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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.
| 351. |
Assertion: Polaroids are used to polarise as well as analyse place polarised light. Reason: Polaroids reduce the intensity of light to zero. |
| 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/assertion-384238" style="font-weight:bold;" target="_blank" title="Click to know more about ASSERTION">ASSERTION</a> is <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> but reason is false. A polaroid allows those light <a href="https://interviewquestions.tuteehub.com/tag/vibrations-726480" style="font-weight:bold;" target="_blank" title="Click to know more about VIBRATIONS">VIBRATIONS</a> to pass through it which are parallel to its <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a>.</body></html> | |
| 352. |
Statement-I: Average of sinusoidal A.C. can never be zero for half cycle. Statement-II: Impedance given by inductance does not depends on frequency |
| Answer» <html><body><p>If both <a href="https://interviewquestions.tuteehub.com/tag/statement-16478" style="font-weight:bold;" target="_blank" title="Click to know more about STATEMENT">STATEMENT</a>- I and Statement- <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</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 Statement - II is the correct <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of Statement– I. <br/>If both Statement - I and Statement-II are true but Statement - II is not the correct explanation of Statement – I.<br/>If Statement-I is true but Statement-II 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/> If Statement-I is false but Statement-II is true.</p>Answer :D</body></html> | |
| 353. |
Which combination of resistance gives maximum current ? |
| 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/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> <a href="https://interviewquestions.tuteehub.com/tag/combination-922669" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINATION">COMBINATION</a>.</body></html> | |
| 354. |
The de Broglie wavelength associated with an electron accelerated by a potential of 64 V is |
| Answer» <html><body><p>a)1.227 <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</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>) 0.613 nm <br/><a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) 0.302 nm <br/>d) 0.153 nm </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> | |
| 355. |
When a dielectric medium is introduced between the plates has a capacitance 'C', If the oil is removed, then its capacitor becomes |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> constant<br/>increases<br/><a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a><br/>first decreases and then increases</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> | |
| 356. |
Find the wavlength of the K_(alpha) line in copper (Z=29) if the wavelength of the K_(alpha) line in iron (Z=26) is known to be equal to 193 p m. |
| 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> :From Moseley's <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> <br/> `omegak_(alpha)=(3)/(4)<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>(Ƶ-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)^(2)` <br/> or `lambdak_(alpha)=(4)/(3R)(1)/((Ƶ-1)^(2))` <br/> Thus `(lambdak_(alpha)(Cu))/(lambdak_(alpha)(Fe))=((25)/(<a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>))^(2)=((Ƶ_(Fe)-1)/(Ƶ_(cu)-1))^(2)` <br/> Substitution gives <br/> `lambdak_(alpha)(Cu)= 153.9p m`</body></html> | |
| 357. |
A wave travelling along the x-axis is described by the equation y(x,t) = 0.005cos(alphax - betat) . If the wavelength and the time period of the wave are 0.08m and 2.0s, respectively, then alpha/beta = n xx 5, then the value of n is ……………….. (where alpha, beta are in appropriate units). |
| 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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a></body></html> | |
| 358. |
The electric field for p-n junction is 1xx10^(6) V/m and depletion region is 5000 Å wide then the potential barrier = …… V. |
| Answer» <html><body><p>0.05<br/>0.005<br/>0.5<br/>5</p>Solution :0.5 <br/> `<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</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>)/(d)` <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> V=Ed=1xx10^(+<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)xx5000xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)` <br/> `therefore V=0.5V`</body></html> | |
| 359. |
A block of mass 0.1 kg is connected to an elastic spring of spring constant 640" Nm"^(-1) and oscillates in a medium of constant 10^(-2)" kg s"^(-1). The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to : |
| 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> s<br/>3.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> s <br/>5 s<br/>7 s</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> | |
| 360. |
Find out the value of current through 2Omega resistance for the given circuit. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/resistance-1186351" style="font-weight:bold;" target="_blank" title="Click to know more about RESISTANCE">RESISTANCE</a> of `2Omega` does not belong to any <a href="https://interviewquestions.tuteehub.com/tag/circuit-916593" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCUIT">CIRCUIT</a> where in a <a href="https://interviewquestions.tuteehub.com/tag/source-1219297" style="font-weight:bold;" target="_blank" title="Click to know more about SOURCE">SOURCE</a> of emf is <a href="https://interviewquestions.tuteehub.com/tag/connected-929389" style="font-weight:bold;" target="_blank" title="Click to know more about CONNECTED">CONNECTED</a>. Hence no current <a href="https://interviewquestions.tuteehub.com/tag/flows-993448" style="font-weight:bold;" target="_blank" title="Click to know more about FLOWS">FLOWS</a> through it.</body></html> | |
| 361. |
(a) Write two distinguishing features of nuclear forces. (b) Complete the following nuclear reactions for a and B decay: (i) " "_(92)^(238)Uto ? +" "_(2)^(4)He +Q (ii) " "_(11)^(22)Na to " "_(10)^(22)Ne +?+nu |
| Answer» <html><body><p></p>Solution :(a) Two distinguishing features of nuclear forces are as follows : <br/> 1. Nuclear forces are extremely short range forces. <br/> 2. Nuclear forces are charge independent. <br/> (b) The <a href="https://interviewquestions.tuteehub.com/tag/completed-409685" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETED">COMPLETED</a> nuclear reactions are as given below : <br/> (i) `" "_(92)^(238)Uoverset(-<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>)(to) " "_(90)^(234)Th +" "_(2)^(4)He +Q` (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) `" "_(11)^(22)Na overset(-<a href="https://interviewquestions.tuteehub.com/tag/beta-2557" style="font-weight:bold;" target="_blank" title="Click to know more about BETA">BETA</a>^(+))(to) " "_(10)^(22)Ne +" "_(+1)^(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)e+nu`</body></html> | |
| 362. |
The energy of a photon of light of wavelength 66 eV is |
| Answer» <html><body><p>`4.4 <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^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)`<br/>`2.5 xx 10^(-19) J`<br/>`1.25 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>) J`<br/>`2.5 xx 10^(-17) 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> | |
| 363. |
When you have learned to integrate derive the formula to calculate the work of expansion of a gas at constant temperature. |
| 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>></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> :The work <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a> by a gas expanding by a gas expanding at constat temperature is <br/> `W = int_(v_1)^(v_2) p <a href="https://interviewquestions.tuteehub.com/tag/dv-433533" style="font-weight:bold;" target="_blank" title="Click to know more about DV">DV</a> = m/M RT int_(v_1)^(v_2) (dV)/(V) = m/M RT ln (V_2)/(V_1) = p_1V_1 ln (V_2)/(V_1)`.</body></html> | |
| 364. |
e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece ? |
| Answer» <html><body><p></p>Solution :e) The <a href="https://interviewquestions.tuteehub.com/tag/image-11684" style="font-weight:bold;" target="_blank" title="Click to know more about IMAGE">IMAGE</a> of objective lens in eye pie is called 'eye ring' All the rays from the object <a href="https://interviewquestions.tuteehub.com/tag/refracted-7709131" style="font-weight:bold;" target="_blank" title="Click to know more about REFRACTED">REFRACTED</a> by the objective go through the eye ring. Therefore, ideal position for our eyes for <a href="https://interviewquestions.tuteehub.com/tag/viewing-3262137" style="font-weight:bold;" target="_blank" title="Click to know more about VIEWING">VIEWING</a> is this eye ring only. <br/> When eye is too close to the eye piece, field of view reduces and eyes do not collect much of the light. The precise. location of the eye ring <a href="https://interviewquestions.tuteehub.com/tag/would-3285927" style="font-weight:bold;" target="_blank" title="Click to know more about WOULD">WOULD</a> depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.</body></html> | |
| 365. |
Draw an OR gate using two diode and explain its operation. Write the truth table and logic symbol of OR gate. |
| Answer» <html><body><p></p>Solution :OR Gate `:` It has two input terminals and one output terminal.If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of thegate is high.The truth tables of OR gate.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C15_E03_006_S01.png" width="80%"/><br/> In truth table logicfunction is written as A or B 'OR' logic function is represented by the symbol 'plus' <br/> `Q = A+B ` <br/> Logic gate 'OR' is shown given below . <br/> Implementation of OR gate using diode `:` <br/>Let `D_(1)` and `D_(2)` be two <a href="https://interviewquestions.tuteehub.com/tag/diodes-954346" style="font-weight:bold;" target="_blank" title="Click to know more about DIODES">DIODES</a>. <br/> A potential of 5V represents the logical value 1 A potential of 0V <a href="https://interviewquestions.tuteehub.com/tag/representes-2987440" style="font-weight:bold;" target="_blank" title="Click to know more about REPRESENTES">REPRESENTES</a> the logicalvalue <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/> When A=,B=0 both the diodes are reverse <a href="https://interviewquestions.tuteehub.com/tag/biased-896564" style="font-weight:bold;" target="_blank" title="Click to know more about BIASED">BIASED</a> and there is no current through the resistance .So, the potential at Q is zero i.e., Q = 0 . When A=0 or B =0 and the other equal to a potential of 5Vthe diode whose anode is at a potential of 5V is forward-biasedand that diode behaves like a closed switch. The output potential then becomes5V i.e., Q =1.When both A and B are 1, both the diodes are forward -biased and the potential at Q is same as that at A and Bwhcih is 5V i.e., Q =1 . The output is same as that of the OR gate.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C15_E03_006_S02.png" width="80%"/> <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C15_E03_006_S03.png" width="80%"/></body></html> | |
| 366. |
Which is more efficient mode of transmission FM or AM? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/fm-457706" style="font-weight:bold;" target="_blank" title="Click to know more about FM">FM</a> <a href="https://interviewquestions.tuteehub.com/tag/transmission-1426202" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSMISSION">TRANSMISSION</a> is more efficient because all the transmitted power is <a href="https://interviewquestions.tuteehub.com/tag/useful-1441152" style="font-weight:bold;" target="_blank" title="Click to know more about USEFUL">USEFUL</a> but in AM transmission most of the power <a href="https://interviewquestions.tuteehub.com/tag/goes-2676180" style="font-weight:bold;" target="_blank" title="Click to know more about GOES">GOES</a> waste in transmitting the carrier <a href="https://interviewquestions.tuteehub.com/tag/alone-2412120" style="font-weight:bold;" target="_blank" title="Click to know more about ALONE">ALONE</a>.</body></html> | |
| 367. |
Is ohm's law true for all conductors ? |
| 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> :No. it is only true for metallic <a href="https://interviewquestions.tuteehub.com/tag/conductors-14781" style="font-weight:bold;" target="_blank" title="Click to know more about CONDUCTORS">CONDUCTORS</a> provided physical <a href="https://interviewquestions.tuteehub.com/tag/conditions-424384" style="font-weight:bold;" target="_blank" title="Click to know more about CONDITIONS">CONDITIONS</a> does not <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a>.</body></html> | |
| 368. |
The relation between mu_r andX_m is mu_r=1+X_m How will you arrive at this relation ? Explain. |
| Answer» <html><body><p></p>Solution :We have total <a href="https://interviewquestions.tuteehub.com/tag/magnetic-551115" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNETIC">MAGNETIC</a> <a href="https://interviewquestions.tuteehub.com/tag/induction-1042578" style="font-weight:bold;" target="_blank" title="Click to know more about INDUCTION">INDUCTION</a> `B=B_0+B_m` <br/> where `B_0` - magnetic induction in <a href="https://interviewquestions.tuteehub.com/tag/free-465311" style="font-weight:bold;" target="_blank" title="Click to know more about FREE">FREE</a> <a href="https://interviewquestions.tuteehub.com/tag/space-649515" style="font-weight:bold;" target="_blank" title="Click to know more about SPACE">SPACE</a>. <br/> `B_0=mu_0H,B_m=mu_0I and B=muH "":. muH=mu_0H+mu_0I` <br/> But `<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a> =mu_rmu_0 "":.mu_rmu_0H=mu_0H+mu_0I` <br/> `mu_rH=H+I=(H+I)/H=1+I/H=1+X_m:.mu_r=1+X_m`</body></html> | |
| 369. |
A constant horizontal force of 20N acts on a body on a smooth horizontal plane. The body starts from rest and is observed to move 20 m in two seconds. The mass of the body is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> kg <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> kg <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> kg <br/>0.5 kg </p>Answer :B</body></html> | |
| 370. |
Which of the following are polynomials- |
| 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>(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)=X^2-3X+2`<br/>`f(x)=X^2-3X+2)X`<br/>`f(x)=X^2-3X+2sqrtx`<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</body></html> | |
| 371. |
In a magnetic field vecB=hati+yhatj+3hatk a charge particle (q, m) is moving with velocity vecV=2hati+3hatj+zhatk experiences a force vecF=-hati+2hatj+hatk. The value of y and z may be |
| Answer» <html><body><p>y=4, z=2<br/>y=2, z=4<br/>y=3, z=6<br/>y=6, z=3</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> | |
| 372. |
Arrange the following electromagnetic radiations per quantum in the order of increasing energy :A : Blue lightB : Yellow lightC : X - rayD : Radiowave |
| Answer» <html><body><p>D, B, A, C<br/>A, B, D, C<br/>C, A, B, D<br/>B, A, D, C</p>Solution :Quantam energy = hf<br/>`= (hc)/(<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>)` (hc = <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a>)<br/>`prop (1)/(lambda)`<br/>Arranging in decreasing order of wavelength Radiowaves `gt` Yellow light `gt` Blue light `gt` X - rays<br/>Hence quantam energy<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>`Arrange quantum energy in increasing order.<br/>Radiowaves `gt` Yellow light `gt` Blue light `gt` X - rays<br/>`D gt B gt A gt C`.</body></html> | |
| 373. |
The current in the primary circuit of a potentiometer wire is 0.5 A. Specific resistance of wire is 4 xx 10^(-7) Omega m and area of cross section of wire is 8 xx 10^(-6) m^(2). The potential gradient on the wire would be |
| Answer» <html><body><p>2.5 m V/m<br/>25 mV/m<br/>25 V/m<br/>10 V/m </p>Solution :25 mV/m <br/> <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> gradient `sigma= (I rho)/(A) = (0.5 xx <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>))/(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> xx 10^(-6))` <br/>`therefore sigma = 0.25 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = 25 xx 10^(-3)` <br/> `therefore sigma = 25 ` m V/m</body></html> | |
| 374. |
When the electrical conductivity of a semi-conductor is only due to the breaking of its covalent bonds, then the semiconductor is said to be |
| Answer» <html><body><p>Donor<br/>Acceptor<br/>Intrinsic<br/>Extrinsic</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> | |
| 375. |
The following fusion reaction take place 2_1^2Ararr_2^3B+n+3.27MeV. If 2 kg of ._1^2Ais subjected to the above reaction,the energy released is used to light a100 Wlight a lamp, how long will the lamp glow ? |
| Answer» <html><body><p>`7xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` years <br/>`3xx10^(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)` years <br/>`5xx10^(4)` years <br/>`2xx10^(6)` years </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> | |
| 376. |
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density lambda without using Gauss's law. |
| Answer» <html><body><p></p>Solution :Let us consider a long thin wire of linear charge density `lambda`. We have to find the <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> electric field due to this wire at point P. <br/> Now, consider a very small element of length dx at a distance x from C. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C04_E05_048_S01.png" width="80%"/> <br/> The charge on this <a href="https://interviewquestions.tuteehub.com/tag/elementary-450205" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENTARY">ELEMENTARY</a> portion of length dx <br/> `q = lambda dx` ...(1) <br/> Electric field intensity at point P due to the elementary portion <br/> `<a href="https://interviewquestions.tuteehub.com/tag/de-945063" style="font-weight:bold;" target="_blank" title="Click to know more about DE">DE</a>=1/(4pi epsi_(0)). q/((OP)^(2))=1/(4pi epsi_(0)) (lambda dx)/((OP)^(2))` [` :'` from (1)] <br/> Now, in `Delta PCO""(PO)^(2)=(<a href="https://interviewquestions.tuteehub.com/tag/pc-590329" style="font-weight:bold;" target="_blank" title="Click to know more about PC">PC</a>)^(2)+(CO)^(2)` <br/> `(OP)^(2)=r^(2)+x^(2)` <br/> `dE=1/(4pi epsi_(0)) (lambda dx)/((x^(2)+r^(2)))` ...(2) <br/> The components of dE are `dE cos theta` along PD and `dE sin theta` along PF. <br/> Here, there are so many elementary portion. So all the `dE sin theta` components balance each other. the resultant electric field at P is due to only `dE cos theta` components. <br/> The resultant electric field due to elementary component, `dE'=dE cos theta` <br/> `dE'=1/(4pi epsi_(0)). (lambda d x)/((x^(2)+r^(2))) cos theta` ...(3) <br/> In `Delta OCP tan theta=x/r implies x =r tan theta` <br/> Differentiation with respect to `theta`, we get <br/> `dx= r sec^(2) theta d theta` <br/> Putting in equation (3), we get <br/> `dE'=1/(4pi epsi_(0)) (lambda.r sec^(2) theta d theta cos theta)/((r^(2)+r^(2) tan^(2) theta))` <br/> `dE'=1/(4pi epsi_(0)) (lambda.r sec^(2) theta d theta cos theta)/(r^(2) sec^(2) theta)` <br/> `dE'=1/(4pi epsi_(0)). lambda/r cos theta d theta` <br/> As the wire os of infinite length, so integrate <a href="https://interviewquestions.tuteehub.com/tag/within-732414" style="font-weight:bold;" target="_blank" title="Click to know more about WITHIN">WITHIN</a> the limits `-pi/2` to `pi/2`, we get <br/> `E'=int dE'=1/(4pi epsi_(0))lambda/r underset(-pi//2)overset(pi//2)(int)cos theta d theta` <br/> `E'=1/(4pi epsi_(0)).lambda/r[sin theta]_(-pi//2)^(pi//2)=1/(4pi epsi_(0)) lambda/r ["sin"pi/20sin(-pi/2)]` <br/> `E'=1/(4pi epsi_(0)) lambda/r [1+1]` <br/> `E'=(2 lambda)/(4pi epsi_(0) r)` <br/> `:. E'=lambda/(2pi epsi_(0) r)`</body></html> | |
| 377. |
Show thatn=(sin((A+D)/(2)))/(sin((A)/(2))) where symbols have their usual notations. |
| Answer» <html><body><p></p>Solution :BC - Base of the prism <br/> AB and AC - Refracting sides representing the planes <br/> ABC - Principle section of prism. <br/> PQ - incident ray. <br/> RS - emergent ray <br/> `d_(1)` - angle of deviation `d_(1)=i_(1)-r_(1)` <br/> `d_(2) ` - angle of deviation`d_(2)=i_(2)-r_(2)` <br/> `u hatTS `- d - total angle of deviation or net deviation <br/> i.e., `d=d_(1)+d_(2)` (exterior angle = sum of two opposite interior angles)<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_ABT_PHY_QB_XII_C09_E03_010_S01.png" width="80%"/> <br/> NM is normal to AB at <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> <br/> N.M is normal to AC at R <br/> `i_(1)`- angle of incidence <br/> `i_(2)` - angle of emergence.<br/> D = angle of minimum deviation, <br/> From figure (1) <br/> From the quadrilateral AQMR, <br/> `A hatQM~=A hatRM=90^(@)` (Nm, N.M are normal to AB and BC) <br/> So that `A+M=180^(@)""` ......(1) <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> `AQMR is a cyclic quadrilateral. From the triangle QMR <br/> `r_(1)+r_(2)+hatM=180^(@)("Property of a "Delta) "" `......(2) <br/> From (1) and (2) <br/> `A+M=r_(1)+r_(2)+M` <br/> i.e.,`A=r_(1)+r_(2)""`.......(3) <br/> For non-symmetric condition `i_(1) ne i` <br/> `d_(1) ne d_(2)` also net deviationd. <br/> `d=d_(1)+d_(2)`<br/> `d=i_(1)-r_(1)+i_(2)-r_(2)`<br/> `therefored=i_(1)+I_(2)-(r_(1)+r_(2))` <br/> using (3) we write, <br/> `d=i_(1)+i_(2)-A""`.......(4)<br/>from the fig., we note that for non-symmetric condition for two angles of incidence, angle of net deviation is the same. However for a symmetric condition, `i_(1)=i_(2)=i and r_(1)=r_(2)=r` net deviation becomes the minimum angle of deviation. In this case the refracted ray will be parallel to the base. <br/> `therefore` for symmetric condition, <br/> (4)can be written as `D=2i-A` <br/> i.e., `i=((A+D)/(2))""` ......(5) <br/> (3) can be written as, <br/> `A=2r` <br/> or`r=((A)/(2))"" ` ......(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) <br/> From Snell.s law of refraction at AB,</body></html> | |
| 378. |
A battery of emf 10 V internal resistance 3 Omega is connected to a resistor R. (i) If the current in the circuit is 0.5 A. calculate the value of R. (ii) What is the terminal voltage of the battery when the cirucit is closed. |
| 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> :`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>)/(<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>+r)or` <br/> `R=(E)/(I)-r-(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)/(0.5)-3=17Omega` <br/> `V=IR=0.5xx17=8.5V`.</body></html> | |
| 379. |
A radio active isotope has a half-life of T years. How long will it take the activity to reduce to 3.125% |
| 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) <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> T <a href="https://interviewquestions.tuteehub.com/tag/years-1464846" style="font-weight:bold;" target="_blank" title="Click to know more about YEARS">YEARS</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>) 6.65 T years</body></html> | |
| 380. |
The quantities of heat required to raise the temperatures of two copper spheres of radii r_(1) and r_(2) (_(1)=1.5 _(2)) through 1 Kare in the ratio of |
| Answer» <html><body><p>`(27)/(8)`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)/(4)`<br/>`(3)/(<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</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> heat, `Q prop ^(2)` <br/> where <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> is the radius. <br/> `therefore (Q_(1))/(Q_(2))=(r_(1)^(2))/(r_(2)^(2))` <br/> `(Q_(1))/(Q_(2))=((1.5)^(2))/((1)^(2))` <br/> `(Q_(1))/(Q_(2))=(9)/(4)`. <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> choice is (b).</body></html> | |
| 381. |
A vessel of capacity V_0 contains n molecules. Calculate the probability of all the molecules assembling in a part of the vessel V lt V_0. |
| Answer» <html><body><p><br/></p>Solution :First method. Since both parts of the vessel are equivalent (this is the consequence of the homogeneity and the isotropy of space) the probability of finding a particle <a href="https://interviewquestions.tuteehub.com/tag/inside-1045864" style="font-weight:bold;" target="_blank" title="Click to know more about INSIDE">INSIDE</a> a <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> `V lt V_0` is equal to the ratio of volumes: <br/> `p = V/(V_0) , q = 1 - p = (V_0 - V)/(V_0)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/ARG_AAP_PIP_PHY_C18_E01_007_S01.png" width="80%"/> <br/> In the present problem k = n, i.e. all the molecules are contained inside V. The answer is obtained from the solution of Problem 18.5. For a large number of molecules the probability of such an event is negligible. <br/> Second method. The problem may be solved without the binomial distribution, bul directly on the basis of the <a href="https://interviewquestions.tuteehub.com/tag/theorem-706777" style="font-weight:bold;" target="_blank" title="Click to know more about THEOREM">THEOREM</a> for compound probability. Spocificully, the probability of finding a molecule inside the given volume is `p = V//V_0)`. The probability of finding simultaneously all the n molecules in it is the <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a> of the <a href="https://interviewquestions.tuteehub.com/tag/individual-1041871" style="font-weight:bold;" target="_blank" title="Click to know more about INDIVIDUAL">INDIVIDUAL</a> probabilities : `w = p^n = (V//V_0)^n`</body></html> | |
| 382. |
The magnetic flux linked with a coil varies with time as phi=5t^2-4t+16 weber The induced emf in the circuit of t=0.2s is |
| Answer» <html><body><p>0.4V<br/>`0.4V`<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a>`<br/>2V</p>Answer :D</body></html> | |
| 383. |
Draw circuit diagram for a full wave rectifier and show input output voltages. |
| Answer» <html><body><p></p>Solution :Rectification. Rectification is the process of converting alternating voltage/current into direct voltage/current. A device used for this purpose is called rectifier and this phenomenon is called rectification. <br/> Principle. It is based on the principle that a p-n junction diode conducts when it is forward biased and does not <a href="https://interviewquestions.tuteehub.com/tag/conduct-928781" style="font-weight:bold;" target="_blank" title="Click to know more about CONDUCT">CONDUCT</a> when it is reverse biased. <br/> Construction. The apparatus for rectification consists of <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> diodes `D_(1)` and `D_(2)` connected to two ends of the secondary of a step down transformer. Output is <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> out from mid point of the secondary and common point N of two diodes. Ouput is taken out from ends of the load R. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MBD_EM_PHY_XII_C09_E02_022_S01.png" width="80%"/> <br/> Working. During the positive <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a> of input `AC,D_(1)` is forward biased and `D_(2)` reverse biased. Hence the current <a href="https://interviewquestions.tuteehub.com/tag/flows-993448" style="font-weight:bold;" target="_blank" title="Click to know more about FLOWS">FLOWS</a> through the upper circuit as shown. During negative half, lower portion `(D_(2))` is forward biased and upper `(D_(1))` is reverse biased. <br/> Thus during each half, we get the current either from `D_(1)` or from `D_(2)`. <br/> The output voltage is unidirectional having ripple contents. <br/> Ripple factor of a rectifier `=("r.m.s of a.c.component")/("value of d.c. component")` <br/> To smoothen the output electric filters are used. The electric filters are combination of inductors are capacitors. Some of the useful filters are L-filter and `pi`-filter.</body></html> | |
| 384. |
A shell fired a cannon with speed v m/s at angle theta with horizontal explodes into three pieces of equal masses at the highest point of trajectory. One piece falls down vertically while the other retraces its path. What is speed of the third piece ? |
| Answer» <html><body><p></p>Solution :If .m. is mass of <a href="https://interviewquestions.tuteehub.com/tag/shell-11505" style="font-weight:bold;" target="_blank" title="Click to know more about SHELL">SHELL</a>, at highest point, <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> momentum<br/>`= m(v <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> theta)`, along horizontal<br/> given, momentum of one fragment `=(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>)(-v cos theta)`<br/>momentum of second along horizontal = 0<br/>If `.v_(1).` is speed of third fragment, from momentum conservation<br/>`-(m)/(3)v cos theta +(m)/(3)v_(1)+0=m(v cos theta)`,<br/>`v_(1)=4v cos theta`</body></html> | |
| 385. |
What is the effect on the current measuring range of a galvanometer when it is shunted? |
| 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/increased-1040402" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASED">INCREASED</a>.</body></html> | |
| 386. |
If vecP and vecQ denote the sides of parallelogram and its area is (1)/(2)PQ, then the angle between vecP and vecQ is |
| Answer» <html><body><p>`0^(@)` <br/>`30^(@)` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a>^(@)` <br/>`60^(@)` </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>)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) PQ =|vecPxxvecQ| = PQ <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> theta` <br/> or sin `theta=(1)/(2) or theta =30^(@)`</body></html> | |
| 387. |
A dynamo converts |
| Answer» <html><body><p>a. <a href="https://interviewquestions.tuteehub.com/tag/mechanical-2780" style="font-weight:bold;" target="_blank" title="Click to know more about MECHANICAL">MECHANICAL</a> <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> into thermal energy <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>. <a href="https://interviewquestions.tuteehub.com/tag/electrical-968148" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRICAL">ELECTRICAL</a> energy into thermal energy<br/>c. thermal energy into electrical energy <br/>d. mechanical energy into electrical energy</p>Answer :D</body></html> | |
| 388. |
A proton and a deutron are acclerated through the same accelerating potential. Which one of the two has (a) Greater value of de-Broglie wavelength associated with it, and (b) Less momentum ? Given reason to justify your answer. |
| Answer» <html><body><p></p>Solution :For an <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> potential V the de-Broglie <a href="https://interviewquestions.tuteehub.com/tag/wavelength-1450414" style="font-weight:bold;" target="_blank" title="Click to know more about WAVELENGTH">WAVELENGTH</a> of a charged particle of <a href="https://interviewquestions.tuteehub.com/tag/charge-914384" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGE">CHARGE</a> q and mass m is given by <br/> `lamda=(h)/(sqrt((2mqV)))` and momentum of particle `p=sqrt(2mqV)`. <br/> As mass of <a href="https://interviewquestions.tuteehub.com/tag/deutron-950001" style="font-weight:bold;" target="_blank" title="Click to know more about DEUTRON">DEUTRON</a> is greater than that of proton i.e., `m_(D) gt m_(P)`, hence we conclude that (a) de-Broglie wavelength of proton is greater than that of deutron, and (b) momentum of proton is less than that of deutron.</body></html> | |
| 389. |
The electrical resistance of pure platinum increases linearly with increasingtemperature over a small range of temperature . This property is used in a Platinum resistance thermometer. The relation between R_(theta) (Resistance at theta K) and R_(0) (Resistance at theta_(0)K) is given by R_(theta)=R_(0)[1+a(theta-theta_(0))], where alpha= temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 0^(@)C when its resistance is 80Omega and 100^(@)C when its resistance is 90 Omega find the temperature at which its resistance is 86Omega. |
| Answer» <html><body><p></p>Solution :Using the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> relationship <br/> we have `(90-80)<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>=alpha(80 Omega)(100)`………..(i) <br/> `(86-80)Omega=alpha(80 Omega)(theta)`………..(ii) <br/> where `theta` is the desired <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a>. <br/> <a href="https://interviewquestions.tuteehub.com/tag/taking-1238585" style="font-weight:bold;" target="_blank" title="Click to know more about TAKING">TAKING</a> the <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of (i) and (ii) <br/> `10/6=100/(theta)impliestheta=60K`</body></html> | |
| 390. |
The maximum kinetic energy of emitted photoelectrons from a surface depends on the ____ of incident radiation as well as the _____of the given surface. |
| 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/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a>, <a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> <a href="https://interviewquestions.tuteehub.com/tag/function-11303" style="font-weight:bold;" target="_blank" title="Click to know more about FUNCTION">FUNCTION</a> (or <a href="https://interviewquestions.tuteehub.com/tag/nature-1112013" style="font-weight:bold;" target="_blank" title="Click to know more about NATURE">NATURE</a>).</body></html> | |
| 391. |
Find the potential difference between and N in the given branch of a circuit figure. |
| Answer» <html><body><p></p>Solution :Let us have a closed <a href="https://interviewquestions.tuteehub.com/tag/circuit-916593" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCUIT">CIRCUIT</a> by joining to <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> with dotted lines. Let the potentials at M and N be `V_(M)` and `V_(N)` respectively. Following counter clockwise direction and applying Kirchhoff.s <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a>, we get <br/> `(V_(N)-V_(M))-4+(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.4xx1)+(0.4xx2)-6(0.4xx1)+(0.4xx6)=0` <br/> `(V_(N)-V_(M))-10+0.4+0.8+0.4+2.4=0` <br/> `V_(N)-V_(M)=6V`</body></html> | |
| 392. |
For any particular medium the refractive index is greater if the wavelength of light is _________ [Fill in blank]. |
| 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/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a></body></html> | |
| 393. |
A message signal of frequency 10 kHz and peak voltage of 10 volts is used to modulate a carrier of frequency 1 mHz and peak voltage of 20 volts Determine (a) modulation index, (b) the side bands produced. |
| Answer» <html><body><p></p>Solution :(a) modulation <a href="https://interviewquestions.tuteehub.com/tag/index-505893" style="font-weight:bold;" target="_blank" title="Click to know more about INDEX">INDEX</a> = `10//20=0.5` <br/> (b) The side <a href="https://interviewquestions.tuteehub.com/tag/bands-892049" style="font-weight:bold;" target="_blank" title="Click to know more about BANDS">BANDS</a> are at (1000+10 <a href="https://interviewquestions.tuteehub.com/tag/khz-1063029" style="font-weight:bold;" target="_blank" title="Click to know more about KHZ">KHZ</a>) <br/> = 1010 kHz and (1000-10 kHz) <br/> = <a href="https://interviewquestions.tuteehub.com/tag/990-1937263" style="font-weight:bold;" target="_blank" title="Click to know more about 990">990</a> kHz.</body></html> | |
| 394. |
Two beams P and Q of light of the same wavelength fall upon the same metal surface causing photoemission of electrons. The photoelectric current produced by P is four times that produced by Q. Which of the following gives the ratio ("wave amplitude of beam P")/("wave amplitude of beam Q")? |
| 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>>`(1)/(<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/>`(1)/(2)`<br/>2<br/>4</p>Solution :`("<a href="https://interviewquestions.tuteehub.com/tag/wave-22399" style="font-weight:bold;" target="_blank" title="Click to know more about WAVE">WAVE</a> amplitude of beamP")/("Wave amplitude of beam Q")=(("<a href="https://interviewquestions.tuteehub.com/tag/current-940804" style="font-weight:bold;" target="_blank" title="Click to know more about CURRENT">CURRENT</a> of P")^(1//2))/(("Current of Q")^(1//2))=((4)/(1))^((1//2))=2`</body></html> | |
| 395. |
Establish a relation for electric potential due to short dipole at a point distance r from the dipole along a line inclined at an angle theta from the dipole axis . Hence obtain value of electric potential at a point lying along (i) axial line , (ii) equatoral line of the dipole . |
| 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>></p>Solution :Consider a short electric dipole <a href="https://interviewquestions.tuteehub.com/tag/ab-360636" style="font-weight:bold;" target="_blank" title="Click to know more about AB">AB</a> consisting of two charges of `-<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>` and `+q` separated by a small distance AB = 2a . The dipole has a dipole moment `vecp` whose magnitude is q (2a) and which points along AB . <br/> Let P be a point at a distance r from mid-point O of electric dipole and line OP subtends an angle `theta` from the dipole axis . Then electric potential at point P is given by <br/> `V = V_(A) + V_(B) = (q)/(4 pi in_(0) (AP)) + (q)/(4 pi in_(0) (BP))` <br/> Draw normals BC and AD from points B and A respectively on extended line PO. As dipole is a short dipole , hence as shown in fig. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/U_LIK_SP_PHY_XII_C02_E10_015_S01.png" width="80%"/> <br/> `BP = CP = OP - OC = r - a cos theta` and `AP = <a href="https://interviewquestions.tuteehub.com/tag/dp-959208" style="font-weight:bold;" target="_blank" title="Click to know more about DP">DP</a> = OP + OD = r + a cos theta` <br/> `therefore V = (q)/(4 pi in_(0) (r + a cos theta)) + (q)/(4 pi in_(0) (r - a cos theta)) = (q)/(4pi in_(0)) [ (1)/((r - a cos theta)) - (1)/((r + a cos theta))]` <br/> `= (q)/(4pi in_(0)) (2 a cos theta)/((r^(2) - a^(2) cos^(2) theta)) = (p cos theta)/(4 pi in_(0) (r^(2) - a^(2) cos^(2) theta)) = (vecp * vecr)/(4 pi in_(0) (r^(2) - a^(2) cos^(2) theta))` <br/> As `a lt lt r` hence the <a href="https://interviewquestions.tuteehub.com/tag/term-1241851" style="font-weight:bold;" target="_blank" title="Click to know more about TERM">TERM</a> `a^(2) cos^(2) theta` may be neglected as compared to `r^(2)` and so we have `V= (p cos theta)/(4 pi in_(0) * r^(2)) = (vecp * vecr)/(4pi in_(0) r^(2))` <br/> Important cases to be remembered are <br/> (i) If point P lies along axial line of dipole towards `+q` charge , then `theta = 0^(@)` and hence `V= (p)/(4 pi in_(0) * r^(2))` <br/> (ii) If point P lies along axial line of dipole towards `-q` charge , then `theta = 180^(@)` and hence `V = (p)/(4 pi in_(0) r^(2))` <br/> (iii) If point P lies along equatoral line of dipole then `theta = 90^(@)` and hence V = 0</body></html> | |
| 396. |
The S.I. unit of latent heat is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/jkg-2139511" style="font-weight:bold;" target="_blank" title="Click to know more about JKG">JKG</a>^(-1)`<br/>`Jmol^(-1)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/nkg-3746303" style="font-weight:bold;" target="_blank" title="Click to know more about NKG">NKG</a>^(-1)`<br/>`"<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/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>"^(-1)`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>=(Q)/(m)=(J)/(kg)=Jkg^(-1)`</body></html> | |
| 397. |
How we define wave optics ? |
| Answer» <html><body><p></p>Solution :The branch of <a href="https://interviewquestions.tuteehub.com/tag/optics-22452" style="font-weight:bold;" target="_blank" title="Click to know more about OPTICS">OPTICS</a> concerned with the nature of light and primarily the <a href="https://interviewquestions.tuteehub.com/tag/theory-706806" style="font-weight:bold;" target="_blank" title="Click to know more about THEORY">THEORY</a> of <a href="https://interviewquestions.tuteehub.com/tag/waves-13979" style="font-weight:bold;" target="_blank" title="Click to know more about WAVES">WAVES</a> is called <a href="https://interviewquestions.tuteehub.com/tag/wave-22399" style="font-weight:bold;" target="_blank" title="Click to know more about WAVE">WAVE</a> optics.</body></html> | |
| 398. |
A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength lamda equals to 6000 Å and the angular width of the central maxima in the resulting diffraction pattern is measured. When the slit is next illuminated by light of wavelength lamda, the angular width decreases by 30%. calculate the value of the wavelength lamda. |
| 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> :4200Å</body></html> | |
| 399. |
When a coil is connected to a D.C source of emf 12V, a current of 4 amp flows in it. If same coil is connected to 12V, 50Hz AC source, the current is 2.4 A. The self inductance of the coil is |
| Answer» <html><body><p>`(1)/(20pi)`<br/>`(1)/(10pi)`<br/>`(1)/(25pi)`<br/>`(1)/(<a href="https://interviewquestions.tuteehub.com/tag/5pi-1900435" style="font-weight:bold;" target="_blank" title="Click to know more about 5PI">5PI</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> :<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> | |
| 400. |
Which radiation in sunlight, causes heating effect |
| Answer» <html><body><p>Ultraviolet<br/><a href="https://interviewquestions.tuteehub.com/tag/infrared-515961" style="font-weight:bold;" target="_blank" title="Click to know more about INFRARED">INFRARED</a><br/>Visible light<br/>All of these.</p>Solution :Infrared <a href="https://interviewquestions.tuteehub.com/tag/causes-910982" style="font-weight:bold;" target="_blank" title="Click to know more about CAUSES">CAUSES</a> <a href="https://interviewquestions.tuteehub.com/tag/heating-1017297" style="font-weight:bold;" target="_blank" title="Click to know more about HEATING">HEATING</a> <a href="https://interviewquestions.tuteehub.com/tag/effect-966056" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECT">EFFECT</a>.</body></html> | |