Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 40451. |
A short bar magnetis arrangedwithits north polepointinggergraphical north. It is found that thehorizontal componentof earth'smagneticinduction(B_(H)) is balacedby themagnetic induction of the magnetat a point whichis at a distanceof 20 cmfrom its centre .The magnetic momentofthe magnet is ( if H = 4 xx 10^(-5) Wbm^(-2)) |
|
Answer» |
|
| 40452. |
A: Red substance seems to be black in presence of yellow light.R: Scattering of red colour is least. |
|
Answer» Both ASSERTION and REASON are TRUE and the reason is CORRECT EXPLANATION of the assertion. |
|
| 40453. |
Activity of radioactive element decreased to one third of original activity I_0 in 9 years. After further 9 years, its activity will be |
|
Answer» `I_0` |
|
| 40454. |
Angle of prism is A and its one surface is silvered. Light rays falling at an angle of incidence 2A on first surface return back through the same path after suffering reflection at second silvered surface. Refractive index of the material of the prism is |
|
Answer» `2SINA` |
|
| 40455. |
In the figure shown , , a parallel beamof light is incident on the plane of the slits of a Young.s doubleslit experiment. Lightincident on the slit S_(1)passes througha mediumof variablerefractiveindexmu =+ax (where .x. is the distnce from tehplaneof slitsas shown )uptoa distance.l. before fallingon S_(1) . Restof the space is filled with air .If at .O. minima is formed , then the minimum value of the positive constant a ( in terms of l and wavelength lambda in aire is ) is : |
|
Answer» `LAMBDA/l ` |
|
| 40456. |
(I): In a beta^(-1) decay in a nucleus, a daughter nucleus that has discrete energy states is produced. The daughter nucleus reaches ground state from excited state by emitting gamma-rays. (II)- The binding energy of hydrogen nucleus is far less than the binding energy of helium nucleus: |
|
Answer» 1 FALSE, II False |
|
| 40457. |
Statement I: A closed solenoid is placed in an external magnetic field. Both of the magnetic field and axis of the solenoid are directed along z-axis. No electromotive force will be induced if the solenoid is rotated along its own axis. Statement II: Electromagnetic induction in a conducting coil takes place only when the magnetic flux linked with the coil changes with time. |
|
Answer» |
|
| 40458. |
The acceleration of an oscillating particle of mass 0.1 kg described by the equation d^2y/dt^2=7-3y. If the amplitude of motion is squrt2 cm. The total energy is |
| Answer» ANSWER :D | |
| 40459. |
The conical pendulum is in steady circular motion with constant angular velocity omega as shown in Fig. Mass of the particle is M and string makes angle alpha with vertical. (a) Find angular momentum of the particle about the center C of the circle and the hinge O about which the string is attached. (b) Also check whether L is changing or is constant. |
|
Answer» Solution :(a) Conceptualize/Classify: We start by finding `L_(C)`, angular momentum of mass M about point C. We can see from Fig. that velocity of the particle is always going to be perpendicular to the line joining particle with the center C. Also, both vectors r and v are lying on horizontal (xy) plane, so the angular momentum, `L_(C)` will be in the positive z direction. It has magnitude `|r_(BOT)||p|=|r||p|=rp`, where r is the radius of the circular motion as we can see from Fig. Calculation: The magnitude of momentum is given by `|p|=Mv=Mromega` The angular momentum about the center C of the circle `L_(C)=rxxp` = `Mr^(2)omegahatk` Now LET us evaluate the angular momentum `L_(O)` about the point O located at the hinge. From Fig. `vecL_(0)=vecrxxvecp` Velocity of the particle is perpendicular to the line OA as SHOWN in the figure. `|L_(O)|=|r.xxp||r.||p|=L|p|` or `|L_(O)|=Mlromega` where `|r.|=l`, is the length of the string. (b) Calculation: We must note that `L_(C)` is constant, both in magnitude and direction. The magnitude of `L_(O)` is constant, but its direction is not constant as can be seen from Fig. `L_(O)` is always perpendicular to the plane containing r. and p. As the particle moves, the plane containing r. and p vector hinges, thus changing the direction of `L_(O)` at different times. As the bob swings around, `L_(O)` rotates as shown in fig. The vertical component of `L_(O)` is not moving, but the horizontal component rotates in the circle with the bob. We can draw the angular momentum at any position we CHOOSE, only the magnitude and direction of L have to be same. 
|
|
| 40460. |
While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonace condition at a column length of 18 cm duiing winter. Repeating the same experiment during summer, she measures the column be x cm for the second resonance. Then: |
|
Answer» `18 GT x` |
|
| 40461. |
Location of image after multiple events Find the location of final image as shown in Fig.34-13. Take the refractive index of the slab as 4/3. |
|
Answer» Solution :(1) This is again an example of multiple events before final image is FORMED. The rays coming from O will pass through the slab, get refracted by it and then reflected by the mirror, and again refracted by the slab. This refraction will from the final image. Whenever we refer to the image of the system, it refers to this image. (2) The mirror will not see the object at 15cm but shifted toward it. After reflection, image will not be directly formed at the point where the mirror focuses the light but will be shifted in the direction of light incident on the slab. Calculation : The very first event is the refraction from the glass slab. It causes a SHIFT in the direction of the incident rays. This normal shift is given by Eq. 34-8. `x=t(1-(1)/(n))` `=4xx(1-(3)/(4))=1cm` Thus, the object will seem to be shifted toward the mirror (in the direction of the incident rays). So, the mirror will see the object at `15-1=14cm`. So, for the mirror `u=-14cm` . After reflection from the mirror, `v=(14xx10)/(14-10)=-35cm` One more time reflection will occur as light after reflecting from mirror will pass through the slab. Again, normal shift will be 1cm, but in the direction of propagation of light. Now the light rays are propagating in the LEFT direction. Therefore, final image will be at a DISTANCE of `36cm` from the mirror. Learn : If there are more than one slabs, the total normal shift as Normal shift =`t_(1)[1-(n)/(n_(1))]+t_(2)[1-(n)/(n_(2))]+......` 
|
|
| 40462. |
Matter inside a ''white dwarf'' is in a state of degeneracy, and the dependence of the pressure on the density is of the form P=Arho^(5//3), where P is the pressure and rho is the density. Find the expression for the constant A and show that the pressure is due to the electron gas, the pressure due to the heavy particles being negligible. |
|
Answer» `PV^(5//3)=H^(2)/(5m_(e))(3/(8pi))^(2//3)N_(e)^(5//3)` There are two electrons to each helium nucleus, so `N_(e)=2N_(alpha)=2M//m_(alpha)` where M is the mass of the star, and `m_(alpha)=4.002xx1.66xx10^(-27)kg` is the mass of a helium nucleus. Hence `PV^(5//3)=AM^(5//3),orP=Arho^(5//3)`, where `A=h^(2)/(5m_(e)(m_(alpha)//2)^(5//3))(3/(8pi))^(2//3)=3.2xx10^(5)Pa*m^(5)*kg^(5//3)` |
|
| 40463. |
Two capacitors A and B with capacities C_1 and C_2 are charged to potential difference of V_1 and V_2 respectively. The plates of the capacitors are connected as shown in the figure with one wire free from each capacitor. The upper plate of A is positive and that of B is negative. An uncharged capacitor of capacitance C_3 and lead wires falls on the free ends to complete circuit, then |
|
Answer» the final charge on each capacitor are same to each other Charge on negative plate of `B` is `-C_(2)V_(3)`. When `d` plate of capacitor `C` is connected with the plate of ``, then the total charge of `(d,a)` plate system will be `C_(1)V_(1)` (conservation of charge). SIMILARTY on `(c,s)` plates, total charge will be `-C_(2)V_(2)`. The total charge on `(h,g)` plate system will be `+C_(2)V_(2)-C_(1)V_(1)`. 
|
|
| 40464. |
A train starting from rest is moving along a straight track with a constant acceleration fo 2.5m//s^(2). A passenger at rest in the train observes a particle of mass 2kg to be at rest on the floor with which it has a coefficient of friction mu_(s) = mu_(k) = 0.5 .Six seconds after the starting of the train , a horizontal force F = 13N is applied to the particle for two seconds duration. The passenger now observes the particle to move perpendicular to the direction of the train. (a) calculate the kinetic energy of the particle with respect to the passenger at the end of 8 seconds after starting of the train. ( b) repeat the calculate of ( a) for an observer on the ground. |
|
Answer» |
|
| 40465. |
A wire of length l carries a current I along the Y direction and mgnetic field is given by vec(B) = (beta)/(sqrt(3)) ( hat(i) + hat(j) + hat(k)) T. The magnitude of Lorentz force acting on the wire is ................. . |
|
Answer» `sqrt((2)/(sqrt(3))) beta`IL |
|
| 40466. |
What was Noodle's suggestion for the crew to read books: |
|
Answer» He SUGGESTED to THROW the BOOKS away |
|
| 40467. |
(a) Using the necessary ray diagram, derive the mirror formula for a concave mirror. (b) In the magnified image of a measuring scale (with equidistant markings) lying along the principal axis of a concave mirror, the markings are not equidistant. Explain. |
|
Answer» Let us consider RIGHT angled triangles A.B.F and MPF, which are similar triangles. Therefore, `(B.A.)/(PM)=(B.F)/(FP)` But as PM= AB, hence, we have `(B.A.)/(BA)=(B.F)/(FP)` ....(i) Again triangles APB and A.PB. are also similar triangles and we have `(B.A.)/(BA)=(B.P)/(BP)`  Comparing (i) and (ii), we get `(B.P)/(BP)=(B.F)/(FP)=(B.P-FP)/(FP)` As per sign convention followed: BP = -u, B.P =-v, and FP =-f Therefore, we get `((-v))/((-u))=((-v)-(-f))/((-f))rArrv/u=(v-f)/f` `rArrvf=uv-uf` or `uf+vf=uv` DIVIDING throughout by uvf, we get `1/v+1/u=1/f`, which is the requisite mirror formula. (b) Magnification of a mirror is given as: `m=-v/u=f/(f-u)` As different markings of a measuring scale lying along the principal axis of a concave mirror are at different positions (i.e., u is different for them), we get magnified images of these marks but magnification differs from one mark to another. As a result, the markings in image are not equidistant. |
|
| 40468. |
A: Like currents repel and unlike currents attract eachother (in conductor). R: Magnetic force acts in the direction of current. |
|
Answer» If both Assertion & REASON are true and the reason is the correct explanation of the assertion then mark (1) |
|
| 40469. |
The technology used for stopping the brain from processing pain is |
|
Answer» PRECISION medicine |
|
| 40470. |
A hemisphere is uniformaly charged positively. The electric field at a point on a diameter away from the centre is directed |
|
Answer» PERPENDICULAR to the DIAMETER |
|
| 40471. |
A rope of length 5m, is kept on frictionless surface and a force of 5N is applied to one of its end. Find tension in the rope at 1 m from this end |
|
Answer» 1N |
|
| 40472. |
A system displaying a hysteresis loop such as a ferromagnet is a device for storing memory '. Explain the meaning of the statement . |
| Answer» Solution :magnetisation of a ferrognetic MATERIAL indicates the number of MAGNETIC CYCLES undergone by the samplein aappliedmagnetic field . Thus it can be said that the sample stores the history of its magnetisation as its MEMORY . | |
| 40473. |
If vec A and vec B are two vectors given by vec A = 2 hati +3 hat j and vec B = hati + hat j. The magnitude of the component of vec A along vec B is |
|
Answer» `5/sqrt 2` |
|
| 40474. |
There are two bodies of mass 100 kg, 10000 kg separated by a distance of 1m. At what distance from the smaller body the intensity of gravitational field is zero: |
|
Answer» `(1)/(9)m`  `rArr (GM_(1))/(X^(2))=(GM_(2))/((1-x)^(2)) rArr (100)/(x^(2))=(10,000)/((1-x)^(2))` `rArr (1-x)/(x)=10 rArr 11x=1` or `x=(1)/(11)m` Hence correct CHOICE is (c ). |
|
| 40475. |
A stone is projected from the top of a tower with velocity 20ms^(-1) making an angle 30^@ with the horizontal. If the total time of flight is 5s and g=10ms^(-2 ) |
|
Answer» the HEIGHT of the tower is 75 m. |
|
| 40476. |
In which situation is there a displacement current but no conduction current: |
| Answer» Solution :During charging or discharging there is a DISPLACEMENT CURRENT, but no conduction current between the PLATES of CAPACITOR. | |
| 40477. |
Name the phenomenon in which an emf is induced in a coil due to the change of current in the same coil. |
| Answer» Solution :SELF induction is the PHENOMENON by which an emf is induced in a coil DUE to the CHANGE of currentin the same coil. | |
| 40478. |
In Q. 103, the radiation force on the roof will be : |
|
Answer» `8.53 xx 10^(-5) W` `=(1.6 xx 10^(5))/(3 xx 10^(8))=5.32 xx 10^(-4)N` |
|
| 40479. |
Explain why photoelectrons come out with different kinetic energies even though the incident radiation is monochromatic. |
| Answer» Solution :In a metal, different ELECTRONS are with different ENERGY levels. Hence the electrons will COME out with different energies. | |
| 40480. |
We know that the Sun is fundamental source of all energy that we use. Huge amount of energy is being produced in the Sun and this energy is radiated all around in the form of electromagnetic waves of several possible wavelengths. We can treat the Sun as a point source because it radiates energy uniformly in all directions. Intensity of wave at a point is defined as amount of energy passing that point per unit time and per unit area. We know that Earth is at a distance approximately 1.5 xx10^(11) m from the Sun and assume that intensity of radiation of the Sun reaching Earth's surface is 10^3 W/m^2. Howmuchforceis applied by the radiationon aroofis size9 m xx 10 m.Assumethat matericalof roofis perfectlyabsorbing . |
|
Answer» `2XX 10^(-4)N` ` F= (U )/( c) = (IA )/(c ) = ( 10 ^3 xx 90 ) /( 3xx 10^8 ) = 3xx 10^(-4) N` |
|
| 40481. |
In positive logic, logic state 1 corresponds to |
|
Answer» POSITIVE VOLTAGE |
|
| 40482. |
Calculate, period of one complete cycle of intensity being received by the detector |
|
Answer» `2sec` |
|
| 40483. |
Assertion: If two tuning forks are sonuded together, they produce 5 beats/s. If by putting wax on one tuning fork, beat frequency increases to 6 beats/s, then one can conclude that wax is put on tuning fork with lesser frequency. |
|
Answer» ASSERTION is TRUE and REASON is CORRECT explanation of Assertion. |
|
| 40484. |
In the secondary circuit of a potentiometer , a cell of internal resistance 1.5 Omega gives a balancing length of 52 cm. To get a balancing length of 40 cm , how much resistance is to be connected across the cell ? |
|
Answer» Solution :`R= 1.5 Omegal_1= 52 CM , l_2 = 40 cm` `r/R = (l_1 - l_2)/(l_2)` Resistance `R=(r.l_2)/((l_1 - l_2)) =(1.5xx40)/((52-40))` `=(1.5xx40)/(12)= 5 Omega` |
|
| 40485. |
The velocity of a body moving vertically up is 49 ms^(-1)at half the maximum height. The height to which it could further rise is |
|
Answer» 245 m |
|
| 40486. |
Moment of inertia of a circular wire of mass M and radius R about its diameter is : |
|
Answer» `MR^(2)//2` M.I. of a ring about ONE of diameter = `(MR^(2))/(2)` 
|
|
| 40487. |
A metal sheet is placed in a variable magnetic field which is increasing from zero to maximum. Induced current flows in the directions as shown in figure . The direction of magnetic field will be __ |
|
Answer» NORMAL to the paper, inwards |
|
| 40488. |
If a simple harmonic motion is represented by (d^(2)X)/(dt^(2))+alphax=0 its time period is : |
|
Answer» `2PI//alpha` Comparing it with STANDARD equation of SHM. `(d^(2)X)/(dt^(2))+W^(2)X=0` we have `w^(2)=a` Time period `T=(2pi)/(w)=(2pi)/(sqrt(a))` So the CORRECT choice is (b). |
|
| 40489. |
The potential energy U=3ax^(3)-2bx^(2). The force constant is represented by |
|
Answer» 8b We know force constant, `k=(d^2 U)/(dx^2)` `THEREFORE (dU)/(dx)= 9ax^(2) - 4 bx`. At EQUILIBRIUM, `(dU)/(dx)=0` `therefore 9ax^(2) - 4 bx =0` or `x=(4b)/(9a)`. `therefore (d^2 U)/(dx^2)=18 ax-4b`. At `x=4b//9a`. `k=(d^2 U)/(dx^2) = (18a xx 4b)/(9a) - 4b-=4b`. |
|
| 40490. |
A train is moving in an elliptical orbit in anticlockwise sense with a speed of 110ms-1. Gaurd is also moving in the given direction with same speed as that of train. The ratio of lengths of major and minor axes is 4/3. Driver blows a whistle of 1900Hz at P, which is received by guard at .S.. The frequency received by gaurd is (Velocity of sound V = 330ms-1, OP = b and OS = a) |
| Answer» ANSWER :B | |
| 40491. |
What is the power factor of a series LCR circuit in which X_(L)gtX_(C) ? |
| Answer» SOLUTION :`cosphi=(R)/(Z)` where `Z=sqrt(R^(2)+(X_(L))^(2))` for `X_(L)gtX_(C)` | |
| 40492. |
What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building 'memory stores' in a modern computer? |
| Answer» Solution :CERAMICS, (specially treated barium IRON oxides) also called FERRITES, are used for coating magnetic tapes in a cassette recorder or for BUILDING memory stores in a MODERN computer system. | |
| 40493. |
for the network shown in figure, determine the value of R and the current through it, if the current through the branch AO is zero. |
|
Answer» |
|
| 40494. |
A charge is distributed uniformly over a ring of radius r. obtain an expression for the electric intensity vecE at a point on the axis of the ring. Hence show that for points at large distances, the ring behaves as a point charge. |
|
Answer» Solution :Let us consider a circular loop of radius a held perpendicular to the plane of paper. Let whole of loop carries a charge q. charge on a small element of loop MN of LENGTH dl is `DQ=(q)/(2pia)dl`. . . (i)  Electric field at P due to charge element MN is `dE=(1)/(4piepsi_(0))(dq)/(r^(2))` `=(1)/(4piepsi_(0))(dq)/((x^(2)+a^(2)))` along PX dE can be resolved into two rectangular components. dE cos `theta` acting horizontally along PZ and `dE sin theta` acting along PY. For a pair diagrammatically opposite element M., N., dE cos `theta` acts along PX and hence ADDD up but `dE sin theta` along PY. cancel with dE sin `theta` along PY. `therefore sumdE sin theta=0` for full loop and dE cos `theta` addd up `therefore E=sumdEcostheta` `=sum(1)/(4piepsi_(0))(dq)/((x^(2)+a^(2)))xx(x)/((x^(2)+a^(2))^(1//2))` or `E=int(1)/(4piepsi_(0))xx(x)/((x^(2)+a^(2))^(3//2))xx(q)/(2pia)dl` [Substituting the value of dq from (i)] or `E=(1)/(4piepsi_(0))xx(x)/((x^(2)+a^(2))^(3//2))xx(q)/(2pia)intdl` But `intdl=l=2pia` `therefore E=(1)/(4piepsi_(0))xx(x)/((x^(1)+a^(2))^(3//2))xx(q)/(2pia)xx2pia` or `E=(qx)/(4piepsi_(0)(x^(2)+a^(2))^(3//2))`. . (ii) Special case. when `x gt gt a ` i.e., P lies far off, then `a^(2) lt lt x^(2)` and hence can be neglected. `therefore E=(qx)/(4piepsi_(0)(x^(2))^(3//2))=(qx)/(4piepsi_(0)x^(3))` or `E=(1)/(4piepsi_(0))xx(q)/(x^(2))` which is the expression for E at a distance x from a charge q. therefore circular loop of charge behaves as a point when P is at a very very LARGE distance from the loop. |
|
| 40495. |
Consider a zener diode with the breakdown voltage 6.2V.How is zener diode different from an ordinary diode? |
| Answer» Solution :ZENER is HEAVILY doped compared to ordinary DIODE (or) Zener diode is used in reverse BIAS whereas ordinary diode works in forward bias. | |
| 40496. |
Which one of the following is a possible nuclear reaction? |
|
Answer» `""_(5)^(10)B+""_(2)^(4)He rarr""_(7)^(13)N+""_(1)^(1)H` |
|
| 40497. |
The following observations were taken for determining surface tension T of water by capillary method : diameter of capillary, D=1.25xx10^(-2)m rise of water, h=45xx10^(-2)m. Using g=9.80m//s^(2) and the simplified relation T=(rhg)/(2)xx10^(3)N//m, the possible error in surface tension is closest to : |
|
Answer» 0.1 |
|
| 40498. |
Electric field due to an infinite sheet of charge having surface density sigma is E Electric field due to an infinite conducting sheet of same surface density of charge is |
|
Answer» `E//2` |
|
| 40499. |
If number of turns in primary and secondary coils is increased to two times each, the mutual inductane ..... |
|
Answer» becomes FOUR TIMES MUTUAL INDUCTANCE =`M=phi/I=(mu_0 N_1N_2A)/L` `therefore M prop N_1N_2` `therefore M_2/M_1=(N._1N._2)/(N_1N_2)=((2N_1)(2N_2))/(N_1N_2)=4` `therefore M_2=4M_1` `therefore` Mutual inductance becomes four times. |
|
| 40500. |
If diffraction pattern obtained by using violet light instead of patterns obtained by red light, then ... |
|
Answer» diffraction pattern will disappear. `:.sin theta lambdaprop lambda and lambda_(v) lt lambda_(R)` sin FUNCTION increases in first quadrant. `:. theta_(v) lt theta_(R)` `:.` The pattern of interference FRINGES OBTAINED closer. |
|