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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 6151. |
Which of the following statement is true for the forward bias of a p-n junction? |
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Answer» The positive TERMINAL of the battery is connected to p-side and the DEPLETION REGION becomes thin. Thepositive terminal p-side and negative terminal n-side of the battery are connected by the forward in the p-n junction. |
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| 6152. |
Generally, we consider that during slowing of a moving automobile, its retardation is constant, but in practice this is seldom the case. Under many circumstances, especially at high speed, we usually apply the brakes slowly at first and then more strongly as the car slows. The braking force therefore depends on the time during the interval over which the car is slowing and acceleration changes with time. a_(x)(t)=(v_(A)-v_(x))/(dt) =(F_(x)(t))/(m) int_(v_(0_(x)))^(v_(x))dV_(x)=int_(0)^(1)(F_(x)(t))/(m)dt therefore V_(x)=V_(0x)+(1)/(m) int_(0)^(1)F_(x) (t) dt x(t)=x_(0)+int_(0)^(1) V_(x) (t) dt An example for the same is discussed here. A car of mass m = 1000 kg is moving with 25 m/s. The driver begins to apply the brakes so that the magnitude of the braking force increases linearly with time at a rate of 2000 N/s. Read the above passage carefully and answer the following questions. How much time passes before the car comes to rest? |
| Answer» ANSWER :A | |
| 6153. |
Distinguish between unpolarised and plane polarised light. An unpolarised light is incident on the boundary between two transparent media. State the condition when the reflected wave is totally plane polarised. Find out the expression for the angle of incidence in this case. |
Answer» SOLUTION :DISTINCTION between UNPOLARISED and PLANE polarised light :  For condition of polarisation of reflection ETC. |
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| 6154. |
A uniform bar of square cross-section is lying along a frictionless horizontal surface. A horizontal force is applied to pull it from one of its ends then |
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Answer» The bar I sunder same stress throughout its length |
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| 6155. |
A block of solid onk with dimensions 10cmxx20cmxx20cm floats in water Fig. The block is submerged a little and let go. Find the frequency and the period of vibrations. |
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| 6156. |
The radionuclide .^(11)C decays according to ._(6)^(11)C to ._(5)^(11)B+ e^(+)+v : T_(1//2)=20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values , m(._(6)^(11)C)=11.011434 u and m(._(6)^(11)B)=11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted. |
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Answer» Solution :Mass defect in the given reaction is `Delta m = m(._(6)C^(11))` `= [m (._(5)B^(11))+Me]` This is in terms of NUCLEAR masses. If we EXPRESS the Q VALUE interms of atomic masses we have to subtract `6m_(e)` from atomic mass of carbon and `5 m_(e)` from that of boron to GET the corresponding nuclear masses Therefore, we have `Delta m = [m(._(6)C^(11))-6 m_(e)-m(._(5)B^(11))+5m_(e)-m_(e)]` `=[m(._(6)C^(1))-m(._(5)B^(11))-2m_(e)]` ` = [11.011434-11.009305-2xx0.000548]u` `= 0.001033 u` As, 1u = 931 MeV `Q = 0.001033xx931 MeV = 0.961 MeV` Which is the maximum energy of emitted POSITION. |
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| 6157. |
Two beams of red and violet colours are made to pass separately through a prism of A = 60^(@). In the minimum deviation position, the angle of refraction inside the prism will be |
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Answer» greater for red colour For minimum deviation, A = 2r or `"" R = (A)/(2) = (60^(@))/(2) = 30^(@)` for both colours. |
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| 6158. |
Two charge +q and -q are placed at distance r. Match the following two columns when distance between them is charged to r'. |
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| 6159. |
A glass tube of 1.0 m length is filled with water. The water can be drained out slowly at the bottomof the tube. If a vibrating tuning fork of frequency 500 c/s is brought at the upper end of the tube and the velocity of sound is 330 m/s, then the total number of resonances obtained will be |
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Answer» 4 |
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| 6160. |
A river is of width 120m which flows at a speed of 8 ms(-1). If a man swims with a speed of 5ms^(-1) at an angle of 127^@ with the stream, his drift on reaching other bank is |
| Answer» ANSWER :B | |
| 6161. |
The correct value of 0^@C on the Kelvin scale is |
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Answer» 273.15 K |
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| 6162. |
A rope ladder of length L is attached to ballon of mass m. As the man of mass m clims the ladder in to the lballon basket the ballon comes donws by a vertical distance s. Then increase in potential energy of man divided by increase inn potentialenergy of ballon is |
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Answer» `(L-s)/(s)` mg L = mg (L-s)++mg.s where .mg(L-s) = increases in P.E of man mg s = increases in P.Eof man mg s= increases in P.Eballoon so (increase in P.E of man)/(increase of P.E of BALLON)=(mg(L-s)/(mgs)` `=(L-s)/(s)` |
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| 6163. |
The magnetic flux linked with a coil satisfies the relation phi=4t^(2)+6t+9 Wb, where t is the time in second. The em.f. induced in the coil at t = 2 second is |
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Answer» 22 V `therefore` Induced e.m.f. `|EPSILON|=(d phi)/(DT)=(d)/(dt)(4t^(2)+6t+9)=8t+6 V` At `t=2 s, |epsilon|=8xx2+6 V= 22V` |
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| 6164. |
What is the difference between virtual image formed by a convex lens and that formed by a concave lens ? |
| Answer» Solution :Virtual IMAGE FORMED by a convex lens is erect and magnified but the virtual image formed by a CONCAVE lens is ALWAYS erect and diminished ONE. | |
| 6165. |
असम और उत्तरी बंगाल तथा कर्नाटक में कौन- सी रोपण फसलें हैं? |
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Answer» चाय व कॉफी |
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| 6166. |
When a galvanometer is shunted with 4Omega the deflection is reduced to (1/5)^th of original deflection .The resistance og galvanometer is : |
| Answer» Answer :D | |
| 6167. |
From the graph between currentI and voltageV shown below, identify the portion corresponding to negative resistanc |
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Answer» AB |
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| 6168. |
For steady interference pattern in biprism experiment, which of the following is true? |
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Answer» Distance between slit and biprims should be less. |
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| 6169. |
A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are |
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Answer» 300 V, 15 A |
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| 6170. |
What is the resulting intensity after interference of two coherent waves given as y_1=a_1 cos omegat and y_2=a_2 cos ((pi)/2-omegat)? |
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Answer» `a_1+a_2` |
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| 6171. |
Which of the following statements are correct, a) Electric lines of force are just imaginary lines b) Electric lines of force will be parallel to the surface of conductor c) If the lines of force are crowded then field is strong d) Electric lines of force are closed loops |
| Answer» Answer :A | |
| 6172. |
A vertical U tube contains a liquid of density rho, length of the liquid in the tube is 2H. The liquid is displaced slightly and released. Show that the liquid will execute SHM. Also find its time period. Take liquid to be ideal. |
| Answer» SOLUTION :`T = 2PI SQRT((H)/(G))` | |
| 6173. |
Quality of sound depends upon : |
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Answer» frequency CORRECT CHOICE is (B) . |
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| 6174. |
A body is completely submerged inside the liquid. It is in equilibrium and in rest condition at certain temperature. It gamma_(L) volumetric expansion coefficient of liquid alpha_(s) = linear expansion coefficient by of body. It we increases temperature by Delta theta amount than find (a) New thrust force if initial volume of body is V_(0) and density of liquid is d_(0). (b) Relation between alpha_(s) and gamma_(L) so body will (i) move upward (ii) down ward (iii) remains are rest |
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Answer» Solution :(a) `v_(0)d_(0) g((1+3alpha_(s)DELTA THETA)/(1+gamma_(L) Delta theta))` (b) (i) `gamma_(L) lt 3 alpha_(s)` (ii) `gamma_(L) gt 3alpha_(s)` (iii) `gamma_(L) =3alpha_(s)` |
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| 6175. |
A simple pendulum having cahargeq_1 mass ma and effective length 1, is suspended from a rigid support between the plates of charged horizontal capacitor. Electric field in the capacitor is perpendicular to the plates and directed vertically downwards . If the plates are oriented such thet plates made an angeltheta with the horizontal then the new time period of the oscillation of the pendulum is give by . |
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Answer» ` T= 2 pi SQRT(l/sqrt (g^2 + ((QE)/m)^2 + 2g ((QQ)/m) SIN theta)` |
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| 6176. |
Determine the relationship between the torque N and the torsion angle varphi for (a) the tube whose wall thickness Deltar is considerably less than the tube radius, (b) for the solid rod of circular cross-section. Their lengh l, radius r, and shear modulus G are supposed to be known. |
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Answer» Solution :(a) Consider a hollow CYLINDER of length l, outer radius `r+Deltar` inner radius r, fixed at one END and twisted at the other by means of a couple of moment N. The angular displacement `varphi`, at a distance l from the fixed end, is proportional to both l and N. Consider an element of length `dx` at the twisted end. It is moved by an ANGLE `varphi` as shown. A vertical SECTION is also shown and the twisting of the parallelopipe of length l and area `Deltardx` under the action of the twisting couple can be discussed by elementary means. If f is the tangential force generated then shearing stress is `f//Deltardx` and this must equal `Gtheta=G(rvarphi)/(l)`, since `theta=(rvarphi)/(l)`. Hence, `f=GDeltardx(rvarphi)/(l)`. The force f has moment `f r` about the axis and so the total moment is `N=GDeltar(varphi)/(l)r^2intdx=(2pir^3Deltarvarphi)/(l)G` (b) For a solid cylinder we must integrate over r. Thus `N=overset(0)overset(r)INT(2pir^3drvarphiG)/(l)=(pir^4Gvarphi)/(2l)` 
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| 6177. |
A simple pendulum having cahargeq_1 mass ma and effective length 1, is suspended from a rigid support between the plates of charged horizontal capacitor. Electric field in the capacitor is perpendicular to the plates and directed vertically downwards . the anglealpha between the equilibrium position of the pendulum and the vertical is given by. |
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Answer» ` ALPHA = SIN^(-1) [ ((qQE//m)sin THETA)/(h + (qE//m) VOD theta) ]` |
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| 6178. |
Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists |
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Answer» |
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| 6179. |
A physical quantity Q=(a^(2)b^(3)c^(5/2))/(x^(2)).If the percentage errors in the measurement of 'a','b','c' and x are 1%,2%,3%,4% reperctively, What is the percentage error in the measurement of Q ? |
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Answer» Solution :Here `Q=(a^(2)b^(3)c^(5//2))/(x^(2))` `:.(dQ)/(Q)=(2da)/(a)+(3db)/(b)+(5)/(2)(DC)/(c )+2(dx)/(x)` or `(dQ)/(Q)xx100=[2(da)/(a)+3(db)/(b)+(5)/(2)(dc)/(c )+2(dx)/(x)]xx100` or percentage errors `Q` is `=[2xx1+3xx2+(5)/(2)xx3+2xx4]` `=23.5%~~24%` `:.(a)` is correct. |
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| 6180. |
Which of the following types of electromagnetic radiation has the longest wavelength ? |
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Answer» Gamma rays |
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| 6181. |
What are conductors and insulators? |
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Answer» Solution : Conductors: Substances which allow electricity to pass through them are known as conductors. All METALS, human body and the EARTH are good conductors of electricity. The charge carries in metal conductors are free electrons. Insulators: Substances which do not allow electricity to pass through them are known insulators. Glass, rubber, PLASTIC etc., are insulators. |
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| 6182. |
a. A closed loop is held stationary int eh magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets? b. A closed loop move normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. c. A rectangular loop and a circular loop are moving out of a uniform magnetic field (region). (see fig.) to field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. d. Predict the polarity of the capacitor in the situation described by the figure below. |
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Answer» Solution :a. No. However strong the magnet may be, current can be induced only by changing the magnetic FLUX through the loop. b. No current is induced in either case. Current cannot to induced by changing the ELECTRIC flux. c. The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field REGION is not constant, hence induced emf will vary accordingly. d. The POLARITY of plate .A. will be positive with RESPECT to plate .B. in the capacitor. |
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| 6184. |
In the following nuclear reaction, how many alpha and beta particles are emitted ? _92X^235 to _82Y^207 |
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Answer» `3 alpha-particle` and `2 alpha-particle` |
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| 6185. |
Whena compact disc. Is illuminated by a source of white light, coloured lines are observed. This is due to |
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Answer» dispersion |
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| 6186. |
In the circuit shown in the Figure calculate the quantity of charge that flows through the switch after it is closed. Give your answer for following two cases- (a) C_(1)=C_(2)=2 muF (b) C_(1)=2 mu F , C_(2) = 1 muF |
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Answer» (B) `20 MUC` |
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| 6187. |
What type of impurity is added to obtain n-type semicondutor? |
| Answer» SOLUTION :PENTAVALENT IMPURITY i.e. As or SB. | |
| 6188. |
Explain construction of cyclotron with diagram. |
Answer» Solution :1. Construction of cyclotron is as shown in diagram.  2. It consists of two small hollow, metallic half cylinders `D_(1)andD_(2)` called dees as they are in the shape of D. 3. They are mounted inside a vacuum chamber between the poles of a powerful electromagnet. 4.The dees are connected to the source of HIGH frequency ALTERNATING voltage of few hundred kilovolts. 5. The beam of charged particles to be accelerated is injected into the dees near their centre P, in a plane perpendicular to the MAGNETIC field. 6. The charged particles are pulled out of the dees by a deflecting plate which is negatively charged through a WINDOW W. 7. The whole device is in high vacuum so that air MOLECULES may not collide with the charged particles. |
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| 6189. |
How it can be said that light posses wave nature? |
| Answer» SOLUTION :From maxwell.s EQUATION for electromagnestism and Hertz experiment to produce (detect) ELECTROMAGNETIC waves.It can be SAID that light posses wave NATURE. | |
| 6190. |
The equivalent resistance of two resistors in parallel is always |
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Answer» the SUM of resistances |
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| 6191. |
The velocity of a 2 kg body is changed from (4hati +3hatj) ms^(-1) to 6hat ms^(-1). The work done on the body is |
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Answer» 9 J `v_1=sqrt((4)^2+(3)^2)=5 ms^(-1)` and velocity `v_2=6 ms^(-1)` NowW=increase in K.E. `=1/2m(v_2^2-v_1^2)` =`1/2xx2(36-25)` =11 J |
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| 6192. |
In the study of Geiger-Marsdon experiment on scattering of a-particles by a thin foil of ... trajectory of alpha-particles in the coulomb field of target nucleus. Explain briefly how one gets the … on the size of the nucleus from this study. From the relationR=R_(0)A^(1//3), where R_(0)is constant and A is the mass number of the nucleus, nuclear matter density is independent of A. |
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Answer» Solution :Elnstein’s photoelectric EQUATION: Where v = incident FREQUENCY,`v_(0 =`threshold frequency, `V_(0) =` stopping potential (i)Incident energy of photon is used in to (a) to liberate ELECTRON from the metal surface (b) rest of the energy appears as maximum energy electron. (ii)Only one electron can absorb energy of one photon. Hence increasing intensity increases the numbers of electrons hence current. (iii)If incident energy is less than WORK function, no emission of electron will take place. (iv)Increasing v (incident frequency) will increase maximum kinetic energy of electrons but number of electrons emitted will remain same. For wavelength `(HC)/(lambda_(1))=phi_(0)+K` `= phi_(0)+ ._(e )v_(0) ""` .... (i) where `K = ._(e )v_(0)` Form wavelength `lambda_(2)` `(hc)/(lambda_(1))=phi_(0)+2_(e )v_(0)` ....(ii)(because KE is doubled) Form equartion (i) and (ii), we get `(hc)/(lambda_(2))=f_(0)+2((hc)/(lambda_(1))-phi_(0))` `= phi_(0)+(2hc)/(lambda_(1))-2phi_(0)` `rArr "" lambda_(0)=(2hc)/(lambda_(1))-(hc)/(lambda_(2))` For thershold wavelength `lambda_(0)`, kinetic energy, K = 0 and work function`lambda_(0)=(hc)/(lambda_(0))` `therefore(hc)/(lambda_(0))=(2hc)/(lambda_(1))-(hc)/(lambda_(0))` `rArr "" (1)/(lambda_(0))=(2)/(lambda_(1))-(1)/(lambda_(2))` `rArr "" lambda_(0)=(lambda_(1)lambda_(2))/(2lambda_(2)-lambda_(1))` Work function, `lambda_(0)=(hc(2lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))` |
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| 6193. |
A ray of white light passes through a rectantular glass slab, entering and emerging at parallel faces. The angle of incidence, measured from the normal to the glaass surface, is large. |
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Answer» White light will EMERGE from the slab. |
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| 6194. |
In p-n junction diode, holes diffuse from p-region to n-region because |
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Answer» the FREE electrons in the n-region attract them. |
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| 6195. |
The current flowing through an inductor of self inductance L is continuously increasing. Plot agraph showing the variation of (i) Magnetic flux versus the current (ii) Induced emf versus dI//dt (iii) Magnetic potential energy stored versus the current. |
Answer» SOLUTION :(i) MAGNETIC FLUX VERSUS CURRENT |
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| 6196. |
Sin(180^(@) +alpaha) +sin(90^(@) -alpaha)= |
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Answer» `COS ALPAHA` |
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| 6197. |
A block is sliding on a rough horizontal surface. If the contact force on the block is sqrt2 times the frictional force, the coefficient of friction is |
| Answer» ANSWER :D | |
| 6198. |
In optical fibres, propagation of light is due to |
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Answer» diffraction |
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| 6199. |
See the digram, area of each plate is 2.0 m^(2) and d=2xx10^(-3)m. A charge of 8.85xx10^(-8) C is given to Q. Then, the potential of Q becomes |
| Answer» ANSWER :C | |
| 6200. |
रेत का पानी में विलयन..........का उदाहरण है |
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Answer» तत्व का |
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