InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1. |
A wire is bent to form a semicircle of the radius a. The wire rotates about its one end with angular velocity omega . Axis of rotation is perpendicular to the plane of the semicircle . In the space , a uniform magnetic field of induction B exists along the aixs of rotation as shown in the figure . Then - |
| 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>>potential <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> between p and <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> is equal to` 2 B omega a^2`<br/> potential difference betweenp andQ is equal to`2 pi ^2B omega a^2`<br/>p is at <a href="https://interviewquestions.tuteehub.com/tag/higher-1022060" style="font-weight:bold;" target="_blank" title="Click to know more about HIGHER">HIGHER</a> potential than Q<br/>potential difference betweenP and Q iszero.</p>Solution :`E=1/2 Bl^2 omega =1/2B(<a href="https://interviewquestions.tuteehub.com/tag/2a-300249" style="font-weight:bold;" target="_blank" title="Click to know more about 2A">2A</a>)^2 omega =2 B omega a^2`</body></html> | |
| 2. |
A massless non conducting rod AB of length 2l is placed in uniform time varying magnetic field confined in a cylindrical region of radius (R gt l) as shown in the figure. The center of the rod coincides with the centre of the cylin- drical region. The rod can freely rotate in the plane of the Figure about an axis coinciding with the axis of the cylinder. Two particles, each of mass m and charge q are attached to the ends A and B of the rod. The time varying magnetic field in this cylindrical region is given by B = B_(0) [1-(t)/(2)] where B_(0) is a constant. The field is switched on at time t = 0. Consider B_(0) = 100T, l = 4 cm(q)/(m) = (4pi)/(100) C//kg. Calculate the time in which the rod will reach position CD shown in the figure for th first time. Will end A be at C or D at this instant ? |
| 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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>,s <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> A</body></html> | |
| 3. |
A concave lens with equal radius of curvature both sides has a focal length of 12 cm. The refractive index of the lens is 1.5. How will the focal length of the lens change if it is immersed in the liquid of refractive index 1.8 ? |
| Answer» <html><body><p></p>Solution :Focal <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> will <a href="https://interviewquestions.tuteehub.com/tag/become-389953" style="font-weight:bold;" target="_blank" title="Click to know more about BECOME">BECOME</a> `+<a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a> <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>`</body></html> | |
| 4. |
If the tempearture of black body is raised by 5%, the heat energy radiated would increases by : |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>%`<br/>`20%`<br/>`21.51%`<br/>`41.61%`</p>Solution :Acc. Of Stefan.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/> `(E_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(E_(1))=((T_(2))/(T_(1)))^(4)` <br/> `(E_(2))/(E_(1))=((105)/(100))^(4)=1*2158` <br/> `:.""(E_(2))/(E_(1))-1=1*2155-1` <br/> `rArr(E_(2)-E_(1))/(E_(1))xx100=*2155xx100=21*<a href="https://interviewquestions.tuteehub.com/tag/55-325247" style="font-weight:bold;" target="_blank" title="Click to know more about 55">55</a>%` <br/> Thus <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 (c ).</body></html> | |
| 5. |
What are the co-ordinates of the image of S formed by a plane mirror as shown in figure? |
| 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> <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</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/>(-4 cm, 0) <br/>`(4sqrt2 cm , 0)`<br/>(0, <a href="https://interviewquestions.tuteehub.com/tag/4cm-318906" style="font-weight:bold;" target="_blank" title="Click to know more about 4CM">4CM</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> | |
| 6. |
The direction of ray of light incident on a concave mirror is shown by PQ in Fig. The direction in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4. Which of the four rays correctly shows the direction of reflected ray? |
| Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :The ray <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> 2 is the correct reflected ray because a ray passing through focus <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> F after reflection, <a href="https://interviewquestions.tuteehub.com/tag/travels-1426842" style="font-weight:bold;" target="_blank" title="Click to know more about TRAVELS">TRAVELS</a> parallel to the <a href="https://interviewquestions.tuteehub.com/tag/principal-1166121" style="font-weight:bold;" target="_blank" title="Click to know more about PRINCIPAL">PRINCIPAL</a> axis of <a href="https://interviewquestions.tuteehub.com/tag/mirror-1098189" style="font-weight:bold;" target="_blank" title="Click to know more about MIRROR">MIRROR</a>.</body></html> | |
| 7. |
What is meant by polarisation ? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/polarisation-1157519" style="font-weight:bold;" target="_blank" title="Click to know more about POLARISATION">POLARISATION</a> is the phenomenon of restricting the <a href="https://interviewquestions.tuteehub.com/tag/waves-13979" style="font-weight:bold;" target="_blank" title="Click to know more about WAVES">WAVES</a> in <a href="https://interviewquestions.tuteehub.com/tag/oscillations-22507" style="font-weight:bold;" target="_blank" title="Click to know more about OSCILLATIONS">OSCILLATIONS</a> to <a href="https://interviewquestions.tuteehub.com/tag/particular-1147539" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICULAR">PARTICULAR</a> plane.</body></html> | |
| 8. |
Two concentric coils each of radius equal to 2πcm are placed right angles to each other. If 3 A and 4 A are the currents flowing through the two coils respectively. The magnetic induction( in Wb m^(-2) )at the center of the coils will be |
| Answer» <html><body><p>`7 xx 10^(-5)`<br/>`5 xx 10^(-5)`<br/>`10^(-5)`<br/>`12 xx 10^(-5)`</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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>= (mu_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)I)/(<a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a>)` <br/> `B_(1)= (mu_(0))/(2) (3)/(2pi xx 10^(-2))= 3 xx 10^(-5)T` <br/> `B_(2)= (mu_(0))/(2) (4)/(2pi xx 10^(-2))= 4 xx 10^(-5)T` <br/> `B= sqrt(B_(1)^(2))+ B_(2)^(2)= 5 xx 10^(-5)T`</body></html> | |
| 9. |
Assertion: Out of ""_(1)He^(3) and ""_(7)He^(3), the binding energy of ""_(1)He^(3)is greater than ""_(2)He^(8). Reason: Inside the nucleus of""_(1)H^(3), there is more repulsion than inside the nucleus of ""_(2)He^(4). |
| Answer» <html><body><p> If both assertion and reason are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> and reason is a 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> o <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> the assertion. <br/>If both assertion and reason are correct but reason is not the correct explanation o f assertion. <br/>If assertion is correct but reason is incorrect. <br/>If assertion is incorrect but reason is correct</p>Answer :B</body></html> | |
| 10. |
In which accelerated motion, K.E of the particle is constant |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/non-580180" style="font-weight:bold;" target="_blank" title="Click to know more about NON">NON</a>. U.C.M.<br/>U.C.M.<br/>circular motion<br/>motion</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> | |
| 11. |
60 W, 100 W and 1000 W bulbs are connected in parallel to the mains. The bulb that will be brightest is (all of them have same specified voltage) |
| Answer» <html><body><p>60W <br/><a href="https://interviewquestions.tuteehub.com/tag/100w-265920" style="font-weight:bold;" target="_blank" title="Click to know more about 100W">100W</a> <br/>1000W <br/>all are <a href="https://interviewquestions.tuteehub.com/tag/equally-2065515" style="font-weight:bold;" target="_blank" title="Click to know more about EQUALLY">EQUALLY</a> <a href="https://interviewquestions.tuteehub.com/tag/bright-391129" style="font-weight:bold;" target="_blank" title="Click to know more about BRIGHT">BRIGHT</a> </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> | |
| 12. |
Demonstarate that electronsmove in a betatronalong around orbitof constantradiusprovidedthe magnetic induction on the orbitof constantradiusprovidedthe magnetic inductionon the orbitis equalto halfthe meanvalue of that indide the orbit (the betatron condition). |
| Answer» <html><body><p></p>Solution :On the one hand, <br/> `(dp)/(dt) = eE = (e)/(2pi r) (d Phi)/(dt) = (e)/(2pi r) (d)/(dt) int_(0)^(r) 2pi r' <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> (r') dr'` <br/> On the other, <br/> `p = B (r) er, r` = constant. <br/> so, `(dp)/(dt) = er (d)/(dt) B(r) = er dotB (r)` <br/> Hence, `er dotB (r) = (e)/(2 pi r) pi r^(2) (d)/(dt) <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> B gt` <br/>So, `dotB (r) = (<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) (d)/(dt) lt B gt` <br/> This is most <a href="https://interviewquestions.tuteehub.com/tag/easily-964537" style="font-weight:bold;" target="_blank" title="Click to know more about EASILY">EASILY</a> <a href="https://interviewquestions.tuteehub.com/tag/satisfied-636887" style="font-weight:bold;" target="_blank" title="Click to know more about SATISFIED">SATISFIED</a> by taking`B(r_(0)) = (1)/(2) lt B gt` .</body></html> | |
| 13. |
A wire of length L is fixed at both ends such that F is tension in it. Its mass per unit length varies from one end to other end as mu=mu_0x, where mu_0 is constant. Find time taken by a transverse pulse to move from lighter end to its mid point. |
| 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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/(3sqrt2)<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((mu_0)/<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/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>)^(3//2)`</body></html> | |
| 14. |
A light ray travels through four adjacent media of refractive indices mu_(1), mu_(2), mu_(3), mu_(4) as shown in the Fig. 2.83. The bases of the media are parallel. If emergent ray is parallel to the incident ray AB then |
| Answer» <html><body><p>`mu_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = mu_(<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/>`mu_(2) = mu_(3)`<br/>`mu_(3) = mu_(4)`<br/>`mu_(4) = mu_(1)`</p>Answer :D</body></html> | |
| 15. |
Combination of two convex lenses works always as ....... |
| Answer» <html><body><p>concave lens<br/> convex lens<br/>rectangular glass slab <br/>plane mirror</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> focal length of combination,<br/>In `<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>=(f_1f_2)/(f_1+f_2)` for concave lens.<br/>`f=((f_1)(f_2))/(f_1+f_2)`<br/>`therefore f=`positive `therefore` <a href="https://interviewquestions.tuteehub.com/tag/works-17618" style="font-weight:bold;" target="_blank" title="Click to know more about WORKS">WORKS</a> as convex lens.</body></html> | |
| 16. |
Consider the following reactions, The end product 'S' is - |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CP_JM_MT_10_E01_050_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CP_JM_MT_10_E01_050_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CP_JM_MT_10_E01_050_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CP_JM_MT_10_E01_050_O04.png" width="30%"/></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> :By <a href="https://interviewquestions.tuteehub.com/tag/diazotisation-951049" style="font-weight:bold;" target="_blank" title="Click to know more about DIAZOTISATION">DIAZOTISATION</a> <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>.</body></html> | |
| 17. |
Uranium ._(92)^(238)U is an ustable nucleus. It decays to Thorium ._(92)^(238)Th, which is again an unstable nucleus which further decays to ._(91)^(234)Pa. Let ._(92)^(238)U be called A of decay constant lambda_(1) and ._(90)^(234)Th is called as B of decay constant lambda_(2) and stable nuclei ._(91)^(234)Pa be called as C. Here A is called parent nucleus and B is called daughter nucleus of A. Any two adjacent nuclei may be consider parent or daughter nuclei A, B and C respectively at time 't'. Then we can write Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC Rate of disintergration of A=(dN_(1))/(dt)=lambda_(1)N_(1) Rate of disintergration of B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2) Rate of formation of nuclei C is equal to (dN_(3))/(dt)=lambda_(2)N_(2) If at t=0, there are N_(0) number of nuclei of A where as nuclei B and C are absent in the sample Answer the following questions The graph N_(1), N_(2) and N_(3) with time can be best represent by |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_PHY_XII_V05_C03_E01_376_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_PHY_XII_V05_C03_E01_376_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_PHY_XII_V05_C03_E01_376_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_PHY_XII_V05_C03_E01_376_O04.png" width="30%"/></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/successivebrbr-2284089" style="font-weight:bold;" target="_blank" title="Click to know more about SUCCESSIVE">SUCCESSIVE</a> <a href="https://interviewquestions.tuteehub.com/tag/disintegration-440598" style="font-weight:bold;" target="_blank" title="Click to know more about DISINTEGRATION">DISINTEGRATION</a></body></html> | |
| 18. |
Explain the laws of photoelectric effect from the Einstein photoelectric equation. |
| Answer» <html><body><p></p>Solution :Einstein’s explanation of photoelectric effect. <br/> Photoelectric effect is the phenomenon of emission of <a href="https://interviewquestions.tuteehub.com/tag/electrons-969138" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRONS">ELECTRONS</a> by certain substances, chiefly metals, when these are illuminated by radiations like X-rays, ultraviolet rays and even visible light. The electrons ejected from a. substance in this manner are called photoelectrons. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MBD_EM_PHY_XII_C07_E02_017_S01.png" width="80%"/> <br/> According to quantum theory, light consists of bundles of energy, called photons, each of energy hv, where h is the Planck’s constant and V is the frequency of the light. <br/> When a photon of energy hv is incident on the surface of a <a href="https://interviewquestions.tuteehub.com/tag/photosensitive-2221004" style="font-weight:bold;" target="_blank" title="Click to know more about PHOTOSENSITIVE">PHOTOSENSITIVE</a> substance, a part of its energy `W_(0)` is used up in liberating the electrons from the surface and the rest of the energy is given to the ejected electron to impart it is a velocity v and hence a kinetic energy `1/2 mv^(2)`.<br/> Work Function. The part of the incident photon WQ which is used up in liberating an electron from the surface is called the work function and depends upon the nature of the surface. <br/> Einstein’s photoelectric equation <br/> Einstein assumed that one incident photon can eject only one electron. The photon energy is utilised in two ways : <br/> (i) A part of energy of photon is used for electron free from the atom and away from the metal surface. This energy is called photoelectric work function WQ. <br/> (ii) The balance of the photon energy is used up in giving the electron a kinetic energy `1/2 mv^(2)`. <br/> `therefore`Energy of incident photon = Work function + K.E. of electron <br/> or `hv =W_(0) +1/2mv^(2)`...........(i) <br/> If `v_(0)`is the threshold frequency which is just sufficient to eject an electron with zero velocity, i.e. v = 0. <br/> So, `W_(0) =hv_(0)`...............(ii) <br/> Putting (ii) in (i) we get <br/> `hv = hv_(0) + 1/2mv^(2)` <br/> or `1/2 mv^(2) = h(v-v_(0))`...........(iii) <br/> This equation is called Einstein.s photoelectric equation.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MBD_EM_PHY_XII_C07_E02_017_S02.png" width="80%"/><br/> From Eqn. (iii), it is clear that <br/> (i) v depends only upon v, m and h are constants. Thus the velocity of the photoelectron ejected depends upon the frequency of light. From equation (iii), it is seen that greater is the frequency v, greater is the velocity of the photoelectrons ejected. This is the first <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of photoelectric emission. <br/> (ii) If the intensity of light is increased, the number of incident photons increases which <a href="https://interviewquestions.tuteehub.com/tag/results-2643" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTS">RESULTS</a> in an increase in the number of photoelectrons ejected. This is the second law. <br/> (iii) If `v lt v_(0), v^(2)`is negative, i.e., v is imaginary. In other words, if the frequency of the incident photon is less than `v_(0)` no photoelectric effect 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>. This is the third law of photoelectric emission. <br/> (iv) The phenomenon of photoelectric emission has been conceived as an effect of collision between a photon and an electron in metal. Hence there should be no time lag between the incidence of photon and ejection of the photoelectrons. This is the fourth law.</body></html> | |
| 19. |
Blue light can eject photo electron from a photo sensitive surface while orange light cannot. Then |
| Answer» <html><body><p>Violet light can <a href="https://interviewquestions.tuteehub.com/tag/eject-7375683" style="font-weight:bold;" target="_blank" title="Click to know more about EJECT">EJECT</a> electron<br/>Red light can not eject electron<br/>Green and <a href="https://interviewquestions.tuteehub.com/tag/yellow-1464953" style="font-weight:bold;" target="_blank" title="Click to know more about YELLOW">YELLOW</a> <a href="https://interviewquestions.tuteehub.com/tag/lights-1073792" style="font-weight:bold;" target="_blank" title="Click to know more about LIGHTS">LIGHTS</a> may or may not eject electron<br/>All the above</p>Answer :D</body></html> | |
| 20. |
A photocell is illuminated by a small bright source placed 1m away. When the same source of light is placed (1/2)m away, the number of electrons emitted by photocathode would |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> by a <a href="https://interviewquestions.tuteehub.com/tag/factor-458892" style="font-weight:bold;" target="_blank" title="Click to know more about FACTOR">FACTOR</a> 2<br/>decrease by a factor 2<br/>increase by a factor <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/>decrease by a factor 4</p></body></html> | |
| 21. |
A current carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon ______ . |
| Answer» <html><body><p>Shape of the <a href="https://interviewquestions.tuteehub.com/tag/loop-1079198" style="font-weight:bold;" target="_blank" title="Click to know more about LOOP">LOOP</a><br/>Area of the loop<br/>Value of the current <br/>Magnetic field</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to torque `tau=NIABsintheta` equation `vectau`, doesnot depend <a href="https://interviewquestions.tuteehub.com/tag/upon-2315586" style="font-weight:bold;" target="_blank" title="Click to know more about UPON">UPON</a> the shape of the loop.</body></html> | |
| 22. |
A sensor to find liquid level in a tank is a cylindrical capacitor of length with outer and conductor radii R_(0)and R_(i) respectively. When a liquid fills the tank up to height h, (h le l)tank.s bottom the dielectric in lower part is liquid (K_l)and dielectric in upper part is vapour liquid (K_r) Value of capacitance of capacitor when tank is completely empty is |
| Answer» <html><body><p>`2.5 xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>`<br/>`3.5 xx10^(-9)F`<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> xx10^(-9)F`<br/>`1.1 xx10^(-9)F`</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> | |
| 23. |
The position - time graph for a particle is shown in fig. Then in the time interval of (0, 6sec) |
| Answer» <html><body><p>Particle <a href="https://interviewquestions.tuteehub.com/tag/receives-1179339" style="font-weight:bold;" target="_blank" title="Click to know more about RECEIVES">RECEIVES</a> two identical impulses<br/>Particle receives two impulses in same direction, but of different <a href="https://interviewquestions.tuteehub.com/tag/magnitudes-1083155" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNITUDES">MAGNITUDES</a><br/>Particle receives two impulses of same <a href="https://interviewquestions.tuteehub.com/tag/magnitude-1083080" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNITUDE">MAGNITUDE</a> and opposite in directions.<br/>Particle receives two impulses of different magnitudes and in opposite directions</p>Answer :A</body></html> | |
| 24. |
Name the experiment in which nucleus was first discovered. |
| Answer» <html><body><p> <br/> <br/></p>Solution :Lord Rutherford <a href="https://interviewquestions.tuteehub.com/tag/discovered-440478" style="font-weight:bold;" target="_blank" title="Click to know more about DISCOVERED">DISCOVERED</a> the nucleus from his <a href="https://interviewquestions.tuteehub.com/tag/experiment-980358" style="font-weight:bold;" target="_blank" title="Click to know more about EXPERIMENT">EXPERIMENT</a> on scattering of `<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>` particles from thin <a href="https://interviewquestions.tuteehub.com/tag/metallic-547634" style="font-weight:bold;" target="_blank" title="Click to know more about METALLIC">METALLIC</a> <a href="https://interviewquestions.tuteehub.com/tag/foils-994068" style="font-weight:bold;" target="_blank" title="Click to know more about FOILS">FOILS</a>.</body></html> | |
| 25. |
A nucleus with Z =92 emits the following in a sequence alpha, beta, alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha. The Z of the resulting nucleus is : |
| Answer» <html><body><p>78<br/>82<br/>74<br/>76</p>Solution :Here, `<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> <a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>, 4 beta^(-) and <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> beta^(+)` <a href="https://interviewquestions.tuteehub.com/tag/radiation-1175811" style="font-weight:bold;" target="_blank" title="Click to know more about RADIATION">RADIATION</a> take <a href="https://interviewquestions.tuteehub.com/tag/place-1155224" style="font-weight:bold;" target="_blank" title="Click to know more about PLACE">PLACE</a>, so change in atomic no. is `=2 xx 8 -4 xx 1+2 xx 1=14` <br/> New atomic no. =92-14=78</body></html> | |
| 26. |
A wire of resistance 8R is bent in the form of a circle. What is the effective resistance between the ends of a diameter AB ? |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/U_LIK_SP_PHY_XII_C03_E08_010_S01.png" width="80%"/> <br/><a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> <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 wire bent in the form of a circle = <a href="https://interviewquestions.tuteehub.com/tag/8r-1930864" style="font-weight:bold;" target="_blank" title="Click to know more about 8R">8R</a><br/>Resistance of each half ACB or ADB =`(8R)/(2) = 4R`<br/> The two half lengths are connected in <a href="https://interviewquestions.tuteehub.com/tag/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> across AB<br/>`R_(eff) = (4R xx 4R)/(4R + 4R) = 2R`</body></html> | |
| 27. |
For steady interference, the two sources of light must be : |
| Answer» <html><body><p>coherent<br/>monochromatic<br/>equality bright<br/>all of these</p>Answer :D</body></html> | |
| 28. |
Which of the following can be expressed in coulomb : |
| Answer» <html><body><p>`oint vecB . <a href="https://interviewquestions.tuteehub.com/tag/vec-723433" style="font-weight:bold;" target="_blank" title="Click to know more about VEC">VEC</a>(dl)`<br/>`oint vecE.vec(dl)`<br/>`oint_s epsilon_(0) vecE.vec(<a href="https://interviewquestions.tuteehub.com/tag/ds-960390" style="font-weight:bold;" target="_blank" title="Click to know more about DS">DS</a>)`<br/>`oint_s (vecB)/(mu_(0)).vec(ds)`</p>Answer :C</body></html> | |
| 29. |
A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m^(-1). If the mass of the drop is 1.6 xx 10^(-3)g, the number of electrons carried by the drop is (g = 10 ms^(-2)) |
| Answer» <html><body><p>`10^(<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>)`<br/>`10^(<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)`<br/>`10^(12)`<br/>`10^(9)`</p>Solution :For the drop to be stationary, <br/> <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> on the drop <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to electric field = Weight of the drop <br/> qE = mg <br/> `:. q = (1.6 xx 10^(-6) xx10)/(100) = 1.6 xx 10^(-7) C` <br/> Number of electrons carried by the drop is <br/> `N = (q)/(e) = (1.6 xx 10^(-7)C)/(1.6 xx 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>)C) = 10^(12)`</body></html> | |
| 30. |
A wavefront AB moving in air is incident on a plane glass surface XY as given in figure.Its position CD after refraction through the glass slab is shown along with normals A to D. The refractive index of glass w.r.t. Air will be equal to : |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/bd-389836" style="font-weight:bold;" target="_blank" title="Click to know more about BD">BD</a>)/(AC)`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/ab-360636" style="font-weight:bold;" target="_blank" title="Click to know more about AB">AB</a>)/(CD)`<br/>`(BD)/(AD)`<br/>`(AC)/(AD)`</p>Solution :`mu = (<a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> i)/(sin <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)` , here i = `theta` and r = `<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/> `therefore mu= (sin theta)/(sin phi)`</body></html> | |
| 31. |
A wavefront AB moving in air is incident on a plane glass surface XY as given in figure.Its position CD after refraction through the glass slab is shown along with normals A to D. The refractive index of glass w.r.t.air is given by : |
| Answer» <html><body><p>`(BD)/(AC)`<br/>`(sin theta)/(sin phi)`<br/>`(sin phi)/(sin theta)`<br/>`(AB)/(<a href="https://interviewquestions.tuteehub.com/tag/cd-407381" style="font-weight:bold;" target="_blank" title="Click to know more about CD">CD</a>)`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a> = (sin i)/(sin <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)` , here i = `theta` and r = `phi`<br/> `therefore mu= (sin theta)/(sin phi)`</body></html> | |
| 32. |
A sensor to find liquid level in a tank is a cylindrical capacitor of length with outer and conductor radii R_(0)and R_(i) respectively. When a liquid fills the tank up to height h, (h le l)tank.s bottom the dielectric in lower part is liquid (K_l)and dielectric in upper part is vapour liquid (K_r)When liquid fills the tank up to height h, the capacity of capacitor is |
| Answer» <html><body><p>`(2piepsilon_0l)/(<a href="https://interviewquestions.tuteehub.com/tag/ln-1076444" style="font-weight:bold;" target="_blank" title="Click to know more about LN">LN</a>(R_0lR_i)){(K_f-K_v)h/l+K_v}` <br/>`(2piepsilon_0l)/(ln(R_0lR_i)){(K_v-K_f)h/l+K_i}` <br/>`(2piepsilon_0l)/(ln(R_ilR_0)){(K_v-K_l)h/l+K_v}` <br/>`(2piepsilon_0l)/(ln(R_ilR_0)){(K_v-K_l)h/l+K_v}` </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> | |
| 33. |
A sensor to find liquid level in a tank is a cylindrical capacitor of length with outer and conductor radii R_(0)and R_(i) respectively. When a liquid fills the tank up to height h, (h le l)tank.s bottom the dielectric in lower part is liquid (K_l)and dielectric in upper part is vapour liquid (K_r)If l=2m , R_(0) = 5mm, K_l=1.4K_v =1.0 . Then , capacitance of capacitor when tank is fully filled is |
| Answer» <html><body><p>`4.5 xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>` <br/>`7.5 xx10^(-9)F` <br/>`3.5 xx10^(-9)F` <br/>`1.5 xx10^(-9)F` </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> | |
| 34. |
A uniform resistance wire of resistance 10Omega is bent in the form of a circle.The resistance between any two diameterically opposite points will be |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/5omega-1899383" style="font-weight:bold;" target="_blank" title="Click to know more about 5OMEGA">5OMEGA</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/20omega-1821725" style="font-weight:bold;" target="_blank" title="Click to know more about 20OMEGA">20OMEGA</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/10omega-1774700" style="font-weight:bold;" target="_blank" title="Click to know more about 10OMEGA">10OMEGA</a>`<br/>`2.5Omega`</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> | |
| 35. |
A particle of mass 3 xx 10^(-6) g has the same wavelength as an electron moving with a velocity 6 xx 10^(6) ms^(-1). The velocity of the particle is…………………… |
| Answer» <html><body><p>`1.82 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>^(-18) ms^(-1)` <br/>`9 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) ms^(-1)` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> xx 10^(-31) ms^(-1)` <br/>`1.82 xx 10^(-15) ms^(-1)` </p>Solution :de-Broglie wavelength of electron <br/> `lambda = (h)/(mv) = (6.63 xx 10^(-34))/(9.1 xx 10^(-31) xx 6 xx 10^(6)) = 1.214 xx 10^(-10)`m <br/> Velocity of the <a href="https://interviewquestions.tuteehub.com/tag/particle-1147478" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICLE">PARTICLE</a> <br/> `v = (h)/(m lambda) = (6.63 xx 10^(-34))/(3 xx 10^(-9) xx 1.214 xx 10^(-10)) = 1.8204 xx 10^(-15) ms^(-1)`</body></html> | |
| 36. |
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is |
| Answer» <html><body><p>1.8 <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a><br/>1.3 V<br/>0.5 V<br/>2.3 V</p>Solution :0.5 V</body></html> | |
| 37. |
A conducting rod of length l is higed at point O. It is free to rotate in a vertical plane. There exists a uniform magnetic field vec(B) in horizontal direction. The rod is released from the position shown in figure. Potential difference between the two ends of the rod is proportional to |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(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/>`l^(2)`<br/>`sin theta`<br/>`(sin theta)^(1//2)` </p>Solution :Potential difference `=(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> omega l^(2)`<br/> From energy conservation : <br/>`<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>.(1)/(2)sin theta =(1)/(2)I omega^(2)`, where `I=(ml^(2))/(3)`<br/>`rArr omega prop (sin theta)^(1//2)rArr <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a> prop omega rArr e prop (sin theta)^(1//2)` <br/> Also `e prop l^(2)`.</body></html> | |
| 38. |
In the above problem . What is the least distance from the central maximum where the bright fringe due to both the wavelength coincide ? |
| Answer» <html><body><p>0.32 cm<br/>0.48 cm<br/>0.16 cm<br/>0.08 cm</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> | |
| 39. |
An iron block of mass m = 500 kg is kept at the back of a truck moving at a speed v_(0)=90km h^(-1). The driver applies the brakes and shows down to a speed of v=54km h^(-1) in 10s. What constant force acts on the block during this time if the block does not slide on the truck-bed? |
| Answer» <html><body><p></p>Solution :The acceleration, `a=(v-v_(0))/(t)` <br/> `=(54 km h^(-1)-90kmh^(-1))/(10s)` <br/> `=(15ms^(-1)-25ms^(-1))/(10s)` <br/> `=-1m//s^(2)` <br/> or, Force F = ma <br/> `=500kg (-1 ms^(-2))` <br/> `=-500N` <br/> -ve sign <a href="https://interviewquestions.tuteehub.com/tag/indicates-1041501" style="font-weight:bold;" target="_blank" title="Click to know more about INDICATES">INDICATES</a> that the force <a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</a> opposite to the <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of the block. The magnitude of force is <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> N.</body></html> | |
| 40. |
The magnetic fields at two pointslying at same distance from an isolated poleare |
| Answer» <html><body><p>the samebothin magnitude and direction <br/>differentboth <a href="https://interviewquestions.tuteehub.com/tag/inmagnitude-1044956" style="font-weight:bold;" target="_blank" title="Click to know more about INMAGNITUDE">INMAGNITUDE</a> and direction <br/>the samein magnitude naddifferent in <a href="https://interviewquestions.tuteehub.com/tag/directions-955033" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTIONS">DIRECTIONS</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/different-951434" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENT">DIFFERENT</a> in magnitude but samein direction </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> | |
| 41. |
Define Snell's Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism. |
| Answer» <html><body><p></p>Solution :Snell's law : <br/> The ratio of the sine of the <a href="https://interviewquestions.tuteehub.com/tag/angle-875388" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLE">ANGLE</a> of incidence to the sine of angle of refraction is constant, called the refractive index of the medium. <br/> `(sini)/(sinr)=mu"(constant)".` <br/> Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is `mu`. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = <a href="https://interviewquestions.tuteehub.com/tag/emergent-2610413" style="font-weight:bold;" target="_blank" title="Click to know more about EMERGENT">EMERGENT</a> ray. <br/> `"Let angle of incidence "= r_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)"angle of emergence "=i_(2)` <br/> `"angle of refraction "=r_(1)"angle of refraction at R"=r_(2)` <br/> After travelling through the prism it falls on AC and emerges as RS. <br/> The D = angle of deviation. <br/> From the `DeltaQRT` <br/> `r_(1)+r_(2)+angleT=<a href="https://interviewquestions.tuteehub.com/tag/180-279527" style="font-weight:bold;" target="_blank" title="Click to know more about 180">180</a>^(@)"..................(1)"` <br/> From the quadrilateral AQTR <br/> `angleA+angleT=180^(@)` <br/> `angleT=180^(@)-A."..............(2)"` <br/> From the equations (1) and (2) <br/> `r_(1)+r_(2)+angleT=180^(@)" we get "` <br/> `r_(1)+r_(2)+180^(@)-A=180^(@)` <br/> `r_(1)+r_(2)=A."...............(3)"` <br/> from the `DeltaQUR` <br/> `i_(1)-r_(1)+i_(2)-r_(2)+180^(@)-D=180^(@)` <br/> `i_(1)+i_(2)-(r_(1)+r_(2))=D` <br/> `i_(1)+i_(2)-A=D""[because r_(1)+r_(2)=A]` <br/> `i_(1)+i_(2)=A+D"....................(4)"`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C02_E03_008_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/minimum-561095" style="font-weight:bold;" target="_blank" title="Click to know more about MINIMUM">MINIMUM</a> deviation : Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation `'delta'` as shown in the fig. <br/> When D decreases the two angles `i_(1) and i_(2)` become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e., `i_(1)=i_(2)` <br/> As `i_(1)=i_(2), r_(1)=r_(2)` <br/> `therefore i_(1)=i_(2)=i, r_(1)=r_(2)=r,` <br/> substituting this in (1) and (2) we get<br/> `"2r = A"rArr r=A//2` <br/> `i+i=A+delta rArr i=(A+delta)/(2)` <br/> According to Snell's law `mu=("Sin i")/("Sin r")` <br/> `mu=("sin"((A+delta)/(2)))/("Sin A/2")` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C02_E03_008_S02.png" width="80%"/></body></html> | |
| 42. |
A conductor has got different areas of cross sections. Are the currents same at different cross sections? What conservation law helps you to decide the answer? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/conductor-928880" style="font-weight:bold;" target="_blank" title="Click to know more about CONDUCTOR">CONDUCTOR</a> may be having different cross <a href="https://interviewquestions.tuteehub.com/tag/sections-1198817" style="font-weight:bold;" target="_blank" title="Click to know more about SECTIONS">SECTIONS</a> at different <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a> along its <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a>. But the current i will be the same for all cross sections of the conductor. This is a consequence of the law 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.</body></html> | |
| 43. |
The capacity of each condenser in the following fig. is .C.. Then the equivalent capacitance acrossA and B is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>/4<br/>3C/4<br/>4C/3<br/>3C</p>Answer :C</body></html> | |
| 44. |
How many times in succession should the light fraction be separated in the centrifuge to obtain a mixture containing 80% of light uranium isotope? |
| 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> :20times</body></html> | |
| 45. |
Non-transfer of pollen from anther to stigma of the came flower due to a mechanical barrier is |
| Answer» <html><body><p>Dichogamy<br/>Herkogamy<br/>Heterostyly<br/>Cleistogamy</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> | |
| 46. |
A solid sphere rolls down on an inclined plane of 30^(@) inclination. Ratio of acceleration when it rolls and slides is : |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>:<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>`<br/>`7:5`<br/>`2:5`<br/>`3:5`</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> :Here `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>)=(5)/(7)gsinthetaanda_(s)=gsintheta` <br/> `a_(r):a_(s)::5:7`</body></html> | |
| 47. |
A carbon resistor is shown in the Fig. Using colour code, write the value of the resistance. |
| Answer» <html><body><p></p>Solution : The <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of <a href="https://interviewquestions.tuteehub.com/tag/resistance-1186351" style="font-weight:bold;" target="_blank" title="Click to know more about RESISTANCE">RESISTANCE</a> `47 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^2 Omega = 4.7 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> Omega` .</body></html> | |
| 48. |
Write the condition of destructive interference in terms of path difference. |
| Answer» <html><body><p>path difference =`n <a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>`<br/>path difference `=(n+(1)/(2)) lambda`<br/>path difference `=<a href="https://interviewquestions.tuteehub.com/tag/2n-300431" style="font-weight:bold;" target="_blank" title="Click to know more about 2N">2N</a> lambda`<br/>path difference `=(2n+1) lambda`</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> | |
| 49. |
Obtain the expression for the equivalent resistance for 3 resistors connected in parallel and also write the expression of equivalent resistance for connection of' n' resistors. |
| Answer» <html><body><p></p>Solution :`rArr "" R_(1) ,R_(2) and R_(3)` are connected between a and b points as shown in figure. <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XII_P1_C03_E01_035_S01.png" width="80%"/><br/> `rArr` By connecting terminals of <a href="https://interviewquestions.tuteehub.com/tag/battery-893842" style="font-weight:bold;" target="_blank" title="Click to know more about BATTERY">BATTERY</a> of voltage V with a and b, I current will be passed through `R_(1) , R_(2), R_(3) " are " I_(1), I_(2) , I_(3)` respectively. <br/> `rArr` According to Ohm.s law, p.d. <a href="https://interviewquestions.tuteehub.com/tag/across-367070" style="font-weight:bold;" target="_blank" title="Click to know more about ACROSS">ACROSS</a> `R_(1), R_(2), R_(3)` is V. <br/> `rArr V = I_(1) R_(I) rArr I_(1) = (V)/(R_(1)) "" `....(1) <br/> `V = I_(2) R_(2) rArr I_(2) = (V)/(R_(2)) "" ` ....(2) <br/>and V = `I_(3) R_(3) rArr I_(3) = (V)/(R_(3)) ` ... (3) <br/> `rArr` At point .a. , <br/> `I = I_(1) + I_(2) + I_(3) ""` ... (4) <br/> By substituting values of equation (1), (2) and (3) in equation (4), <br/> `I = (V)/(R_(1)) + (V)/(R_(2)) + (V)/(R_(3))` <br/> `therefore (I)/(V) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) "" `[ Dividing by V] <br/> `rArr " If"(I)/(V) = (1)/(R_(eq)) , ` then <br/> `(1)/(R_(eq) ) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) ` <br/> If `R_(eq) ` is represented as `R_(p)`, then `(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))`<br/> `rArr` For <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> resistance of .n. unequal resistors in parallel,<br/> `(1)/(R_(eq))= (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) + ...+ (1)/(R_(n))`<br/> `rArr` For equivalent resistance of .n. equal resistors of resistance <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> in parallel,<br/> `(1)/(R_(eq)) = (n)/(R)` <br/> `therefore R_(eq) = (R)/(n)`</body></html> | |
| 50. |
A milliammeter of range 10 mA has a coil of resistance 1Omega. To use it as a voltmeter ofrange 10 V, the resistance that must be connected in series with it is |
| Answer» <html><body><p>`999Omega` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/1000omega-1771215" style="font-weight:bold;" target="_blank" title="Click to know more about 1000OMEGA">1000OMEGA</a>` <br/>`9Omega` <br/>`99Omega` </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> :`R=(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>)/(I_(g))-R_(g)=(10)/(10xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))-1=999Omega`</body></html> | |