
| 44. |
Consider a hypothetical situation where we are comparing the properties of two crystals made of atom A and atom B. Potential energy (U) v//s interatomic separation (r) graph for atom A and atom B is shown in figure (i) and (ii) and respectively. Choose correct statement |
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Answer» VOLUME of `A` and `B` EXPAND on heating |
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| 45. |
Consider a hypothetical situation where we are comparing the properties of two crystals made of atom A and atom B. Potential energy (U) v//s interatomic separation (r) graph for atom A and atom B is shown in figure (i) and (ii) and respectively. When we heat the crystal of either atoms ,the atom undergo oscillation. Choose correct statement for atoms of crystal A |
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Answer» Their EQUILIBRIUM POSITION remains unchanged but average SEPARATION decreases |
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| 46. |
A student uses the circuit diagram of a potentiometer as shown in Fig. (a) For a steady current I passing through thepotentiometer wire, he gets a null point for the cell epsi_1and not for epsi_2 . Give reason for this observation and suggest how thisdifficulty can be resolved. (b) What is the function of resistance R used inthe circuit? How will the change in its value affect the null point ?(c ) How can the sensitivity of the potentiometer be increased? |
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Answer» Solution :(a) The student does not get a null point for the cell `epsi_2` because cell `epsi_2`is connected in the circuit wrongly. The difficulty can beresolved by connecting + ve terminal of `epsi_2`to point A and - ve terminal to key .b.. (b) The resistance R reduces the current PASSING through the galvanometer when the jockey contactis FAR away from the null point. As a result, the galvanometer is protected from the damage likely to be caused by a STRONG current. A change in value of resistance R does not affect the null point. (c) Sensitivity of the potentiometer can be increased by lowering the value of potential gradient along the potentiometer wire. It can be achieved (i) by INCREASING the total length of potentiometer,(ii) by placing a resistance in the auxiliary circuit of DRIVER cell in series of the potentiometer. |
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| 47. |
What is diffraction? |
| Answer» Solution :Diffraction is BENDING of waves around sharp edges into the geometrically SHADOWED REGION. | |
| 49. |
Due to the application of a force on a body of mass 100 kg that is initially at rest, the body moves with an acceleration of 20ms^(-2)in the direction of the force. Find the magnitude of the force. |
| Answer» ANSWER :B | |
| 50. |
A certain string will resonate to several frequencies, the lowest of which is 200 Hz. What are the next three higher frequency to which it resonates ? |
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Answer» 400, 600, 800 Here v = 200 Hz. `therefore` 2v, 3v, and 4v and 400 Hz, 600 Hz and 800 Hz respectively HENCE the correct choice is (a) . |
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