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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. |
What is the relation between the apparent frequency and true frequency of sound when the source and observer move towards each other along the same line ? |
| Answer» SOLUTION :n (apparent)=n (true) x `((V+v_("observer"))/(v-v_("SOURCE")))`, where v is the speed of sound in still AIR. | |
| 2. |
Directions for Questions 56 and 57: In each question, there is a table having 3 columns and 4 rows. Based on the table, there are 3 questions. Each question has 4 options (a), (b), (c) and (d), ONLY ONE of these four options is correct. Two charges of opposite sign and equal magnitude Q = 2.0 C are held 2.0 m apart. And there is a point Pat 4.0 cm from +Q charge. In the given table there are four values each of electric field (Column I), potential difference each of electric field (Column I), potential difference {:("Column I","Column II", "Column III"),((I) E=1.27 xx 10^(9) V//m,(i) V=1.5 xx 10^(9) V,(J) (##MST_AG_JEE_MA_PHY_V02_C24_E03_060_Q01.png" width="80%"> The combination for maximum electric potential is |
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Answer» (I) (II) (M) |
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| 3. |
A body takes 1(1)/(3) times as much time to slide down a rough inclined plane as it takes to slide down an identical bust smooth inclined plane. If the angle of inclination is 45^(0), the coefficient of friction is |
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Answer» `1/16` |
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| 4. |
Curie temperature is the temperature above which |
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Answer» ferromagnetic material BECOMES PARAMAGNETIC |
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| 5. |
The time of period of a simple pendulum on the earth's surface is T. Its time period at a bottom of a mine will be |
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Answer» Zero |
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| 6. |
उस वृत्त की त्रिज्या जिसका 15 सेमी का चाप केन्द्र पर 3/4 रेडियन का कोण बनाता है, |
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Answer» 10 सेमी |
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| 7. |
In which fields x-rays found extensive use. |
| Answer» Solution :Surgery,RADIO THERAPY,Engg and SCIENTIFIC RESERCH. | |
| 8. |
What are the possible cause of one side deflection in Galvanometer while performing potentiometer experiment ? |
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Answer» Solution :(i) EITHER +ve terminals of all the CELLS are not CONNECTED to same end of potentiometer. or (ii) The total POTENTIAL drop across wire is less than the emf to be measured. |
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| 9. |
Give electronic configuration of silicon. |
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Answer» `1s^(1)" "1s^(2)" "2s^(2)" "3S^(2)" "3P^(2)" "3p^(6)` |
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| 10. |
A massless string passes over a frictionless pulley and carries two masses m_1 and m_2(m_1 > m_2) at it's ends. If g is the acceleration due to gravity, then the thrust on the pulley is (m_1>m_2). |
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Answer» `FRAC{2m_1m_2}{m_1-m_2}` |
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| 11. |
Let h_(0) be the initial height of ball with respect to the earth. The coefficient of restitution is e. |
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Answer» <P> |
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| 12. |
The resistance of all the wires between any two adjacent dots is R. Then equivalent resistance between A and B as shown in figure |
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Answer» `5` `I=3` AMP `3=((15+3))/(1+2+R)xx1=18/(3+R) implies3R=9` `R=3 OMEGA` |
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| 13. |
In a photoelectric experimentanode potential is plotted against plate current.Then |
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Answer» A and B will have DIFFERENT INTENSITIES while B and C will have different frequencies. |
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| 14. |
What will be the answer if string is moved upward with acceleration a ? |
| Answer» SOLUTION :`2sqrt((L)/(g+a))` | |
| 15. |
A transmitter radiates 12 KW power at 80 % modulation index. What is the power in the carrier? |
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Answer» Solution :POWER of CARRIER WAVE`P_"carrier=("P"_(TOTAL)//[1+((m^2a))//2]` |
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| 16. |
The image formed by a concave spherical mirro |
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Answer» is always VIRTUAL |
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| 17. |
A beam of light of wavelength 600 nm from a distant source falls normally on a single-slit, 1 mm wide, and the resulting diffraction pattern is obtained on a screen 2.0 away. The distance between the first dark fringes on either side of the central bright fringe is |
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Answer» 1.2 mm `=(2xx2.0xx(600xx10^(-9)))/((1xx10^(-3)))=2.4xx10^(-3)m or 2.44mm`. |
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| 18. |
Show three points charges +2q. -q and +3q, Two charges +2q and -q are enclosed within a surface 'S' What is the electric flux due to this configuration through the surface 'S' ? |
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Answer» Solution :Electric flux ` phi_in =(1)/( in_0)SUM ` (CHARGED enclosed WITHIN surface) ` "" =(1)/( in_0)[2q-Q] =+(q)/(in_0) ` |
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| 19. |
Lenz's law is statement of _____ |
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Answer» LAW of CONSERVATION of CHARGE |
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| 20. |
Draw the diagram in each case to show the position and nature of the image formed when the object is placed.(i) At the centre of curvature of a concave mirror.(ii) In front of convex lens. (iii) Between the pole P and focus F of a concave mirror. |
Answer» SOLUTION :
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| 21. |
Two persons A and B are standing on a road. A third person riding on a cycle betweenA and B is ringing his bell and mocing towards A at 27 km/h. The frequency of the bell as heard by A is 420 Hz. What is the frequency of the bell as heard by B ? |
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Answer» |
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| 22. |
In a smooth horizontal groove two particles m_(1) and m_(2) collides match the two column under the given condition at t = 0 first collision takes place. Radius of the circular grove is R |
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Answer» P-2, Q-3, R-1, S-4 |
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| 23. |
What has the child hitherto been? |
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Answer» An ideal child |
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| 24. |
(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point? |
| Answer» Solution :(a) An electrostatic field line either originates from a positive charge or terminates on a negative charge. A sudden break in an ELECTRIC line of force means that the electric field intensity has suddenly dropped to zero from a finite value. But we know that electric field intensity GRADUALLY decreases to zero atinfinity. Basically, the end points of the electric LINES must be at the location of the charge or infinity. Consider an example of a metallic object placed inside an electrostatic field. Due to induction the charge of the REQUIRED polarity is INDUCED on its surface so that the electric lines either terminate on its surface or they start from its surface. | |
| 25. |
The graph showing the varation of potential difference V along x-axis and current I along y-axis for a conductor makes an angle 40^@ with the x-axis. The resistance of the conductor will be |
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Answer» `0.84.Omega` |
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| 26. |
Figure shows the orientation of two vectors u and v in the XY plane Ifu= hati+b hatj and v=p hati+q hatj which of the following is correct? |
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Answer» <P>a and p are POSITIVE while B and Q are NEGATIVE |
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| 27. |
Two charges respectively of +3.2xx10^(-19) coublem and -3.2xx10^(-19) coulomb are separated by a distance of 2.4xx10^(-10)m . This dipole is placed in a homogeneous electric field of 4.0xx10^(5)V//m . Find (i) Electric dipole moment (ii)The maximum moment exerted on the dipole by the electric field. (iii)The energy necessary for rotating the dipole from equilibrium position to 180^(@). |
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Answer» Solution :(i)The electric dipole moment `P=qxx21` `P=3.2xx10^(-19)xx2.4xx10^(-10)` `P=7.68xx10^(-29)`coulomb `xx` meter (ii)The moment of the couple ACTING on the dipole `TAU=pEsintheta` for MAXIMUM moment `theta=90` `tau=PE` `=7.86xx10^(-29xx4.0xx10^(5)` `=31.44xx10^(-24)` `tau_("max")=3.14xx10^(-23)`newton `xx` meter (iii)Potential energy of the dipole `U=-PEcos theta`. In theposition of equilibrium `theta=0` `thereforeU=-PE` `U=-7.86xx10^(-29)xx4xx10^(5)` `=-3.14xx10^(-23)J` The energy Spent in rotating the dipole from `theta=0" to "theta=180^(@)`. `W=U_(180)-U_(0)` `=-PECos180-(-(PECos0)` `=PE+PE` `=2PE` `=2xx7.86xx10^(-29)xx4xx10^(5)` `=6.28xx10^(-23)J` |
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| 28. |
The twinkling effect of star light is due to |
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Answer» total INTERNAL reflection |
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| 29. |
The given Fig. 6.39, shows an inductor L and resistor R connected in parallel to a battery B through a switch S. The resistance of R is same as that of the coil that makes L. Two identical bulbs Pand Q are put in each arm of the circuit as shown. (a) When S is closed, which of the two bulbs will light up earlier ? (b) Will the bulbs be equally bright after some time? Justify your answer. |
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Answer» <P> Solution :(a) The bulb P joined in the ARM containing RESISTOR R will light up earlier when switch S is closed. On closing the switch, a self-induced back emf is developed in the INDUCTOR L in a direction opposite to the emf of battery B whereas no such emf is induced in resistor R. As a result, bulb P glows up earlier and the bulb Q glows up a bit late.(b) Yes, after sometime both the bulbs will GLOW equally bright. It is because once the current has reached its maximum value, self-inductance plays no role. As resistance of both R and L is same and the bulbs are identical, both will finally glow with equal brightness. 
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| 30. |
A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval t_1. An observer at the top of the tower see the same ball go up above him and then come back at this level in a time interval t_2 . The height of the tower is |
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Answer» `1/2 G t_1 t_2` |
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| 31. |
Skip zone in radio wave transmission is that range where |
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Answer» there is no RECEPTION of EITHER ground WAVE (or) SKY wave |
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| 32. |
Derive the formula for the electric potential due to an electric dipole at a point from it. |
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Answer» Solution :A dipole is placed with ORIGIN at its midpoint. Its charge -q and +q is separated by a distance 2a. The magnitude of a dipole moment of the dipole is p= 2aq and its direction is from -qto +q. Let point P away from the midpoint of dipole and `r_(1)` and `r_(2)` are the distance of P from +q and q respectively which is shown in the figure .  The potential DUE to the charge +q at point P is `V_(1) (k(+q))/(r_(1))`and The potential due to the charge -q, `V_(2)=(k(-q))/(r_(2))= (kq)/(r_(2))` According to superposition principle, the total potential at point P `V=V_(1)+V_(2)` `=(kq)/(r_(1))-(kq)/(r_(2))` `=kq [(1)/(r_(1))-(1)/(r_(2))]` DRAW qN `bot` OP In `DeltaqON angleqON = theta` `:. ON = r- r_(1)` and `COS theta= (ON)/(Oq) implies ON = Q q cos theta` `:. r- r_(1) = a cos theta implies r_(1) = r - a cos theta ` `=r(1-(a_(r1))/(r) cos theta)` `implies r_(1)^(2)= r^(2) (1-(a)/(r)cos theta)^(2)` `=r^(2)(1-""(2acostheta)/(r)+(a^(2))/(r^(2))cos^(2)theta)` `=r^(2)(1-(2acostheta)/(r))` Putting `rgtgt a implies (a^(2))/(r^(2))=0` and draw OM `bot` (-qp=P) then in `Delta- ` qOM P(-q) M `= theta` and (-q) M `= r_(2) -r ` `:. cos theta((-q)M)/(O-(-q))implies (-q)M =m O (-q) cos theta` `:. r_(2) -r = acostheta` `:> r_(2) = r +a cos theta` `:. r_(2)= r(1+(a)/(r) cos theta)` Similarly `r_(2)^(2) =r^(2) (1+(2acostheta)/(r))` Now from equation (4 ) `:. (1)/(r_(1))=(1)/(r)(1-(2acostheta)/(r))^(-1//2)` `=(1)/(r)(1+(acostheta)/(r))` only two terms be taken from expansion and, `(1)/(r_(2))=(1)/(r)(1-(acostheta)/(r))` Now from equation (5) only two terms be taken from expansion . `implies` Putting the values of equation (6) and (7) in equation (1) , `V=(kq)/(r)[(1+(acostheta)/(r))-(1-(acostheta)/(r))]` `=(kq)/(r)[(2acostheta)/(r)]` `=(k(p)costheta)/(r^(2)) [ because 2aq=p]` `=(kvecP.vecr)/(r^(2))[becausep costheta= vecP.hatr]` `:. V = (p.hatr)/(4piin_(0)r^(2))or (pcostheta)/(4piin_(0)r^(2))or (kpcostheta)/(r^(2))` Hence potential due to a point charge decreases according `(1)/(r)` and potential due to a dipole decreases according to `(1)/(r^(2))` distance . The molecular dipoles are very small and such an approximation is very WELL applicable to them. |
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| 33. |
A magnetic field of 2xx10^(-2)T acts at right angles to a coil of area 100 cm^(2), with 50 turns. The average e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is |
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Answer» 10 s `t=(NBA)/(epsilon)=(50xx2xx10^(-2)xx10^(-2))/(0.1)=0.1 s` |
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| 34. |
The electric field amplitude of an electromagnetic wave is 15V/m. Find the magnetic field amplitude of the wave |
| Answer» SOLUTION :`[B=E/C =15/(3*10^8) = 5*10^-8 T]` | |
| 35. |
A charged particle enters a uniform magnetic field with velocity v_(0) = 4 m//s perpendicular to it, the length of magnetic field is x = ((sqrt(3))/(2)) R, where R is the radius of the circular path of the particle in the field. Find the magnitude of charge in velocity (in m/s) of the particle when it comes out of the field. |
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Answer» `Delta vec(v)=(v_(0)cos 60^(@) hat(i)+v_(0) SIN 60 hat(j)-v_(0)hat(i))IMPLIES |Delta vec(v)|=v_(0)`. 
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| 36. |
Light emitted during the de excitation of electron from n = 3 to n = 2, when incident on a metal, photoelectrons are just emitted from that metal. In which of the following de excitations photoelectric effect is not possible ? |
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Answer» From n=2 to n=1 `(hv)_(4-3) lt (hv)_(3-2)` |
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| 37. |
The difference in the number of wavelengths, when yellow light (of wavelength 6000 overset(*)A in vacuum) propa- gates through air and vacuum columns of the same thickness is one. If the refractive index of air is 1.0003, the thickness of the air column is: |
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Answer» 1.8 mm |
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| 38. |
By using only two resistance ciols,either in single or in series or in parallel one should be to obtain resistance of 3Omega,4Omega,12Omegaand 16Omega. The indivdual resistance of the coils are |
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Answer» `3Omega`and `4OMEGA` |
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| 39. |
A Carnot engine operating between temperatures T_(1) and T_(2) has efficiency (1)/(6). When T_(2) is lowered by 62 K its efficiency increase to (1)/(3). Then T_(1) and T_(2) are, resectively : |
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Answer» 372 K and 330 K `rArr (T_(2))/(T_(1))=(5)/(6)"….(i)"` `eta_(2)=1-(T_(2)-62)/(T_(1)=(1)/(3)"….(ii)"` Solving we GET, `T_(1)=372 K and T_(2)=(5)/(6)xx372=310 K` Correct choice : (d). |
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| 40. |
A magnetic vector needs three quantities for it's speciffication . Name the three quantities conventionally used to specify earth's magnetic fields. |
| Answer» SOLUTION :(a) MAGNETIC declination , (B) angle of dip, © horizontal COMPONENT of earth.s magnetic field. | |
| 41. |
Name the type of waves which are used for the 5 line of slight (LOS) communication. What is the range of their frequencies ? A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 xx 10^(6)m) |
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Answer» Solution : SPACE wave are used for the line of sight (LOS) communication. The range of their frequencies is 40 MHZ and above. We have, height of transmitting antenna, `h_T = 20 m` and height of RECEIVING antenna, `h_R = 45 m`. Then, maximum distance between the two ANTENNAS, `d_m= SQRT(2Rh_T) + sqrt(2Rh_R)` `implies d_m= sqrt( 2 xx 6.4 xx 10^(6) xx 20) + sqrt( 2 xx 6.4 xx 10^(6) xx 45)` ` 2 xx 8 xx 10^(3) + 3 xx 8 xx 10^(3)` `=40 km` |
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| 42. |
A body is dropped from a height of 125 m. If = 10 ms^(-2) what is the ratio of the distances travelled by it during the first and last second of its motion : |
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Answer» `2:9` During first SECOND the body falls through a height `h_(1)=(1)/(2)g t_(1)^(2)=(1)/(2)xx101^(2)=5M` During first four second the body falls through height `h_2=(1)/(2) "gt"_(2)^(2)=(1)/(2)xx10xx4^(2)=80 m` `:.` height through which the body falls in 5TH second is `h_(3)`=125-80=45 m `:.` `(h_(1))/(h_3)=(5)/(45)=(1)/(9)` |
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| 43. |
What's the expession for torque on a bar magnet in a uniform magnetic field ? |
| Answer» SOLUTION :`TAU = VECM XX VECB` | |
| 44. |
Use Huygens' principIe to show that a point object pIacecd in front of a pIane mirror produces a virtuaI image at the back of the mirror whose distance is equaI to the distance of the object from the mirror. |
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Answer» Solution :LET A be a point source of light a distance y from the plane merror MM . In absence of the mirror let the wavefront travel A ' in time t. But due to the presence of the mirror the REFLECTED wavefront will reach the point A in the same TIMEINTERVAL t. THUS AA' =ct (here , c=velocity of light) i.e AO +OA ' =ct Again AO+OA =ct `:.` OA =OA' 
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| 45. |
Cutting slots in a copper plate, oscillating between the poles of a magnet, reduces the effect of _____. |
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Answer» |
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| 46. |
Discuss about Nicol prism. |
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Answer» SOLUTION :Nicol prism is an optical device incorporated in optical instruments both for producing and ANALYSING plane polarised LIGHT. The constructin of a Nicol prism is based on the phenomemon of Double refraction. One of the most common forms of the Nicol prism is made by taking a CALCITE crystal which is a double refracting crystan with its lenght three times its breadth. It is cut into two halves along the DIAGONAL so that their face angles are `72^(@) and 108^(@)`. Two two halves are joined together by a layer of canada balsam, a transparent cement. |
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| 47. |
Energy in a current carrying coil is stored inthe form of |
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Answer» ELECTRIC field |
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| 48. |
A plane flies 483 km east from city A to city B in 48.0 min and then 966 km south from city B to city C in 1.50 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c ) magnitude and (d) direction of its average velocity, and (e) its average speed? |
| Answer» SOLUTION :(a) `1.08xx10^(3)KM`, (b) `63.4^(@)` south of EAST, or `26.6^(@)` east of south, ( c) 470 km/h 470 km/h `angle-63.4^(@)`, (d) , (E) 630 km/h | |
| 49. |
The adjacent figure shows, the cross-section of a long rod with its length perpendicular to the plane of the paper. It carries constant current flowing along its length. B_(1),B_(2),B_(3)"and"B_(4) respectively respresent the magnetic fields due to the current in the rod at points 1,2,3"and"4 lying at different separations from the centre O as shown in the figure. which of the following shall hold true? |
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Answer» `B_(1)gtB_(2) NE 0` |
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| 50. |
In given figure two coils of self inductance L_1 and L_2 electromagnetically coupled. If mutual inductance between two coils is M, then equivalent self inductance is (##AAK_MCP_32_NEET_PHY_E32_020_Q01##) |
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Answer» `L_1+L_2` |
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