45406 + Interview Questions in Physics in Class 12

1.	स्त्रीकेसर का वह भाग जो परागकण ग्रहण करता है :-
Answer» वर्तिका अंडाशय वर्तिकाग्र अण्ड Answer :C

Discussion

2.	On transmitting a bem of Xrays with wavelength lambda = 17.8pm through a polycrystalline specimen a system of diffraction rings is producedon a screen located at a distance l = 45 cm fom the specimen. Determine the radius of the bright rind corresponding to second order of reflection from the system of planes with interplanar distance d = 155pm
Answer» SOLUTION :In a polycrysyalline specimen, microcystals are orientyed at various angles with respect to one another. The microcrystals which are oriented at certain special angles with respect to the incident beam produce diffraction MAXIMA that appear as rings. The radiqal of these rings are given by `r = l tan 2 ALPHA` where the Bragg's law gives `2D SIN alpha = k lambda` In our case `k = 2, d = 155pm, lambda = 17.8pm` so `alpha = sin^(-1) .(17.8)/(155) = 6.6^(@)` and `r = 3.52 cm`.

Discussion

3.	The potential energy of a body of mass m is given by U=ax+by where x,y are position co-ordinate of the particle. The acceleration of the particle is
Answer» `(a^2+b^2)/m` `((a+b)^(1/2))/m` `(a+b)/m` `((a^2+b^2)^(1/2))/m` ANSWER :D

Discussion

4.	The wavelength of the yellow spectral emission line of sodium is 590 nm. At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?
Answer» SOLUTION :`4.3xx10^(-6)EV`

Discussion

5.	The average power dissipated in A.C. circuit is 2 watt. If a current flowing through a circuit is 2 A and impedance is 1 Omega, what is the power factor of the A.C. circuit ?
Answer» 0.5 1 0 `(1)/(sqrt(2))` SOLUTION :`P = VI cos phi = I^(2) Z cos phi ""(because V = IZ)` `cos phi = (P)/(I^(2)Z) = (2)/(4 xx 1) = 0.5`

Discussion

6.	A right-angled prism of refractive index mu_(1), is placed in a rectangular block of refractive index mu_(2), which is surrounded by a medium of refractive index mu_(3), as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between mu_(1), mu_(2) and mu_(3), it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'. Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists:
Answer» `{:(P,Q,R,S),(2,3,1,4):}` `{:(P,Q,R,S),(1,2,4,3):}` `{:(P,Q,R,S),(4,1,2,3):}` `{:(P,Q,R,S),(2,3,4,1):}` Solution :For PATH e - f, angle in medium `mu_(2)`, is less hence `mu_(2)gtmu_(1)`. Moreover, angle in medium `mu_(3)`is ALSO increasing hence `mu_(3)ltmu_(2)`. Hence P can be MATCHED with 2. For path e-g, there is no deviation while emerging from prism, hence `mu_(1)=mu_(2)`. Hence Q can be matched with 3. For path e-h, angle in medium `mu_(2)` is greater than in `mu_(1)` and hence `mu_(2)ltmu_(1)`. Moreover, there is no internal reflection so `mu_(2)ltsqrt(2)` `mu_(2), mu_(2)gtmu_(3)`is also correct according to RAY DIAGRAM. Hence R can be matched with 4. For path e-i there is total internal reflection taking place and hence `mu_(1)gtsqrt(2)``mu_(2)`. Hence S can be matched with 1. `P-2, Q-3, R-4, S-1` is correct. Hence, option (d) is correct.

Discussion

7.	If 'm' is the mass of an electron and 'c' is the speed of light, the ratio of the wavelength of a photon of energy E to that of the electron of the same energy is
Answer» a. `c sqrt((2M)/(E ))` b. `sqrt((2m)/(E ))` c. `sqrt((2m)/(CE))` d. `sqrt((m)/(E ))` ANSWER :A

Discussion

8.	An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to
Answer» `B/v` `v/R` `sqrt(v/B)` `sqrt(B/v)` Solution :Equivalent CURRENT DUE to motion of electron in a circular orbit of radius R is `I = (EV)/(2 pi R)` and magnetic FIELD B at the centre of the circular orbit `B = (mu_0 I)/(2R) = (mu_0)/(2R) cdot (ev)/(2 pi R) = (mu_0 ev)/(4pi R^2)` `:. R = sqrt((mu_0 ev)/(4 pi B)) "or" R prop sqrt((v)/(B))`

Discussion

9.	If the impedance of an L-C-R circuit is 40 Omega, then the admittance of the circuit will be :
Answer» 0.1 siemen 0.025 siemen 0.05 siemen 0.075 siemen Answer :B

Discussion

10.	What is mean life of nucleus? Give the expression.
Answer» Solution :The MEAN life TIME of the nucleus is the ratio of sum or INTEGRATION of life of all nuclei to the total NUMBER nuclei present initially. The expression for mean life time, `tau = 1/lambda`

Discussion

11.	In radiation heat energy can be transferred
Answer» a)In presence of MATERIAL medium B)without any material medium C)In presence of any material medium or without any material medium d)All of the above. Answer :C

Discussion

12.	A vessel of depth 2d is half filled with a liquid of refractive index n_(1) and the upper half with a liquid of refractive index n_(2). The apparent depth of the vessel when seen normally from the top is
Answer» `d((n_(1)n_(2))/(n_(1)+n_(2)))` `d((1)/(n_(1))+(1)/(n_(2)))` `2D((1)/(n_(1))+(1)/(n_(2)))` `(2d)/(n_(1)n_(2))` SOLUTION :Apparent depth `=(d)/(n_(1))+(d)/(n_(2))=d((1)/(n_(1))+(1)/(n_(2)))`

Discussion

13.	YDSE is carried out in a liquid of refractive index mu=1.3 and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. Find the positions of the fourth maxima. The wavelength of light is air is lambda_0 = 0.78 mum and D//d = 1000.
Answer» Solution :(a) `1-(1/mu)t = ((3lambda)/mu)` `:. t=((3lambda)/(mu-1))` ` = ((3xx0.78)/(1.3-1))` `=7.8 mu m` (b) UPWARDS `(yd)/D-1-(1/mu)t = (4LAMBDA)/mu` ` Solving, we get y=4.2 mm` Downwards `T1-(1/mu) +(yd)/D=(4lambda)/mu` Solving, we get `y=06` mm.

Discussion

14.	Magnetic induction is a
Answer» Scalar QUANTITY VECTOR quantity Both (a) and (B) None of the above ANSWER :B

Discussion

15.	Calculate the longest and shortest wavelength in the Balmer series of hydrogen atom. Given Rydberg constant =1.0987xx10^7m^-1.
Answer» SOLUTION :`lambda_l =6553 A^@ , lambda_s=3640A^@`

Discussion

16.	A child is sitting on a swing.Its minimum and maximum heights from the ground are 0.75 m and 2m respectively. Its maximum speed will be :
Answer» 10 m/s 8 m/s 5 m/s 15 m/s Solution :Here effective height `h=2-0.75=1.25` m. Now `(1)/(2)MV^(2)="mgh"impliesv=SQRT(2gh)` `=sqrt(2xx10xx1.25)=5` m/s. `:.` CORRECT CHOICE is ( c )

Discussion

17.	A long straight cable of length l is placed symmetrically along z-axis and has radius a(ltltl). The cable consists of a thin wire and co-axial conductingtube. An alternating current I(t)=I_0sin (2pivt) flows down the central thin wire and return along the co-axial conducting tube. The induced elelctric field at a distance s form the wire inside the cable isvecE(s,t)=mu_0I_0 v cos (2pivt)log_e(s/a)hatk (i) Calculate the displacement current density inside the cable. (ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current I^d. (iii) Compare the conduction current I_0 with the displacement current I_0^d.
Answer» Solution :(i) Given, the induced electric field at a distance r from the wire INSIDE the cable is `vecE(s,t)=mu_0I_0 v cos (2pivt)log_e(s/a)hatk` DISPLACEMENT current density, `vecJ_D=in_0 (vec(dE))/(dt)=in_0 d/(dt)[mu_0I_0 v cos (2pivt)log_e(s/a)hatk]` `=in_0 mu_0 I_0 v d/(dt)[cos 2pivt]log_e (s/a)hatk=1/(c^2) I_0 v^2 2pi[-sin 2pivt]loe_e (s/a)hatk` `=(v^2)/(c^2)2piI_0 sin 2pivt log_e (a/s)hatk[ :' log_e(s/a)=-log_e(a/s)]` `=1/(lambda^2) 2pi I_0 log_e(a/s) sin 2pivthatk=(2piI_0)/(lambda^2) log_e a/s sin 2pivt hatk` (ii) `I_D=int J_D s DS d theta=int_(s=0)^a int_0^(2pi) d theta=int_(s=0)^a J_Dsdsxx2pi` `=int_0^a[(2pi)/(lambda^2)I_0log_e(a/s)s ds sin 2pivt]xx2pi=((2pi)/lambda)^2I_0 int_(s=0)^alog_e(a/s)s ds sin 2pivt` `((2pi)/lambda)^2I_0 int_(s=0)^alog_e(a/s) 1/2d(s^2).sin 2pivt=(a^2)/2((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2` ` (a^2)/4((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2=-(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivt int_(s=0)^a log_e(a/s). d(s/a)^2` `=-(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivtxx(-1)[ :' int_(s=0)^a log_e(a/s). d(s/a)^2=-1]` `I_D=(a^2)/4 ((2pi)/lambda)^2I_0 sin 2pivt=((2pia)/(2lambda))^2I_0sin 2pivt` (iii) The diplacementcurrent, `I_D=((2pia)/(2lambda))^2I_0sin 2pivt=I_(0D)sin 2pivt` where `I_(0D)=((2pia)/(2lambda))^2I_0=((pia)/(lambda))^2I_0 :. (I_0D)/(I_0)=((API)/(lambda))^2`

Discussion

18.	In Bohr's model of an atom, which of the following is an integral multiple of (h)/(2pi)?
Answer» Kinetic energy Radius of an atom Potential energy Angular momentum Solution :`L = MVR = (NH)/(2 PI)`

Discussion

19.	To make the central fringeat the centreO , a miea - sheet index 1.5 is itroduced.Choose the correct statement (s)
Answer» The THICKNESS of sheet is `2(SQRT(2)-1)` d infront of `S_(1)` The thickness of sheet is `(sqrt(2)-1)` d infront of `S_(2)` The thickness of sheet is `2sqrt(2)d` infront of `S_(1)` The thickness of sheet is `(2sqrt(2)-1)` d FRONT of `S_(1)` ANSWER :A

Discussion

20.	A p - type semiconductor has acceptor level 57 meV above the valence band. The maximum wavelength of light required to create a hole is :
Answer» `57Å` `57xx10^(-3)Å` `217100 Å` `11.61 XX 10^(-33)Å` ANSWER :C

Discussion

21.	How is the speed of electromagnetic waves in vacuum determined by electric and magnetic fields ?
Answer» Solution :`E_(0)=cB_(0) or c=(E_(0))/(B_(0))`

Discussion

22.	Name the phenomenon which justifies the transverse nature of em waves.
Answer» SOLUTION :POLARIZATION.

Discussion

23.	पाँच जगत वर्गीकरण प्रस्तुत किया
Answer» जीनियस ने हक्सले ने व्हिटेकर ने हचिन्सन ने Answer :C

Discussion

24.	The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations show whether it can detect light of wavelength 400 nm.
Answer» Solution :Here energy gap of photodiode `E_(g)` = 1.2 eV, and WAVELENGTH of incident light `lambda` = 400 nm = ` 400 XX 10^(-9)` m. `THEREFORE ` Energy of incident light photon E = `(hc)/(e lambda) eV = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx 400 xx 10^(-9)) `eV = 3.1 eV As energy of incident light photon is greater than the energy gap `(E gt E_(g))` of given photodiode, the photodiode will definitely DETECT the light.

Discussion

25.	A signal emitted by an antenna from a certain point can be received at another point of the surface
Answer» SKY WAVE ground wave SEA wave Both (1) and (2) ANSWER :D

Discussion

26.	(A): Increasing the number of observations minimizes random errors (R): positive and negative random errors occur with equal probability.
Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false ANSWER :A

Discussion

27.	How much energy must a gamma ray photon have, if it is to materialize into a pair of electron and positron with each particle having a K.E. of 1 MeV?
Answer» 2MeV 3.02 MEV 1.0MeV 0.51 MeV ANSWER :B

Discussion

28.	Find the distance at which the magnetic field on axis as compared to the magnetic field at the centre of the coil carrying current I and radius R is 1/8.
Answer» R `sqrt2R` 2R `sqrt3R` Solution :`B_("centre")/B_("axis")=[1+x^(2)/R^(2)]^(3/2),"But "B_("axis")=(B_("centre"))/8` `thereforeB_("centre")/B_("axis")XX8=[1+x^(2)/R^(2)]^(3/2)` `THEREFORE(2^(3))^(2/3)=1+x^(2)/R^(2)` `therefore(2^(2))=[1+x^(2)/R^(2)]` `therefore4=1+x^(2)/R^(2)" "thereforex^(2)/R^(2)=3rArrx=sqrt3R`

Discussion

29.	An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.
Answer» Solution :The focal LENGTH `f=-15//2cm=-7.5cm` (i) The object distance `u=-10cm.` Then Eq. (9.7) GIVES `(1)/(v)+(1)/(-10)=(1)/(-7.5)` `"or"v=(10xx7.5)/(-2.5)=-30CM` The image is 30 cm from the mirror on the same side as the object. `" Also, magnification m "=-(v)/(u)=-((-30))/(-10)=-3` The image is MAGNIFIED, real and INVERTED. (ii) The object distance `u = -5 cm`. Then from Eq. (9.7), `(1)/(v)+(1)/(-5)=(1)/(-7.5)` `or v=(5xx7.5)/((7.5-5))=15cm` This image is formed at 15 cm behind the mirror. It is a virtual image. Magnification `m=-(v)/(u)=-(15)/((-5))=3` The image is magnified, virtual and erect.

Discussion

30.	Which of the following is not a simple harmonic motion?
Answer» Motion of needle of SEWING machine Vibrations of ATOMS in molecules Motion of tip of hands of a clock Motion of PENDULUM clock Answer :C

Discussion

31.	Induced electric field due to to changing magnetic flux are more readily observed than induced magnetic field due to changing electric field. Why?
Answer» SOLUTION :The CHANGING electric FIELD PRODUCES displacement current, which is very small and hence the magnetic field set up by it is also small, the same cannot the be observed easily. In an a.c. circuit displacement current can be increased by increasing the ANGULAR frequency of current. This would increase the iduced electric field due to changing magnetic flux can be increased by taking more number of turns of the coil. The induced e.m.f. in different turns of the same coil are added up, resulting in induced electric field which is easily observed.

Discussion

32.	Give the reason for the twinkling effect of stars.
Answer» Solution :Light from STAR undergoes REFRACTION continuously before it REACHES earth. DUE to this, the apparent position of the star is slightly different from its actual position. The apparent position varies due to the change in temperature and density of atmospheric layers. The fluctuating apparent position of the star GIVES rise to twinkling effect.

Discussion

33.	If the intensity ratio of two coherent sources used in Young's double slit experiemnt is 49:1 then the ratio between the maximum and minimum intensities in the interference pattern is
Answer» a. `1:9` B. `9:16` C. `25:16` d. `16:9` ANSWER :D

Discussion

34.	a. If the fundamental source wave is a vibrating object, what is the fundamental source of an electromagnetic wave ?b.Give the expression for velocity of E.M waves in vacuum.c.On what factors does its velocity in vaccum depend ?
Answer» Solution :a. The varying electric and MAGNETIC FIELDS can act as sources of each other and electromagnetic wave is produced. b.Velocity `=c=(1)/(sqrt(mu_(0)epsilon_(0)))` Permeability `(mu_(0))` and permittivity `(epsilon_(0))` of free space

Discussion

35.	A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with 80 km/h, what is the least stopping distance ?
Answer» 8 m 2 m 4 m 6 m Solution :STOPPING distance `PROP` `(velocity)^(2)` `(s_(2))/(2)=((80)/(40))^(2)=4` `:. S_2 =8`

Discussion

36.	A coherent parallel beam of microwaves of wavelength A = 0.5 mm fails on a Young's double slit apparatus. The separation between maxima is measured on a screen placed paraiiei to the plane, of the slits at a distance of 1.0 m from it as shown in figtird. The separation between the slits is 2d - 1mm. (a) If the incident beam fails normally on the double slit apparatus, find the y-coordinates of all the interface minima of the screen. (b) If the incident beam makes an angle of 30° with the x-axis (as shown in fig. find the y-coordinates of the first minima on either side of the central maximum.
Answer» Solution :(a) Givenseperation between two slits = 2d 1.0mm, D=1.0m, Path diffenrece = `2dsin theta` For minima `2 d SINTHETA = (2n -1 ) lambda/2` `therafore 1 xx 10^(-3) = ((2n-1)xx(0.5xx10^(-3)))/2` as `theta=90^@` for the HIGHEST possible order of minimum. or `(2n-1) = (2xx10^(-3))/(0.5xx10^(-3)) implies n=2.5` Thus, only two minima are possible on either side of central maxima. Thus total number of minima = 2+2 =4 For minima `y= pm ((2n-1)lambdaD)/(4d) = (2n-1)0.25` `y_(1)=0.25m , y_(2)=0.75m` and `y_(4)= -0.75m` (b) Initial path DIFFERENCE = `2d sin 30^@` For the central maxima `therefore y/D xx2d= 2d sin30^@` `y=Dsin30^@=0.5m` For first minima `y/D xx2d -2dsin 30^@ = om lambda/2` or `y= pm 0.25 + 0.5` or `y_(1)0.75m ` and `y_(2) = 0.25 m`.

Discussion

37.	A rod of mass m and length l is rotating about a fixed point in the ceiling with an angular velocity omega as shown in the figure. The rod maintains a constant angle theta with the vertical. What is the rate of change of angular momentum of the rod ?
Answer» `(momega^(2)L^(2)sintheta)/(6)` `(momega^(2)l^(2)sin2theta)/(6)` `(momega^(2)l^(2)sin^(2)theta)/(6)` `(momega^(2)l^(2)cos^(2)theta)/(6)` Solution :The vertical component of the ANGULAR momentum remains constant while its HORIZONTAL component keeps changing the direction. `DL=(momegalsin(2theta))/(6)xxomegadtrArr(dL)/(dt)=(momega^(2)l^(2))/(6)sin(20)`

Discussion

38.	A rod of mass m and length l is rotating about a fixed point in the ceiling with an angular velocity omega as shown in the figure. The rod maintains a constant angle theta with the vertical. What angle will the rod make with the vertical
Answer» `cos^(-1)((g)/(omega^(2)l))` `cos^(-1)((2g)/(3omega^(2)l))` `cos^(-1)((3g)/(2omega^(2)l))` `cos^(-1)((3g)/(4omega^(2)l))` SOLUTION :`TAU=(dL)/(DT)` `Mg.(L)/(2)sintheta=(mu omega^(2)l^(2))/(6)sin(2THETA)` `rArrcostheta=((3g)/(2omega^(2)l))` or, `theta=cos^(-1)((3g)/(2omega^(2)l))`

Discussion

39.	The phenomenon of interference is possible in the case of
Answer» LONGITUDINAL WAVES transverse waves both none Answer :C

Discussion

40.	In order that a body of 15 kg weighs zero at the equator, then the angular speed of earth is : (g=10ms^(-1))
Answer» `(1)/(80) " RAD s"^(-1)` `(1)/(400)" rad s"^(-1)` `(1)/(100)" rad s"^(-1)` `(1)/(1600)" rad s"^(-1)` ANSWER :C

Discussion

41.	In a large bullding there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :
Answer» `12 `A `14`A `8A` `10A` Solution :`12A` Total power = 40 `xx 15 + 100 xx 5 + 80 xx 5 + 1000` = `600 + 500 + 400 + 1000 ` `therefore VI = 2500` W `therefore I = (2500)/(220) = 11.36`A `therefore I = 12 A `

Discussion

42.	If photoelectrons are to be emitted from a potassium surface with a speed 6 xx 10^(6) ms^(-1), what frequency of radiation must be used ? (Threshold frequency for potassium is 4.22 xx 10^(14) Hz, h = 6.6 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg)
Answer» SOLUTION :Here, `V = 6 xx 10^(6) ms^(-1)` `V_(0) = 4.22 xx 10^(14)` Hz From Einstein.s photoelectric equation, `K.E = (1)/(2) mv^(2) = H(upsilon - upsilon_(0))` `upsilon = (1)/(2) (mv^(2))/(h) + upsilon_(0)` `= (1)/(2) xx (9.1 xx 10^(-31) + (6 xx 10^(6))^(2))/(6.6 xx 10^(-34)) + 4.22 xx 10^(14)` `= (2.48 xx 10^(14)) + (4.22 xx 10^(14))` `upsilon = 6.7 xx 10^(14)` Hz

Discussion

43.	A cyclotron, when being used to accelerate protons (mass = 1.67 xx 10^(-27) kg, charge = 1.6 xx 10^(-19) C) has a magnetic field of (pi)/2 tesla applied normal to the plane of its does. What must be the value of the frequency of the applied alternating electric field to be used in it?
Answer» Solution :Here mass of protons `m = 1.67 xx 10^(-27)KG`, charge on a proton `q = +1.6 xx10^(-19) C`, magnetic field `B = pi/2 T`. `:.` FREQUENCY of applied alternating electric field `v = (QB)/(2 pi m) = (1.6 xx 10^(-19) xx (pi//2))/(2pi xx 1.67 xx 10^(-27)) = 2.4 xx 10^(7) HZ` .

Discussion

44.	Two projectile P and Q are fired form same distance with same velocity at angle 30^@ and 60^@ The horizontal range is
Answer» P is EQUAL to Q P is LESS than Q P is GREATER than Q P can be lesser or greater than Q Answer :A

Discussion

45.	In the above question, the ratio of the initial rate of fall of temperature is
Answer» `((PI)/(6))^(1//3):1` `1:1` `((4PI)/(3))^(1//3):1` `((3)/(4pi))^(2//3):1` ANSWER :A

Discussion

46.	Find the pressure of a pulsar's magnetic field and compare it with the presseure of gravitational forces
Answer» SOLUTION :The pressure of the magnetic field is equal to the energy density: `p_(m)=w_(m)=B^(2)//2mu_(0)` `P_(g)rav=barrhobargh=(3M)/(4piR^(3)),(gammaM)/(2R^(2)).(R)/(2)=(3gammaM^(2))/(16piR^(4))` Substituting the values of the pulsar.s mass and radius (see Problem 14.21) we OBTAIN the VALUE of the pressure EXERTED by the gravitational forces.

Discussion

47.	An aircraft loops the loop of radius R=500m with a constant velocity v=360km per hour. Find the weight of the flyer of mass m=70kg in the lower, upper, and middle points of the loop.
Answer» SOLUTION :While moving in a loop, NORMAL reaction exerted by the flyer on the loop at different points and uncompensated weight if any contribute to the weight of flyer at those points. (a) When the aircraft is at the lowermost point, Newton's second LAW of motion in projection form `F_n=mw_n` gives `N-mg=(mv^2)/(R)` or, `N=mg+(mv^2)/(R)=209kN` (b) When it is at the upper most point, again from `F_n=mw_n` we get `N^(' ')+mg=(mv^2)/(R)` `N^('')=(mv^2)/(R)-mg=07kN` (C) When the aircraft is at the middle point of the loop, again from `F_n=mw_n` `N^'=(mv^2)/(R)=14kN` The uncompensated weight is `mg`. Thus effective weight `=sqrt(N^('^2)+m^2g^2)=156kN` acts obliquely.

Discussion

48.	An object of height 1.5 cm is placed in the axis of convex lens of faces length 25 cm . A real image is formed at a distance of 75 cm from the lens. The size of image will be :
Answer» 4.5 cm 5 cm 3.0 cm 0.75 cm Answer :A

Discussion

49.	A transformer has an efficiency of 80%. It works at 100V and 4kW. If secondary voltage is 240 V, the current in primary coil is ……
Answer» 0.4 A `4 A` `10 A` `40 A` Solution :Power of 100 V and I ampere for transformer is`4xx10^3` W `therefore P_1=V_1 I_1` `therefore I_1=P_1/V_1` `=(4xx10^3)/100`=40 A

Discussion

50.	Three equal mass satellites A,B and C are in coplaner orbits around a planet as shown in the figure. The magnitudes of the angular momentum of the satellites as measured about the planet are L_(A),L_(B), and L_(C).Which of the following statements is correct?
Answer» `L_(A)gtL_(B)gtL_(C)` `L_(C)gtL_(B)gtL_(A)` `L_(B)gtL_(C)gtL_(A)` `L_(B)gtL_(A)gtL_(C)` ANSWER :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

स्त्रीकेसर का वह भाग जो परागकण ग्रहण करता है :-

The potential energy of a body of mass m is given by U=ax+by where x,y are position co-ordinate of the particle. The acceleration of the particle is

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An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to

If the impedance of an L-C-R circuit is 40 Omega, then the admittance of the circuit will be :

What is mean life of nucleus? Give the expression.

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A vessel of depth 2d is half filled with a liquid of refractive index n_(1) and the upper half with a liquid of refractive index n_(2). The apparent depth of the vessel when seen normally from the top is

Magnetic induction is a

Calculate the longest and shortest wavelength in the Balmer series of hydrogen atom. Given Rydberg constant =1.0987xx10^7m^-1.

A child is sitting on a swing.Its minimum and maximum heights from the ground are 0.75 m and 2m respectively. Its maximum speed will be :

In Bohr's model of an atom, which of the following is an integral multiple of (h)/(2pi)?

To make the central fringeat the centreO , a miea - sheet index 1.5 is itroduced.Choose the correct statement (s)

A p - type semiconductor has acceptor level 57 meV above the valence band. The maximum wavelength of light required to create a hole is :

How is the speed of electromagnetic waves in vacuum determined by electric and magnetic fields ?

Name the phenomenon which justifies the transverse nature of em waves.

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The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations show whether it can detect light of wavelength 400 nm.

A signal emitted by an antenna from a certain point can be received at another point of the surface

(A): Increasing the number of observations minimizes random errors (R): positive and negative random errors occur with equal probability.

How much energy must a gamma ray photon have, if it is to materialize into a pair of electron and positron with each particle having a K.E. of 1 MeV?

Find the distance at which the magnetic field on axis as compared to the magnetic field at the centre of the coil carrying current I and radius R is 1/8.

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Which of the following is not a simple harmonic motion?

Induced electric field due to to changing magnetic flux are more readily observed than induced magnetic field due to changing electric field. Why?

Give the reason for the twinkling effect of stars.

If the intensity ratio of two coherent sources used in Young's double slit experiemnt is 49:1 then the ratio between the maximum and minimum intensities in the interference pattern is

a. If the fundamental source wave is a vibrating object, what is the fundamental source of an electromagnetic wave ?b.Give the expression for velocity of E.M waves in vacuum.c.On what factors does its velocity in vaccum depend ?

A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with 80 km/h, what is the least stopping distance ?

A rod of mass m and length l is rotating about a fixed point in the ceiling with an angular velocity omega as shown in the figure. The rod maintains a constant angle theta with the vertical. What is the rate of change of angular momentum of the rod ?

A rod of mass m and length l is rotating about a fixed point in the ceiling with an angular velocity omega as shown in the figure. The rod maintains a constant angle theta with the vertical. What angle will the rod make with the vertical

The phenomenon of interference is possible in the case of

In order that a body of 15 kg weighs zero at the equator, then the angular speed of earth is : (g=10ms^(-1))

In a large bullding there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

If photoelectrons are to be emitted from a potassium surface with a speed 6 xx 10^(6) ms^(-1), what frequency of radiation must be used ? (Threshold frequency for potassium is 4.22 xx 10^(14) Hz, h = 6.6 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg)

A cyclotron, when being used to accelerate protons (mass = 1.67 xx 10^(-27) kg, charge = 1.6 xx 10^(-19) C) has a magnetic field of (pi)/2 tesla applied normal to the plane of its does. What must be the value of the frequency of the applied alternating electric field to be used in it?

Two projectile P and Q are fired form same distance with same velocity at angle 30^@ and 60^@ The horizontal range is

In the above question, the ratio of the initial rate of fall of temperature is

Find the pressure of a pulsar's magnetic field and compare it with the presseure of gravitational forces

An aircraft loops the loop of radius R=500m with a constant velocity v=360km per hour. Find the weight of the flyer of mass m=70kg in the lower, upper, and middle points of the loop.

An object of height 1.5 cm is placed in the axis of convex lens of faces length 25 cm . A real image is formed at a distance of 75 cm from the lens. The size of image will be :

A transformer has an efficiency of 80%. It works at 100V and 4kW. If secondary voltage is 240 V, the current in primary coil is ……

Three equal mass satellites A,B and C are in coplaner orbits around a planet as shown in the figure. The magnitudes of the angular momentum of the satellites as measured about the planet are L_(A),L_(B), and L_(C).Which of the following statements is correct?