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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. |
Object of small length 1 is placed on the axis of concave mirror of focal length f, then size of image will be ...... if object is at distance d from pole. |
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Answer» `(If)/(d-F)` when u=d+1 Image size=`v_1-v_2=l((f)/(d-f))^2` |
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| 2. |
Binding energy per Nucleon of 3^(Li^7)and 2^(He^4) are 5.6MeV and 7.06 MeV. Then find the energy of proton in the following reaction._3Li^7 + P to 2 ._2He^4 |
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Answer» 1.728 MeV |
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| 3. |
If the frequency of incident light on a certain metal is 8.2xx10^(14) Hz, having threshold frequency3.3xx10^(14) Hz, then cut off potentialis : |
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Answer» 3.0 V |
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| 4. |
An observer is moving with a constant speed of 20m/s on a circular track of radius 50m. A source kept at centre of track emits a sound of frequency 200 Hz. Then the frequency received by the observer isx xx10^2Hz, what is value of x. (V= 340m/s) |
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Answer» |
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| 5. |
A double convex lens is made of glass which has refractive index 1.55 for violet rays and 1.50 for red rays. If the focal length for violet rays is 20 cm, the focal length for red rays will be |
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Answer» 9cm |
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| 6. |
Frequency of a wave . |
| Answer» SOLUTION :FREQUENCY of a wave is defined as the number of waves that PASS per unit TIME acrossa given point in the MEDIUM . | |
| 7. |
An inductor of inductance 2.0mH is connected across a charged capacitor of capacitance 5.0muF and the resulting L-C circuit is set oscillating at its natural frequency. Let Q denote the instantaneous change on the capacitor and i the current in the circuit. It is found that the maximum value of Q is 200muC . (a) When Q=100muC , what is the value of |di//dt| ? (b) When Q=200muC , what is the value of i ? (c)Find the maximum value of i (d) When i is equal to one-half its maximum value, what is the value of |Q| ? |
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Answer» 10000 |
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| 8. |
What is a Bohr magneton? |
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Answer» SOLUTION :Bor magneton is the minimum amount of magneitc moment associated with an ORBITING electron and its value is `mu_B = (eh)/(4 PI m_e) = 9.27 xx 10^(-24) A m^2`. |
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| 9. |
In S.I. system, the unit for epsilon_0 is |
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Answer» `C^2//N - m^2` |
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| 10. |
A hollow sphere has a small hole in it. On lowering the sphere in a tank of water, it is observed that water enters into the hollow sphere at a depth of 40 cm below the surface. Surface tension of water is7 xx 10 ^(-2) N//m. The diameter of the hole is approximately: |
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Answer» 0.21 MM `1000xx10xx(40)/(100)=(2sigma)/(R)=(2xx7xx10^(-2))/(R)Rightarrow2R=0.07mm` |
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| 11. |
डाइकोगेमी (विषमकालपकवता) , जो पर परागण मेसहायता करती है , जिसमे : |
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Answer» परागकोश और वर्तिकाग्र भिन्न स्तरों पर होते है |
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| 12. |
At what angle of incidence should a light beam strike a light beam strike a glass slab of refractive index sqrt3, such that the reflected and refracted rays are perpendicular to each other ? |
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Answer» <P> Solution :The REFLECTED and the REFRACTED rays are PERPENDICULAR to each other, when angle of incidence `i` is given by `n=tani_(p)`,`because n=sqrt(3) ` hence `sqrt(3)=taini_(p)impliesi=60^(@)`. |
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| 13. |
परागकोष तथा वर्तिकाग्र के एक ही समय परिपक्व होने को कहते है |
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Answer» समकालपकवता |
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| 14. |
What will be the diffraction angle of first order maxima in diffraction obtained due to light of wavelength 55nm and width of slit 0.55 mm ? |
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Answer» `0.0015` radian `d sin theta_(1)=(3lambda)/(2) "" lambda=5xx10^(-7)m` `d=55xx10^(-5)m` `:. sintheta_(1)=(3lambda)/(2D)=(3)/(2)XX(55xx10^(-9))/(55xx10^(-5))` `=(3)/(2)xx10^(-4)=0.00015` radian |
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| 15. |
A ball is dropped from a height of 20 cm above the water surface of a lake. Refractive index of water = (4)/(3). A fish situated below the surface of water along the path of motion of the ball is observing the ball. When the ball reaches a depth of 12.8cm from the surface of water, then what will be the velocity of the ball relative to the fish? |
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Answer» |
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| 16. |
A circular ring made of copper wire of 0.1 mm diameter and 60 cm long is connected as shown in Fig. 26.1. Find the resistance of the circuit. What should the length of the shorter section AB = x be for the resistance of the circuit to be 0.2 ohm? |
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| 17. |
A stationary wave on a stretched string has an equation given by y = 10 sin (2pi x)/(3) cos 8 pit, where x and y are in cm and t is in second. Then separation between the two adjacent nodes is given by : |
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Answer» 1.5 cm y = a sin kx cos `omega`t y= 10 sin `(2pi x)/(3) ` cos 8`pi` t sok = `(2pi)/(3) and omega = 8 pi `. `(2pi)/(lambda) = (2pi)/(3)rArr lambda = 3` cm. So DISTANCE between two CONSECUTIVE NODES ` = (lambda)/(2) = 1.5` cm . correct CHOICE is a. |
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| 18. |
State Nweton's law of net loss of heat. Hence, show that (dtheta)/(dt) prop(theta-theta_(@)). |
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Answer» Solution :Newton's law of cooling : The rate of loss of heat by a BODY is directly proportional to its excess of temperature over the surroundings, provided the excess is small. Let m, c and `theta` be the mass, speciffic heat capacity and temperature of a body RESPECTIVELY and let `theta_(@)`, assumed to be less than `theta`, be the temperature of the surroundings. As `theta gt theta_(@)`, the body loses heat to the surroundings. ACCORDINGS to Nweton's law of cooling, the rate of loss of heat by the body, `(dQ)/(dt) PROP(theta-theta_(@)` `:. (dQ)/(dt)=K(theta-theta_(@))` where K is a constant that depends on the body and the surroundings. But `(dQ)/(dt)=mc(dtheta)/(dt)` where `dtheta//dt` is the rate of fall of temperature of the body. ltBRgt `:. MC(DTHETA)/(DT)=k(theta-theta_(@)) :. (dtheta)/(dt)=(K)/(mc)(theta-theta_(@))` `:. (dtheta)/(dt)=k(theta-theta_(@))`, where `k=K//mc` is a constant. `:.` Rate of cooling of the body, `(dtheta)/(dt) prop(theta-theta_( |
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| 19. |
The weber m^(-2) is equal to : |
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Answer» tesla |
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| 20. |
Four resistances of 15Omega,12Omega,4Omega and 10Omega respectively are connected in cyclic order to form a wheatstone bridge. Is the network balanced? If not, calculate the resistance to be connected in parallel with the resistance of 10Omega to balance the network |
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| 21. |
Degree of modulation : |
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Answer» can take any value |
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| 22. |
(a) Definethe term 'self-inductance' and write theS.I. unit. (b) Obtainthe expression for the mutualinductance of twolong co-axialsolenoidsS_(1) and S_(2)wound one over the other, each of length L and radii r_(1) and r_(2) andn_(1) and n_(2)number of turnsper unit length , when acurrent I is set up in the outersolenoid S_(2). |
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Answer» Solution :(a) Self induction is the PROPERTY held by the coil byvirtue of which it opposes any changein strengthof the current flowingthrough it(an emf is inducedin it)S.I., is called inertiaof electricity, is its unit is Henery, `1H = (1V)/(1Am//s) = 1 W//A`. (b) `S_(1)` and `S_(2)` are the solenoidwith same length las shown`n_(1)` and `n_(2)` through `S_(1)phi_(2)1Flu`lined with `S_(2)`DUE to currentthrough S. `phi_(21) prop I_(1)`. `phi_(21) = m_(2), I_(1)` Where, `M_(21)` is co-efficientof mutual inductionof the two solenoid. As the current pass through `S_(1)`, emf is inducedin `S_(2)`. Magnecticfield inside `S_(1) = B_(1) = mu_(0)n_(1)O_(1)` Magneticflux linkedwith each turn of `S_(2) = B_(1)A` So total magneticflux linkedwith `S_(2)phi_(21) = B_(1)An_(2)I` `= mu_(0)n_(1)I_(1) xx A xx n_(2)l` `= phi_(21) = mu_(0)n_(1)n_(2) AI_(1)` `M_(21) = mu_(0)n_(1)n_(2)Al` Mutual inductancebetweenthe twosolenoidswhen currentis passed through `S_(2)`and inducedemf is producedin `S_(1)` `M_(12) = mu_(0)n_(1)n_(2)Al` Mutual inductance betweenthe two solenoidsthe two solenoidswhen current is passed through`S_(2)`and indcued emf isproduced in `S_(1)`. `M_(12) = mu_(0)n_(1)n_(2)Al` So, `M_(12)= M_(21) = M`(say) Hence, co-efficient of mutual inductionbetween the long solenoids `M= mu_(0)n_(1)n_(2)Al`. 
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| 23. |
(a) State the working principle of a potentiometer with the help of a circuit diagram.(b) Explain how the internal resistance of a cell is determined. (c ) How do the following affect the potentiometer circuit when(i) the internal resistance of the driver cell increases, (ii) the series resistor connected to the driver cell is reduced ? Justify your answer. |
| Answer» SOLUTION :( C) (i) If internal resistance of the driver cell increases then current I flowing through potentiometer wire and HENCE potential gradient decreases. (ii) If series resistor CONNECTED to the driver cell is reduced, the current I flowing through potentiometer wire as well as potential gradient across the potentiometer wire increase. | |
| 24. |
Give uses of cyclotron. |
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Answer» Solution :The uses of cyclotron as following : (1) The high energy particles produced in a cyclotron are used to bombard nuclei and STUDY the resulting nuclear REACTIONS and hence investigate nuclear structure. (2) It is used to implant IONS into SOLIDS and modify their properties or even synthesis new materials. (3) It is used to produce radioactive isotopes which are used in hospitals for diagnosis and treatment. |
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| 25. |
हाइड्रोजन की ऑक्सीजन से क्रिया करने पे जल बनता इसमे उत्पाद क्या है |
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Answer» हाइड्रोजन |
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| 26. |
A mobile phone lies along the principal axis of a concave mirror, as shown in figure. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror ? |
| Answer» Solution :The FORMATION of the image of the PHONE is shown in figure. The image of the part which is on the plane perpendicular to principal axis will be on the same plane . It will be of the same SIZE. i.e, B.C = BC . You can yourself why the image is distorted. | |
| 27. |
A TV tower has a height of 100m. The average population density around the tower is 1000 per km^(2). The radius of the earth is 6.4xx10^(6)m. The population covered by the tower is |
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Answer» `2XX10^(6)` |
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| 28. |
A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then: |
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Answer» some charge from the CAPACITOR will flow BACK into the source. |
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| 30. |
The ratio of magnetisation I to the magnetic field intensity H is |
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Answer» PERMEABILITY |
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| 31. |
Assertion (A) : Isobars are the nuclei having same mass number but different atomic number. Reason (R) : Neutrons andprotons are present inside a nucleus . |
| Answer» SOLUTION :In ISOBARS number of PROTONS are DIFFERENT but total number of nucleons is the same. | |
| 32. |
The maximum kinetic energy of electrons emitted by a photocell is 2-8 V. What is the stopping potential ? |
| Answer» SOLUTION :STOPPING POTENTIAL will be - 2-8 V | |
| 33. |
What is the magnification when the object is placed at 2f from the pole of a convex mirror ? |
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Answer» 1/3 |
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| 34. |
State Gauss's theoremin electrostatics.Apply this theorem to derive an expression for electric field intensityat a point outside a uniformlycharged thinsphericalshell. |
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Answer» Solution :Gauss's Theorem in electrostatics : It states that the flux of electric field through any closed surface `delta` in the vacuum is `(1)/(epsilon_(0))` TIMES the total charges enclosed by S. `:. phi = oint vec(E).dvec(s)=(q)/(epsilon_(0))` Electric field due to a uniform charged spherical shell : Suppose a thin spherical shell of Radius R and centre O. Let the charge +q is distributed over the surface of sphere. Electric field intensity `vec(E)` is same at every POINT on the surface of sphere directly outward. Let a point P outside the shell with radius vector `vec(r)` and small area element `vec(dS) = hat(n) dS`. According to Gauss's law `oint vec(E).vec(dS)=(q)/(epsilon_(0))RARR oint E dS = (q)/(epsilon_(0))` Since `vec(E) and hat(n) ` arein the same direction `E 4 pi r^(2) = (q)/(epsilon_(0))rArr E = (1)/(4 pi epsilon_(0)) (q)/(r^(2))` Vectorially,`vec(E)=(1)/(4 pi epsilon_(0))(q)/(r_(0))hat(r)` Special cases (i) At the point on the surface of the shell, `r=R " " :. " " E=(1)/(4pi epsilon_(0))(q)/(R^(2))` (ii) If `sigma` is the surface chargedensity on theshell then `q=4 pi r^(2) sigma` `:. E=(1)/(4pi epsilon_(0)) (4 pi R^(2)sigma)/(R^(2))=(sigma)/(epsilon_(0))` (iii) If the point P lies INSIDE hte spherical shell then the Gaussian surface encloses no charge `:. q=0` Hence E=0 `##SB_PHY_XII_08_DB_E01_025_S01.png" width="80%"> |
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| 35. |
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 xx 10^(-7) m^2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 xx 10^28 m^(-3) . |
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Answer» SOLUTION :Here `A= 1.0 xx 10^(-7) m^2 , I = 1.5 A , n = 9 xx 10^28 m^(-3) " and " e = 1.6 xx 10^(-19) C` `I = nA ev_d` ` therefore ` AVERAGE drift speed `v_d = (I)/(nAe) = (1.5)/(9 xx 10^28 xx 1.0 xx 10^(-7) xx 1.6 xx 10^(-19) ) ` ` = 1.04 xx 10^(-3) MS^(-1)" or" 1.04 mm s^(-1)` |
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| 36. |
Two waves are said to be coherent if they have |
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Answer» DIFFERENT FREQUENCY, and same phase |
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| 37. |
A shell of mass 15 kg, initially a rest, explodes into three fragments of masses in the ratio 1:1: 3. The fragments with equal masses fly off in mutually perpendicular directions with a speed of 6 ms^(-1) The speed of the heaviest fragment will be : |
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Answer» `12ms^(-1)` `:. m_(1) = m_(2) = 3 kg` and `m_(3) = 9 kg` Now ACCORDING to law of CONSERVATION of momentum, momentum `(p_(3))` of heavier part must be equal and OPPOSITE to resultants of `P_(1)` and `P_(2)`  `|p_(3)|=sqrt(p_(1)^(2)+p_(2)^(2))` `:.9xxupsilon_(3)=sqrt((3xx6^(2)+3xx6^(2)))` `upsilon_(3)=(6sqrt6)/(9)=(2sqrt6)/(3)m//s` Hence the correct CHOICE is (d) |
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| 38. |
What is Huygens's principle ? |
| Answer» Solution :EVERY point on a given WAVEFRONT is a source of SECONDARY WAVELETS. | |
| 39. |
A body is released from tower of height 1000 m and falls freely. Another body is released from the same height exactly one second later. Then the separation between the two bodies 3 second after the release of Ist body is : |
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Answer» 24.5 m |
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| 40. |
If the equation of a progressive wave is Y=Asin2pi(nt-x/lamda) , maximum particle velocity is |
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Answer» `nlamda` |
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| 41. |
A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.15 g? (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than or equal to 0.15 g? Why? |
| Answer» Solution :(a) `12^(@)`, (b) The acceleration would be more. We can look at this in TERMS of forces or in terms of energy. In terms of forces, the uphill static friction would then be ABSENT, so the downhill acceleration would be due only to the downhill gravitational pull. In terms of energy, the ROTATIONAL term in Eq. 11-5 would be absent so that the potential energy it STARTED with would simply BECOME `1//2mv^(2)`? (without it being "shared" with another term) resulting in a greater speed (and, because of Eq. 2-16, greater acceleration) | |
| 42. |
The vector vec P =hati+ 12hatj + hatk and vec Q= 2hati -a hatj + 4hatk are perpendicular, then value of a is |
| Answer» ANSWER :B | |
| 43. |
Thereddish appearance of rising and setting sun is due to |
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Answer» REFLECTION of light |
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| 44. |
The power of a heater is 750 W at 1000^@C.What will be it's powerat 200^@C if alpha =4xx10^(-4)per^@C? |
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Answer» 400 W |
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| 45. |
The intensity of plane electromagnetic wave with B_(0)=1.0xx10^(-4)T is ….. Wm^(-2).[c=3xx10^(8)ms^(-1), mu_(0)=4pi xx 10^(-7)NA^(-2)] |
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Answer» `2.38xx10^(6)` `c=3XX10^(8)m//s, B_(0)=1xx10^(-4)T` `mu_(0)=4pi xx10^(-7)N//A^(2)` Intensity `I=(cB_(rms)^(2))/(mu_(0))` `therefore I=(cB_(0)^(2))/(2mu_(0))` `therefore I=(3xx10^(8)xx10^(-8))/(2xx12.56xx10^(-7))=1.19xx10^(6)Wm^(-2)` |
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| 46. |
A galvanometer of resistance 25Omega is connected to a battery of 2V along with a resistance of 3000Omega. In this case, a full - scale deflection of 30 units is obtained in the galvanometer. In order or reduce this deflection to 10 units, how much more resistance ("in "Omega) should be added to the circuit in series? |
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Answer» |
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| 47. |
A T.V. transmitting antenna is 128 m tall . If the receiving antenna is at the ground level , the maximum distance between them for satisfactory communication in L.O.S. mode is (Radius of the earth = 6.4 xx 10^(6) m ) |
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Answer» `64 XX sqrt(10)` km |
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| 48. |
Derive equation of potential energy of an electric dipole in a uniform electric field . |
Answer» Solution :Consider a dipole with charges -q and +q placed in a uniform electric field `vecE` as SHOWN in figure.  Dipole placed in a uniform electric field `vecE` at an angle `theta` Equal and opposite forces `+qvecE` and `-qvecE` are acting on charges -q and +q . The forces acting on dipole constitute moment of force and hence dipole experiences a torque `tau= vecpxx vecE` where `vecp=(2veca)q` which will tend to rotate it ( unless `vecp` is parallel or antiparallel to `vecE`). Suppose an external torque acting on dipole and it ROTATES with small angle `Deltatheta = theta_(1) - theta_(0)` the small work done, , `DeltaW= tau Deltatheta = p E sin theta d theta` Total work done during ROTATION from `theta_(0)` to `theta_(1)` `W= int_(theta0)^(theta) p E sin thetad theta` `= p E[ -cos theta]_(theta0)^(theta1)` `:. W = p E [ cos theta_(0)- cos theta_(1)]` This work is stored as the potential energy of the SYSTEM . `:.` Potential energy of dipole, `U = pE[ cos theta_(0)- cos theta_(1)]` Initially a dipole placed at an angle `theta_(0)=(pi)/(2)` and if it makes angle 6, = 0 from that position by ROTATING then the potential energy of dipole is `U=pE[cos""(pi)/(2)-costheta]` `=pE[0-cos theta]` `:. U=-pEcostheta` `:. U = -(vecp.vecE)` |
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| 49. |
Thebackemf inaDC motorismaximumwhen, |
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Answer»
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| 50. |
What's the pentrating power of gamma-rays ? |
| Answer» SOLUTION :HIGHLY PENETRATING | |