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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.
| 40801. |
If current changes at a unit rate through a closed coil, the induced emf produced in the coil is equal ______. |
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Answer» |
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| 40802. |
Find the magnetic flux linked with a rectangular coil of size 6 cmxx8 cm placed at right angle to a magnetic field of 0.5 Wbm^(-2) |
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Answer» |
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| 40803. |
A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is 1/4 that of first, the rate at which ice melts in g//s will be |
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Answer» 3.2 |
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| 40804. |
A boy of mass 25kg slides down a rope hanging from the branch of a tree. If the force of friction against him is 50N, the boy's acceleration is (g=10 ms^(-2)) |
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Answer» `10 MS^(-2)` |
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| 40805. |
If in hydrogen atom, radius ofn^(th)Bohr orbit is , n_(r)frequency of revolution of electron in n^(th) orbit is f_(n)choose the correct option |
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Answer»
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| 40806. |
What is modulation? |
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Answer» Solution :(a) For long distance transmission, the low frequency baseband signal ( input Signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified. They are (a) amplitude modulation, (b) frequency modulation, and (c ) phase modulation. (a) Amplitude Modulation If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.  (b) Frequency modulation (i) The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. (II) Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier SIGNALAND vice versa. (iii) This leads to compressions and rarefactions in the frequency SPECTRUM of the modulated WAVE. Louder signal leads to compressions and relatively weaker signals to rarefactions. (iv) When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C) (vi) The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (Figure (C )).  (c ) Phase Modulation (PM) (i) The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. (ii) The carrierphase changes according to increase or decrease in the amplitude of the baseband signal  (iii) When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. (iv) Due to this, the carrier signal is compressed or its frequency is increased. (v) The neagative half cycle of the baseband signal produces a phase lag in the carrier signal. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged. (VII) the frequency shift depends on (i) amplitude of the modulating signal and (ii) the frequency of the signal. |
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| 40807. |
A 1.0kg block oscillates with a frequency of 10Hz at the end of a certain spring. The spring is then cut into two halves. The 1.0kg block is then made to oscillate at the end of one of the halves. What is the frequency of oscillation of the block? |
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Answer» 5Hz |
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| 40808. |
What is the condition for first minimum in case of diffraction due to a single slit ? |
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Answer» Solution :The ANGULAR position of first minimum, `theta_(1)=(LAMBDA)/(d)` |
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| 40809. |
The electric current in a circuit is given by i=3t Here, t is in second and I in ampere. The rms current for the period to=0 to t=1 s is |
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Answer» 3A |
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| 40810. |
(#FIITJEE_PHY_MB_02_C01_SLV_052_Q01.png" width="80%"> |
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Answer» <P> SOLUTION :A `rarr` (p, t) `""` B `rarr` (Q) , `""` C `rarr` (q,R), `""` D `rarr` (s) |
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| 40811. |
The effective capacitances between points A and B is (the capacitance of each of the capacitors is C) |
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Answer» C |
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| 40812. |
A cell of emf E has an internal resistance r & is connected to rheostat. When resistance R of rheostat is changed correct graph of potential difference across it is |
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Answer»
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| 40813. |
When researchers find a reasonably complete fossil of a dinosaur, they can determine the mass and weight of the living dinosaur with a scale model sculpted from plastic and based on the dimensions of the fossil bones. The scale of the model is 1/20, that is, lengths are 1/20 actual length, areas are (1//20)^(2) actual areas, and volumes are (1//20)^(3) actual volumes. First, the model is suspended from One arm of a balance and weights are added to the other arm until equilibrium is reached. Then the model is fully submerged in water and enough weights are removed from the second arm to reestablish equilibrium (Fig. ). For a model of a particular T. rex fossil791.10 g had to be removed to a reestablish equilibrium . what was the volume of (a) the model and (b) the actual T . rex ? (c) if the density of T.rex was approximately the density of water , what was its mass ? |
| Answer» Solution :(a) `791.1 cm^(3) , (B) 6.329 m^(3) , (C) 6.329 xx 10^(3)` kg | |
| 40814. |
A Cs plate is irradiated with a light of wavelength lamda=(hc)/(phi),phi being the work function of the plate, h Plank's constant, and c the velocity of light in vacuum. Assume all the photoelectron are moving perpendicular to the plate toward a YDSE setup when accelerated through a potential difference V. Take charge on a proton =e and moss of an electron =m. Read the paragraph carefully and answer the following question: Q. Instead of moving perpendicular to the plate, if the electrons were moving randomly, then the central maximum would shift |
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Answer» upward |
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| 40815. |
What is 'retentivity' in magnetism? |
| Answer» SOLUTION :The PROPERTY of a material to retain the magnetism EVEN after the removal of the magnetising field is known as RETENTIVITY. | |
| 40816. |
An element X decays into element Z by two-step process. XrarrY+""_(2)^(4)He YrarrZ+2bare then |
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Answer» X & Z are isobars `""_(Z)^(A)Xoverset(alpha-"decay")rarr""_(Z-2)^(A-4)Y+""_(2)^(4)He` `""_(Z-2)^(A-4)Yoverset(beta-"decay")rarr""_(Z)^(A-4)Z+2_(-1)e^(0)` `therefore XandZ` are isotopes. |
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| 40817. |
A coll of radius r = 25 cm wound of a thin a copper wireof lengthl = 500 mrotates with an angularvelocityomega = 300 rad//sabout itsaxis. The coll is connectedtoa baliistic galvanometerby meansof slidingcontacts. Thetotal resistanace of thecircuitis equalt R = 21 Omega. Find teh specific totalresistanceof the circuit is equalto R = 21 Omega. Find the specfiic chargeof currentcarriesin copperif a sudden stoppageof thecoil makesa charge q = 10 nC flow through the galvanometer. |
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Answer» Solution :In a rotating route, to first order in `omega`, the main effect is a colotis froce `2 m vec(v) xx vec(omega)`. This unbalanced froce will cause electronsto REACT by setting up a magnetic field `vec(B)` so that the magnitude force `e vec(v) xx vec(B)` balances the coritolis force. Thus `- (epsilon)/(2M) vec(B) -vec(omega)` or,`vec(B) = - (2m)/(epsilon) vec(omega)` The flux assocatied with this is `Phi = N pi r^(2) B = N pi r^(2) (2m)/(e) omega` where `N = (l)/(2 pi r)` is the number of thrus of the RING. If `omega` CHANGES(and then is time for electron to rearrange) then `B` ALSO chargesand so does`Phi`. An `emf` will beinduced and a currentwill flow. This is `I = N pi r^(2) (2m)/(e) Phi//R` The total charge flowingthrough the ballastic galvanometer as the ringis stopped, is `q = N pi r^(2)// (2m)/(e) omega//R` So,`(e)/(m) = (2N pi r^(2) omega)/(q R) = (l omega r)/(q R)` |
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| 40818. |
इनमें से गंगा की सबसे बड़ी सहायक नदी कौन सी है? |
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Answer» घाघरा |
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| 40819. |
In a.c. circuit, the instantaneous value of e.m.f. and current are |
| Answer» Solution :`P_(AV)= (E_(0)I_(0)cos THETA)/2 = (100 xx 2)/2 xx cos 60^(@) = 50 W`. | |
| 40820. |
A beam of monochromatic light of wavelength 6000 A in air enters water of refractive index = 4/3. What is its wavelength in water ? |
| Answer» Answer :C | |
| 40821. |
A point A is located at a distance r - 1.5 m from a point source of sound of frequency 600 Hz.Thepower 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 (DeltaP)_("in") |
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Answer» `4.98 N//m^(2)` |
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| 40822. |
Explain the variation of average binding energy with the mass number by graph and discuss its features. |
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Answer» SOLUTION :We can find the AVERAGE binding energy per nucleon `overline(BE)`. It is given by `overline(BE) = ([Zm_(H) + Nm_(n) -M_(A)]c^(2))/(A)` The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. `overline(BE)` is plotted against A of all known nuclei.  Important interferences from of the average binding energy curve: (i)The value of `overline(BE)` rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (IRON) and then it slowly decreases. (ii)The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive. (iii) For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radio active.If two light nuclei with `A lt 28` combine with a nucleus with `A lt 56`. the binding energy per nucleon is more for final nucleus that initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb. (IV) If a nucleus of heavy ELEMENT is split (fission) intotwo or more nuclei of medium value A, the energy released would again be a large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled. |
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| 40823. |
The image of a candle formed by a convex lens is obtained on a screen. Will full size of the image be obtained if the lower half of the lens is painted completely opaque? |
Answer» Solution :The FULL size of the image will be obtained. But the intensity of image will be REDUCED. This is because the NUMBER of RAYS of light refracted through different parts of the lens will be reduced. 
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| 40824. |
Using mirror formula, explain why does a convex mirror always produce a virtual image. |
| Answer» SOLUTION :a.Yes. A VIRTUAL OBJECT can PRODUCE a REAL image. | |
| 40825. |
The frequencies for series limit of Balmer and Paschen series respectively are v_(1) and v_(3). If frequency of first line of Balmer series is v_(2) then the relation between v_(1), v_(2) and v_(3) is |
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Answer» `v_(1)-v_(2) = v_(3)` |
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| 40826. |
A sinusoidal carrier voltage of 80 volts amplitude and 1 MHz frequency is amplitude modulated by a sinusoidal voltage of frequency 5kHz producing 50% modulation. Calculate the amplitude and frequency of lower and upper side bands. |
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Answer» Solution :Amplitude of both LBS and USB are equal and GIVEN by `(MEC)/(2)=(0.5xx80)/(2)=20`volts Now FREQUENCY of LSB =`f_(C)-f_(s)=(1000-5)kHz` |
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| 40827. |
Two point charges placed at a distance of 0.20m in air repel each other with a certain force. When a dielectric slab of thickness 0.08m is introduced in between the charges, the force of interaction is half of its previous value. Find the dielectric constant. |
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Answer» Solution :r= 0.2m, x= 0.08m `F= (1)/(4PI in_(0)) (q_(1)q_(2))/((0.2)^(2))` …….in air `F^(1)= (1)/(4pi in_(0))(q_(1)q_(2))/([(0.2-0.08) + 0.08 sqrtK]^(2)) " Here " F^(1)= (1)/(2)F` SUBSTITUTING we get `K ~= 4` |
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| 40828. |
A current carrying loop is placed in a uniform magnetic field. The Torque acting on it does not depend upon |
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Answer» shape of the loop |
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| 40829. |
Which test did Beinkensopp miss? |
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Answer» Maths |
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| 40830. |
A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x=A to x=A//2 is : |
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Answer» `(T)/(6)` |
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| 40831. |
A spherical conductor of radius 2 m is charged to a potential of 120 V. It is now placed inside another hollow spherical conductor of radius 6 m. Calculate the potential to which the bigger sphere would be raised |
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Answer» 120 V |
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| 40832. |
Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm ). The distance of the light source from the glass surface is 100 cm. At what position the image is formed ? |
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Answer» Solution :`(n_(2))/(V) - (n_(1))/(u) = (n_(2) - n_(1))/(R ) ` u = - 100 CM , v = ? , R = + 20 cm , `n_(1) = 1 and n_(2) = 1.5`. We then have `(1.5)/(u) + (1)/(100) = (0.5)/(20) or v = + 100 ` cm The image is at a distance of 100 cm from the glass surface , TOWARDS RIGHT. |
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| 40833. |
Water is flowing continuously from a tap having an internal diameter 8 xx 10^(-3) m. The water velocity as it leaves the tap is 0.4 ms^(-1). The diameter of the water stream at a distance 2 xx 10^(-1) m below the tap is close to: |
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Answer» `7.5 XX 10^(-3) m` `P_(0)+1/2pv_(1)^(2)+pgh=P_(0)+1/2pv_(2)^(2)+0` `v_(2)=SQRT(v_(1)^(2)+2gh)=sqrt(0.16+2xx10xx0.2)=2.03m//s` From continuity equation `A_(2)v_(2)=A_(1)v_(1)` `pi(D_(2)^(2))/4v_(2)=pi(D_(1)^(2))/4v_(1)` `D_(2)D_(1)sqrt((v_(1))/(v_(2)))3.5510^(-3)m` CORRECT choice : ( c) |
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| 40834. |
In a vernier calliper, vernier scale has 20 division which match with 19^(th) division of main scale. The least count of main scale is 1 mm. When two jaws of vernier calliper is in contact, zero of main scale is visible and 14^(th) vernier scale division match with a main scale division. In measuring the length of iron rod by this instrument, main scale reads 35 division and 8^(th) division of vernier scale coincides with a main scale division. Find length of rod in cm. |
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Answer» 20 VSD = 19 MSD LEAST count = MSD - VSD = 0.005cm +ve zero error = `14 xx LC = .070 cm` MEASURED LENGTH `= MSR + VSR xx LC - (+ve "error")` `= 3.5 xx 0.1 + 8 xx 0.005 - 0.070` `= 3.5 + 0.040 - 0.070` `=3.500 - 0.030 = 3.47 cm`. |
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| 40835. |
A particle is thrown with a velocity u at an angle theta from the horizontal. Another particle is thrown with the same velocity at an angle a from the vertical. The ratio of times of flight of the two particles will be |
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Answer» `TAN 2 THETA :1` |
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| 40836. |
When the concentration of alkyl halide is triple and concentration of OH reduced to half , the rate of S_(n) rectio increased by : |
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Answer» 3 times |
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| 40837. |
A 20 volts 5 watt Jamp is used on a.c. mains of 200 volts 50 c.p.s. Calculate the value of (a) capacitance, (b) inductance to be put in series to run the lamp. (c) how much pure resistance should be included in place of the above device so that lamp can run on its voltage |
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Answer» Solution :For LAMP`i=(5)/(20)=0.25A` `R=(20)/(0.25)=80Omega` Current through the lamp should be 0.25 A (a) When condenser C is placed in series`i=(200)/sqrt(R^(2)+(1)/(omegaC)^(2))=0.25` putting the value of `omega=2pixx50` `thereforeC=4.0muF` (B) When inductor is USED`I=(200)/(sqrt(R^(2)+(omegaL)^(2)=0.25impliesL=2.53H` (c) When resistance is used`I=(200)/(R+r)=0.25` `impiiesr=720Omega` |
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| 40838. |
A wave pulse starts propagating in the x direction along a non uniform wire of length 10m with mass per unit length is given by mu = mu_0 + ax and under a tension of 100N. The time taken by a pulse to travel from the lighter end to heavier end(mu_0 = 10^(-2) kg//m " and " a = 9 xx 10^(-3) kg//m^2) is |
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Answer» 22.27 SEC |
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| 40839. |
Explain the refraction of plane wavefront of light through a prism. |
Answer» SOLUTION :The secondary WAVELETS FORM B will travel almost the whole distance inside the prism to REACH the POINT C. But the secondary wavelets from K will travel the whole distance in air before it is incident on the prism at A. The points K and B are phase, and so do L and C. LC and KB represent the plane wavefronts. 
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| 40840. |
The angle of the prism is 6^@ and it's refractive index is 1.5 w.r.t. green colour. The deviation of green ray is : |
| Answer» ANSWER :D | |
| 40841. |
A system has net charge zero. Can it have magnetic moment ? |
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Answer» yes |
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| 40842. |
The gate for which output is high, if atleast one input is low? |
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Answer» NAND |
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| 40843. |
Assertion:A bar magnet is dropped into a long vertical copper tube. Even taking air resistance as negligible, the magnet attains a constant terminal velocity. If the tube is heated, the terminal velocity gets increased. Reason: The terminal velocity depends on eddy current produced in bar magnet. |
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Answer» If both ASSERTION and REASON are t rue and the reason is the correct explanation of the assertion. |
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| 40844. |
How you can decreases noise in FM receivers? |
| Answer» Solution :Noise can be DECREASED in FM RECEIVERS by increasing the frequency of DEVIATION. | |
| 40845. |
The wavelength of the yellow doublet components of the resonance Na line caused by the transition 3p rarr 3S are equal to 589.00 and 589.56nm. Find the splitting of the 3P term in eV units. |
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Answer» Solution :The splitting of the `NA` LINES is due to the fine structure splitting of `3p` lines (The `3S` state in nearly single except for possible hyperfine effects). The splitting of the `3p` level then equals the energy difference `DELTAE=(2piħc)/(lambda_(1))-(2piħc)/(lambda_(2))=(2piħc(lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))~~(3piħcDelta lambda)/(lambda^(2))` Here `Delta lambda`= WAVELENGTH difference & `lambda=` average wavelength. Substitution GIVES `DeltaE=2.0meV` |
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| 40846. |
Compare the rate of radiation of heat of a body at 327^@C with the rate of emission of heat of same body at 27^@C |
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Answer» `(6/3)^4` |
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| 40847. |
A moving coil galvanometer 'A' has 200 turns and a resistance of 100Omega. Another meter 'B' has 100 turns resistance 400 . All the other quantities are same in both the cases. The current sensitivity of A is |
| Answer» Answer :A | |
| 40848. |
In the doble slit experiment the distance of the second dark fringe from the central line is 3 mm. The distance of the fourth bright fringe from the line is: |
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Answer» 8 mm |
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| 40849. |
Electrostatic field lines of force do not form close loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force? |
| Answer» Solution :It is a conservative FORCE. SINCE WORK done by the conservation force in moving a body along a CLOSED path is zero. | |
| 40850. |
A uniformly charged conductingsphereof 2.4 m diameter has a(a)find the charge on the sphere (b) what is the total eletricfluxd leaving the surface of the sphere |
| Answer» SOLUTION :`-6.67 NC` | |
