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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 12 knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Obtain mirror equation for the real image obtained by concave mirror. |
| Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XII_P2_C09_E01_012_S01.png" width="80%"/><br/>An object AB is perpendicular to principal axis away from C of a concave mirror.<br/>AM ray from point A incidents on mirror at M and reflected ray passes through principal <a href="https://interviewquestions.tuteehub.com/tag/focus-25840" style="font-weight:bold;" target="_blank" title="Click to know more about FOCUS">FOCUS</a> F.<br/>AP ray from point A incidents on pole P and reflects back in form of PA.<br/>These both reflected rays intersect at A., hence A. is real image of A.<br/>A.B. is image of object AB due to reflection of rays from mirror.<br/>Let FP = focal length f<br/>CP = radius of curvature R<br/>BP = object distance u<br/> B.P= image distance v<br/>For paraxial rays, MP can be considered to be a straight <a href="https://interviewquestions.tuteehub.com/tag/line-1074199" style="font-weight:bold;" target="_blank" title="Click to know more about LINE">LINE</a> perpendicular to CP.<br/>The two right-angled triangles A.B.F and MPF are similar.<br/>`therefore(B.A.)/(PM)=(B.F)/(FP)`<br/>But PM = BA<br/>`therefore (B.A.)/(BA)=(B.F)/(FP)`....(1)<br/>Since `angleAPB = angleA.PB.`, the right angled triangles A.B.P and ABP are also similar. Therefore,<br/>`therefore(B.A.)/(BA)=(B.P)/(BP)`...(2)<br/>Comparing Equations (1) and (2), <br/>`(B.F)/(FP)=(B.P)/(BP)`<br/>but B.F = B.P - FP<br/>`therefore (B.P-FP)/(FP)=(B.P)/(BP)` ....(3)<br/>Now, <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to sign <a href="https://interviewquestions.tuteehub.com/tag/convention-933064" style="font-weight:bold;" target="_blank" title="Click to know more about CONVENTION">CONVENTION</a>, B.P = - v, FP = - f, BP = - u<br/>From equation (3),<br/>`(-v+f)/(-f)=(-v)/(-u)`<br/> `therefore (v-f)/(f)=v/u` <br/> `therefore v/f-1=v/u`<br/>`therefore v/f=1+v/u`<br/>Now,dividing by v,<br/>`therefore 1/f=1/v+1/u`<br/>Which is mirror equation and it is called Gaussian equation as it was given by scientist Gauss.</body></html> | |
| 252. |
Gren house effect is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/black-898687" style="font-weight:bold;" target="_blank" title="Click to know more about BLACK">BLACK</a> radiations from both the <a href="https://interviewquestions.tuteehub.com/tag/types-15918" style="font-weight:bold;" target="_blank" title="Click to know more about TYPES">TYPES</a> of <a href="https://interviewquestions.tuteehub.com/tag/sources-1219409" style="font-weight:bold;" target="_blank" title="Click to know more about SOURCES">SOURCES</a> at high and low temperature<br/>transmit radiations from both the types of source<br/>transmit radiations from a source at low source at high temperature<br/>transmit radiates from a source at high temperature and block radiations from a source at low temperature</p>Answer :D</body></html> | |
| 253. |
Three identical objects each of mass M move along circle of radius R under the action of their mutual gravitational attraction, the speed of each is : |
| Answer» <html><body><p>`((<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/( <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> ))^(1//<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <br/>`((GM)/( R ))^(1//2)`<br/>`((GM)/(sqrt(3)R))^(1//2)`<br/>`((GM)/(sqrt(2)R))^(1//2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :C</body></html> | |
| 254. |
Mention the way of producing induced emf. |
| Answer» <html><body><p></p>Solution :Induced <a href="https://interviewquestions.tuteehub.com/tag/emf-970036" style="font-weight:bold;" target="_blank" title="Click to know more about EMF">EMF</a> can be produced by changing <a href="https://interviewquestions.tuteehub.com/tag/magnetic-551115" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNETIC">MAGNETIC</a> flux in any of the <a href="https://interviewquestions.tuteehub.com/tag/following-463335" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWING">FOLLOWING</a> ways: (i) By changing the magnetic <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a> B <br/> (ii) By changing the area of the coil and <br/> (iii) By changing the relative orientation of the coil with magnetic field</body></html> | |
| 255. |
An stronmical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> <a href="https://interviewquestions.tuteehub.com/tag/focal-2648683" style="font-weight:bold;" target="_blank" title="Click to know more about FOCAL">FOCAL</a> <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> and <a href="https://interviewquestions.tuteehub.com/tag/large-1066424" style="font-weight:bold;" target="_blank" title="Click to know more about LARGE">LARGE</a> diameter<br/>large focal length and small diameter<br/>small focal length and small diameter<br/>large focal length and large diameter</p>Answer :D</body></html> | |
| 256. |
If the work function of a certain metal is 3.2xx10^(-19)J and is it illuminated with light of frequency 8xx10^(14) Hz, the maximum kinetic energy of photo electrons would be (h=6.6xx10^(-34)Js) |
| Answer» <html><body><p>`2.1xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)<a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>`<br/>`8.5xx10^(-19)J`<br/>`5.3xx10^(-19)J`<br/>`3.2xx10^(-19)J`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 257. |
A circular coil of radius 8cm, 400 turns and resistance 2Omega is placed with its plane perpendicular to the horizantal component of the earth's magnetic fiedl. It is rotated about its vertical diameter through 180^(@) in 0.30 s. Horizontal component of earth magnitude of current induced in the coil is approximately |
| Answer» <html><body><p>`4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)A`<br/>`8xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)A`<br/>`8xx10^(-2)A`<br/>`1.92xx10^(-3)A`</p>Solution :Initial flux through the coil <br/> `phi_(i)=BAcos theta` <br/> `=3xx10^(-5)xxpixx(8xx10^(-2))xxcos 0^(@)` <br/> `=192pixx10^(-9)Wb` <br/> <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> flux after the rotation <br/> `phi_(<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>)=3xx10^(-5)xxpi(8xx10^(-2))^(2)cos 180^(@)` <br/> `=-192pixx10Wb` <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/teh-1240626" style="font-weight:bold;" target="_blank" title="Click to know more about TEH">TEH</a> magnitude of induced emf is <br/> `epsi=N|(dphi|)/(dt)=(N|phi_(f)phi_(i)|)/(dt)` <br/> `=(400xxx(384pxx10^(-9)))/(0.30)=1.6xx10^(-3)V` <br/> Current `I=(epsi)/(R)=(1.6xx10^(-3))/(2)=8xx10^(-4)A`</body></html> | |
| 258. |
According to the Bohr-Sommerfeld postulate the periodic motion of a particle in a potential field must satisfy the following qyantization rule: ointp dq= 2piħn, where q and p are generalized coordinate and momenum of the particle, n are integers. Making use of this rule, find the permitted values of energy for a particle of mass m moving (a) In a uniimensional rectangular potential well of width l (b) along a circule of radius r, (c ) in a unidimentional potential field U=alphax^(2)//2, where alpha is a positive constant: (d) along a round orbit in a central field, where the potential enargy of the particle is equal to U= -alpha//r(alpha is a positve constant) |
| Answer» <html><body><p></p>Solution :(a) if we measure energy from the bottom of the well, then `V(x) = 0` inside the walls. Then the quantization condition reads `oint p d x = 2 l p = 2 pi nħ` <br/> or `p = pi ħ//l`<br/> Hence`E_(n) = (p^(2))/(2 m) = (pi^(2) n^(2) ħ)/(2 m l)`. <br/> `oint p d x = 2 l p` because we have to consider the <a href="https://interviewquestions.tuteehub.com/tag/integral-516833" style="font-weight:bold;" target="_blank" title="Click to know more about INTEGRAL">INTEGRAL</a> form `- (1)/(2)` to `(1)/(2)` and then back to `-(1)/(2)`. <br/> (b) Here, `oint p d x = 2 pi r p = 2 pi nħ` <br/> or `p = (n ħ)/(r )` <br/> Hence`E_(n) = (n^(2)ħ^(2))/(2 m r^(2))` <br/> (c ) By energy <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a>`(p^(2))/(2 m) + (1)/(2) alpha x^(2) = E` <br/> so`p = sqrt(2m E - m alpha x^(2))` <br/> Then `oint p d x = oint sqrt(2m E - m alpha x^(2) dx)` <br/> `= 2sqrt(m alpha) int_(-(sqrt( 2E))/(alpha))^(sqrt(2E)/(alpha)) sqrt((2 E)/(alpha) - x^(2)) dx` <br/> The integral is`int_(-a)^(a)sqrt(a^(2) - x^(2)) dx = a^(2) int_(-x//2)^(x//2) cos^(2) theta d theta` <br/> `= (a^(2))/(2) int_(-x//2)^(x//2)(1 + cos 2 theta) d theta = a^(2)(pi)/(2)`. <br/> Thus `oint p d x = pi sqrt(m a). (2 E)/(alpha) = E.2 pi sqrt((m)/(alpha)) = 2 pi n ħ` <br/> Hence `E_(n) = n ħsqrt((alpha)/(m))`. <br/> ( b) It is required to find the energy levels of the <a href="https://interviewquestions.tuteehub.com/tag/circular-916697" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCULAR">CIRCULAR</a> orbit for the rotential <br/> `U( r) = -(alpha)/(r )` <br/> In a circular orbit, the particle only has <a href="https://interviewquestions.tuteehub.com/tag/tangible-663303" style="font-weight:bold;" target="_blank" title="Click to know more about TANGIBLE">TANGIBLE</a> velocity and the qunatization condition reads `ointp d x = m v. 2 pi r = 2 pi nħ` <br/> so `m v r = M = n ħ` <br/> The energy of the particle is <br/> `E = (n^(2) ħ^(2))/(2 m r^(2)) - (alpha)/(r )` <br/> Equilibrium requires that the energy as a function of `r` be minimum. Thus <br/> `(n^(2) ħ^(2))/(mr^(3))=(alpha)/(r^(2)) or r=(n^(2) ħ^(2))/(m alpha)` <br/> Hence `E_(n)= -(malpha^(2))/(2n^(2) ħ^(2))`</body></html> | |
| 259. |
The displacement-time graphs of two moving particles makes anlges of 30^(@) and 45^(@)with the X-axis.The ratio of their velocities is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>:2)`<br/>`1:1`<br/>`1:2`<br/>`:sqrt(3)`</p>Solution :`(V_A)/(V_B)=(<a href="https://interviewquestions.tuteehub.com/tag/tan30-1238909" style="font-weight:bold;" target="_blank" title="Click to know more about TAN30">TAN30</a>^(@))/(<a href="https://interviewquestions.tuteehub.com/tag/tan45-3097696" style="font-weight:bold;" target="_blank" title="Click to know more about TAN45">TAN45</a>^(@))=(1)((sqrt(3))/(1))=(1)/(sqrt(3))`</body></html> | |
| 260. |
A ray of light travelling in a medium of refractive index mu is incident at an angle theta ona composite transparent plate consisting of 50 plates of R.I. 1.0mu, 1.02mu, 1.03mu,"……",1.50mu. The ray emerges from the composite plate into a medium of refractive index 1.6 mu at angle 'x'. Then |
| Answer» <html><body><p>`sin x (1.01/1.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)^(50) sin <a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sinx-1210519" style="font-weight:bold;" target="_blank" title="Click to know more about SINX">SINX</a> = 5/8 sin theta`<br/>`sinx= (8)/(5) sin theta`<br/>`sinx = (1.5/1.01)^(50) sin theta`</p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XII_C09_E01_102_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/accordingto-847003" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDINGTO">ACCORDINGTO</a> <a href="https://interviewquestions.tuteehub.com/tag/snell-2274380" style="font-weight:bold;" target="_blank" title="Click to know more about SNELL">SNELL</a>'s law, `musintheta= 1.6 mu sin x` <br/> `:. sin x = 5/8 sin theta`</body></html> | |
| 261. |
The rate of emission of heat energy per unit area of an iron ball of radius 10 cm is 10J//m^2s s, then rate ·of emission of heat energy per unit area by a copper ball of radius 5 cm at same temperature will be ( emissivity of both the balls is same) |
| Answer» <html><body><p>`2J//m^2s`<br/>`2.5J//m^2s`<br/>`2cal//m^2s`<br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> of these</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 262. |
A massless elastic cord (that obeys Hooke's law) break if the tension in the cord exceeds T_("max"). One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m as shown in the figure. If a second, smaller object ofmass m moving at an initial speed v_(0) strikes the larger mass and the two stick together the cord will stretch and break but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one dimensional collision the cord will still break and the larger mass will move off with a final speed of v_(f). All motion occurs on a horizontal frictionless surface. (Assume that Hooke's law is obeyed throughout untill the cord breaks) Find (v_(f))/(v_(0)) |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(sqrt(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>))` <br/>`(1)/(sqrt(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))` <br/>`(1)/(sqrt(6))` <br/>`(1)/(sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :C</body></html> | |
| 263. |
A massless elastic cord (that obeys Hooke's law) break if the tension in the cord exceeds T_("max"). One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m as shown in the figure. If a second, smaller object ofmass m moving at an initial speed v_(0) strikes the larger mass and the two stick together the cord will stretch and break but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one dimensional collision the cord will still break and the larger mass will move off with a final speed of v_(f). All motion occurs on a horizontal frictionless surface. (Assume that Hooke's law is obeyed throughout untill the cord breaks) Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial knetic energy of the smaller mass prior to the collision |
| Answer» <html><body><p>`1//4`<br/>`1//3`<br/>`1//2`<br/>`3//4` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 264. |
Maximum value of photo electrons V_max = ____. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :2hc(`lambda_0 - <a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>`)/`mlambda lambda_0`</body></html> | |
| 265. |
The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length L_A=1.5m and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is |
| Answer» <html><body><p></p>Solution :(a)Only odd <a href="https://interviewquestions.tuteehub.com/tag/harmonics-1016029" style="font-weight:bold;" target="_blank" title="Click to know more about HARMONICS">HARMONICS</a> can exist in a pipe with one <a href="https://interviewquestions.tuteehub.com/tag/open-586655" style="font-weight:bold;" target="_blank" title="Click to know more about OPEN">OPEN</a> end. The resonant frequencies are given by f = nv/4L for harmonic numbers n = 1,<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>,5,....<br/>Because resonant frequency increases with increasing n, the second lowest resonant frequency <a href="https://interviewquestions.tuteehub.com/tag/corresponds-935737" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDS">CORRESPONDS</a> to n = 3, the second lowest <a href="https://interviewquestions.tuteehub.com/tag/choice-915996" style="font-weight:bold;" target="_blank" title="Click to know more about CHOICE">CHOICE</a> of n.<br/>(b) Any harmonic can exist in a pipe with two open ends. The resonant frequencies are given by f = n/2L for harmonic numbers n = 1,2,3,... <br/>Because resonant frequency increases with increasing n, the second lowest resonant frequency corresponds to n = 2, the second lowest choice of n. Therefore, we can write<br/>`f= (2v)/(2L_B)` <br/> Solving for `L_B` and substituting known data yield<br/>`L_B = v/f = (1522 m//s)/(761 Hz) = 2.00 m `</body></html> | |
| 266. |
Rutherford's model suggested that electron revolves round the nucleus in ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/circular-916697" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCULAR">CIRCULAR</a> <a href="https://interviewquestions.tuteehub.com/tag/orbits-1138199" style="font-weight:bold;" target="_blank" title="Click to know more about ORBITS">ORBITS</a></body></html> | |
| 267. |
A uniform wire of length L, diameter D ad densityP is stretched under a tension T. The correct relation between its fundamental freqency 'f', the length L and the diameter D is : |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> prop (<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>)/(D^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))` <br/>`f prop (1)/(D^(2))` <br/>`f prop (1)/(LD)`<br/>`f prop (1)/(L sqrt(D))` </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>= `(1)/(2l ) sqrt((T)/(m))` <br/> Now mass per unit length m `rho . A= rho. (pi D^(2))/(4 ) ` <br/> v `= (1)/(2l) sqrt((T)/((pi D^(2))/(4))) = (1)/(l) sqrt((T)/(pi D^(2))) ` <br/> For <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> tension T, v `prop (1)/(l)` <br/> So correct choice is c.</body></html> | |
| 268. |
A convex lens of power 0.04 dioptre produces an image which is double the size of object placed in front of it. Find the position of the object. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> supplied, <br/> P = 0.04 dioptre,`"" ` m = 2,<a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a> = ?<br/> `<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> = (1)/(p) = (1)/(0.04) = (100)/(4) = 2.5 ` m<br/>` m = (f)/(u + f)2 =(<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>)/(u + 25)2u + <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> = 252u = - 25u = - 12.5 m `</body></html> | |
| 269. |
STATEMENT-1: It is not possible for a charged particle to move in a circular path around a long straight conductor carrying current. STATEMENT-2: The electromagntic force on a moving particle is normal to its plane of rotation. |
| Answer» <html><body><p>Statement-1 is <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a>, Statement-2 is True, Statement-2 is a <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a>, for Statement-1.<br/>Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.<br/>Statement-1 is True, Statement-2 is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a>.<br/>Statement-1 is a False, Statement-2 is True.</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 270. |
A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity, the impulse received by the ball due to gravity force during its flight is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> m^(2) gh)`<br/>`sqrt(4 m^(2) gh)`<br/>`sqrt(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> m^(2) gh)`<br/>`4 sqrt(m^(2) gh)`</p>Answer :C</body></html> | |
| 271. |
Two shells are fired from a canon successively with speed u each at angles of projection alphaandbeta, respectively. If the time interval between the firing of shells is dt and they collide in mid-air after a time t from the firing of the first shell. Then |
| Answer» <html><body><p>`tcosalpha=(t-dt)<a href="https://interviewquestions.tuteehub.com/tag/cosbeta-2549507" style="font-weight:bold;" target="_blank" title="Click to know more about COSBETA">COSBETA</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/alphagtbeta-2412697" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHAGTBETA">ALPHAGTBETA</a>`<br/>`(t-dt)cosalpha=dtcosbeta`<br/>`(usinalpha)t-1/2gdt^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(usinbeta)t-dt-1/2g(t-dt)^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A::<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>::D</body></html> | |
| 272. |
A variable frequency alternating voltage of constant magnitude is applied across a capacitor. Which of the following graph shows the variation of current set up in the circuit with frequency v? |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/U_LIK_SP_PHY_XII_C07_E12_005_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/U_LIK_SP_PHY_XII_C07_E12_005_O02.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/U_LIK_SP_PHY_XII_C07_E12_005_O03.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/U_LIK_SP_PHY_XII_C07_E12_005_O04.png" width="30%"/> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 273. |
An error of 3% is committed in the measurement of volume of a cube. What is the percentage of error committed in the measurement of each of its sides. |
| Answer» <html><body><p></p>Solution :The volume of a cube `V=<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>`<br/> l=length of each side <br/> so `long(V) = 3 log l` <br/> Differentiating both sides `DeltaV/V=3(Deltal)/l` so, `(Deltal)/l=l//3xx(DeltaV)/V=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>//3xx3%=1%`</body></html> | |
| 274. |
A capacitor of capacitance 10mu F is charged to a potential of 50V with a battery. The battery is disconnected and an additional charge 200 mu C is given to positive plate of the capacitor. The potential difference across the capacitor will be |
| Answer» <html><body><p>50V<br/>80 V<br/>70 V<br/>60 V</p>Answer :D</body></html> | |
| 275. |
Two plane wavespropagate in a homogeneous elasticmedium, one along the x axis and the other along the y axis : xi _(1) = a cos ( omega t - kx), xi _(2) = a cos ( omega t - ky ) .Find the motion patternfo particles in the plane xy if both waves. (a) are tansverse and their oscillation directions coincide, (b) are longitudinal. |
| Answer» <html><body><p></p>Solution :`(a)` Equation of the resultant wave, <br/> `xi=xi_(1)+xi_(2)=2 a cos k ((y-x)/(2))<a href="https://interviewquestions.tuteehub.com/tag/xoa-3887560" style="font-weight:bold;" target="_blank" title="Click to know more about XOA">XOA</a> {omegat-(k(x+y))/(2)}, ` <br/> `=a'cos { omegat-(k(x+y))/2}`,where `a^(') = 2 a cos k^(') ((y-x)/2)` <br/>Now, the equation of wave pattern is, <br/> `x+ y=k`, (aConst.) <br/> For sought plots see the answer `-` sheet of the problem book. <br/> For antinodes, i.e. <a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> intensity <br/> `cos((k(y-x))/(2))=+-1= cos n pi` <br/> or, `+-(x-y)=(2npi)/(k) = n lambda` <br/> or `y=x+- nlambda, n=0,1,2,.....` <br/> Hence, the particles of the medium at the <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a>, lying ono the solide straight lines `(y=x+-n lambda)` , oscillate with maximum amplitude.ltbr. For nodes, i.e. minimum intensity, <br/> `cos ((k(y-x))/(2))=0` <br/> or `+- (k(y-x))/( 2)=(2n+1) ( pi)/(2)` <br/> or , `y=x+- ( 2n +1) lambda//2`, <br/> and hence the particles at the points, lying on dotted lines do not oscillate. <br/> `(b)` When the waves are longitudinal, <br/> For sought plots see the answer `-` sheet of the problem book. <br/> `k(y-x)=cos^(-1)((xi_(1))/(a))- cos^(-1)((xi_(2))/( a))` <br/> or, `(xi_(1))/( a)= cos { k(y-x)+cos ^(-1) ((xi_(2))/(a))}` <br/> `=(xi_(2))/( a) cos k(y-x) - <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> k y( y-x) sin(cos ^(-1)((xi_(2))/(a)))` <br/> `=( xi_(2))/(a) cos k ( y-x)- sin k ( y-x) <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(1-(xi_(2)^(2))/( a^(2)))...(1)` <br/> from `(1)`,<br/> if `sin k ( y-x) = 0 =cos ( 2n +1) ( pi)/(2)` <br/> `(xi_(1)^(2))/(a)=1-xi_(2)^(2)//a^(2)`, acircle. <br/> Thus the particles, at the points, where `y= x+-(n+- 1//4) lambda` , will oscillate along circles, <br/> In general, all other particles will move along ellipse.</body></html> | |
| 276. |
Which of the following determine shortest and longest wavelengths in hydrogen atom spectrum ? |
| Answer» <html><body><p>`(1)/(lamda)=R((1)/(1^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(6^(2)))`<br/>`(1)/(lamda)=R((1)/(2^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(oo^(2)))`<br/>`(1)/(lamda)=R((1)/(3^(2))-(1)/(4^(2))),(1)/(lamda)=R((1)/(4^(2))-(1)/(5^(2)))`<br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> of these </p>Solution :`(1)/(lamda)=R((1)/(1^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(6^(2)))` <br/> First equation of option (A) shows <a href="https://interviewquestions.tuteehub.com/tag/series-1201802" style="font-weight:bold;" target="_blank" title="Click to know more about SERIES">SERIES</a> limit of Lyman series, which is in ultraviolet region. So `lamda` is minimum, second equation of option (A) shows first line of Pfund series, which is in <a href="https://interviewquestions.tuteehub.com/tag/infra-515954" style="font-weight:bold;" target="_blank" title="Click to know more about INFRA">INFRA</a> <a href="https://interviewquestions.tuteehub.com/tag/red-613739" style="font-weight:bold;" target="_blank" title="Click to know more about RED">RED</a> region and shows maximum wavelength.</body></html> | |
| 277. |
An equilateral prism has a refractive index equal to 1.56 for the light of wavelength 450 nm. Determine the minimum angle of deviation for light of 450 nm. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`42.52^(@)`</body></html> | |
| 278. |
For a television network, 5 xx 10^(5) channels are granted. If the central frequency of the microwave link is 25 GHz and the alloted bandwidth for each channel is 2 KHz, then how much percentage of the link is used for the network ? |
| Answer» <html><body><p>`4%`<br/>`10%`<br/>`25%`<br/>`5%`</p>Solution :Let `x%` of link is used for network. Given, central frequency = 25 GHz <br/> `= 25 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(9) Hz` <br/> Bandwidth of channels `= x%` of 25 GHz <br/> `= (x)/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>) xx 25 xx 10^(9) = 25x xx 10^(7)` <br/> Number of channel <br/> `= ("Total Bandwidth")/("Bandwidth needed per channel")` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/5xx-1901301" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX">5XX</a> 10 ^(5) = (25 xx 10 ^(7))/(2 xx 10^(3))` <br/> `10xx 10 ^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) = 25 x xx 10 ^(7)` <br/> `implies x = 4%`</body></html> | |
| 279. |
The height of a T.V. tower is 100m. If radius of earth is 6400km, then what is the maximum distance of transmission from it ? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/100km-265959" style="font-weight:bold;" target="_blank" title="Click to know more about 100KM">100KM</a>`<br/>`64sqrt(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)<a href="https://interviewquestions.tuteehub.com/tag/km-1064498" style="font-weight:bold;" target="_blank" title="Click to know more about KM">KM</a>`<br/>`6.4sqrt(10)km`<br/>`8sqrt(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)km.`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 280. |
Give the expression for the wavelength of a charged particle in terms of accelerating potential and explain the meaning of the symbols used. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a> = (h)/( sqrt(<a href="https://interviewquestions.tuteehub.com/tag/2me-1837831" style="font-weight:bold;" target="_blank" title="Click to know more about 2ME">2ME</a> <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>))`</body></html> | |
| 281. |
Consider an ideal juction diode. Find the value of current flowing through AB is |
| Answer» <html><body><p></p>Solution :The barrier <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> of the diode is neglected as it is an ideal diode. The value of <a href="https://interviewquestions.tuteehub.com/tag/current-940804" style="font-weight:bold;" target="_blank" title="Click to know more about CURRENT">CURRENT</a> flowing through AB can be <a href="https://interviewquestions.tuteehub.com/tag/obtained-7273275" style="font-weight:bold;" target="_blank" title="Click to know more about OBTAINED">OBTAINED</a> by using Ohm.s law<br/> `I=V/R = (3-(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>))/(1 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^3)= 10/(10^3) = 10^(-2)A = 10 m A`</body></html> | |
| 282. |
If 13.6 eV energy is required to 10ise the hydrogen atom, then energy required to remove an electron from n = 2 is |
| Answer» <html><body><p>`10.2` eV<br/>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>` eV<br/>`3.4` eV<br/>`6.8` eV</p>Solution :`E_(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>) = (<a href="https://interviewquestions.tuteehub.com/tag/136-272658" style="font-weight:bold;" target="_blank" title="Click to know more about 136">136</a>)/(n^(2))eV` <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> E = E_(<a href="https://interviewquestions.tuteehub.com/tag/infty-2738147" style="font-weight:bold;" target="_blank" title="Click to know more about INFTY">INFTY</a> - E_(2) = 0 + (13.6)/(2^(2)) = 3.4 eV`</body></html> | |
| 283. |
A thermos bottle containing coffee is vigorously shaken. Consider coffee as a system: Has work been done on it ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Yes, work has been <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a> on <a href="https://interviewquestions.tuteehub.com/tag/coffee-421753" style="font-weight:bold;" target="_blank" title="Click to know more about COFFEE">COFFEE</a> against <a href="https://interviewquestions.tuteehub.com/tag/viscous-1447063" style="font-weight:bold;" target="_blank" title="Click to know more about VISCOUS">VISCOUS</a> <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a></body></html> | |
| 284. |
If the bar magnet in exercise 13 is turned around by 180^@ where will the new null points be located ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :On the equatorial line . <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> d. be the corresponding distance . Since the two fields are equal to earth.s field <br/> `10^(-7)xx(<a href="https://interviewquestions.tuteehub.com/tag/2m-300757" style="font-weight:bold;" target="_blank" title="Click to know more about 2M">2M</a>)/d^3=10^(-7)xxm/(d.^3)"":.d.^3=(d^3)/2`<br/> `d.="d"xx2^(-1/3)=14xx2^(-1/3)=11.1cm`</body></html> | |
| 285. |
Two particles having position vectors vec(r_1) = (3veci+5vecj)m and vec(r_2) =(-5veci +3vecj)m are moving with velocities vec(V_1) =(4veci -4vecj)ms^(-1) and vec(V_2) = (aveci – 3vecj)ms^(-1). If they collide after 2 seconds, the value of 'a' is |
| Answer» <html><body><p>2<br/>4<br/>6<br/>8</p>Answer :D</body></html> | |
| 286. |
Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> (i.<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>., `n_(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>) = n_(e) = n_(i)` ).</body></html> | |
| 287. |
A source of sound gives five beats per second, when sounded with another source of frequency 100 s^(-1). The second harmonic of the source, together with a source of frequency 205 s^(-1) gives five beats per second. What is the frequency of the source ? |
| Answer» <html><body><p>`105 s^(-1)` <br/>205 `s^(-1)` <br/><a href="https://interviewquestions.tuteehub.com/tag/95-342378" style="font-weight:bold;" target="_blank" title="Click to know more about 95">95</a> `s^(-1)` <br/>100 `s^(-1)` .</p>Solution :The freq. of the <a href="https://interviewquestions.tuteehub.com/tag/source-1219297" style="font-weight:bold;" target="_blank" title="Click to know more about SOURCE">SOURCE</a> is `(100 pm 5) ` i.e., 105 or 95 Hz. The second harmonic is 210 or 190 . As is <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> 5 beats / sec with 205 , <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, its freq. is 105 and harmonic 210 . <br/> correct choice is (a). <br/> correct choice is (c).</body></html> | |
| 288. |
A current flowing through a galvanometer is 10(-6)A and produce a deflection of 1 scale devision. If the resistance of moving coil is 400Omega and galvanometer has 30 scale division, the voltage across the coil is : |
| Answer» <html><body><p>`4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>) V`<br/>`4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) V`<br/>`4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) V`<br/>`4xx10^(-3) V`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 289. |
Figure shows a rectangular 20-trun coil of wire of dimensions 10 cm by 5.0 cm. It carries a current of 0.10 A and is hinged along one long side. It is mounted in the xy. Plane, at angle theta=30^(@) to the direction of a uniform magnetic field of magnitude 0.50 T. In unit-vector notation, what is the torque acting on the coil about the hinge line? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`(-4.3xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)N.m)<a href="https://interviewquestions.tuteehub.com/tag/hatj-2693584" style="font-weight:bold;" target="_blank" title="Click to know more about HATJ">HATJ</a>`</body></html> | |
| 290. |
Which of the following is an example of point to point communication mode ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/radio-4998" style="font-weight:bold;" target="_blank" title="Click to know more about RADIO">RADIO</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/television-1240846" style="font-weight:bold;" target="_blank" title="Click to know more about TELEVISION">TELEVISION</a> <br/>telephony<br/>All of these</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 291. |
A charged particle enters in a magnetic field B with its initial velocity making an angle of 45° with B. The path of the particle will be |
| Answer» <html><body><p>an ellipse<br/>straight line<br/>a circle<br/>a heilical</p>Answer :D</body></html> | |
| 292. |
An electric lamp designed for operation on 110 V AC is connected to a220 V AC supply, through a choke coil of inductance 2H, for proper operation. The angular frequency of the AC is 100sqrt(10) rad //s. If a capacitor is to be used in the place of the choke coil, its capacitance must be |
| Answer» <html><body><p>`1muF` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <a href="https://interviewquestions.tuteehub.com/tag/muf-549966" style="font-weight:bold;" target="_blank" title="Click to know more about MUF">MUF</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> muF`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> muF`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :C</body></html> | |
| 293. |
Three point-mases m_(1),m_(2)andm_(3) are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m_(1)? |
| Answer» <html><body><p>`(m_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)+m_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))(a^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` <br/>`(m_(1)+m_(2)+m_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))(a^(2))/(4)` <br/>`(m_(2)+m_(3))(a^(2))/(4)` <br/>`(m_(1)+m_(3))(a^(2))/(4)` </p>Solution :`I_(AD)=m_(1)xx0^(2)+m_(2)xxBD^(2)+m_(3)xxCD^(2)` <br/> `=0+m_(2)((a)/(2))^(2)+m_(3)((a)/(2))^(2)` <br/> `I_(AD)=(m_(2)+m_(3))(a^(2))/(4)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_RPA_OBJ_PHY_C06_E01_057_S01.png" width="80%"/></body></html> | |
| 294. |
At a place horizontal component of the earths magnetic field is B and angle of dip at the place is 60°. What is the value of horizontal component of the earths magnetic field. (i) at Equator, (ii) at a place where dip angle is 30° |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/2b-300274" style="font-weight:bold;" target="_blank" title="Click to know more about 2B">2B</a>, Bsqrt3`.</body></html> | |
| 295. |
For a series L-C-R circuit the power loss at resonance is : |
| Answer» <html><body><p>`I^2omegaL`<br/>`I_omegaC`<br/>`I^2R`<br/>`(E^<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)/<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(R^2+(omegaL-1/(<a href="https://interviewquestions.tuteehub.com/tag/omegac-2889227" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGAC">OMEGAC</a>))^2`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 296. |
Bodies in non-inertial frames disobey "___________". |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/newton-11212" style="font-weight:bold;" target="_blank" title="Click to know more about NEWTON">NEWTON</a>'s law<br/>Theory of <a href="https://interviewquestions.tuteehub.com/tag/relativity-1183718" style="font-weight:bold;" target="_blank" title="Click to know more about RELATIVITY">RELATIVITY</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/general-471730" style="font-weight:bold;" target="_blank" title="Click to know more about GENERAL">GENERAL</a> Relativity<br/>Einstein's theories</p>Answer :A</body></html> | |
| 297. |
Match the following |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/JYT_AJP_AIO_PHY_XII_C13_E03_007_S01.png" width="80%"/></body></html> | |
| 298. |
{:((i)"Quantisation of charges",(a) int_(infty)^(P)- vec(E).d vec(r) ),((ii)"Electric field",(b)"ne"),((iii)"Electric dipole moment",(c) (F)/(q)),("(iv) Electric potential",(d)2qa):} |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :(i)`<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>` (d); (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>)`rarr`(a); (iii) `rarr` (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>); (iv) `rarr`(c)</body></html> | |
| 299. |
What is the value of magnetic susceptibility for diamagnetic materials ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> and <a href="https://interviewquestions.tuteehub.com/tag/positive-1159908" style="font-weight:bold;" target="_blank" title="Click to know more about POSITIVE">POSITIVE</a></body></html> | |
| 300. |
Electromagnetic waves can be propagated throught conductor. In an electromagnatic wave, what is the phase difference between electric and magnetic field variation ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :They are in <a href="https://interviewquestions.tuteehub.com/tag/phase-22748" style="font-weight:bold;" target="_blank" title="Click to know more about PHASE">PHASE</a></body></html> | |