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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.
| 45001. |
Name c.g.s and S.I. units of charges. |
| 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> :The c.g.s units of charge is 1 <a href="https://interviewquestions.tuteehub.com/tag/stat-632937" style="font-weight:bold;" target="_blank" title="Click to know more about STAT">STAT</a> <a href="https://interviewquestions.tuteehub.com/tag/coulomb-426789" style="font-weight:bold;" target="_blank" title="Click to know more about COULOMB">COULOMB</a> or 1 e.s.u. of charge. <a href="https://interviewquestions.tuteehub.com/tag/si-630826" style="font-weight:bold;" target="_blank" title="Click to know more about SI">SI</a> <a href="https://interviewquestions.tuteehub.com/tag/unit-1438166" style="font-weight:bold;" target="_blank" title="Click to know more about UNIT">UNIT</a> of charge is 1 coulomb <br/> 1 Coulomb = `3xx10^9` stat coulomb</body></html> | |
| 45002. |
If wavelengths of light in liquid A and B are 3500 Å and 7000 Å respectively, then critical angle of liquid A with respect to liquid B is ..... |
| Answer» <html><body><p>`15^@`<br/>`30^@`<br/>`45^@`<br/>`60^@`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/refractive-2247079" style="font-weight:bold;" target="_blank" title="Click to know more about REFRACTIVE">REFRACTIVE</a> index of A w.r.t. B, `(n_B)/(n_A)=n`<br/> but`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/n=(v_A)/(v_B)=(lambda_Af)/(lambda_Bf)`[`because` <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> is constant]<br/>`therefore1/n=(lambda_A)/(lambda_B)=(<a href="https://interviewquestions.tuteehub.com/tag/3500-1858022" style="font-weight:bold;" target="_blank" title="Click to know more about 3500">3500</a>)/(7000)`<br/> `1/n=1/2`<br/> sinC=`1/2`<br/>`therefore C=sin^(-1)(1/2)`<br/>`thereforeC=30^@`</body></html> | |
| 45003. |
What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? |
| Answer» <html><body><p></p>Solution :We knowthat radius of nith orbit in a hydrogen atom 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/> `r_(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>) = r_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)n^(2)`, where `r_(0)` = radiusof orbitin stablein stablestatecorrespondington = 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> `radiusof hydrogenorbitin firstexcitedstate`(n = 2) r_(2) = r_(0) (2)^(2) = 4r_(0)` <br/>`rArr "" (r_(2))/(r_(1)) =(r_(2))/(r_(0))= (4)/(1)`</body></html> | |
| 45004. |
The variation of the intensity of magnetisation with respect to the magnetisign field Hin a diamagnetic substance is described by the graph |
| Answer» <html><body><p>od <br/><a href="https://interviewquestions.tuteehub.com/tag/oc-582976" style="font-weight:bold;" target="_blank" title="Click to know more about OC">OC</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/ob-582890" style="font-weight:bold;" target="_blank" title="Click to know more about OB">OB</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/oa-582846" style="font-weight:bold;" target="_blank" title="Click to know more about OA">OA</a> </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> | |
| 45005. |
A missile is launched at an angle of 60^(@) to the vertical with a velocity sqrt( 0.75gR) from the surface of the earth ( R is the radius of the earth). Find the maximum height from the surface of earth. (Neglect air resistance and rotating of earth). |
| 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> :0.25R</body></html> | |
| 45006. |
Conservatios of charge is the propery by virtue of whichtotal charge of an isolatedsystem always remainsconstant or conserved. For example, whenwe rub two inslatingbodies, A and B, suchthat n electrons fromA transfer to B, then chargeacquiredby A = + ne and chargeacquired byB = -ne, where e is magnitudeof chargeon electron. The net chargeon isolated systemof bodiesA and B =n e -ne - 0, which was the chargebefore rubbing. Thus, itis not possibleto create or destroynet charge carriedby any isolated system. It also imples that chargescan be created or destroyedin equaland unlike pairs only. (i) What is the basic causeof conservationof charge ?(ii) Name any other fundamentalpropertyof electric charge. (iii) At a time, can you createtwo likecharges of magnitudeq =n e each ? (iv) How is the property of conservation of charge reflected in day to day life ? |
| 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) As is known, <a href="https://interviewquestions.tuteehub.com/tag/charging-914495" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGING">CHARGING</a> is dueto actualtransfer of electrons from one day to the other. Assuming that there is no stay lossloss of electrons number of electrons gainedby one body is equalto number of electrons lost by the other body. This is the basiccause of <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of charge. <br/> (ii) Another fundamentalof electric charge is'Quantization of charge'. <br/> (iii) No, as it is against the conservation of charge. Creationof a charge `+q` is <a href="https://interviewquestions.tuteehub.com/tag/accompained-1960605" style="font-weight:bold;" target="_blank" title="Click to know more about ACCOMPAINED">ACCOMPAINED</a> by creations of charge `-q`. <br/> (iv)In day to daylife, conservation of charge impliesthat gainof oneis alwaysequal to lossof the other. It also imples that sumtotal of gain in lifeis equalto sumtotal of losses in <a href="https://interviewquestions.tuteehub.com/tag/life-541672" style="font-weight:bold;" target="_blank" title="Click to know more about LIFE">LIFE</a> infact, the sum total of gainsand losses in lifeis constant.</body></html> | |
| 45007. |
A two wire transmission line has a capacitance of 20 pF/m and a characteristic impedance of 50 Omega (a) What is the inductance per metre of this cable ? (b) Determine the impedance of an infinitely long section of such cable. |
| Answer» <html><body><p></p>Solution :(a) The <a href="https://interviewquestions.tuteehub.com/tag/characteristic-914243" style="font-weight:bold;" target="_blank" title="Click to know more about CHARACTERISTIC">CHARACTERISTIC</a> impendance `<a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a>=sqrt(<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>//C)L=(Z^(2))(C)=(50)^(2)(20xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>))H=0.05H`. <br/> (b) The characteristic impedance of a trasnmission line is the impedance that an infinite length of line would <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> to a power supply at the input end of the line. Thus, `Z_(oo)=Z_(0)=50Omega`.</body></html> | |
| 45008. |
A beam of light is incident normally upon a polaroid and the intensity of the emergent beam is found to be unchanged when the polaroid is rotated about an axiis perpendicular to its pass axis. The incident beam is ____in nature. |
| 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/unpolarised-3248101" style="font-weight:bold;" target="_blank" title="Click to know more about UNPOLARISED">UNPOLARISED</a>.</body></html> | |
| 45009. |
Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel. |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_PHY_HND_BOK_MAR_18_E01_029_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> `I_1,I_2` represent branch currents.<br/> Let V be the commonpotential ,so that ,<br/> `I_1=(E_A-V)/r_1, I_2=(E_2-V)/r_2` <br/>hence, `I=(E_1-V)/r_1 + (E_2-V)/r_2` <br/>i.e., `I=(E_1/r_1 + E_2/r_2)-V (1/r_1+1/r_2)` <br/>Comparingthis with the terminalpotentialdifference, <br/> `V_E_(eq)-Ir_(eq)` <br/> i.e.,`V=E_(eq)-1/((1/r_(eq)))` <br/>For two <a href="https://interviewquestions.tuteehub.com/tag/cells-25681" style="font-weight:bold;" target="_blank" title="Click to know more about CELLS">CELLS</a> in parallel <a href="https://interviewquestions.tuteehub.com/tag/combination-922669" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINATION">COMBINATION</a> , <br/> `(E_(eq))_p=(E_1/r_1 +E_2/r_2)/(1/r_1+1/r_2)=(E_1r_2 + E_2r_1)/(r_1+r_2)` <br/> `1/(r_(eq))_p =1/r_1+1/r_2`and main current=`i=E_(eq)/(R+r_(eq))` <br/> Note : (i) For .n. cells in parallel<br/> `V=[(sum_(r=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_1)]-[I/(sum_(i=1)^n 1/r_1)]` ,<br/> `(E_(eq))_"parallel" =[(sum_(r=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_1)]-[I/(sum_(i=1)^n 1/r_1)]` and `(1/r_(eq))_(p)=sum_(i-1)^n (1/r_i)`. <br/> (ii) For .n. number of identicalcells <br/> E-emf of each <a href="https://interviewquestions.tuteehub.com/tag/cell-25680" style="font-weight:bold;" target="_blank" title="Click to know more about CELL">CELL</a> <br/>r-internalresistance of each cell. <br/> `(E_(eq))_p =(n(E/r))/(n(1/r))` <br/>i.e.,`(E_(eq))_p=E` <br/> and `(1/r_(eq))_p=n(1/r)` <br/>i.e. `(r_(eq))_p =r/n` <br/>Main current, `I=(nE)/(R+n/r)` and terminalpotential difference across cells V=IR.</body></html> | |
| 45010. |
How much work has to be done to isolate the two charges and take them infinitely away from each other? |
| 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> :0.7 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a></body></html> | |
| 45011. |
Ruby lases at a wavelength of 694 nm. A certain ruby crystal has 4.00xx10^(19)Cr ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting 1.50mus. As the pulse begins, 60.0% of the Cr ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? |
| 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> :`1.5xx10^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)J//s=1.5xx10^(6)<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>`</body></html> | |
| 45012. |
A body of mass 5 kg collides elastically with a stationary body of mass 2.5 kg. After the collision, the 2.5 kg body begins to move with a kinetic energy of 8). Assuming the collision to be one-dimensional, the kinetic energy of the 5 kg body before collision is: |
| Answer» <html><body><p>3 J<br/>6 J<br/>9 J<br/>11J</p>Solution :Here we know that the fraction of K.E. <a href="https://interviewquestions.tuteehub.com/tag/transferred-2310410" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSFERRED">TRANSFERRED</a><br/>=`(<a href="https://interviewquestions.tuteehub.com/tag/4x-319255" style="font-weight:bold;" target="_blank" title="Click to know more about 4X">4X</a>)/((1+x)^<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(4xx1/2)/((1+1//2)^2)` <br/>=8/9<br/>This means that<br/>`("K.E of 2.5 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> after collison")/("K.E of 2.5 Kg before collison")=8//9` <br/>`:.` K.E of 5 kg before collison is<br/>`E_k=9//8xx8=9 J`</body></html> | |
| 45013. |
A particle of mass m is attached to one end of missless rigid non-conducting rod of length l. Another particle of the same mass is attached to the other end of the rod. Two particles carry charges +q and -q . Thisarrangement is held in the region of uniform electric field E such that the rod makes and angle theta (lt 5^(@)) withthe fielddirection. Find an expression for the minimum time that is needed for the rod to become parallel to the field after it is set free. |
| Answer» <html><body><p></p>Solution :As the rod has twocharges of <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> <a href="https://interviewquestions.tuteehub.com/tag/magnitude-1083080" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNITUDE">MAGNITUDE</a>(be opposite in nature), the torqueacting on the rod, <br/> `tau=qE xx l sin theta` <br/> or, `tau =qE l theta""[ becausetheta <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> 5^(@)]` <br/> We know, `tau=I alpha` <br/> [where I = moment ofinertia and `alpha` = angular acceleration] <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XII_P1_U01_C02_E02_021_S01.png" width="80%"/> <br/> Hence, `I alpha=q E l theta or, alpha=(q E l theta)/(I)` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> period, `T=2pi sqrt(("angular displacement")/("angular acceleration"))=2pi sqrt((I)/(qE l))` <br/> Now, moment ofinertiaof the <a href="https://interviewquestions.tuteehub.com/tag/dipole-440057" style="font-weight:bold;" target="_blank" title="Click to know more about DIPOLE">DIPOLE</a>, <br/> `I=[((l)/(2))^(2)+((l)/(2))^(2)]m=(1)/(2) ml^(2)` <br/> `therefore T=2pisqrt((ml^(2))/(2qEl))=2pi sqrt((ml)/(2qE))` <br/> `therefore ` Required time, `t=(1)/(4)xx "time period"=(T)/(4)=(pi)/(2) sqrt((ml)/(2qE))`</body></html> | |
| 45014. |
A uniform solid sphere has a radius R and density p. The M.I. of sphere passing through its centre is |
| Answer» <html><body><p>28/15`<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a> R^5rho`<br/>15/28`pi R^5rho`<br/>15/8`pi R^5roh`<br/>8/15`pi R^5rho`</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> | |
| 45015. |
An laternativing voltage of 350 V, 60 Hz is applied on a full wave rectifier. The internal resistance of each diode is 200Omega. If R_(L)=5kOmega, then find (i) the peak value of output current. (ii) the value of output direct current. (iii) theoutput dc power. (iv) the rms value of output current. (v) the efficiency of rectifier. (vi) the value of peak inverse voltage (P.I.V.). |
| 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 :(i) `I_("<a href="https://interviewquestions.tuteehub.com/tag/peak-1149480" style="font-weight:bold;" target="_blank" title="Click to know more about PEAK">PEAK</a>")=I_("rms")xxsqrt2=(V_("rms")xxsqrt2)/((R_(L)+2r_(p)))` <br/> `or I_(0)=(350xxsqrt5)/(5400)=0.092A` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) `I_(DC)=(2I_(0))/(pi)=(2xx0.092)/(3.14)=0.058A` <br/> (iii) `p_(DC)=I_(DC)^(2)xxR_(L)=(0.058)^(2)xx(5000)=17W` <br/> (iv) `I_("rms")=(I_(0))/(sqrt2)=(0.092)/(1.41)=0.065A` <br/> (v) Efficiency of reaction `eta=(81.6)/(1+(r_(p))/(R_(L)))` <br/> `or eta=(81.6)/(1+(200)/(5000))=(81.6xx25)/(26) or eta=78%` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/vi-723586" style="font-weight:bold;" target="_blank" title="Click to know more about VI">VI</a>) `P.I.V.=2E_(0)` <br/> `or P.I.V.=2sqrt2E_("rms")=2sqrt2=350` <br/> `or P.I.V.=1000V`.</body></html> | |
| 45016. |
The cube in figure has edge length sqrt2m and is oriented as shown in a region of uniform electric field , in newtons per coulomb, is given by (a)6.00hati, (b) -2.00 hatj, and (c) -3.00 hatj+ 4.00 hatk . (d) what is the total flux through the cubefor each field ? |
| 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) <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</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>) -4.0N.m^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)//<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` © zero (d) zero</body></html> | |
| 45017. |
Optical fibers work on the principle of ___________. |
| 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/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> intermal <a href="https://interviewquestions.tuteehub.com/tag/reflection-13747" style="font-weight:bold;" target="_blank" title="Click to know more about REFLECTION">REFLECTION</a></body></html> | |
| 45018. |
Find the capacitance of the capacitor that would have a reactance of 100 Omega when used with an a.c. source of frequency 5/pi kHz. |
| Answer» <html><body><p></p>Solution :Here, `X_(C) = 100 Omega, v= 5/pi kHz = 5000/pi Hz` <br/> `therefore C = 1/(X_(C).2pi) = (1/(100 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2pi xx (500)/pi)) = 10^(-4) <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> = 1 muF`</body></html> | |
| 45019. |
What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm^(2) . Would you be able to see the squares distinctly with your eyes very close to the magnifier? |
| Answer» <html><body><p></p>Solution :Magnification `=sqrt((6.<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>//1))=2.5` <br/> `v=+2.5u` <br/> `+(1)/(2.5u)-(1)/(u)=(1)/(<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/> i.e., `u=-6cm` <br/> `|v|=15cm` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/virtual-673" style="font-weight:bold;" target="_blank" title="Click to know more about VIRTUAL">VIRTUAL</a> image is <a href="https://interviewquestions.tuteehub.com/tag/closer-7270227" style="font-weight:bold;" target="_blank" title="Click to know more about CLOSER">CLOSER</a> than the normal <a href="https://interviewquestions.tuteehub.com/tag/near-1112277" style="font-weight:bold;" target="_blank" title="Click to know more about NEAR">NEAR</a> point (25 cm) and cannot be seen by the eye distinctly.</body></html> | |
| 45020. |
In a young ,s double slit experiment distance the slits is 1mm. The fringe width is found to be 0.mm. when the screen is moved through a distance of 0.25m away from the plane of the slit used . |
| Answer» <html><body><p></p>Solution : `<a href="https://interviewquestions.tuteehub.com/tag/beta-2557" style="font-weight:bold;" target="_blank" title="Click to know more about BETA">BETA</a>= 0.6xx10^(-3)m,beta_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=0.75xx10^(-3)m`<br/> `d= 1xx10^(-3)m, D_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = Dm, D_(2) = (D+ 0.25 ) m, lamda= ?` <br/> `beta_(1) = (lamdaD_(1))/(d) = (lamdaD)/(1xx10^(-3)) rightarrow(1)` <br/> `beta_(2)= ( lamda(D_(2)))/(d) = (lamda(D+0.25))/(1xx10^(-3)) rightarrow(2)`<br/> `beta-(1))/(beta_(2))= (0.6xx10^(-3))/(0.75xx10^(-3))= (lamdaD)/(lamda(D + 0.25))` <br/> `(D+ 0.25)0.6xx10^(-3)= 0.75xx10^(-3)D` <br/> `RightarrowD=1m` <br/> Subsituting the value of D in eqaution (1) or (2) <br/> `lamda= (beta_(1)d)/(D)= (0.6xx10^(-3)xx10^(-3))/(1)` <br/> `lamda=6xx10^(-7)m= 6000A^(@)`</body></html> | |
| 45021. |
A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The e.m.f. induced between the two ends of the conductor is |
| Answer» <html><body><p>0.10 V <br/>0.15 V <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>.50 V <br/>15.00 V </p>Solution :Assuming magnetic <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>), conductor length (l) and its velocity (v) are mutually perpendicular so, <a href="https://interviewquestions.tuteehub.com/tag/emf-970036" style="font-weight:bold;" target="_blank" title="Click to know more about EMF">EMF</a> induced between the two <a href="https://interviewquestions.tuteehub.com/tag/ends-971392" style="font-weight:bold;" target="_blank" title="Click to know more about ENDS">ENDS</a> of the conductor<br/> `epsilon = Blv`<br/>Here, `B=0.1 T, l=0.1 m, v = 15 ms^(-1), epsilon = ?`<br/>`epsilon =0.1xx0.1xx15=0.15 V`</body></html> | |
| 45022. |
Where gammarays are emited ? |
| Answer» <html><body><p></p>Solution :The `gamma` rays are emitted from the nuclei of some <a href="https://interviewquestions.tuteehub.com/tag/radioactive-2970069" style="font-weight:bold;" target="_blank" title="Click to know more about RADIOACTIVE">RADIOACTIVE</a> materials such as <a href="https://interviewquestions.tuteehub.com/tag/uranium-1440633" style="font-weight:bold;" target="_blank" title="Click to know more about URANIUM">URANIUM</a>,<a href="https://interviewquestions.tuteehub.com/tag/radium-1176224" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUM">RADIUM</a> etc. It was <a href="https://interviewquestions.tuteehub.com/tag/discovered-440478" style="font-weight:bold;" target="_blank" title="Click to know more about DISCOVERED">DISCOVERED</a> by Paul <a href="https://interviewquestions.tuteehub.com/tag/ulrich-2316447" style="font-weight:bold;" target="_blank" title="Click to know more about ULRICH">ULRICH</a> Villard.</body></html> | |
| 45023. |
On metal surface radiation of 2000Å and 5000 Å is incident sequentially .Change in kinetic energy of photo-electric emitted during this will be…….h=6.6xx10^(-34)Js . |
| Answer» <html><body><p>3.71 eV<br/>5.94 eV<br/>7.42 eV<br/>2.97 eV</p>Solution :`(1)/(2)mv_(max)^(2)=hv-phi_(<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/>`K_(max)=(hc)/(lambda)-phi_(0)` <br/>`[because (1)/(2)mv_(max)^(2)=K_(max)` and `v=(c )/(lambda)]` <br/>In first case `(K_(max))_(1)=(hc)/(lambda_(1))-phi_(0)` <br/>In <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> case `(K_(max))_(2)=(hc)/(lambda_(2))-phi_(0)` <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> DeltaK=(K_(max))_(1)-(K_(max))_(2)` <brge>`therefore DeltaK=hc[(lambda_(2)-lambda_(1))/(lambda_(1)lambda_(2))]` <br/>`=6.6xx10^(-34)xx3xx10^(8)` <br/>`[(5xx10^(-7)-2xx10^(-7))/(5xx10^(-7)xx2xx10^(-7))]` <br/>`=6.6xx3xx10^(-26)xx(3xx10^(-7))/(10xx10^(-14))` <br/>`=59.4xx10^(-20)J` <br/>`=(5.94xx10^(-19))/(1.6xx10^(-19))eV` <br/>3.7125 eV <br/>`~~3.71` eV</brge></body></html> | |
| 45024. |
A boat which has a speed 5 km/h in still water crosses a river 100 m along the shortest possible path in 1.5 min. The velocity of river water in km/h is : |
| Answer» <html><body><p>1<br/>3<br/>4<br/>4.1</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> man crosses along <a href="https://interviewquestions.tuteehub.com/tag/shortest-642834" style="font-weight:bold;" target="_blank" title="Click to know more about SHORTEST">SHORTEST</a> path `vec(OB)` represent tesultant velocity of man w.r.t. <a href="https://interviewquestions.tuteehub.com/tag/river-1189947" style="font-weight:bold;" target="_blank" title="Click to know more about RIVER">RIVER</a><br/>`|vec(V_(R))|=(OB)/t=0.1/(1.5/60)=60/15`=4km/h<br/>`vec(OA)` repesents velocity of man<br/>`:. AB=<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(OA^(2)-OB^(2))=sqrt(25-16)=3`<br/>`:.` Velocity of river =3km/h<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_RPA_OBJ_PHY_C03_E01_030_S01.png" width="80%"/></body></html> | |
| 45025. |
In a choke coil, the values of reactance X_L and resistance R are such that |
| Answer» <html><body><p>`X_L = <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>`<br/>`X_L <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> gt R`<br/>`X_L <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> lt R`<br/>`X_L = infty`</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 45026. |
Given a +b+c+d=0 which of the following statement is incorrect |
| Answer» <html><body><p>a,b,<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> and d must each be a null vector.<br/>The magnitude of (a+c) equals the <a href="https://interviewquestions.tuteehub.com/tag/magni-2166213" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNI">MAGNI</a> tude of (b+d)<br/>The magnitude of a can <a href="https://interviewquestions.tuteehub.com/tag/never-570518" style="font-weight:bold;" target="_blank" title="Click to know more about NEVER">NEVER</a> be <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than the sum of the magnitudes of b, c and d<br/> b + c must lie in the plane of a and b if a and d are not collinear, and in the line of a and d, if they are collinear </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> | |
| 45027. |
The range of a projectile, when launched at an angle of 15^@ with the horizontal is 1.5 km. The additional horizontal distance the projectile would cover when projected with same velocity at 45^@ is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> <a href="https://interviewquestions.tuteehub.com/tag/km-1064498" style="font-weight:bold;" target="_blank" title="Click to know more about KM">KM</a> d<br/>4.5 km <br/>1.5 km <br/>2.5 km </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<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> | |
| 45028. |
In the following circuit a 10m long potentiometer wire withresistance 1.2 ohm/m , a resistance R 1 and an accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection in galvanometer is zero. The current supplied by the accumulator will be |
| Answer» <html><body><p>`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^(–4) A`<br/>`<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^(–4)A`<br/>`4 xx 10^(-3) A`<br/>`8 xx10^(-3) A`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 45029. |
Find the position of the image formed by the lens combination given in the figure. |
| 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> :Image <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> by the fist lens `(1)/(v_(1)) -(1)/(u_(1)) = (1)/(f_(1))` <br/> `(1)/(v_(1)) - (1)/(-30) = (1)/(10) or v_(1) = 15 cm ` <br/> The image formed by the first lens serves as the object for the second. This is at a <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> of (15 -5) cm = 10 cm to the right of the second lens. though the image is real, it serves as a virtual object for the second lens, Which means that the rays appear to come from it for the second lens.<br/> `(1)/(v_(2)) - (1)/(10) = (1)/(-10) or v_(2) = infty`<br/> The virtual image is formed at an infinite distance to the left of the second lens. This <a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</a> as an object for the <a href="https://interviewquestions.tuteehub.com/tag/third-1414358" style="font-weight:bold;" target="_blank" title="Click to know more about THIRD">THIRD</a> lens.<br/> `(1)/(v_(3)) - (1)/(u_(3)) = (1)/(f_(3))"" or "" (1)/(v_(3)) = (1)/(infty) + (1)/(30) "or" v_(3) = 30` cm <br/> The final image is formed 30 cm to the right of the third lens.</body></html> | |
| 45030. |
Two identical charged spheres are suspended in air from a ceiling using strings of equal length. The point of suspension is common to both the spheres. Neglect the density of air. The density of material of spheres is 2 "g/cm"^3.Both spheres repel each other and the strings are inclined at a certain angle with the vertical when the system is in equilibrium. Now both the sphere are dipped into a liquid of specific gravity 1 and dielectric constant K, but angle that the strings are making with the vertical remains unchanged. Let T_1 be the tension in both the strings when the spheres are in air and T_2, be the tension in strings when spores are dipped in liquid. Let F_1 be the electrostatic force between charges when they were kept in air and F_2, be the electrostatic force when spheres are dipped in liquid. The magnitude of T_2//T_1 is |
| Answer» <html><body><p>1<br/>2<br/>`1//2`<br/>None </p>Solution :On <a href="https://interviewquestions.tuteehub.com/tag/comparing-2531631" style="font-weight:bold;" target="_blank" title="Click to know more about COMPARING">COMPARING</a> equations (i) and (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>) we get `T_2//T_1`=1/2</body></html> | |
| 45031. |
A string 25 cm long and having a mass of 2.5g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its firsy overtone and the air in the pipe in its fundamental frequency, 8 beats per sec are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string. |
| Answer» <html><body><p></p>Solution :For first overtone (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>., second harmonic) of thestring `f_(s)=<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> sqrt(T)` and for <a href="https://interviewquestions.tuteehub.com/tag/fundamental-466867" style="font-weight:bold;" target="_blank" title="Click to know more about FUNDAMENTAL">FUNDAMENTAL</a> frequency of closed organ pipe, `f_(c)=200 <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/> So `200-40 sqrt(T)=8 or 40 sqrt(T)-200=8` <br/> But decreasing the tension <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> the beat frequency. <br/> So `40 sqrt(T)=208 i.,e T=[(208)/(40)]^(2)=(5.2)^(2)=27N`</body></html> | |
| 45032. |
A simple pendulum of length varphi oscillates with the time period T . Then which of the following graph correctly represents the relation? |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RNK_SM_FIITJEE_PHY_P3_E02_067_O01.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RNK_SM_FIITJEE_PHY_P3_E02_067_O02.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RNK_SM_FIITJEE_PHY_P3_E02_067_O03.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RNK_SM_FIITJEE_PHY_P3_E02_067_O04.png" width="30%"/> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a 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> | |
| 45033. |
Two identical charged spheres are suspended in air from a ceiling using strings of equal length. The point of suspension is common to both the spheres. Neglect the density of air. The density of material of spheres is 2 "g/cm"^3.Both spheres repel each other and the strings are inclined at a certain angle with the vertical when the system is in equilibrium. Now both the sphere are dipped into a liquid of specific gravity 1 and dielectric constant K, but angle that the strings are making with the vertical remains unchanged. Let T_1 be the tension in both the strings when the spheres are in air and T_2, be the tension in strings when spores are dipped in liquid. Let F_1 be the electrostatic force between charges when they were kept in air and F_2, be the electrostatic force when spheres are dipped in liquid. The magnitude of F_2//F_1 is |
| Answer» <html><body><p>3<br/>2<br/>`1//2`<br/>`1//3`</p>Solution :The magnitude of `F_2//F_1`=1/2 as we have <a href="https://interviewquestions.tuteehub.com/tag/calculated-907694" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATED">CALCULATED</a> the dielectric <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> K=2</body></html> | |
| 45034. |
A person using a lens as a sample microscope sees an |
| Answer» <html><body><p>inverted virtual image<br/>inverted <a href="https://interviewquestions.tuteehub.com/tag/real-1178490" style="font-weight:bold;" target="_blank" title="Click to know more about REAL">REAL</a> magnified image<br/><a href="https://interviewquestions.tuteehub.com/tag/upright-1440445" style="font-weight:bold;" target="_blank" title="Click to know more about UPRIGHT">UPRIGHT</a> virtual image<br/>upright real magnified image</p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/person-25481" style="font-weight:bold;" target="_blank" title="Click to know more about PERSON">PERSON</a> <a href="https://interviewquestions.tuteehub.com/tag/see-630247" style="font-weight:bold;" target="_blank" title="Click to know more about SEE">SEE</a> an upright virtual image in a simple microscope.</body></html> | |
| 45035. |
What is half-life of nucleus? Give the expression. |
| 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> :The half - life `T_(1/2)` is <a href="https://interviewquestions.tuteehub.com/tag/defined-947013" style="font-weight:bold;" target="_blank" title="Click to know more about DEFINED">DEFINED</a> as the time <a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> for the number of <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> initiallypresent toreduce to <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> half of the initial amount. <br/> `T_(1/2) = (In 2)/(lambda) = (0.6931)/(lambda)`.</body></html> | |
| 45036. |
The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively. |
| Answer» <html><body><p>112.5 m and 22.5 s<br/>337.5 m and <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> s<br/>112.5 m and <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> s<br/>225.5 m and <a href="https://interviewquestions.tuteehub.com/tag/10s-267218" style="font-weight:bold;" target="_blank" title="Click to know more about 10S">10S</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</body></html> | |
| 45037. |
In a Young's double slit experiment the intensity at a point where the path difference is (lambda)/(6)(lambda being the wavelength of the light used) is I. If I_(0) denotes the maximum intensity then (I)/(I_(0))=… |
| Answer» <html><body><p>`(1)/(sqrt(2))`<br/>`(sqrt(3))/(2)`<br/>`(1)/(2)`<br/>`(3)/(4)`</p>Solution :Maximumintensity `I_(0)=I.+I+2 sqrt(I.I) cos0^(@)`<br/> `=I.+2I. [ :. cos 0^(@)=1]` <br/> `=4I.` <br/> and intensity at any <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a>, <br/> `I=I.+I.+2sqrt(I.I) cos <a href="https://interviewquestions.tuteehub.com/tag/phi-599602" style="font-weight:bold;" target="_blank" title="Click to know more about PHI">PHI</a>` <br/> where `phi=(<a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a>)/(lambda)xx` path <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> <br/> `=(2pi)/(lambda)xx(lambda)/(6)=(pi)/(3)` rad <br/> `:.I=2I.+2I"cos"(pi)/(3)` <br/> `=2I.+I.=<a href="https://interviewquestions.tuteehub.com/tag/3i-310966" style="font-weight:bold;" target="_blank" title="Click to know more about 3I">3I</a>.` <br/> `:.(I)/(I_(0))=(3I.)/(4I.)=(3)/(4)`</body></html> | |
| 45038. |
Molecular mass of amonia is.............. |
| Answer» <html><body><p>15u<br/>24u<br/>17u<br/>36u</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> | |
| 45039. |
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free space. It is under continuous illumination of 200 nm wavelength of light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is Axx10^(z) (where 1ltAlt10). The value of 'Z' is |
| Answer» <html><body><p>2<br/>3<br/>4<br/>7</p>Answer :D</body></html> | |
| 45040. |
Name any two characteristics of light which do not change on polarisation |
| 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/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> and <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a></body></html> | |
| 45041. |
The slopes of anode and mutual characteristics of a triode are 0.02 mA V^(-1) and 1 mA V^(-1) respectively. What is the amplification factor of the valve |
| Answer» <html><body><p>5<br/>50<br/>500<br/>`0.5`</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 45042. |
A thin stick of length f/5 placed along the principal axis of a concave mirror of focal length f such that its image is real and elongated just touch the stick. What is the magnification? |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NEET_PHY_OBJ_V02_C09_S01_016_S01.png" width="80%"/> <br/> One end of image concides with one end of stick i.e., either A or B is centre of curvature. <br/> Since image is elongated i.e., point B is centre ofcurvature. Image of B is formed on B itself. <br/> Image of A <br/> `u=-(2f-<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>//5)=(-8f)/3,f=-f,v=?` <br/> `1/v+1/u=1/f <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 1/v+1/(-8f//5)=1/(-f)` <br/> `1/v=-1/f+5/(8f)=(-8+5)/(8f)=(-3)/(8f)rArrv=-(8f)/3` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NEET_PHY_OBJ_V02_C09_S01_016_S02.png" width="80%"/> <br/> Image length, A'B'=`(8f)/5-2f=-2f//5` <br/> Longitudinal <a href="https://interviewquestions.tuteehub.com/tag/magnification-1083057" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNIFICATION">MAGNIFICATION</a>, <br/> `m_(<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>)=(-A'B')/(AB)=(2f//5)/(f//5)=<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>`</body></html> | |
| 45043. |
Long distance radio broadcasts use short - wave bands. Why ? |
| 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> :Because in SW band, radio <a href="https://interviewquestions.tuteehub.com/tag/waves-13979" style="font-weight:bold;" target="_blank" title="Click to know more about WAVES">WAVES</a> of frequency 4.75 MHz to 9.9 MHz whose very <a href="https://interviewquestions.tuteehub.com/tag/long-537592" style="font-weight:bold;" target="_blank" title="Click to know more about LONG">LONG</a> distance transmission is <a href="https://interviewquestions.tuteehub.com/tag/possible-592355" style="font-weight:bold;" target="_blank" title="Click to know more about POSSIBLE">POSSIBLE</a> by ionosphere.</body></html> | |
| 45044. |
Condition for maximum diffraction is |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>)/(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>`<br/>`(lambda)/(a)=<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/>`(lambda)/(a)=oo`<br/>`(lambda)/(a)=(1)/(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 45045. |
Mention the three types of energy loss in a transformer. |
| Answer» <html><body><p></p>Solution :(a) <a href="https://interviewquestions.tuteehub.com/tag/copper-933790" style="font-weight:bold;" target="_blank" title="Click to know more about COPPER">COPPER</a> loss (<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/magnetic-551115" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNETIC">MAGNETIC</a> flux <a href="https://interviewquestions.tuteehub.com/tag/linkage-1074960" style="font-weight:bold;" target="_blank" title="Click to know more about LINKAGE">LINKAGE</a> loss (c) <a href="https://interviewquestions.tuteehub.com/tag/eddy-965645" style="font-weight:bold;" target="_blank" title="Click to know more about EDDY">EDDY</a> current loss.</body></html> | |
| 45046. |
Let us suppose that the earth had a net surface charge density of 1 electron per m^(2). What would its potential be? Also what is the field just outside the earth's 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/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> supplied, <br/> `sigma=1e//m^(2) =1.6 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-19)C//m^(2), " <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> of the earth , "R=6.4 xx 10^(6)m` <br/> Potential `V=(1)/(4pi epsi_(0) ) q/R=(1)/(4pi epsi_(0)). (sigma 4pi R^(2))/(R) =(sigma R)/(epsi_(0)) =(1.6 xx 10^(-19) xx 6.4 xx 10^(6))/(8.854 xx 10^(-12)) =0.1156V` <br/> Field, `E=(1)/(4pi epsi_(0)).q/R^(2)=(1)/(4pi epsi_(0)) xx (sigma 4piR^(2))/(R^(2)) =(sigma)/(epsi_(0)) =(1.6 xx 10^(-19))/(8.854 xx 10^(-12))=1.81 xx 10^(-8) N//C`</body></html> | |
| 45047. |
A Neutron moving with velocity v collides with a stationary alpha particle. The velocity of the neutron after collision is |
| Answer» <html><body><p>(A) `-3v//5`<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>) `3V//5`<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>) `2V//5`<br/>(D) `-2V//5`</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> | |
| 45048. |
A uniform magnetic field B exists in a circular region of radius R. the field is perpendicular to the plane and is increasing at a constant rate of (dB)/(dt) = alpha. There is a straight conducing rod AB of length 2R. Find the emf induced in the rod when it is placed as shown in figure (a) & (b). Point C is midpoint of the rod in figure (b) |
| 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) <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> (b) `(piR^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) alpha`</body></html> | |
| 45049. |
Ifv = c, the mass of the moving body, of rest mass 'm_(0)' is "_________________". |
| Answer» <html><body><p>zero<br/>infinity<br/>m0<br/>`(m)/(m_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 45050. |
Monochromatic light falls on a right-angled prism at an angle of incidence 45^@. The emergent light is found to slide along the face AC. Find the refractive index of material of prism |
| Answer» <html><body><p></p>Solution :Since the emergent light slides along the <a href="https://interviewquestions.tuteehub.com/tag/face-982590" style="font-weight:bold;" target="_blank" title="Click to know more about FACE">FACE</a> AC, angle of emergence is `<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a>^@`, 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>. <br/> It implies that angle of <a href="https://interviewquestions.tuteehub.com/tag/incidence-1039767" style="font-weight:bold;" target="_blank" title="Click to know more about INCIDENCE">INCIDENCE</a> of the ray that falls on face AC is equal to the critical angle `theta_C`. <br/> `thereforer_2= theta_C ...(1)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XII_V02_C_C02_SLV_045_S01.png" width="80%"/> <br/> From the prism theory, we know <br/> ` r_1+r_2= A = 90^@ ` <br/> ` thereforer_2 = 90^@- r_1 ......(2)` <br/> fromequaiton(1) and (2)`,90^@ - r_1 =theta_c `<br/> ` thereforesin( 90^@ -r_1) = sintheta_C(or)cos r_1= sintheta_C` <br/>Butsin ` theta_C = (1)/(mu )thereforecos r_1 = (1)/(mu)`<br/>Applyingsnell.slawat theboundaryAB , ` 1 xx sin45^@= mu sinr_1= mu sqrt(1- (1)/( mu^2))`<br/>` therefore(1)/( sqrt(2)) = sqrt(mu^2 -1)or mu^2 = 3//2 = 1.5mu = sqrt(1.5)`</body></html> | |