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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.
| 5951. |
दो परिमेय संख्याओ के बीच अधिकतम कितनी परिमेय संख्याए हो सकती है |
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Answer» 1 |
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| 5952. |
Wavelength of matter particle is independent of which of the quantities? |
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Answer» Mass |
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| 5953. |
The radius of la circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be1/(2sqrt(2)) times that at the centre? |
| Answer» ANSWER :C | |
| 5954. |
In the above question, its length equivalent to a simple pendulum will be : |
| Answer» Answer :C | |
| 5955. |
Through what did the boy regain his motor ability? |
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Answer» IRON persistence |
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| 5956. |
Two bodies of equal masses are some distance apart. If 20 % of mass is transferred from the first body to the second body, then the gravitational force between them |
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Answer» increases by 4% `F_(1)=(Gm^(2))/r^(2)` When 20% MASS is transferred from one BODY to another, then their masses are `m_(1)=80/100mandm_(2)=120/100m` So, force is finally `F_(2)=(Gm^(2))/r^(2)xx(24/25)` Hence, reduction in `F_(2)" is "(F_(1)-F_(2))/F_(1)XX100%=-4%` |
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| 5957. |
A pitcher contains 1 kg water at 40^(@)C. It is given that the rate of evaporation of water from the surface of pitcher is 50 gm/s. Find the time it will take to cool down the water inside to 30^(@)C. Given that latent heat of vaporization of water is 540 cal /g and specific heat of water is 1 "cal"//g.^(@)C. |
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Answer» |
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| 5958. |
Figure 6.36 shows a potentiometer with a cell of emf 2.0 V and internal resistance 0.4 Omega maintaining a potential drop across the resistor wire AB. A standard cell that maintains a constant emf of 1.02 V (for very moderate current up to emf mu A) gives a balance point at 67.3 cm length of the wire. To ensure very low current is drawn the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found, similary, turns out to be at 82.3 cm length of the wire. a. What is the value of epsilon? b.What purpose does the high resistance of 600 k Omega have? c. Is the balance point affected by this high resistance? d. Is the balance point affected by internal resistance of the driver cell? e. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V? f. Would the circuit work well for determining an extermely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ? |
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Answer» or `epsilon= (l xx epsilon_(standard))/(l_(standard)) = (82.3 xx 1.02)/(67.3) = 1.2474 V` (b) To reduce the current through the galvanometer when the MOVABLE constant is FAR from the balance point. (c ) No. (d) Yes. (e) No, if `epsilon` as it is, would be unsuited, because the balance point (for `e` of the order of a few `mV`) will be very close to the end `A` and the percentage error in measurement will be very large. The circuit is modified by putting a suitable RESISTOR `r` in series with the wire `AB` so that potential drop across `AB` is only slightly GREATER than the emf to be measured. Then the balance point will be at larger length of the wire and the percentage error will be much smaller. |
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| 5959. |
A p photon having 10.2 eV energy collides with a hydrogen atom in ground state inelastically. After few microseconds, one other photon having 15 eV collides with the same hydrogen atom, then a suitable detector can detect : |
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Answer» 1 photon of 3.4 EV and electron of 1.4 eV IONISATION energy of `H_(2)` atom =13.6eV So SECOND photon will ionise the atom and additional energy =15-13.6=1.4eV is carried by electron as its K.E. So photon of energy 10.2eV and electron of energy 1.4eV is ejected. |
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| 5960. |
The intrinsic semi-conductor has: |
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Answer» A FINITE RESISTANCE which does not CHANGE with temperature |
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| 5961. |
A Planet is revolvingaround a star in an elliptic orbit. The ratio of the fathest distance to the closest distance of the planet from the star is 4. The ratio of kinetic energies of the planet at eh farthest to the closest position is |
| Answer» SOLUTION :N/A | |
| 5962. |
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ? |
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Answer» 7.2 J Length hanging verticallly =`(60)/(100)=0.6 m` The w.t of hanging part is acting at C.G. of this part i.e at a distance =`(0.6)/(2)=0.3m` from FREE end. Mass of hanging part `=(4/2xx0.6)=1.2kg`. `:.` Work done =`1.2xx10xx0.3` (`:.` Work done = change in P.E. MGH) |
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| 5963. |
In Fig. the switch is closed and steady-state conditions are established. The switch is thrown open at t = 0. a. Find the initial voltage E_(0) across L just after t = 0. Which end of the coil is at the heigher potential : a or b? b. Makefreehand graphs of the current in R_(1) and R_(2) as a funtion of time, treating the steady-state directions as positive. Show values before and after t = 0. c. How long after t = 0 does the current in R_(2) have the value 2 mA? |
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Answer» SOLUTION :a. `b` shell be at a higher potential `E_(0) = I_(0)(R_(1) + R_(2)) = 9 xx 8 = 72 V` b. `I = I_(0)e^(-((R_(1)+R_(2))t)/(L))`for `t = gt 0` `I` through `R_(1)`  `I` through `R_(2)`  c. `I = 9 e^((-8t)/(0.4))` `2 xx 10^(-3) = 9 e^(-20t) rArr e^(20t) = (9 xx 10^(3))/(2)` `20t = In (4.5 xx 10^(3))` `t = (1)/(20) In (4.5 xx 10^(3))s` |
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| 5964. |
A galvanometer having a resistance of son, gives a full scale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross section 2.97 xx 10^(-2) cm^(2) that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is ( specific resistance of the wire = 5 xx 10^(-7) Omega - m |
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Answer» 9 |
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| 5965. |
Two inductors each of inductance L are connected in parallel. One more inductor of value 5 mH is connected in series of this configuration then the effective inductance is 15 mH . The value of L is ………..mH. |
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Answer» 10 |
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| 5966. |
A wire of length L and three identical cells of negligible internal resistances are connected in seires. Due to the current, the temperature of the wire is raised by Delta T in a time t. N number of similar cells are now connected in series with a wire of the same material and crosssection but of length 2L. The temperature of the wire is raised by the same amount Delta Tin the same time t. The value of 'N' is |
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Answer» 4 |
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| 5967. |
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz, what is its wavelength ? |
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Answer» Solution :`vec(E )` and `vec(B)` are in x-y plane and are MUTUALLY PERPENDICULAR. `v = upsilon LAMBDA` `lambda = (v)/(upsilon)=(3xx10^(8))/(30xx10^(6))=(3)/(30)xx10^(2)=10 m` |
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| 5968. |
Tuning fork P of frequency 260 Hz gives 8 beats per sec with a tuning fork R. prongs of R are filed a little . When P and R are again sounded 6 beats per second are heard. Frequencies of R before and after filing respectively are |
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Answer» 252 HZ and 254 Hz. |
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| 5969. |
In the following question a statement of assertion (A) is followed by a statement of reason (R ) A : An equipotential surface is normal to electric field lines. R : Potential increases in direction of electric field . |
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Answer» If both Assertion & Reason are true and the reason is the correct EXPLANATION of the assertion , then mark (1). |
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| 5970. |
the followingtwo reactionsare carried outseparately . CH_(3)CH_(2)-o-CH_(3) +HI (1 mol ) overset("heat")to "products" (CH_(3))_(3)C-O-CH_(3)+HI(1 mol ) overset("heat")to "products" the pairof majorproductsobtained in thefirstand second reactions are respectively - |
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Answer» `CH_(3)CH_(2)OH and CH_(3)I, (CH_(3))_(3)C-I andCH_(3)OH` `(CH_(3))_(3)overset(I-H)C-O-CH_(3)underset(SN_(1))underset(1 mol)overset(HI)""to (CH_(3))_(3)C-I +CH_(3)-OH` |
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| 5971. |
Is electric intensity a scalar or vector quantity ? |
| Answer» SOLUTION :VECTOR, `NC^(-1) or VM^(-1)` | |
| 5972. |
The displacement bar(r ) of a charge Q in an electric field E= e_(1) hat(i) + e_(2) hat(j) + e_(3) hat(k) " is " bar(r )= a hat(i) + b hat(j). The work done is |
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Answer» `Q(a e_(1) + B e_(2))` |
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| 5973. |
The distance of the closest approach of an alpha particle fired at a nucleus with kinetic of an alpha particle fired at a nucleus with kinetic energy K is r_(0). The distance of the closest approach when the alpha particle is fired at the same nucleus with kinetic energy 2K will be |
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Answer» `4 r_(0)` |
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| 5974. |
Consider a vertical tube open at bothh ends. The cube consists of two parts, each of different cross-sections and each parth having a piston which can move smoothly in respective tubes. The two pistons are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of cross-section of the upper piston is 10 cm^(2) greater than that of the lower piston. Amount of gas enclosed by the pistons is one mole. When the gas is heated slowly, pistons move by 50 cm. Find rise in the temperature of the gas, in the form 25X//R K where R is universal gas constant. Use g=10m//s^(2) and outside pressure =10^(5)N//m^(2)). Fill value of X in the answer sheet. |
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Answer» <P>  `P(A_(1)-A_(2))=P_(0)(A_(1)-A_(2))+(m_(1)+m_(2))g` `impliesP/_\V=(P_(0)/_\A+mg)L` `L` is displacement of the piston. `P./_\V=nR/_\T` `/_\T=(P/_\V)/(nR)=((P_(0)/_\V+mg)L)/(nR)` `=((10^(5)P_(a)xx10^(-3)m^(2)+5xx10)(50xx10^(-2)))/(1xxR)=75/R K` |
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| 5975. |
For an object moving with uniform acceleration, travelling 50m in 5th sec, 70m in 7th sec. (a)Its initial velocity is 5 m/s (b) Its acceleration is 20 "m/s"^2 (c)Its travels 100 m in 9th sec (d) Its average velocity during 9th sec is 90 m/s |
| Answer» Answer :B | |
| 5976. |
There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r lt R) ? |
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Answer» `Epropr^(-2)` |
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| 5977. |
Find the potential due to an infinite sheet of charge at a distance x from it. Assume that the potential is zero at a distance a from it. |
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Answer» Solution :ALTHOUGH the infinite sheet can be treated as the special case of a disk, yet there are points that need be discussed. Let US do this by the FORMULA we derived (Eq. 23-19), because the electric field of an infinite disk is known and is simple to handle. Here we can use the formula we found before `Delta V= - INT overset(to) (E). d overset(to)(x)=-int_(a)^(x) (sigma)/( 2 epsilon_(0)). dx.` Here, we have assumed the potential to be zero at a distance of a from the sheet. `V_(x) = (sigma (a-x) )/(2 epsilon_(0))`. We can note here that we cannot take the potential to be zero at infinity. If we do so, the potential will be infinite. This is because the electric field is also extending to infinity. We can also note that a is just a arbitrary number we chose according to our convenience. It can have any fixed value and still it will not affect our calculations because in any situation, we want to find the change in potential energy. For that, we need to find only the change in potential. It will be the same for every value of constant a. |
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| 5978. |
In fig. a plastic rod in the form of circular are with charge +Q unformly distributed on it produces an electric fieldof magnitude E at the centre of curvature (at the origin). In fig., more circular rods with identical umform charges +Q are addes until the circle is complete. A fifth arrangement (which would be labeled e) is like taht in d except that the roed in the fourth quadrant has charge-Q. Rank all the five arrangements according to the magnitudes of the electric field at the center of curvature, greatest first. |
Answer» Solution :Make use of SYMMETRY property.  ,It is clear from DIAGRAMS that order of fields will be FIRST): (e ),(b), (a,c),(d),(2E, sqrt(2)E, and ZERO, repectively). |
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| 5979. |
The energy flux of sunlight reaching the surface of the earth is 1.388xx10^(3)W//m^(2) .How many photons (nearly) per square meter are incident on the Earth per second?Assume that the photons in the sunlight have an average wavelength of 550 nm. |
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Answer» Solution :Energy FLUX of the sun =`1.388xx10^(3)W//m^(2)` `lambda=550 nm=55xx10^(-8)m` `h=6.63xx10^(-34)JS` `c=3xx10^(8)m//s` `implies` Energy of each photon, `E=(hc)/(lambda)` `therefore E=(6.63xx10^(-34xx3xx10^(8)))/(55xx10^(-8))` `thereforeE=0.3616xx10^(-18) therefore E~~3.62xx10^(-19)J` `implies` No. Of photon incident on surface of the earth per unit area per unit time `=("Energy flux")/("Energy of each photon")` `(1.388xx10^(3))/(3.62xx10^(-19))` `=0.3834xx10^(22)` `~~3.8xx10^(21)("photon")/(m^(2)xxs)` SECOND Method: If N number of photon incident per unit area the NE=P `N=(P)/(E)=(P)/((hc)/(lambda))=(Plambda)/(hc)` `=(1.388xx10^(3)xx5.5xx10^(-7))/(6.63xx10^(-34xx3xx10^(8)))=3.8xx10^(21)` |
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| 5980. |
A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the water surface, the radius of this circle (in cm) is ..... |
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Answer» `36root_2`  `SINC=(R)/(sqrt(R^2+D^2))` `1/n=(R)/(sqrt(R^2+D^2))` `(1)/(n^2)=(R^2)/(R^2+D^2)` `thereforeR^2+D^2=n^2R^2` `thereforeD^2=R^2(n^2-1)` `therefore(D^2)/(n^2-1)=R^2` `thereforeR=(D)/(sqrt(n^2-1))` `thereforeR=(12)/(sqrt((4/3)^2)-1)=(12)/(sqrt((16)/(9)-1))` `thereforeR=(36)/(sqrt7)` cm |
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| 5981. |
Why a common emitter amplifier preferred over common base amplifier ? |
| Answer» Solution :Because in COMMON AMPLIFIER, the current gain, voltage gain and power gain are more as COMPARED to that in case of common BASE amplifier. | |
| 5982. |
According to Einstein's photoelectric equation, the graph of KE of the photoelectron emitted from the metal versus the frequency of the incident radiation gives a straight line graph, whose slope |
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Answer» depends on the intensity of the incident radiation  `KE_("max")=hv-phi_(0)` Compring with the equation of straight LINE |
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| 5983. |
A body is projected at an angle 30° with the horizontal with momentum p. At its highest point the magnitude of the momentum is : |
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Answer» p Velocity at the HIGHEST point = v cos `THETA` Momentum,p. = mv cos `theta` |
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| 5984. |
Derive an expression for the electric potential energy of a system of two point charges in the absence of an external electric field. |
Answer» Solution : Consider the charges `q_(1) and q_(2)`are initially at INFINITY. First, the charge `q_1`is brought from infinity to the point A and no work is done for this. This is because there is no electric field to oppose this charge. Consider another point B at a distance .R. from A. Electric potential at B due to `q_1`is given by `V_(1) = 1 / (4piepsilon_(0))q_(1)/r` Next the charge `q_2`is brought from infinity to the point B. When the charge `q_2`is moved, the electric field due to `q_1`opposes it. Hence work has to be done. Work done in BRINGING the charge `q_2`from infinity to the point B is given by `w=V_1xxq_(2) "" [ :. V=(w)/q_(0)impliesw=q_(0)V]` `w = 1/(4piepsilon_(0))(q_1)/(r)xxq_2` This work done is stored in the charges as potential ENERGY and is given by `PE = 1/(4piepsilon_(0))(q_1q_2)/r` |
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| 5985. |
In Young' s double-slit experiment, separation between the two slits is d and separation between plane of the slits and the screen is D. A small hole is made on the screen at a point directly in front of one of the slit. Some wavelengths are missing from the light coming out from hole on the other side. These wavelengths are |
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Answer» `(d^2)/D` `Deltax = (yd)/D` For the point directly in front of ONE of the slits `y = d//2`, hence on substituting we get path difference as follows: `Delta x = (yd)/D = ((d/2)d)/D` `implies Delta x = (d^2)/(2D)` There will be destructive interference of those wavelengths which are missing, at this point and condition for destructive interference can be written as follows: `Delta x = (2n + 1)lambda/2` Here, `lambda` is wavelength od light `implies (d^2)/(2D) = (2n + 1) lambda/2` `implies lambda = (d^2)/((2n + 1)D)` We can substitute `n = 0, 1, 2, 3,............` `implies lambda = (d^2)/(D), (d^2)/(3D) , (d^2)/(5D),.....` |
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| 5986. |
Choose the word or phrase that is most closely associated with the given word. You may use a choice more than once or not at all. Eliminate those choices that you think to be incorrect and mark the letter of your choice on the answer sheet (1)Graham's law of diffusion (2)Charles's law (3)Ohm's law (4) Gresham's law I=E/R |
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| 5987. |
(A):Acceleration is the rate of change of velocity . (R ):Abody having nonzero acceleration can have a constant velocity. |
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Answer» |
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| 5988. |
Choose the word or phrase that is most closely associated with the given word. You may use a choice more than once or not at all. Eliminate those choices that you think to be incorrect and mark the letter of your choice on the answer sheet (1)Graham's law of diffusion (2)Charles's law (3)Ohm's law (4) Gresham's law V/V_1=T/T_1 |
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| 5989. |
Give reason why trasnsmission of TV signals via sky waves is not possible. |
| Answer» Solution :TV frequencies lie in the range 100-220 MHz which cannot be REFLECTED by the ionosphere. So, Sky WAVE PROPAGATION method is not used in TV transmission. | |
| 5990. |
Choose the word or phrase that is most closely associated with the given word. You may use a choice more than once or not at all. Eliminate those choices that you think to be incorrect and mark the letter of your choice on the answer sheet (1)Graham's law of diffusion (2)Charles's law (3)Ohm's law (4) Gresham's law R_1//R_2=sqrt(M_2//M_1) |
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| 5991. |
भाग लेमा a=bq+r में r=0 होगा जब |
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Answer» B,a को विभाजित करें |
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| 5992. |
Two horizontal wires PQ and RS of resistance 10 Omega and 20 Omega are separated by a distance or 10 cm and connected in parallel in vertical plane across a cell of emf 200 V and negligible internal resistance. A wire AB of mass 1 g and length 1 cm in balanced exactly,midway between them. What must be the current in it ? |
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Answer» Solution :AB experience a force of attraction force PQ and RS. Apply V = iR ` 200 i_1 xx 10 RARR i_1 = 20 A` `200 =I_2 xx 20 rArr i_1 = 10 A` AB will be in equilibrium if `(mu_0)/( 2pia) i_2 i_3 l + MG = (mu_0 i_1 i_3)/(2 pi a)` `(mu_0)/(2pi a) xx (i_1 i_3 - i_2i_3) l = mg `  `(4 pi xx 10^(-7))/(2 pi xx 5 xx 10^(-2))(20 i_3 -10 i_3) xx (1)/(100) =10^(-3) xx 9.8` ` (2 xx 10^(-5) xx 10 i_3)/(5) = 9.8 xx 10^(-3)` `i_3 = 245A.` |
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| 5993. |
A uniform ball, of mass M = 6.00 kg and radius R, rolls smoothly from rest down a ramp at angle theta = 30.0^(@) . (a) The ball descends a vertical height h = 1.20 m to reach the bottom of the ramp. What is its speed at the bottom? |
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Answer» Solution :The mechanical energy E of the ball-Earth system is con served as the ball rolls down the ramp. The reason is that the only force doing work on the ball is the gravitational force, a conservative force. The normal force on the ball from the ramp does zero work because it is perpendicular to the ball.s path. The frictional force on the ball from the ramp does not transfer any energy to thermal energy because the ball does not slide (it rolls smoothly). Thus, we conserve mechanical energy (`E_(F)= E_(i)`): `K_(f)+ U_(f)= K_(i) + U_(i)` (11-11) where subscripts f and i refer to the FINAL values (at the bottom) and initial values (at rest), respectively. The gravitational POTENTIAL energy is initially `U_(i) = Mgh` (where M is the ball.s mass) and finally `U_(f) = 0`. The kinetic energy is initially `K_(i) = 0`. For the final kinetic energy K, we need an additional idea: Because the ball rolls, the kinetic energy involves both translation and rotation, so we include them both by using the right side of eq 11-5. Calculations: Substituting into Eq. 11-11 gives us `((1)/(2)I_("com")omega^(2)+ (1)/(2)Mv_("com")^(2))+ 0= 0+ Mgh` (11-20) where `I_("com")` is the ball.s rotational inertia about an axis com through its center of mass, `v_("com")` is the requested speed at the bottom, and `omega` is the angular speed there. Because the ball rolls smoothly, we can use Eq. 11-2 to substitute `v_("com")R` for `omega` to reduce the unknowns in com Eq. 11-12. Doing so, substituting `2//5MR^(2)` for `I_("com")` ( from Table 10-2f), and then solving for `v_("com")` com give us `v_("com")= sqrt((10)/(7)gh)= sqrt((10)/(7)(9.8m//s^(2))(1.20m))` `= 4.10 m//s` Note that the answer does not depend on Mor R. |
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| 5994. |
How much energy must be spent to pull the satellite in Q. No. 304 out of the earth's gravitational field? |
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Answer» `(2GmM)/((R+h)^(2))` |
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| 5995. |
As shown in figure, a plano-concave lens is placed in such a way that it becomes completely fit with plano-convex lens. Their plane surfaces are parallel. If their refractive indices are 1.6 and 1.5 respectively and radius of curvature is R, then focal length of combination is ..... |
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Answer» `(R)/(6.2)` For first LENS,` (1)/(f_1)=(n-1)((1)/(R_1)-(1)/(R_2))` `R_1= INFTY,R_2=-R,n=mu_1` `(1)/(f_1)=(mu_1-1)((1)/(infty)+(1)/(R))` `therefore (1)/(f_1)=(mu-1)/(R)` … (1) For SECOND lens, `(1)/(f_2)=(n-1)((1)/(R_1)-(1)/(R_2))` `R_1=-R,R_2=infty,n-mu_2` `(1)/(f_2)=(mu_2-1)(-(1)/(R)-(1)/(infty))` `infty (1)/(f_2)=(1-mu_2)/(R)` ... (2) Focal LENGTH of combination,`1/f=(1)/(f_1)+(1)/(f_2)=(mu_1-1)/(R)+(1-mu_2)/(R)` `therefore 1/f=(mu_1-mu_2)/(R)` `therefore f=(R)/(mu_1-mu_2)=(R)/(1.6-1.5)=(R)/(0-1)` |
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| 5996. |
Why alkali metals most suited as photo-sensitive metals ? |
| Answer» Solution :Alkali metals have LOW WORK function . EVEN VISIBLE radiation can eject out ELECTRONS from them. So, alkali metals are most suitable photo sensitive metals. | |
| 5997. |
How did the grandmother react to her illness? |
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Answer» She SAID her END was near |
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| 5998. |
How much energy must be spent to pull the satellite in Q. No. 304 out of the earth's gravitational field? (If the earthshrink suddenly to half its present size.) |
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Answer» `(GmM)/(2(R+H)^(2))` `r_(1)=h_(1)+R` 
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| 5999. |
The minimum number of NAND gates are used to form AND gate is |
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Answer» 1 |
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| 6000. |
Variation of photoelectric current with intensity of light is shown by graph……. |
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Answer»
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