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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.
| 6301. |
Assertion: Thermionic emission occurs in metals when they are heated to a very high temperature. Reason: Electrons emitted due to the electromagnetic radiation. |
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Answer» |
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| 6302. |
In Fig. 30-44, after switch S is closed at time t = 0, the emf of the source is automatically adjusted to maintain a constant current i through S. (a) Find the current through the inductor as a function of time. (b) At what time is the current through the resistor equal to the current through the inductor? |
| Answer» SOLUTION :i[1-exp(- Rt/L)],(B)(L/R)ln3 | |
| 6303. |
A TV signal, which contains both voice and picture information, is allocated a bandwidth of |
| Answer» Answer :B | |
| 6304. |
Briefly discuss the observations of Herts, Hallwachs and Lenard. |
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Answer» Solution :Hertz `:` (i) Heinrich Hertz generate and detct electromagnetic wave with high voltage induction coil to cause a spark discharge between two metallic spheres. (ii) When a sparkis formed, the charge will oscillate backand forth rapidly and the electromagnetic waves are produced. (iii) The electromagnetic waves thus produced were detected by a detector that has a copper wirebent in the shape of a circle. (iv) Although the detection of waves is successful, there is a problem in observing the tiny spark produced in the detector. (v) In small dectector spark becamemore vigorous whenit was exposed to ultraviolet light. (vi) The reason for this behaviour of the spark was not known at that time. (vii) Later it was found that it isdue to the photoelectric emission. (viii) Whenever ultraviolet light is incident on the metallic sphere, the electrons on the OUTER surface are emitted which caused the spark to be more vigorous. Hallwachs' observation `:` (i) Wilhelm Hallwachs, a German physicist confirmed that the strange behaviour of the spark is due to the action of ultraviolet light with his simple experiment. (ii) A clean circular plate of zinc is mounted on an insulatingstand and it attached to a gold leaf electroscope by a wire. (iii) When the uncharged zinc plate is irradiated by ultraviolet light from an arc lamp, it becomes positviely charged and the leaves will open. (iv) Further, if the negatively charged zinc plate is exposedto ultraviolet light, the leaves will close as the charges leaked away quickly. (v) If the plate is positively charged, it becomes more positive upon UV rays irradiation and the leaves will open further. (vi) From these observations, it was concluded that negatively charged electrons were emitted from the zinc plate under the action of ultraviolet light.  LENARD's observation `:` (i) Lenard studied this electron emission phenomenon in DETAIL. His simple EXPERIMENTAL setup is as shown in Figure. (ii) The apparatus consists of two metallic plates A and C placed in an evacuaed quartz bulb B are connected in the circuit.  (iii) When ultraviolet light is incident on the negative plate C, an electric current flows in the circuit that is INDICATED by the deflection in the galvanometer. (iv) On other hand, if the positive plate is irradiated by the ultraviolet light, no current is observed in the circuit. (v) From these observations, it is concluded that when ultraviolet light falls on the negative plate, electrons are ejected from it which are attracted by the positive plate A. (vi) On reaching the positive plate through the evacuated bulb, the circuit is completed and the current flows in it. (vii) Thus, the ultraviolet light falling on the negative plate causes the electron emission from the surface of the plate. |
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| 6305. |
Statement I. The value of acceleration due to gravity does not depend upon the mass of the body on which it acts. Statement II. Acceleration due to gravity (g) is a constant quantity. |
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Answer» STATEMENT-I is TRUE, Statement-II is true and Statement-II is CORRECT explanation for Statement-I. So correct choice is c. |
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| 6306. |
निम्नलिखित में से कौन-सा रिक्त समुच्चय है? |
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Answer» `{x:x € R, x^2–1 = 0}` |
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| 6307. |
____is synthesized top down approach |
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Answer» BALL milling |
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| 6308. |
Obtain the resonant frequency omega_(r ) of a series LCR circuit with L = 2.0H. C = 32 mu F and R = 10 Omega. What is the Q-value of this circuit ? |
| Answer» SOLUTION :`125s^(-1),25` | |
| 6309. |
A capacitor having capacitance 1 mu F with air is filled half -half with two dielectrics as shown . How many times capacitance will become ? |
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Answer» Solution :Answer (2) NEW capacitance will be `(8C)/(2)+(4C)/(2)` |
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| 6310. |
As we move away from the edge into the geometrical shadow of a straight edge, the intensity of illumination |
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Answer» decreases |
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| 6311. |
The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V , what increase in plate potential will keep the plate current unchanged |
| Answer» Answer :B | |
| 6312. |
What is the power of two convex lenses offocal length 1m placed in contact. |
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Answer» <P> SOLUTION :P = `P_1+P_2` = 2D |
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| 6313. |
A body is projected vertically upwards and attains maximum height H. If the time to reach height 'h' and H are in the ratio of 1:3, then the ratioh/H is : |
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Answer» `4/9` |
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| 6314. |
Establish a relation between coverage distance and height of transmitting antenna. |
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Answer» Solution :Let `AB` be a T.V. tower of height `h` above the earth. `T.V.` signal is received within a circle of `CQ=QD` on the surface of earth. Let `R` be the radius of the earth. Let `CQ=QD=d` and `AB=h`. `:. OB=R+h` In rt. `/_d` triangle `OCB,` `OB^(2)=OC^(2)+BC^(2)` `(R+h)^(2)=R^(2)+d^(2)` `d^(2)=(R+h)^(2)-R^(2)` `=R^(2)+h^(2)+2Rh-R^(2)` or `d^(2)=h^(2)+2Rh` SINCE `h^(2)` can be neglected as compared to `2hR` `:.d^(2)=2Rh` or `d=sqrt(2hR)` or `h=(d^(2))/(2R).`  AREA covered by T.V. signal `=pid^(2)` `=pi2hR=2pihR` Population covered = Area covered `XX` Population density Since `dpropsqrt(h,)( :.2R` is constant ) So greater the height of T.V. transmitting antenna, grater is its range. |
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| 6315. |
In this lesson, the narrator talks about how he advises everybody what they should do to protect the: |
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Answer» elderly |
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| 6316. |
Distinguish between HF chokes and LF chokes. |
Answer» SOLUTION :
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| 6317. |
फ्रेंकल दोष के कारण आयनिक ठोसों का घनत्व |
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Answer» घटता है |
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| 6318. |
The difference between the measured value and the true value is known as |
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Answer» RELATIVE error |
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| 6319. |
An athlete completes one round of a circular track of radius R in 20 second. What will be his displacement at the end of 2 minute 20 second? |
| Answer» ANSWER :A | |
| 6320. |
The first factor length f_(1) for refraction at a spherical surface is defined as the value of u corresponding to v=oo (as shown) with refractive indices of two mediums, as n_(1) and n_(2). The second focal length f_(2) is defined as value of v for u=oo. |
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Answer» `f_(2)` is EQUAL to `(n_(2)R)/((n_(2)-n_(1))` |
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| 6321. |
An electron enters a region with uniform velocity and it moves in the region with the same velocity. In the region |
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Answer» there MAY be ELECTRIC field, but no MAGNETIC field |
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| 6322. |
What is the flux through a cube of side a if a point charge q is at one of its corner ? |
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Answer» `Q/(epsilon_0)` |
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| 6323. |
In a meter bridge with a standard resistanceof 15Omega in the right gap, the ratio of balancing length is 3:2. Find the value of the other resistance. |
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Answer» <P> SOLUTION :`Q=15Omega, l_(1):l_(2)=3:2``(l_(1))/(l_(2))=(3)/(2)` `(P)/(Q)=(l_(1))/(l_(2)) rArr P=Q(l_(1))/(l_(2))` `P=15(3)/(2)=22.5Omega` |
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| 6324. |
(A): A bird perches on a high power lines and nothing happends to the bird. (R ): The level of bird is very high from the ground. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct explanation of .A. |
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| 6325. |
A particle is moving in a circular path of given radius with (i) constant (ii) increasing (iii) decreasing number of rotations per second. What happens to work done in three cases. |
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Answer» SOLUTION :We know that the work done is equal to the change in KINETIC energy of the SYSTEM. Hence (i) When speed is constant, K.E. does not change so , work done is zero. (ii) When K.E. is increasing, work is done by EXERTING force. (iii) When K.E. is decreasing, work is done by the body on the force. |
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| 6326. |
In alpha-particle scattering experiment, the perpendicular distance(b) of initial velocity vector of the the alpha-particle from the centre of nucleus is equal to (where E is initial kinetic energy of alpha-particle) (##AAK_MCP_38_NEET_PHY_E38_020_Q01##) |
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Answer» `(Ze^2sin^2(theta/2))/(4πepsilon_0E)` |
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| 6327. |
The critical angle for a ray of the light coming from a medium to air is 30^@. The velocity light in the medium is : |
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Answer» `1.5 XX 10^8 m/s` |
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| 6328. |
In the figure shwon there is non-conducting disc of mass M=2k and radius R=4m. On its upper & lower part of circumference +Q and -Q charge are uniformly attached such that liner charge density is Q/(piR). The disc can freely rotate about an horizontal axis passing through O. There is a uniform electric field barE in the vertical direction such that QE=mg. If the disc is rotated by a small angle it performs S.H.M. its time period is given by T=~~sqrt(( pi)/n) then, P+n, is (Take pi^(2)=g) |
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Answer» `:. T=sqrt(2pi)` `:.P+n` is 3 |
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| 6329. |
An object is placed at 20 cm from a convex mirror of focal length 10 cm. The image formed by a mirror is |
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Answer» REAL and at 20 CM from the mirror |
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| 6330. |
1 amu is equal to ................g |
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Answer» `1.66xx10^(-25)` |
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| 6331. |
A glass tube of 1.0 m length is filled with water, the water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500Hz is brought at the upper end of the tube and the velocity of sound is 330 m/s, then the total number of resonances obtained will be |
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Answer» 4 |
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| 6332. |
The decimal expansion of the rational 14587/1250 number will terminate after |
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Answer» ONE DECIMAL place |
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| 6333. |
An 80muC charge is given to the 4muF capacitor in the circuit shown in the figure so that the upper plate A is positively charged. An unknown resistance R is connected in the left limb. As soon as the switch S in the central limb is closed, a current of 2 A flows through the 2Omega resistor in the central limb. The capacitive time constant for the circuit is |
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Answer» `56mus` |
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| 6334. |
A satellite is moving in an orbit with half the speed required to escape the earth's field. What is the height of the satellite? |
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Answer» 64 KM `therefore (GM)/(R+h)=(1)/(4)*(2GM)/(R ) rArr 2R=R+h rArr h=R` Thus correct choice is (c ). |
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| 6335. |
In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. On shunting the cell with a resistance of 2 Omega, the balancing length becomes 120 cm. The internal resistance of the cell is |
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Answer» `1 Omega` `E-ir=E-(ER)/(2+r) =kl_(2) rArr E = kl_(2) [(2+r)/(2)]`...(ii) From eqn. (i) and (ii), `(2+r)/(2) = (l_(1))/(l_(2)) = (240)/(120)` `rArr r= 2OMEGA` |
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| 6336. |
Which of the following series of H-atom lies in visible region ? |
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Answer» Lyman |
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| 6337. |
(a) Draw a labelled diagram of an a.c.generator. Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field vecB. (b) A horizontal conducting rod 10 m long extending from east to west is falling with a speed 5.0 ms^(-1) at right angles to the horizontal component of the Earth's magnetic field, 0.3 xx 10^(-4) Wb m^(-2). Find the instantaneous value of the emf induced in the rod. |
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Answer» Solution :(a) See Short Answer Question Number 53. (b) Here length of rod l = 10 m, speed of falling of rod `V = 5.0 m s^(-1)` and HORIZONTAL component of earth.s MAGNETIC FIELD `B_(H) = 0.3 xx 10^(-4) WB m^(-2) = 0.3 xx 10^(-4)T` `therefore` Instantaneous value of induced emf `|varepsilon|= B_(H)lv = 0.3 xx 10^(-4) xx 10 xx 5.0 = 15 xx 10^(-4)V = 1.5 mV` |
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| 6338. |
A point A is located at a distance r = 1.5 m from a point source of sound of frequency 600 Hz. The power of the source is 0.8 watt. Speed in air is 340 m/s and density of air is 1.29kg//m^3 . Find the pressure oscillation amplitude(Delta P)_m |
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Answer» `4.98 N//m^2` |
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| 6339. |
Which of the following changes to a double-slit interference experiment with light would increase the widths of the fringes in the diffraction pattern that appears on the screen? |
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Answer» Use light of a shorter wavelength |
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| 6340. |
A 10 Omegaresistance is connected with an electric cell. Now this resistance is replaced by a 20 Omega resistance. The potential difference between two poles of the cell |
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Answer» will INCREASE `epsilon` and r are constant in V = `epsilon` -IR, but connecting 20`Omega` instead of 10`Omega` cannot flowing through the circuit decreses So as per equation V= `epsilon - Ir,` V increases. |
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| 6341. |
A circuit has primary inductance of 2H and secondary of 8H. The mutual inductance of the circuit is: |
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Answer» 4H |
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| 6342. |
A vacuum has been created in a radio tube, i.e. a state of gas where the mean free path of its particles exceeds the characteristic dimensions of the vessel. Assuming the tube's length to be 5 cm and it to be filled with argon, estimate the density and the pressure of the gas at room temperature (20^@C). |
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Answer» |
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| 6343. |
Consider two conducting spheres ofradill R_(1) and R_(2) with R_(1) gt R_(2). If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. |
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Answer» Solution :Here, `V_(1) = V_(2)` `(Q_(1))/(4PI in_(0) R_(1)) = (Q_(2))/(4pi in_(0) R_(2)) :. (Q_(1))/(Q_(2)) = (R_(1))/(R_(2))` …(i) As `R_(1) gt R_(2). :. Q_(1) gt Q_(2)` i.e., larger sphere has more charge than the smaller sphere. Now, `sigma_(1) = (Q_(1))/(4pi R_(1)^(2)) and sigma_(2) = (Q_(2))/(4pi in R_(2)^(2)) :. (sigma_(2))/(sigma_(1)) = (Q_(2))/(R_(2)^(2)) . (R_(1)^(2))/(Q_(1))` USING (i) , `(sigma_(2))/(sigma_(1)) = (R_(2))/(R_(1)) . (R_(1)^(2))/(R_(2)^(2)) = (R_(1))/(R_(2))` As `R_(1) gt R_(2)`, therefore, `sigma_(2) gt sigma_(1)` i.e., charge density of smaller sphere SI more than the charge density of larger sphere. |
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| 6344. |
The graphs below show angular velocity as a function of theta. In which one of these is the magnitude of angular velocity constantly decreasing with time? |
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Answer»
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| 6345. |
The capacity of each condenser in the following fig. is 'C'. Then the equivalent capacitance across A and B is (##AKS_NEO_CAO_PHY_XII_V02_B_APP_E02_085_Q01.png" width="80%"> |
| Answer» ANSWER :C | |
| 6346. |
Two point charges +1 nC and 4 nC are 1 m apart in air. Find the positions along the line joining the two charges at which resultant potential is zero. |
Answer» Solution : At A charge is `1 xx 10^(-9) C` At B charge is `- 4 xx 10^(-9) C ` Now at P RESULTANT potential is zero. ` therefore (1)/(4 r epsi_0) [ (1 xx 10^(-9))/(x ) - (4 xx 10^(-9))/(1 - x) ] = 0` `([(1 xx x) - 4(x)] xx 10^(-9))/(x(1 -x )) = 0` ` + 1 - x -4x = 0` ` 1 = 5x` `x= 1/5 = 0.2m` |
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| 6347. |
The initial and final temperature of water as record by an observer are lettheta_(1)40.6pm0.2^(@)Cdeltatheta_(1)=pm0.2"^(@)C |
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Answer» SOLUTION :LET `theta_(1)40.66"^(@)C,Deltatheta_(1)=pm0.2"^(@)C` `theta_(2)78.3"^(@)C,Deltatheta_(2)=pm0.3"^(@)C` `impliestheta=theta_(2)-theta_(1)=78.3-40.6=37.7"^(@)C Hence rise in TEMPERATURE |
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| 6348. |
Assuming the Earth to be a spherical conductor of radius 6380 km, calculate its capacity. |
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Answer» |
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| 6349. |
TV tower has a height of 80m. The height of the tower to be increased to, so that the coverage area is doubled is |
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Answer» 80m |
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| 6350. |
A counter rate - meter is used to measure the activity of a given sample. At one instant, the meter show 4750 counts per minute. Five minutes later it shows 2700 counts per minute. The half-life of the sample is nearly ______ minute [log_(10)^(1.760)=0.2455] |
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Answer» 15 min |
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