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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.
| 44301. |
Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is : |
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Answer» 36.3eV |
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| 44302. |
What force is required to stretch a Cu wire to double its length ? (A = 0.1 mm^2,Y = 1.28 xx 10^12 "dyne"/(cm^2)) |
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Answer» a)`1.28 XX 10^9` DYNE |
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| 44303. |
When an electric cell is in use, .... Relation holds good. |
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Answer» `EPSILON = V - Ir ` When CURRENT flows AWAY from positive terminal of the cell, it gets discharged and so its tenninal voltage is, V = `epsilon - Ir ` `THEREFORE epsilon = V + Ir ` |
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| 44304. |
In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelenght of light used is 5000A^(0). Find the distance between 7th maxima 11th minima on the screen. |
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Answer» SOLUTION :Given, `d= 0.1 mm, D= 0.5 m` and `LAMBDA= 5000 dotA= 5.0xx10^(-7)m` `triangle y= (y_(11))_("dark")-(y_(7))_("bright")= ((2xx11 -1)lambda D)/(2D)-(7 lambda D)/(d)` `= 8.75xx 10^(-3) m = 8.75 mm` |
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| 44305. |
In a Young's double slit experiment, a monochromatic source of wavelength lambda is used to illuminate the two slits S_(1) and S_(2). The slits S_(1) and S_(2) are identical and source S is placed symmetrical as shown. Interference pattern is observed on a screen at a distance D from the centre of slit. The distance between the slits is d. if the resultant intensity at P is same as that O, then the distance OP cannot be |
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Answer» `(LAMBDA D)/(d)` |
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| 44306. |
If Ca_(3)(PO_(4))_(2) andH_(3)PO_(3) contain same number of 'P' atom then the ratio of oxygen atoms in these compounds respectively is : |
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Answer» `8//3`  `2x=y"(Same 'P' atoms)" RARR x=(y)/(2)` `"Ratio of 'O' atoms" = 8x3y=8xx(y/2)/(3y)=4/3` |
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| 44307. |
If the refractive indices of glass and water with respect to air are (3)/(2) and (4)/(3) respectively, whatis the refractive index of glass with respect to water ? |
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Answer» Solution :Data SUPPLIED, `n_(ga)= (3)/(2), "" n_(wa) = (4)/(3) ` `n_(GW) = (n_(ga))/(n_(wa)) = ((3)/(2))/((4)/(3)) = 1.128` |
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| 44308. |
Assertion An alpha-particle and a deuteron having same kinetic energy enter in a uniform magnetic field perpendicular to the field. Then, radius of circular path of alpha-particle will be more. Reason (q)/(m) ratio of an alpha-particle is more than the (q)/(m) ratio of a deuteron. |
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Answer» If both Assertion and REASON are TRUE and Reason is the correct explanationof Assertion. |
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| 44309. |
During the propagation of emw in a medium : |
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Answer» ELECTRIC ENERGY DENSITY is double of the magnetic density |
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| 44310. |
Satisfactory explanation of the phenomenon of photo electric effect is based on |
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Answer» Planck's QUANTUM theory |
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| 44311. |
The plates of a parallel plate capacitor are charged by a battery and the battery is disconnected after the charging. Now, the plates are placed as shown in the figure. Then (plates are not parallel to each other). . |
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Answer» the surface CHARGE density is GREATER at POINT `A` |
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| 44312. |
A pure capacitor of capacitive reactance of 10Omega is connected to an A.C. source. If the frequency of a source is doubled, the capacitive reactance will be ....... Omega. |
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Answer» 0.5 `X_C=1/(2pifC)` `THEREFORE X_C prop 1/f` Now if the frequency is doubled , `X_C` becomes half `therefore X._C =1/2xxX_C =10/2 "" therefore X._C = 5OMEGA` |
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| 44313. |
Sphere of radius 5 cm at temperatures 127^@C (Stefan's constant = 5.7 xx 10^(-8)S.I. units) . Find the energy. |
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Answer» 1851 J |
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| 44314. |
What is meant by ‘Wattful current’? |
| Answer» SOLUTION :The component of current (IRMS `COS phi`) which is in phase with the voltage is CALLED active component. The power consumed by this current = `V_(RMS) I_(RMS) cos phi`. So that it is also known as .Wattful. current. | |
| 44315. |
A 60 watt bulb is hung over the center of a table 4 m xx 4 m at a height of 3 m . The ratio of the intensities of illumination at a point on the centre of the edge and on the corner of the table is |
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Answer» `(17//13)^(3//2)` |
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| 44316. |
At the surface of the Earth, an object of mass m has weight w. If this object is transported to an altitude that's twice the radius of the Earth, then, at the new location, |
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Answer» its mass is `(m)/(2)` and its weight is `(w)/(2)` |
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| 44317. |
A test charge 'q' is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shows in the figure. (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why ? |
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Answer» Solution :(i) As electric field is a conservative field, work doene in MOVING the TEST charge q is exactly same along the path ABC or along straight path AC. Distance`AC = vec R = - 4 hati`, and force applied against the electric field `vecF_("ext")= - vecE_q= -E_qhati` `:.` Work done `W_(AC) = vec F_("ext").vecr = [-EQ hat i].[-4hati] = + 4 Eq` `:.` Potential difference between C and A `V_C - V_A= (W_(AC))/q = (4Eq)/q = + 4E` (ii) Electric potential at point C is GREATER i.e., `V_C gt V_A`. |
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| 44318. |
A plane wavefront of wavelength lamda is incident on a single slit of width a. The angular width of principal maximum is |
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Answer» `lamda/a` |
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| 44319. |
Four positive point charges Q are placed at the corners of a square of side a. Calculate the charge that should be placed at the centre so that system remains in equilibrium. |
| Answer» SOLUTION :`(Q/4 +Q/sqrt2)` | |
| 44320. |
If the nuclear of radius of O^8 is 2.4 fermi, then radius of Al^27 would be |
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Answer» 6.0 fermi |
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| 44321. |
In a Young's experiment , the slits are 1.5 m from the screen . The width of the fringes observed with light of wavelength 6000 Å is 1.0 nm . What is the separation of the slits . |
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Answer» |
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| 44322. |
One mole of an ideal monatomic gas at temperature T_0 expands slowly according to the law P = kV (k is constant). If the final temperature is 4T_0 then heat supplied to gas is |
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Answer» `2RT_(0)` |
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| 44323. |
The conduction band of an insulator is_____empty. Ans. Practically. |
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Answer» |
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| 44324. |
Column - I gives certain situtations involving two thin conducting shells connected by a conducting wire via a key K. In all situations one sphere has net charge +q and other sphere has no net charge. After the key K is pressed, Column - II gives some resulting effect. Match the figures in Column - I with the statements in Column - II. |
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Answer» |
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| 44325. |
At room temperature, a p-type semiconductorhas |
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Answer» LARGE number of holes and few ELECTRONS |
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| 44326. |
When will the conductivity of a Ge semiconductor decrease? |
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Answer» on dopping donor impurity. Less electrons escape in the covalent BOND and the temperature decreases and less HOLES and produced, so the NUMBER of charge carrier decreases and conductivity decreases. |
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| 44328. |
A parallel bean of monochromatic light is allowed to incident normally on a plane transmission grating having 5000 lines per centimeter. A second order spectral line is found to be diffracted at an angle 30^(@). Find the wavelength of the light. Data : N=5000 lines/cm =5000xx10^(2) lines/m, m=2, theta=30^(@), lambda=? |
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Answer» SOLUTION :`sintheta=Nmlambda` `lambda=(sintheta)/(NM)` `lambda=(SIN30^(@))/(5xx10^(5)xx2)=(0.5)/(5xx10^(5)xx2)` `lambda=5xx10^(-7)m=5000Å`. |
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| 44329. |
A semicircular wire PQS of radius R is connected to a wire bent in the form of a sine curve to form a closed loop as shown in the figure. If the loop carries a current I and is placed in a uniform magnetic field B, then the totla force acting on the sine curve is |
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Answer» 2BiR (DOWNWARD) |
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| 44330. |
A : When the coherent sources are far apart, interference pattern cannot be detected. R : If two point coherent sources are infinitely close to each other, firinges appears very sharp. |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION of A |
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| 44331. |
A beam of fast moving alpha particles were directed towards a thin film pf gold. The parts A'B corresponding to the incident parts A, B and C of the beam, are shown in the adjoining diagram. The number of alpha particles. In- |
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Answer» B' will be minimum and in C' MAXIMUM |
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| 44332. |
A: Optical communication system is more economical than other system of communications. R: The information carrying capacity of a communication system is directly proportional to its band width. |
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Answer» |
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| 44333. |
A uniform insulating rod of length L moves with a velocity bar(upsilon) in a magnetic field B where bar(upsilon) is perpendicular to both L and B. Then the induced EMF at the ends of the rod is given by |
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Answer» BL `UPSILON` |
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| 44334. |
Find the current flowing through the branch AC in the steady state as also the charge on the capacitor C. If the externally applied potentials are now withdrawn, how will the charge on the capacitor vary as a function of time? (R = 1 kOmega, C = 10 muF) |
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Answer» Solution :The current through `BA = (10V-5V)/(1) = 5 mA` Similarly current through `AC = 5 mA, & BC = 10 mA` steady state CHARGE on `C = 5V xx 10 muF = 50 muC`If the potential differences are withdrawn at time t = 0, the charge on the capacitor varies as a function of time as it discharges through the external resistance. The equivalent resistance of the circuit across AC is `(2R.R)/(2R+R)=(2)/(3)xxR=667Omega(approx)` The time CONSTANT `tau=(2//3R)XXC=6.67msec` The charge across the capacitor decreases exponentially`q(t) 50 muC xsx e^(-t//6.67) MS` |
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| 44335. |
A coil having an area A_(0) is placed in a magnetic field which changes from B_(0) to 4B_(0) in a time interval t. The e.m.f. induced in the coil will be |
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Answer» `(3B_(0))/(A_(0)t)` |
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| 44336. |
If the frequency of revolution of electron in orbit in H atom is n then the equivalent current is: |
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Answer» `(2pire)/N` |
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| 44337. |
Explain how energy is released in a fission process. |
| Answer» SOLUTION :TOTAL MASS of the fission products is less than that of the reactants. This mass DIFFERENCE is CONVERTED into energy. | |
| 44338. |
The horizontal component of the earth's magnetic field at a place is 1/(sqrt(3)) times its vertical component there. Find the value of the angle of dip at that place. What is the ratio of the horizontal component to the total magnetic field of the earth at that place? |
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Answer» Solution :As per question `B_H = 1/sqrt3B_V ` or `(B_V)/(V_H) = sqrt3` But `(B_V)/(B_H)= tan DELTA` , where `delta` is the ANGLE of dip at given PLACE. HENCE, `tan delta = sqrt3 ` or ` delta = tan^(-1) sqrt3 = pi/3 ` or `60^@` Again `B_H = D_E cos delta = B_E cos 60^@ = (B_E)/(2) implies(B_H)/(B_E) =1/2`. |
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| 44339. |
Define Current sensitivity ? |
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Answer» Solution :It Is DEFINED as the deflection PRODUCED PER unit current flowing through it. `I_(s) = (THETA)/(I) = ("NAB")/(K) rArr, I_(s) = (l)/(G)` |
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| 44340. |
Along wire carrying a current of 40A as shown in figure. The rectangular loop carries a current of 15A. The resultant force acting on the loop is assume that a = 1cm, b = 80cm and 1 = 30cm ] |
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Answer» `3.6 XX 10^(-3) `N directed towards wire |
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| 44341. |
At absolute zero temperature Si acts as …….. |
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Answer» non metal At ABSOLUTE zero temperature, valence band in the crystal STRUCTUREIS completely filled and there is no free electron in conduction BAN means it is empty in the context of electric charge HENCE at this temperature it ACTS as a insulator. |
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| 44342. |
A body of mass 5 kg explodes into 3 fragments having masses in the ratio of 2:2 : 1. The fragments with equal masses fly in merely far direction with speed 15 ms^(-1). What will be the velocity of lighter one ? |
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Answer» 15 m/s `1xxv=sqrt((2xx15)^2+(2xx15)^2)=30sqrt2 MS^(-1)`. |
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| 44343. |
A signal emitted by an antenna from a certain point can be received at another point of the surface in the form of |
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Answer» sky wave |
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| 44344. |
In a X-ray tube, after an electron has been removed from an inner shell of a target atom, an electron from outer shell falls into the vacancy and the excess energy is usually released in form of a photon. Many a times an atom chooses to release this excess energy by ejecting another electron. The ejected electron is called Auger electron. In one such event, the bombarding electron removed an electron from K shell of a target atom. An electron from L shell falls to occupy the vacant K shell position, and the excess energy is used by the atom to eject an Auger electron from L shell. What will be kinetic energy of such an Auger electron if the ionization energies for K and L shell of the atom is E_(k) and E_(L) respectively. |
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Answer» |
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| 44345. |
How has the poet described the plains? |
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Answer» BARREN with erosion |
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| 44346. |
Find the mean free transit time and the mean free path of the electrons in copper (at room temperature). |
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Answer» |
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| 44347. |
A coil of area 0.05m^2 and 500 turns is placed in a magnetic field of strength 4xx10^-5tesla. If it is rotated through 90^@ in 0.1sec, then what is the magnitude of e.m.f. induced in the coil? |
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Answer» SOLUTION :`E = (dphi)/DT = nAB/t = (500xx5xx10^-2xx4xx10^-5)/10^-1` `THEREFORE e = 10^-2V` = 10MV.` |
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| 44348. |
The drift current in a p-n junction is |
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Answer» from the N- side to the p - side |
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| 44349. |
A rigid circular loop of radius r and mass m lies in the XY plane on a flat table and has a current I flowering in it. At this particular place, the earth's magnetic field is vecB = B_x hati + B_z hatk . The value of I so that one end of the loop lifts from the table is |
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Answer» `(mg)/( pi R SQRT(B_(X)^(2) +B_(Z)^(2)))` |
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| 44350. |
A coil of inductance 40 henry is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is |
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Answer» 40s |
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