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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. |
A n-type semiconductor is formed |
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Answer» when GERMANIUM crystal is doped with an IMPURITY containing three VALENCE electrons. |
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| 2. |
In the given graph betweentwo physical quantities 'a' and 'b' as shown in the figure 'b' is along x-axis while 'a' is along y-axis. The plot describes the motion of a particle in a straight line then |
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Answer» Quantity 'B' may represent time. Also .a. can ACT as the acceleration for uniformly acceleratedmotion. Thus (a),(c),(d) can be taken to be correct (c) and (d) under different CONDITION as stated above. |
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| 3. |
Hydrogen atom emits blue light when it changes from energy level n = 4 to n = 2. The colour light would the atom emit when it goes from n = 5 to n = 2 level is ..... |
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Answer» yellow |
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| 4. |
A transformer connected to 220 V mains is used to light a lamp of rating 100 W and 110 V. If the primary current is 0.5 A, the effeciency of the transformer is (approximately) |
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Answer» a. 6 |
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| 5. |
For a uniform electric field vecE = E_(0)(hati), if the electric potential at x = 0 is zero, then the vaJut of electric potential at x = +x will be ..... . |
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Answer» `xE_(0)` `V_((x))-V_((0))=-int_(0)^(x)E_(0)dx=-E_(0)[int_(0)^(x)dx]` `=-E_(0)[x]_(0)^(x)=-E_(0)x` but `V_((0))=0` `:. V_((x))= -E_(0)x` |
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| 6. |
Three charges q, q, and -2q are fixed on the vertices of an equilateral triangularplate of edge length a. This plate is in equilibrium between two very large plates having surfaces charge density sigma_1 and sigma_2, respectively. Find the time period of small anglular oscillations about an axis passing through its centroid and perpendicular to the plane. Moment of inertia of the system about this axis is l. |
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Answer» `2pi sqrt((epsilon_0l)/(QA|sigma_1-sigma_2|)` `r=-sqrt(3)qa([sigma_(1)-sigma_(2)]theta)/(2epsilon_(0))=1a` `(sqrt(3)qa|sigma_(1)-sigma_(2)])/(2epsilon_(0))=l((2pi)/(T))` `T=2pisqrt(((2epsilon_(0)l))/(sqrt(3)qa|sigma_(1)-sigma_(2)])` |
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| 7. |
The P.D between the points A and B is |
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Answer» 60V |
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| 8. |
A person wants a real image of his own, 3 times enlarged. Where should he stand infront of a concave mirror of radius of curvature 30 cm? |
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Answer» 20cm `m= - 3 = -(v)/(U) rArr v= 3u` `(1)/(v)+ (1)/(u) = (1)/(F) rArr (1)/(3u) + (1)/(u) = (1)/(-15) rArr (4)/(3u)= -(1)/(15) rArr u= - 20m` |
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| 9. |
Why are we getting zero current in the branch EF of the previous example ? |
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Answer» Solution :Let us disconnect the points E and D of the circuit used in the previous example. Then the situation is as follows: Current i flowing from the positive terminal of the 12V battery will continue to flow in the `3Omega` resistance. A loop involving branch EF is OPEN so current will not flow through this branch. For rest of the circuit, current will be `i= (12)/(3+2+1)= 2A` Now we can see that `V_(F) - V_(E )= 6V` `V_(G) - V_(D) = i xx 3=6V`...(ii) Let potential of common points F and G be zero then: `V_(E) = -6V` `V_(D) = -6V` We have proved that the potential of points E and D are same when E is not CONNECTED with D. HENCE, if E and D are joined by a meal wire, then no current will flow through that wire. Now you can understand that the above question was based on the idea of a potentiometer. 
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| 11. |
Angular momentum of an electron in an excited hydrogen atom having energy -3.4eV is : |
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Answer» `1.7 xx 10^(-34)Js` `n^(2)=-(R_(H).ch)/(E_(n)) rArr n^(2)=4 rArr n=2` ALSO, mvr `=(NH)/(2pi)=2.11 xx 10^(-34)Js` |
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| 12. |
What did the grandmother do in her final hours? |
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Answer» TALKED to EVERYONE in the house |
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| 13. |
A capacitor is connected across an inductor. At time t = 0 charge on the capacitor is equal to 1/sqrt2q_"max",where q_"max"is the maximum charge on the capacitor. The time t, atwhich the energy stored in the capacitor is equal to the energy stored in the inductor is (The inductance of the inductor is L and capacitance of the capacitor is C. Resistance of the circuit is zero) |
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Answer» `2pisqrt(LC)` |
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| 14. |
The equation of a progressive wave is y= 2 sinpi(0.5x -200t) meter. The velocity of the wave is |
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Answer» 100 m/s |
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| 15. |
(A): Inductance coils are usually made of thick copper wire. (R) : Induced current is more in wire having less resistance. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT explanation of 'A'. |
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| 16. |
What is rectification and rectifier ? Why p-n junction diode is used as a rectifier? |
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Answer» Solution :The process of converting AC signal (voltage, CURRENT and POWER)into DC signal is called rectification. The CIRCUIT which PERFORMS this process is called a rectifier. The characteristics of p-n junction is that when giving it forward bias it allows the current to be in the same direction but by connecting reverse bias it does not allow the current to flow so the diode has a low resistance during forward bias and is very high at reverse bias. The current flow by this property is called rectifier. Due to this characteristic p-n junction diode is USED as rectifier. There are two types of rectifiers. (1) Half wave rectifier (2) Full wave rectifier |
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| 17. |
When the adhesive force between a liquid and glass is greater than the cohesive force between the liquid molecules, the meniscus of the liquid in a capillary tube is |
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Answer» CONVEX in shape |
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| 18. |
The sun subtends an angle of 0.5^@ at the vertex of a concave mirror of radius of curvature 18.0m. Find the size of the image of the sun. |
| Answer» SOLUTION :`7.85` CM | |
| 20. |
The ionisation potential of mercury is 10.39 volt. To gain energy sufficient enough to ionise mercury, an electron must travel in an electric field of 1.5xx10^(6)Vm^(-1) at distance of : |
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Answer» `(10.39)/(1.5xx10^(6))m` |
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| 21. |
Electromotive force is most closely related to ............ |
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Answer» ELECTRIC FIELD |
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| 22. |
A thin bar magnetic is cut into two equal parts by cutting is perpendicular to ties length. What is the new magnetic moment of each part ? What is the time - period of each part as compared to that of the original magnet if if vibrated in the same field ? |
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Answer» |
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| 23. |
Wavelength of light used in an optical instrument are lamda_1 = 4500 A and lamda_2 = 6000 A. What is the ratio of the resolving powers corresponding to lamda_1 and lamda_2 ? |
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Answer» 3/4 |
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| 24. |
If white light is used instead of sodium light in the Young's experiment...... |
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Answer» all fringes will appear dark. |
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| 25. |
Draw the approximate voltage vector diagrams in the electric circuits shown in figure. a,b. The external voltage V is assumed to be alternating harmonically with frequency omega |
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Answer» SOLUTION :`(b) tan varphi=(OMEGAL-(1)/( omegaC))/( R) =- ve` as `OMEGA^(2)LT(1)/( LC)`  , 
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| 26. |
For the circuit shown, the ammeter reading is initially I. The switch in the circuit then is closed. Consequently (Battery and ammeter are ideal) |
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Answer» the ammeter reading decreases |
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| 27. |
In producing chlorine by electrolysis 100 kw power at 125 V is consumed. How much chlorine per minute is liberated. |
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Answer» `1.76xx10^(-3) KG` |
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| 28. |
An oscillator is nothing but an amplifier with |
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Answer» POSITIVE FEEDBACK |
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| 29. |
How many revolutions does an electron in the n = 2 state of a hydrogen atom make before dropping to the n = 1 state ? The average lifetime of an excited state is 10^(-8) s. Bohr radius 0.53 Å and velocity V_(1)=2.19xx10^(6)m//s |
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Answer» Solution :Velocity of electron in the nth orbit of hydrogen atom `v_(n)=(v_(1))/(n)` `=(2.19xx10^(6))/(n)` `:.v_(2)=(2.19xx10^(6))/(2)=1.095xx10^(6)m//s` and radius in `n^(th)` orbit `r_(n)=n^(2)r_(1)` `=(2)^(2)xx0.53Å` `=4xx0.53Å` `=2.12Å` `rArr` No. of revolution in ONE second `v=(v_(n))/(2pi r_(n))=(v_(2))/(2pi r_(2))` `:.v=(1.095xx10^(6))/(2xx3.14xx2.12xx10^(-10))` `0.0822xx10^(16)s^(-1)` `rArr` No. of revolution in `10^(-8)` second `N=vxx t` `=0.0822xx10^(16)xx10^(-8)` `:.N=8.22xx10^(6)` revolution. |
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| 30. |
A particle executes S.H.M. with amplitude 2 cm. At extreme position theforce is 4 N. The force acting on it at a position mid-way between the mean and extreme is : |
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Answer» 1 N `:.""F=ma=m omega^(2)y`. `implies""F prop y`. Thus when DISPLACEMENT is halved, then force also becomes half i.e., `(1)/(2)(4)=2N`. HENCE correct CHOICE is (b). |
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| 31. |
The dimensional formula of physical quantity is M^(a) L^(b) T^( c) . Then that physical quantity is |
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Answer» SURFACE TENSION if `a=1,b=1,c=-2` `:.` So dimensional formula of angular frequency is `M^(0)L^(0)T^(-1)` So cannot CHOICE is `(c )`. |
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| 32. |
A concave mirror of focal length 15 cm forms an image having twice linear dimensions of the object. The position of the object when the image is virtual will be |
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Answer» 22.5 cm `As (1)/(u) + (1)/(v) = (1)/(F)` `therefore(1)/(u)-(1)/(2u)=(1)/(-15)or(1)/(2u)=(1)/(-15)rArru=-7.5 cm` |
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| 33. |
Photoelectric cells are used to control "__________" of furnaces. |
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Answer» pressure |
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| 34. |
A domain in ferromagnetic iron is in the form of a cube of side length 1mu m. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/(cm)^(3).Assume that each iron atom has a dipole moment of 9.27 xx 10^(-24) "Am"^(2). |
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Answer» Solution :(a ) Density of given substance is `rho = (M.)/( V) … (1)` No. of moles of atoms in above substance is, `mu= (M.)/( M_0) = (N)/( N_A) rArr M. = (NM_(0) )/( N_A) …(2)` Where M. = total mass of atoms in volume V `M_0=` molar mass of atoms `N=` total no. of atoms in volume V `N_A=` Avogadro number From equation (1) and (2), `rho = (NM_0)/( VN_A)` `therefore N= (rho VN_A)/( M_0)` therefore `N= (rho l^(3) N_(A) )/( M_0)` (For a cube of side length l, its volume is V `=I^(3)`) `therefore N= ((7.9 xx 10^(3) ) (10^(-6) )^(3) (6.02 xx 10^(23) ))/( (55 xx 10^(-3) ))` `therefore N= 8.647 xx 10^(10)` atoms (b) If `overset(to) (m_b)` is the induced magnetic dipole moment of ONE atomic dipole then when N such atomic dipoles ALIGN in same direction then we have maximum possible induced magnetic dipole moment, `overset(to)((m_b) )_("max") = Noverset(to)( m_b)` Taking magnitudes, `(m_b))_("max") = Nm_b` `therefore (m_b)_("max") = (8.647xx 10^(10) ) (9.27 xx 10^(-24) )` `=8.016 xx 10^(-13) "Am"^(2)` (c) Corresponding maximum magnetisation obtained will be, `M_("max") = ((m_b)_("max") )/( V)` `= (8.016 xx 10^(-13) )/( 10^(-18) )""( because V- l^(3) = (10^(-6) )^(3) = 10^(-18) m^(3) )` `= 8.016 xx 10^(5) "Am"^(-1)` |
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| 35. |
In the production of X-rays by a Coolidge tube, the changes which can be made in its operation are listed in column A. The possible effects of such changes are listed in column B. |
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Answer» |
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| 36. |
If in a resonance tube, oil of density higher than that ofwater is used, then the resonance frequency would |
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Answer» INCREASE |
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| 37. |
The output power in step-up transformer used in practice is ….. |
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Answer» greater than the input power. |
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| 38. |
Match the nuclear processes given in the column-I with the appropriate option(s) in column-II. |
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Answer» |
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| 39. |
Two charges - q and + q are located at points A(0, 0, - a) and B(0, 0, + a) respectively, work done in moving small test charge from point P(7, 0, 0) to Q(- 3, 0, 0) is ...... . |
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Answer» zero 
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| 40. |
Suppose the initial charge on the capacitor is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time? |
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Answer» SOLUTION :`E = (Q^2)/(2C) = ( (6 XX 10^(-3) )^2)/(2 xx 30 xx 10^(-6) ) = 0.6 J` total energy is same for the later TIME . |
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| 41. |
Add up two vibratory motions analytically and using a vector diagram: s_(1)=3sin(6t+(pi)/(4)) and s_(2)4sin(6t-(pi)/(4)) Find the amplitude of the velocity of the resulting vibrations. |
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Answer» |
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| 42. |
In X-ray coolidge tube 25 keV energy of electron strike from the anode due to this the wavelength of first line of series K is obtained 0.5 Å. If applied potential is increased by 50%.(1)What is the value of wavelength of first line of K–series. (2)Minimum value of wavelength. |
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Answer» Solution :(1) 0.5 Å `lambda_(KA)` - class REST depond on applied POTENTIAL 0.5 Å (2) 0.33 Å |
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| 43. |
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48xx10^(6)m, and the radius of lunar orbit is 3.8xx10^(8)m. |
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Answer» Solution :(a) Here, `f_(0) = 15 m` and `f_(e) =1.0 cm = 10^(-2) m` `THEREFORE` ANGULAR magnification `|m| =f_(0)/f_(e) =15/10^(-2) = 1500` (b) Diameter of MOON `D = 3.48 xx 10^6` m and distance of moon from earth = radius of lunar orbit of moon r = `3.8 xx 10^8 m`. The moon SUBTENDS an angle a at the telescope objective where `alpha = D/r`. The objective lens forms image of moon in its focal plane. If SIZE of image be h, then `alpha =h/f_(0)` `rArr h/f_(0) =D/r, therefore h=D/r f_(0)) =(3.48 xx 10^(6) xx 15)/(3.8 xx 10^(8) = 13.7 xx 10^(-7) m` or `13.7` cm |
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| 44. |
A suspended nano wire is a wire that is produced in………... |
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Answer» AIRMEDIUM |
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| 45. |
The two aspects of work are: |
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Answer» INDIVIDUAL and SOCIAL work |
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| 46. |
The poet wants us to listen to |
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Answer» our own voice |
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| 47. |
Assertion: Two metallic cylinders of different materials and having the same cross section are connected in series with each other. Electric field in both of them would be equal each other. Electric field in both of them would be equal when current passes through them. Reason: When two conductors are connected in series then the same current is passed through them. |
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Answer» If both assertion and reason are CORRECT and reason is a correct EXPLANATION of the assertion |
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| 48. |
A concave shaving mirror is of focal length 50cm. How far the mirror should be held from the face to produce an image 2 fold magnification. |
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Answer» Solution :Here F = 50CM, m = 2 v/u = u = 2U using 1/v+1/u = i/f u = `50xx3/2` = -75CM |
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| 49. |
How is it convenient to obtain image by reflection from spherical mirror ? |
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Answer» Solution :We can take any TWO rays emanating from a point on an object, trace their paths, find their point of intersection and thus, obtain the image of the point due to reflection at a SPHERICAL mirror. In practice, however, it is convenient to choose any two of the following rays. (i) The ray from the point which is parallel to the PRINCIPAL axis. The REFLECTED ray goes through the focus of the mirror.  (ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. The reflected ray simply retraces the path.  (iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through (or directed towards) the focus of a convex mirror. The reflected ray is parallel to the principal axis.  (iv) The ray incident at any ANGLE at the pole. The reflected ray follows laws of reflection. 
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| 50. |
A force of 3 kg wt is just sufficient to pull a block of 4 kg wt over a flat surface. What is the angle of friction? |
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Answer» `36^0 52^1` |
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