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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.
| 45351. |
All bodies, no matter how hot or cold, continuously radiate photons. At a given temperature, the intensities of the electromagnetic waves emitted by an object vary from wavelength to wavelength throughout the visible, infrared, and other regions of the spectrum. Figure illustrates how the intensity per unit wavelength depends on wavelength for a perfect blackbody emitter. Although this figure can strictly be applied only to a black body, yet this will approximately discribe the behavior of many of the self radiating systems. For example, sun has an approximate temperature of 6000K. it is not a black body, it has an emissivity of nearly 0.6 But its peak almost occurs at a wave length that predicted by Wein's law. Suppose we have a bulb of power 100W. It emits only about 5W as visible light, rest is emitted as infrared radiated. Assume that the bulb filament has a surface area of 10mm^(2) (hc=1250 eV-nm) What is the approximate temperature of the filament? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> k<br/>350 k<br/>2500 K<br/>10000 K</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> | |
| 45352. |
Name the pd between terminals of the cell when (i) key K is open(i) K is closed |
| Answer» <html><body><p>emf<br/>lost voltage<br/>terminal voltage<br/>induced voltage</p>Answer :A</body></html> | |
| 45353. |
The ground state energy of hydrogen atom is - 13.6 eV. If an electron makes a transition from an energy level - 0.85 eV to -3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? |
| Answer» <html><body><p></p>Solution :The groundstateenergy of hydrogen `E_(1)`= 13.6 eV . <br/> `therefore ` Energy `E_(i) = -0.85e <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>` <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 a level `n_(i)`where ` - 0.85 = (-13.6)/(n_(i)^(2)) rArr n_(i) = 4` <br/>Againenergy `E_(f) = - 3.4` eV correspondsto a level `n_(f)`suchthat ` -3.4 = - (13.6)/(n_(f)^(2))`, which leads us totheresult `n_(f) = 2` <br/> As transition is takingplace from `n_(i) = 4`level to `n_(f) = 2` levelthewavelenghtbelongs to Balmer series of hydrogen <a href="https://interviewquestions.tuteehub.com/tag/spectrum-1221793" style="font-weight:bold;" target="_blank" title="Click to know more about SPECTRUM">SPECTRUM</a> . <br/>The wavelenghtof spectrall lineis given by therelation. <br/> `hv = (<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(lambda) = E_(i) = E_(f) =- 0.85 eV - (-3.4 eV) = + 2.55 eV`. <br/>`therefore "" lambda = (hc)/(2.55ev) = (6.63 xx 10^(-34) xx 3 xx 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>))/((2.55 xx 1.60 xx 10^(-19))J)= 4.875 xx 10^(-7) m = 487.5` nm</body></html> | |
| 45354. |
All bodies, no matter how hot or cold, continuously radiate photons. At a given temperature, the intensities of the electromagnetic waves emitted by an object vary from wavelength to wavelength throughout the visible, infrared, and other regions of the spectrum. Figure illustrates how the intensity per unit wavelength depends on wavelength for a perfect blackbody emitter. Although this figure can strictly be applied only to a black body, yet this will approximately discribe the behavior of many of the self radiating systems. For example, sun has an approximate temperature of 6000K. it is not a black body, it has an emissivity of nearly 0.6 But its peak almost occurs at a wave length that predicted by Wein's law. Suppose we have a bulb of power 100W. It emits only about 5W as visible light, rest is emitted as infrared radiated. Assume that the bulb filament has a surface area of 10mm^(2) (hc=1250 eV-nm) Which of the following resistances would have maximum surface temperature. All of them have a surface area of10mm^(2)and same emissivity |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</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> Omega`<br/>`2 Omega`<br/>`3 Omega`<br/>`4 Omega`</p>Solution :`(dQ)/(dt)=eAsigma(T^(4)-T_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)^(4))` <br/> `T_(max)Rightarrow(dQ)/(dt) max` <br/> `P_(4)=<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>^(2)/4 P_(1)=(100/2.2)^(2)xx1` <br/> `P_(2)=(100/2.2xx3/5)^(2)xx2` <br/> `P_(3)=(100/2.2xx2/5)^(2)xx3`</body></html> | |
| 45355. |
All bodies, no matter how hot or cold, continuously radiate photons. At a given temperature, the intensities of the electromagnetic waves emitted by an object vary from wavelength to wavelength throughout the visible, infrared, and other regions of the spectrum. Figure illustrates how the intensity per unit wavelength depends on wavelength for a perfect blackbody emitter. Although this figure can strictly be applied only to a black body, yet this will approximately discribe the behavior of many of the self radiating systems. For example, sun has an approximate temperature of 6000K. it is not a black body, it has an emissivity of nearly 0.6 But its peak almost occurs at a wave length that predicted by Wein's law. Suppose we have a bulb of power 100W. It emits only about 5W as visible light, rest is emitted as infrared radiated. Assume that the bulb filament has a surface area of 10mm^(2) (hc=1250 eV-nm) If we want of increase the number of photons emitted by the bulb in the visible region without changing the wattage, which method would be most appropriate? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/increasing-1040633" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASING">INCREASING</a> emissivity by a factor of 2 <br/>increasing the radius of the <a href="https://interviewquestions.tuteehub.com/tag/filament-987780" style="font-weight:bold;" target="_blank" title="Click to know more about FILAMENT">FILAMENT</a> by a factor of 2 and the length by a factor of 4.<br/>decreasing the radius of the filament by a factor of 2 and the length by a factor of 4<br/>doubling the <a href="https://interviewquestions.tuteehub.com/tag/voltage-1447811" style="font-weight:bold;" target="_blank" title="Click to know more about VOLTAGE">VOLTAGE</a> and decreasing the length of the filament by a factor of 2.</p>Solution :`(dN)/(<a href="https://interviewquestions.tuteehub.com/tag/dt-960413" style="font-weight:bold;" target="_blank" title="Click to know more about DT">DT</a>)uparrowifTuparrow` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/dq-959213" style="font-weight:bold;" target="_blank" title="Click to know more about DQ">DQ</a>)/(dt)=eAsigmaT^(4) TuparrowRightarrowAdownarrow`</body></html> | |
| 45356. |
The intrinsic semiconductor becomes an insulator at |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>^(@)`C<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>^(@)`C<br/>300 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a><br/>0 K </p>Solution :At 0 K an <a href="https://interviewquestions.tuteehub.com/tag/intrinsic-499568" style="font-weight:bold;" target="_blank" title="Click to know more about INTRINSIC">INTRINSIC</a> semiconductor behaves as an insulator.</body></html> | |
| 45357. |
A 150 W lamp emits light of mean wavelength of 5500Å. If the efficiency is 12% , find out the number of photons emitted by the lamp in one second. |
| 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/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of photons emitted per second n" = (P lambda)/(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>),,P = 150 W),(,,lambda 5500Å),("If the efficiency is 12%," eta = (12)/(100) = 0.12,,h = 6.6 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-34)Js),(n = (P eta lambda)/(hc),,c = 3 xx 10^(8) ms^(-1)):}` <br/> `= (150 xx 0.12 xx 5500 xx 10^(-10))/(6.6 xx 10^(34) xx 3 xx 10^(8)) = (99000 xx 10^(-10))/(19.8 xx 10^(-26)) = 5000 xx 10^(16)` <br/> `n = 5 xx 10^(19)`</body></html> | |
| 45358. |
The frail woman in Mukesh's house is his |
| Answer» <html><body><p>mother<br/>elder <a href="https://interviewquestions.tuteehub.com/tag/brother-904720" style="font-weight:bold;" target="_blank" title="Click to know more about BROTHER">BROTHER</a>'s wife<br/>wife<br/>niece</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> | |
| 45359. |
Ionosphere can reflect which type of waves ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/microwaves-560348" style="font-weight:bold;" target="_blank" title="Click to know more about MICROWAVES">MICROWAVES</a> <br/>Radiowaves<br/>X-rays<br/>y-rays</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> | |
| 45360. |
Which one of the following has the dimensions of[ML^(-1)T^(-2)]: |
| Answer» <html><body><p>torque<br/>surface tension<br/>viscosity<br/>stress</p>Solution :stress `=("Force")/("area")=(<a href="https://interviewquestions.tuteehub.com/tag/mlt-2163963" style="font-weight:bold;" target="_blank" title="Click to know more about MLT">MLT</a>^(-2))/(L^(2))=[<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>^(-1)T^(-2)]` <br/> So <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/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> is `(d)`</body></html> | |
| 45361. |
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transformer is connected to 220 V 1 A A.C. source, what is output current of the transformer ? |
| Answer» <html><body><p>`1//20 A`<br/>20 A<br/>100 A<br/>`2 A`</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> :`(N_(s))/(N_(p)) = (V_(s))/(V_(p))` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>) = (V_(s))/(<a href="https://interviewquestions.tuteehub.com/tag/220-1824383" style="font-weight:bold;" target="_blank" title="Click to know more about 220">220</a>)` <br/> `V_(s) = 11 V` <br/> `V_(s)I_(s) = V_(p)I_(p)` <br/> `11 xx I_(s) = 220 xx 1` <br/> `I_(s) = 20 A`</body></html> | |
| 45362. |
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> <a href="https://interviewquestions.tuteehub.com/tag/feet-986106" style="font-weight:bold;" target="_blank" title="Click to know more about FEET">FEET</a> <br/>3 feet <br/>16 feet <br/>Any length </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> :B</body></html> | |
| 45363. |
What is the magnetic field along the axis and equatorial line of a bar magnet ? |
| Answer» <html><body><p></p>Solution :Magnetic field at a point along the equatorial line due to a magnetic <a href="https://interviewquestions.tuteehub.com/tag/dipole-440057" style="font-weight:bold;" target="_blank" title="Click to know more about DIPOLE">DIPOLE</a> (bar magnet) <br/> Consider a ber magnet. NS. Let <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> be the north pole and S be the south pole of the bar magnet, each with pole strength `q_(m)` and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line ) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole `(q_(m) C = 1 A m )` at C. The force e`xx`perienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb.s law of magnetism as follows ,<br/> the force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space ) is <br/> `vec(F_(N)) = - F_(N) cos theta hat(i) + F_(N) sin theta hat(j) ` <br/> Where `F_(N) = (mu_(0))/(4pi ) (q_(m))/(r.^(2))` <b> The force of attraction ( in free space ) between southpole of the bar magnet and unit north pole at point C is <br/> `vec(F_(S)) = - F_(S) cos theta hat(i) - F_(S) sin theta hat(j)` <br/> where , `vec(F_(S)) = (mu_(0))/(q_(m))(q_(m))/(r.^(2))` <br/> From equation (1) and equation (2) , the net force at poin C is `vec(<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>) = vec(F_(N)) + F_(S)`. This net force is equal to the magnetic field at the point C . <br/>`vec(B) - (F_(N) + F_(S)) cos theta hat(i)` <br/> since, `F_(N) = F_(S)` <br/> `vec(B) = - (2mu_(0))/(4pi) (q_(m))/(r.^(2)) cos theta hat(i) = (2 mu_(0))/(4pi ) ( q_(m))/((r^(2) + l^(2))) cos theta hati`<br/> In a right angle <a href="https://interviewquestions.tuteehub.com/tag/triangle-1427233" style="font-weight:bold;" target="_blank" title="Click to know more about TRIANGLE">TRIANGLE</a> NOC as shown in the Figure l <br/> cos `theta= ("adjacent")/("hypotenuse") = (l)/(r.) = (l)/((r^(2) + l^(2))^((1)/(2)))`<br/> Substituting equation 4 in equation 3 we get <br/> `vec(B) = - (mu_(0))/(4pi) (q_(m) xx (2l))/((r^(2) + l^(2))^((3)/(2)) )` <br/> Since, magnitude of magnetic dipole moment is `|vec(P_(m))| = P_(m) = q_(m). 2l` and substituting in equation (5 ). The magnetic field at a point C is <br/>`vec(B)_("equatorial") = - (mu_(0))/(4pi) (P_(m))/((r^(2) + l^(2))^((3)/(2)) )` <br/> If the distance between two poles in a bar magnet are small (lools like short magnet ) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r `gt gt ` l, then , <br/> `(r^(2) + l^(2))^((3)/(2)) approx r^(3)` <br/> Therefore, using equation (7) in equation (6), we get<br/> `vec(B)_("equatorial") = - (mu_(0))/(4 pi ) (P_(m))/(r^(3)) hat(i)` <br/> Since, `P_(m) hat(i) = vec(P_(m)) . ` in general, the magnetic field at equatorial point is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by <br/> `vec(B_("equatorial")) = - (mu_(0))/(4 pi ) (P_(m))/(r^(3))` <br/> Note that magnitude of `vec(B_("axial")) `is twice that of magnitude of `B_("equatorial")` and the direction of `B_("axial ")and B_("equantorial") ` are opposite .<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XII_V01_C03_E01_031_S01.png" width="80%"/></b></body></html> | |
| 45364. |
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ? |
| 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> :When <a href="https://interviewquestions.tuteehub.com/tag/dc-430940" style="font-weight:bold;" target="_blank" title="Click to know more about DC">DC</a> source is connected bulb will glow gradually more and more . When AC source is connected , bulb will glow with <a href="https://interviewquestions.tuteehub.com/tag/less-1071906" style="font-weight:bold;" target="_blank" title="Click to know more about LESS">LESS</a> brightness. When inductance is decreased, this brightness will <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a>.</body></html> | |
| 45365. |
Consider an equilateral triangle ABC of side 2a in the plane of the paper as shown. The centroid of the triangle is O. Equal charges (Q) are fixed at the vertices A, B and C In what follows consider all motion and situations to be confined the plane of the paper. (a) A test charge (q), of same sign as Q is placed on the median AD at a point at a distance delta below O. Obtain the force (vec(F)) felt by the test charge. (b) Assuming delta lt lt a discuss the motion of the test charge when it is released. (c) Obtain the force (vec(F)_(D)) on this test charge if it is placed at the point D as shown in the figure. (d) In the figure below mark the approximate locations of the equilibrium point (s) for this system. Justify your answer. (e) Is the equilibrium at O stable or unstable if we displace the test in the direction of OP ? The line PQ is parallel to the base BC. Justify your answer. (f) Consider a rectangle ABCD. Equal charges are fixed at the vertices A, B, C and D. O is the centroid. In the figure below mark the approximate locations of all the neutral points of the system for a test charge with same sign as the charges on the vertices. Dotted lines are drawn for the reference. (g) How many neutral points are possible for a system in which N charges are placed at the N vertices of a regular N sided polygon ? |
| Answer» <html><body><p></p>Solution :(a) `vec(F)=(2KQq(a/sqrt(3)-delta))/((a^(2)+(a/sqrt(3)-delta)^(2))^(3//2))-(KQq)/(((2a)/sqrt(3)+delta))` Here `K=1//4 pi epsi_(0)` and direction is upward (towards A) <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) Using binomial approximation, `vec(F)=KQq (9 sqrt(3))/16 delta/a^(3)` (upward) which is linear in`delta`. Hence charge will oscillate simple harmonically about O when relaesed. <br/> (c) `vec(F)_(D)=(KQq)/(3a^(2))` (downward) <br/> (d) For small `delta` force on the test charge is upwards while for <a href="https://interviewquestions.tuteehub.com/tag/large-1066424" style="font-weight:bold;" target="_blank" title="Click to know more about LARGE">LARGE</a> `delta` (eq. at D) force is downwards. So there is a neutral point between O and D. By symmetry there will be neutral points on other medians also. In figure x. Below all possible (4) neutral points are shown by. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_PHY_ELE_E03_086_S01.png" width="80%"/> <br/> (e) Let the distance along P be x and O to be at (0, 0). electric potential of a test charge along OP can be written as `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>(x)=(Kq)/sqrt(x^(2)+(4//3))+(KQ)/sqrt((x+1)^(2)+(1//3))+(KQ)/sqrt((x-1)^(2)+(1//3))~~KQ sqrt(3/4) (3+9/16 x^(2))` <br/> We can see that `V(x) prop x^(2)`, hence it is <a href="https://interviewquestions.tuteehub.com/tag/stable-1223548" style="font-weight:bold;" target="_blank" title="Click to know more about STABLE">STABLE</a> equilibrium. <br/> (f) Equilibrium points are <a href="https://interviewquestions.tuteehub.com/tag/indicated-7298716" style="font-weight:bold;" target="_blank" title="Click to know more about INDICATED">INDICATED</a> by <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_PHY_ELE_E03_086_S02.png" width="80%"/> <br/> (g) `N+1`</body></html> | |
| 45366. |
Both in interference and diffraction phenomena, alternate dark and bright fringes are obtained on screen I) generally fringe width is same in interference and not same in diffraction II) the central fringe in interference has maximum brightness and the intensity gradually decreases on either side III) in interference the intensity of all bright fringes is same IV) both the phenomena are produced from same coherent sources. |
| Answer» <html><body><p>I only<br/>I and <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a> <br/>I, II and IV<br/>I, <a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a> and IV</p>Answer :D</body></html> | |
| 45367. |
Calculate the potential difference and the energy stored in the capacitor C_2 in the circuit shown in the Fig. Given potential at A is 90 V , C_1 = 20 mu F , C_2 = 30 mu Fand C_3 = 15 mu F |
| Answer» <html><body><p></p>Solution :Let as <a href="https://interviewquestions.tuteehub.com/tag/shown-1206565" style="font-weight:bold;" target="_blank" title="Click to know more about SHOWN">SHOWN</a> in Fig `V_1 , V_2` and `V_3` be the respective potential differences across the capacitors `C_1 , C_2` and `C_3` respectively . In series combination same charge Q exists on each capacitor and `V_1 : V_2 : V_3 = (1)/(C_(1)) : (1)/(C_(2)) : (1)/(C_(3))` <br/> `implies V_(1)= (C_(2))/(C_(1)) V_(2) = (30)/(20) V_(2)` and `V_(3) = (C_(2))/(C_(3)) V_(2) = (30)/(15) V_(2)` <br/> But `V_(1) + V_(2) + V_(3) = 90 V` , hence we have <br/> `(30)/(20) V_(2) + V_(2) + (30)/(15) V_(2) = 90 V implies 45 V_(2) = 90 V implies V_(2) = 20 V ` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/U_LIK_SP_PHY_XII_C02_E10_046_S01.png" width="80%"/> <br/> Moreover , energy stored in capacitor `C_2` is `V_2 = (1)/(2) C_(2) V_(2)^(2) = (1)/(2) xx (30 mu F) xx (20 V)^(2) = <a href="https://interviewquestions.tuteehub.com/tag/6000-329798" style="font-weight:bold;" target="_blank" title="Click to know more about 6000">6000</a> mu <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> = 6.0 mJ`</body></html> | |
| 45368. |
A circuit shown in the figure has resistances 20 Omega and 30 Omega. At what value of resistance R_(x) will the thermal power generated in it be pracitcally independent of small variations of that resistance? The voltage between points A and B is supposed to be constant in this case. |
| Answer» <html><body><p>`3 <a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>`<br/>`7 Omega`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> Omega`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> Omega`</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> | |
| 45369. |
Blood is flowing at a rate of 200 (cm)^3/sec in a capillary of cross sectional area 0.5 m^2. The velocity of flows in (in mm/sec) |
| Answer» <html><body><p>a)0.1<br/>b)0.2<br/>c)0.3<br/>d)0.4</p>Answer :D</body></html> | |
| 45370. |
Explain electric field and also electric field is point charge. |
| Answer» <html><body><p></p>Solution :Suppose, charge Q is placed at origin .O. in free <a href="https://interviewquestions.tuteehub.com/tag/space-649515" style="font-weight:bold;" target="_blank" title="Click to know more about SPACE">SPACE</a>. If another charge q is placed distance r at point P (OP =r), then Coulomb force acts on q. <br/> `vecF = 1/(pi epsilon_(0)).(Qq)/r^(2)hatr` <br/> If Q = 1C, then force acting on unit charge is called electric field E. <br/> `therefore vecF/q = 1/(<a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a> epsilon_(0)).Q/r^(2) hatr` <br/> `therefore vecE = 1/(4pi epsilon_(0)).Q/r^(2)hatr` or `E = (kQ)/r^(2)` <br/> Definition of electric field : The region around the charge in which the effect of electric charge is <a href="https://interviewquestions.tuteehub.com/tag/prevailing-2950879" style="font-weight:bold;" target="_blank" title="Click to know more about PREVAILING">PREVAILING</a> is called the electric field of the charge. <br/> Electric field `vecE` is also called electric fielt intensity. Force acting on charge q of position vector `vecr` is `vecF(vecr) =qvecE(vecr)` <br/> Electric field `vecE` is also called electric field intensity. <br/> Definition of Electric field : .The force acting on a unit positive charge at a given point in an electric field of a point charge of the system at charge is called the electric field or intensity of electric field `vecE`at that point.<br/> SI unit of electric field intensity is `NC^(-1)` or `<a href="https://interviewquestions.tuteehub.com/tag/vm-728443" style="font-weight:bold;" target="_blank" title="Click to know more about VM">VM</a>^(-1)`and <a href="https://interviewquestions.tuteehub.com/tag/dimensional-2582799" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONAL">DIMENSIONAL</a> formula is `[M^(1)L^(1) T^(-3) A^(-1)]`</body></html> | |
| 45371. |
Four equal point charges of 16 muCare kept at vertices of square of side 0.2 m. Find force acting at any one charge. |
| 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> :`<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> = 110.3 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>`</body></html> | |
| 45372. |
In the Q. 122 what will be the velocity of the particle when it returns to the starting point? |
| Answer» <html><body><p>`vecr_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)`<br/>`-vecr_(0)`<br/>`2vecr_(0)`<br/>`-2vecr_(0)`</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> :`vecr=vecr_(0)t-vecr_(0)at^(2)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> <a href="https://interviewquestions.tuteehub.com/tag/vecv-3257988" style="font-weight:bold;" target="_blank" title="Click to know more about VECV">VECV</a>=(dvecr)/(dt)=vecr_(0)-vecr_(0)a2t` <br/> Returning at `t=(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(a)` <br/> `rArr vecv_(1//a)=vecr_(0)-2vecr_(0)axx(1)/(a)` <br/> `rArr vecv=-vecr_(0)`</body></html> | |
| 45373. |
In a common emitter amplifier , output resistance is 500Omega and input resistance is 2000 Omega . If peak value of signal voltage is 10 mV and beta =50 , then the peak value of output voltage is : |
| Answer» <html><body><p>`5 xx10^(-6)<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>`<br/>`2.5 xx10^(-<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/>1.25 V<br/>125 V</p>Answer :C</body></html> | |
| 45374. |
The current in resistance R_(3) in the given circuit is (2)/(x)A Find the value of x. |
| 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> :<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a></body></html> | |
| 45375. |
Magnetic flux passing through a coil varies with time as, phi=(2t^(2)-4) weber, Resistance of the coil is 10Omega. |
| Answer» <html><body><p>At time t=2s, induced <a href="https://interviewquestions.tuteehub.com/tag/current-940804" style="font-weight:bold;" target="_blank" title="Click to know more about CURRENT">CURRENT</a> in the <a href="https://interviewquestions.tuteehub.com/tag/coil-921250" style="font-weight:bold;" target="_blank" title="Click to know more about COIL">COIL</a> is 0.8A <br/>Induced current increases lineralu with time <br/>From t=0 to t=2s,0.8C charge has flown in the coil<br/>in the above time interval net flow of charge is zero</p>Answer :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>::C</body></html> | |
| 45376. |
Two polaroids are placed 90^(@) to each other and the transmitted intensity is zero. What happens when one more polaroid is placed between these two bisecting the angle between them? |
| 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> :`1/4`</body></html> | |
| 45377. |
Draw the voltage-current characteristic of a Zener diode. |
| 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/U_LIK_SP_PHY_XII_C14_E08_022_S01.png" width="80%"/></body></html> | |
| 45378. |
A narrow slit illuminted by light of wavelength 0.64 mu is placed at a distance of 3 m from a straight edge. If the distance between the straight edge and the screen is 6 m calculate the ditance between the first and fourth dark bands. |
| 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> :`4.8xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) m`</body></html> | |
| 45379. |
When NaCl molecule is formed, one electron is transferred from Na atom to CI atom. The equilibrium internuclear distance between Na+ and Cl^(-)ions is 2.75 xx 10^(-10)m. The dipole moment of NaCl molecule is |
| Answer» <html><body><p>`2.75 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <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/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)` cm<br/>`4.4 xx 10^(-29)` cm<br/>`2.75 xx 10^(-29)` cm<br/>`4.4 xx 10^(-19)` cm </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> :Dipole <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> p=q(2a) <br/> `=1.6 xx 10^(19) xx 2.75 xx 10^(-10)` <br/> `=4.4 xx 10^(-29)` cm</body></html> | |
| 45380. |
Prove that the field on the surface of a sphere carrying a uniformly distributed electric charge is equal to that which would have been established, if the entire charge were concentrated in the centre of the sphere. |
| Answer» <html><body><p></p>Solution :Construct a second sphere <a href="https://interviewquestions.tuteehub.com/tag/around-5602275" style="font-weight:bold;" target="_blank" title="Click to know more about AROUND">AROUND</a> the sphere under consideration and suppose it carries a charge <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> in magnitude, but opposite in sign (Fig. 24.14). According to the result <a href="https://interviewquestions.tuteehub.com/tag/obtained-7273275" style="font-weight:bold;" target="_blank" title="Click to know more about OBTAINED">OBTAINED</a> in the <a href="https://interviewquestions.tuteehub.com/tag/previous-592857" style="font-weight:bold;" target="_blank" title="Click to know more about PREVIOUS">PREVIOUS</a> problem, the charge of the outer sphere does not create a field inside it. Therefore the field between the spheres is created only by the charge of the internal sphere. If there is <a href="https://interviewquestions.tuteehub.com/tag/little-1075899" style="font-weight:bold;" target="_blank" title="Click to know more about LITTLE">LITTLE</a> difference between the radii of these spheres (i.c. if `R_(1)-R lt lt R)` , the field in between will be almost homogeneous, and its strength will be (see 37.5) <br/> `E=(sigma)/(epsi_(0))=(Q)/(epsi_(0)S)=(Q)/(4pi epsi_(0)R^(2))`</body></html> | |
| 45381. |
A conductor ABCDE has the shape shown. It lies i the yz plane, with A and E on the y-axis . When it moves with a velocity v in a magnetic field B, an emf e is induced between A and E |
| Answer» <html><body><p>`e=0` if `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>` is in the `y`-direction and `<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>` is in the `x`-direction.<br/>`e=2Bav`, if `v` is in the `y`-direcrtion and `B` is in the `x`-direction<br/>`e=Blamdav`, if `v` is in the`<a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a>`- direction and `b` is in the `x`-direction.<br/>`e=Blamdav`,if `v` is in the `x` -direction and `B` is in the `z`- direction</p>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>::D</body></html> | |
| 45382. |
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case ? |
| Answer» <html><body><p></p>Solution :`f_(0) cm , f_(e)= 6.25` cm <br/> Distance between object lens and eyepiece = <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> cm<br/> `therefore (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))` <br/> - `(1)/(u_(e)) = (1)/(f_(e)) - (1)/(v_(e)) = (1)/(6.25) - (1)/((-25)) = (1)/(5) cm "" therefore u_(e) =- 5 ` cm <br/> Object is on the left of object lens. <br/> Distance of the image from the objective lens = 15 -5 = 10 cm<br/> i.e., `v_(0) = 10` cm <br/> `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) "" - (1)/(u_(0)) = (1)/(f_(0)) - (1)/(v_(0)) = (1)/(2) - (1)/(10) = (5 - 1)/(10) = (4)/(10)`<br/> `u_(0) = -2.5` cm <br/> Magnifying power = M = `(v_(0))/(-u_(0)) [ 1 + (D)/(f_(e)) ]= (10)/(+ 2.5) [ 1 + (25)/(6.25) ] ` = 20<br/> <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>. When final image is at <a href="https://interviewquestions.tuteehub.com/tag/infinity-515780" style="font-weight:bold;" target="_blank" title="Click to know more about INFINITY">INFINITY</a> , then object must be placed at the focus of the eyepiece.<br/> `u_(e)= - f_(e)= - 6.25 ` cm <br/> `v_(0) = <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> - |u_(e)| = 15 - 6.25 = 8.75 ` cm <br/> `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) "" - (1)/(u_(0)) = (1)/(f_(0))- (1)/(v_(0)) = (1)/(2) - (1)/(8.75) = (8.75 - 2)/(8.75 xx 2) = (6.75)/(15.50)`<br/> `u_(e) = - 2.6 ` cm <br/> M = `((-v_(0))/(v_(0))) . (D)/(f_(e)) = (8,75)/(2.6) xx (25)/(6.25) = 13.5`</body></html> | |
| 45383. |
A one metre steel wire of negligible mass and area of cross-section 0.01 cm^(-2) is kept on a smooth horizontal table with one end fixed.A ball of mass 1 kg is attached to the other end. The ball and the wire are rotating with an angular velocity of omega. If the elongation of the wire is 2 mm, then omega is (Young's modulus of steel = 2 xx 10^(11) Nm^(-2) ) |
| Answer» <html><body><p>5 rad `s^(-1)`<br/>10 rad `s^(-1)`<br/>15 rad `s^(-1)`<br/><a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> rad `s^(-1)`</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> :Given, elongation of the wire, `Delta`l= 2mm <br/> = `2 xx 10^(-3)` m<br/> Mass of the ball, m=1 kg <br/> Length of wire, l = 1 m <br/> area of cross - sectional of wire, A= 0.01`cm^(2) = 0.01 xx 10^(-4)` m<br/> Young.s modulus of steel, R = `2 xx 10^(11) Nm^(-2)` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> force in wire , T = m `<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>^(2)`l <br/> `therefore ` Stress = `("Tension")/("Area") = (m omega^(2)l)/(A )` <br/> Strain = `( Delta l)/(l ) = ("stress")/("Young.s modulus")` <br/> ` Delta l=(m omega^(2) l^(2))/(YA)` <br/>or`omega = sqrt((YA Delta)/(<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>^(2))` <br/> putting the given values, we get <br/> `=sqrt((2 x 10^(11) xx 0.01 xx 10^(-4) xx 2 xx 10^(-3))/(1 xx (1)^2))` <br/> `omega = 20 rad/"sec"^(-1)`</body></html> | |
| 45384. |
The magnetic field at the center of a current carrying loop of radius 0.1 m is 5sqrt( 5) times that at a point along its axis. The distance of this point from the centre of the loop is |
| Answer» <html><body><p>0.<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 <br/>0.1 m <br/>0.05 m <br/>0.25 m </p>Solution :We know that,<br/>`(B_("centre"))/(B_("axis"))=(1+(x^(2))/(r2))^(3//2)`<br/>Given that, `B_("centre")=5sqrt(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)B_("axis")`<br/>`(B_("centre"))/(B_("axis"))=5sqrt(5)`<br/>`therefore 5sqrt(5)=[1+(x^(2))/((0.1)^(2))]^(3//2)`<br/>On squaring both sides, we get<br/>`25xx5 = [1+(x^(2))/((0.1)^(2))]`<br/>`root(3)(<a href="https://interviewquestions.tuteehub.com/tag/125-271180" style="font-weight:bold;" target="_blank" title="Click to know more about 125">125</a>)=1+(x^(2))/((0.1)^(2))`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 0.01 + x^(2)=0.05`<br/>`rArr x^(2)=0.005 - 0.01`<br/>`rArr x^(2)=0.04`<br/>`rArr x = 0.2 m`</body></html> | |
| 45385. |
When a train is approaching the stationary observer, the apparent frequency of the whistle observed as 100 Hz, while when it has passed away from the observer with same speed, it is 50 Hz. Calculate the frequency of the whistle when the observer moves with the train (V = 330 m/s) |
| Answer» <html><body><p>33.3 <a href="https://interviewquestions.tuteehub.com/tag/hz-493442" style="font-weight:bold;" target="_blank" title="Click to know more about HZ">HZ</a><br/>50 Hz<br/>66.6 Hz<br/>75 Hz </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> | |
| 45386. |
A thin prism of crown glass (m_r=1.515, m_v=1.525)and a thin prism of flint glass (mu_r=1.612, mu_v=1.632) are placed in contact with each other. Their refracting angles are 5.0^@ each and are similarly directed. Calculate the angular dispersion produced by the combination. |
| 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> :A</body></html> | |
| 45387. |
A long round conductor of cross- sectional area S is made of material whose resistivity depends only on distance r from the axis of the conductor as rho = (alpha)/(r^2) , where alpha is a constant. Find (a) the resistance per unit length of such a conductor. (b) the electric field strength in the conductor due to which a current i flows through it . |
| Answer» <html><body><p></p>Solution :Consider a <a href="https://interviewquestions.tuteehub.com/tag/cylindrical-429099" style="font-weight:bold;" target="_blank" title="Click to know more about CYLINDRICAL">CYLINDRICAL</a> element of radii between r and (r+ <a href="https://interviewquestions.tuteehub.com/tag/dr-959219" style="font-weight:bold;" target="_blank" title="Click to know more about DR">DR</a>) . Its resistance <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XII_V02_A_C04_SLV_006_S01.png" width="80%"/> <br/> `dR = (rhol)/(2pi dr) ` (or)` <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/(dR) = (2pi r dr)/(rhol)""……(i)` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> 1/R = int_0^a 1/(dR) = int_0^a (2pi)/(rhol)r dr` <br/>(where a is the radius of the conductor ) <br/>`int_0^a (2pir dr)/(((alpha)/(r^2))l) = (2pi)/(alphal) int_0^a r^3 dr = (2pi)/(alpha l) ((a^4)/(4)) = ((pia^2)^2)/(2pialphal) = (S^2)/(2pi alpha l)` <br/>`R = (2pi alpha l)/(S^2) "".....(ii)` <br/>The resistance per unit length of wire `R= (2pi alpha)/(S^2)` <br/>(b) Equation (ii) can be written as `R= ((2pialpha)/(S)) ((l)/(S))` <br/><a href="https://interviewquestions.tuteehub.com/tag/compare-11929" style="font-weight:bold;" target="_blank" title="Click to know more about COMPARE">COMPARE</a> with `R = (rhol)/(S) , ` we get `rho = (2pi alpha)/(S)` <br/>By Ohm.s law `E= jrho = i/S xx (2pi alpha)/(S) = (2pi alpha i)/(S^2)`</body></html> | |
| 45388. |
Radius of the nucleus of the atom with A=216 is (R_0 =1.3fm) |
| Answer» <html><body><p>7.2 fm<br/>7.8 fm<br/>280 fm<br/>19 fm</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> | |
| 45389. |
(A): When a battery is short-circuited the terminal voltage is zero (R): In the situation of short circuit, current is zero |
| Answer» <html><body><p>Both 'A' and '<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>' are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and 'R' is the correct explanation of 'A'<br/>Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'<br/>'A' is true and 'R' is false <br/>'A' is false and 'R' is false </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> | |
| 45390. |
An oil droplet of 0.01 mm diameter floats in equilibrium between two horizontal plates the distance between which is 25 mm. What is the charge of the droplet, if the equilibrium corresponds to a voltage of 3.6 xx 10^(4)V across the plates? |
| Answer» <html><body><p></p>Solution :A force of gravity directed <a href="https://interviewquestions.tuteehub.com/tag/downwards-959173" style="font-weight:bold;" target="_blank" title="Click to know more about DOWNWARDS">DOWNWARDS</a> and equal to `mig=4//6pi D^(8)rhog`<a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</a> on the <a href="https://interviewquestions.tuteehub.com/tag/droplet-959857" style="font-weight:bold;" target="_blank" title="Click to know more about DROPLET">DROPLET</a>. It is countorboloncod by on blootrio force `F=qE=q varphi//d` (<a href="https://interviewquestions.tuteehub.com/tag/fig-460913" style="font-weight:bold;" target="_blank" title="Click to know more about FIG">FIG</a>. 24.4). The charge of the droplet is found from the balance of forces. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/ARG_AAP_PIP_PHY_C24_E01_004_S01.png" width="80%"/></body></html> | |
| 45391. |
The current in a coil changes from +10A to -2A in 3 millisecond. hat is the induced e.m.f. in the coil ? The self-inductance of the coil is 2 mH. |
| Answer» <html><body><p>8V<br/>4V<br/>0.8V<br/>0.4V</p>Answer :A</body></html> | |
| 45392. |
State de-Broglie hypothesis. |
| 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> :De-Broglie hypothesis <a href="https://interviewquestions.tuteehub.com/tag/states-1225920" style="font-weight:bold;" target="_blank" title="Click to know more about STATES">STATES</a> that atomic particles of matter moving with a <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a>, (or momentum) can display wave like properties.</body></html> | |
| 45393. |
The kinetic energy K of a particle moving along x axis varies with its position (x) as shown in the figure. The magnitude of force acting on particle at x = 9 m is - |
| Answer» <html><body><p>zero<br/>5 N<br/>20 N<br/>7.5 N</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> | |
| 45394. |
Calculate the ratio of electric and gravitational force between two protons. Charge of each proton is 1.6 xx 10^(-19)C,mass is 1.672 xx 10^(-27) kg and G = 6.67 xx 10^(-11) Nm^(-2)kg^(-2)? |
| 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> :`Q_1 = 1.6 xx 10^(-19) C, Q_2 = 1.6 xx 10^(-19) C` <br/> r is the distance between the two protons <br/> <a href="https://interviewquestions.tuteehub.com/tag/electrostatic-2069468" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTROSTATIC">ELECTROSTATIC</a> force between two protons <br/> `F_1 = 9 xx 10^(9) (Q_1Q_2)/(r^2) m_1 = m_2 = 1.672 xx 10^(-27) kg` <br/> `G = 6.67 xx 10^(-11) <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>^(2)//kg^(2)` <br/> Gravitational force between them <br/> `F_2 = G(m_1 m_2)/(r^2) :. (F_1)/(F_2) = (9 xx 10^9 Q_1Q_2)/(G.m_1 m_2)` <br/> `= (9 xx 10^(9) xx 1.6 xx 10^(-19) xx 1.6 xx 10^(-19))/(6.67 xx 10^(-11) xx 1.672 xx 10^(-27) xx 1.672 xx 10^(-27)) = 1.23 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a>)`</body></html> | |
| 45395. |
The magnetic field of earth can be modelled by that of a pointplaced at the centre of the earth the dipole axis makesan angle of 11.3^(@)with the axis ofearth at mumbaideclination is nearly zero then |
| Answer» <html><body><p>the declination variis between `11.3^(@)` w to `11.3^(@)` e <br/>the least declination is `0^(@)` <br/>the plane defoned by dipole axis andearth axis passes throgh <a href="https://interviewquestions.tuteehub.com/tag/green-476835" style="font-weight:bold;" target="_blank" title="Click to know more about GREEN">GREEN</a> wich <br/>declination <a href="https://interviewquestions.tuteehub.com/tag/averaged-7664694" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGED">AVERAGED</a> over <a href="https://interviewquestions.tuteehub.com/tag/eath-2601631" style="font-weight:bold;" target="_blank" title="Click to know more about EATH">EATH</a> <a href="https://interviewquestions.tuteehub.com/tag/must-2185568" style="font-weight:bold;" target="_blank" title="Click to know more about MUST">MUST</a> be always <a href="https://interviewquestions.tuteehub.com/tag/negative-570381" style="font-weight:bold;" target="_blank" title="Click to know more about NEGATIVE">NEGATIVE</a> </p>Answer :A</body></html> | |
| 45396. |
A monochromatic light source is placed at a distance d from a metal surface. Photo electrons are ejected at a rate n, kinetic energy is being E. If the source is brought nearer to distance d/2, the rate and kinetic energy of emitted photo electrons becomes nearly |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2n-300431" style="font-weight:bold;" target="_blank" title="Click to know more about 2N">2N</a> and 2E<br/>4n and 4E<br/>4n and E<br/>n and 4E</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> | |
| 45397. |
In Young's double slit experimentt. When a glass plate (mu = 1.5) of thickness is introduce path of one of the interfering beams (wavelength = lambda) the intensity at the position where centre maxima occurred previous remains unchanged. The minimum thickness |
| Answer» <html><body><p>`<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/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>`<br/>`lambda`<br/>`(2)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)lambda`<br/>`(lambda)/(3)`</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> | |
| 45398. |
The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3. For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. A personcan see objects clearly fromdistance 10cm to oo. Then, we can say that the person is |
| Answer» <html><body><p>normal sighted <a href="https://interviewquestions.tuteehub.com/tag/person-25481" style="font-weight:bold;" target="_blank" title="Click to know more about PERSON">PERSON</a> <br/>near-sighted person<br/>far-sighted person<br/>a person with <a href="https://interviewquestions.tuteehub.com/tag/exceptional-454608" style="font-weight:bold;" target="_blank" title="Click to know more about EXCEPTIONAL">EXCEPTIONAL</a> eye having no eye defect</p>Solution : Since he can see beyond <a href="https://interviewquestions.tuteehub.com/tag/th-665760" style="font-weight:bold;" target="_blank" title="Click to know more about TH">TH</a> <a href="https://interviewquestions.tuteehub.com/tag/erange-446421" style="font-weight:bold;" target="_blank" title="Click to know more about ERANGE">ERANGE</a> of a normal eye `[25cm,oo]`therefore it is an exceptional eye havingno defect.</body></html> | |
| 45399. |
Solve a similar problem for the case of a tungsten ball. The magnetic susceptibility of tungsten is chi_(m)=1.76 xx 10^(-4). |
| Answer» <html><body><p><br/></p>Answer :`1.5 xx 1.5 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) A.m^2` along the <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a>.</body></html> | |
| 45400. |
The wavelength of de-Broglie wave associated with a thermal neutron of mass m kg at 27^⋅C is (KB is the Boltzmann constant and all quantities are in SI units) |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>/(sqrt100mK_B) <a href="https://interviewquestions.tuteehub.com/tag/metre-1095305" style="font-weight:bold;" target="_blank" title="Click to know more about METRE">METRE</a>`<br/>`h/(sqrt900mK_B) metre`<br/>`h/(sqrt400mK_B) metre`<br/>`h/(sqrt300mK_B) metre`</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> | |