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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.
| 40651. |
A beam of light of wavelength 600 nm from a distance source falls on a single slit 1.00 mm and the resulting diffraction is observed on a screen 2m away. The distance between the first dark fringes on either side of the central is (a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm |
| Answer» SOLUTION :d | |
| 40653. |
A particle performing S.H.M. has an acceleration of 0.2 m//s^2 at a distance of 0.05 m from its mean position then what is it's time period ? |
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Answer» SOLUTION :`AC c^n=omega^2x` ` 0.2=((2PI)/T)^2xx0.05` T^2=(4pi^2xx0.05)/0.2` `T^2=pi^2` `T=pi SEC.` |
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| 40654. |
A container with insulating walls is divided into equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure P and temperature T, whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be: |
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Answer» `P/2,T/2`  It is the FREE expansion So, Tremain CONSTANT `P_(1)V_(1)=P_(2)V_(2)` `PV/2=P_(2)(V)` `P_(2)=(P/2)` So, CORRECT choice is (d) |
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| 40655. |
What is the magnitude direction of the magnetic fields at A, and C. ( ## EXP_SPS_PHY_XII_C04_E03_011_Q01 .png" width="80%"> |
| Answer» Solution :A- outward to the plane of PAPER , C- INWARD to the plane of paper | |
| 40656. |
Two radioactive materials X_1and X_2contain same number of nuclei. If 6lamdas^(-1) and 4lamdas^(-1)are the decay constants of X_1 and X_2respectively, find the time after which ratio of number of nuclei undecayed of X_1to that of X_2will be 1/e |
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Answer» `1/(10lambda)` |
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| 40657. |
Plane microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflection. Between positions of 14 successive maxima, the detector travels a distance 0.14m. If the velocity of light is 3 xx 10^(8) m//s, find the frequency of the transmitter. |
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Answer» `1.5 xx 10^(10)HZ` |
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| 40658. |
Explain the resolving power of microscope. |
Answer» Solution : Image of a point like object by an objective of 3 microscope is shown in figure. Let diameter of the lens be D and its focal length be F. As object distance is kept greater than that of f. Let an image distance be v. The angular width of central maximum due to the effect of diffraction is `theta=(1.22lamda)/(D)` Linear width of central maximum `=vtheta` `:.vtheta=((1.22lamda)/(D))v""......(1)` If image of two point like objects are at a separation LESS than VO, then `vtheta`, it will be seen as a mixed single object. If minimum distance for two IMAGES of two point like objects found to be resolved is `d_(m)` `:.d_(m)=((1.22lamda)/(D))(v)/(m)""......(2)` [ `:.m=(v)/(f)` magnification ] `:.d_(m)=((1.22lamda)/(D))f""[:.(v)/(m)=f]` From figure `(2)/(f)=tanbeta` `:.(D)/(f)=2tanbeta""......(3)` From equation (2) and (3), `d_(m)=(1.22lamda)/(2tanbeta)` If `beta` is too small and is in radian then `tanbeta~~sinbeta` `:.d_(m)=(1.22lamda)/(2sinbeta)""......(4)` If medium ofrefractive index"n" is placed between on object and objective then from equation (4) `:.d_(m)=(1.22lamda)/(2nsinbeta)""......(5)` Where `nsinbeta` is called numerical aperture and `(1)/(d_(m))`is called resolving POWER of microscope. If refractive index "n" of medium is increased, resolving power of microscope can be increased. Value of `sinbeta` is not more than 1, so resolving power (R.P.) of microscope is inversely proportional to the wavelength `lamda`. |
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| 40659. |
Choosethe wrongstatement(s) |
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Answer» Surface TENSION is directly PROPORTIONALTO extension of surface |
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| 40660. |
Deduce the relation between n,u,v, Q, R for refraction at a spherical surface, where the symbols have their usual meaning. |
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Answer» Solution :OM = u = object distance MI = v = image distance MC = R = radius of curvature Angle i = angle of incidence Angle r = angle of refraction ON = INCIDENT RAY NI = refracted ray  NC=normal & `n_1,n_2` are the refractive indices From the figure for SMALL angles tan`angleNOM=(MN)/(OM)`, tan `angleNCM=(MN)/(MC), tan angleNCM=(MN)/(MI)` In the triangle NOC, `hati` = exterior angle = sum of the interior opposite angles `anglei=angleNOC + angleNCO=angleNOM+angleNCM=(MN)/(OM)+(MN)/(MC)` Similarly , `angleNCM = angleCNI + angleNIC = angleCNI+angleNIM` `angler=angleCNI=angleMCN-angleNIM=(MN)/(MC)-(MN)/(MI)` From Snell.s law `n_1` sin i = `n_2` sin r Forsmall angles `n_1 i = n_2 r` Substituting the VALUES of i and r we get `n_1((MN)/(OM)+(MN)/(MC))=n_2((MN)/(MC)-(MN)/(MI))` `n_1/(OM)+n_2/(MI)=(n_2-n_1)/(MC)` APPLYING sign convention, OM=-u , MI=v , MC=R `n_1/v-n_2/u=(n_2-n_1)/R` This is the required expression. |
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| 40661. |
radar sends a radio signal of frequency 9 xx 10^9 Hz towards an aircraft approaching the radar. If the reflected wave shows a frequency shift of 3 xx 10^3 Hz the speed with wich the aircraft is approaching the radar in ms^(-1) (velocity of the rador signal is 3 xx 10^8 ms^(-1) ) |
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Answer» 150 |
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| 40662. |
Is the permeability of a ferromagnetic material independent of the magnetic feld? If not, is it more for lower or higher fields? |
| Answer» Solution :(d) Permeability of ferromagnetic MATERIAL depends on magnetising field. Because in the hysterisis CURVE, we find that as H is increased INITIALLY VALUE of B increases rapidly, which indicates that value of μ must be increasing. | |
| 40663. |
Are matter waves are electromagnetic ? |
| Answer» SOLUTION :No, ELECTROMAGNETIC WAVES are PRODUCED by accelerating CHARGED particles. | |
| 40664. |
Ozone is found in |
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Answer» Stratosphere |
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| 40665. |
Speed of electromagnetic wave is the same : |
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Answer» for all wavelengths |
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| 40666. |
How does the dispersive power of glass prism change when it is dipped in water? |
| Answer» SOLUTION :DECREASES | |
| 40667. |
If a vector (2hat(i) + 3hat(j) + 8hat(k)) is perpendicular to the vector (4hat(i) - 4 hat(i) + alphahat(k)), the value of alpha is : |
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Answer» `1/2` `(2hati + 3hatj + 8hatk).(4hatj -4hati+ ALPHA k)=0` `-8 + 12 + 8alpha=0` or `8 alpha=-4` pr `alpha=-1/2` |
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| 40668. |
The direction (q) of vec(E ) at point p due to uniformly charged rod will be |
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Answer» at ANGLE `30^(@)` from X-axis |
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| 40669. |
For making standard resistance wire of following material is used |
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Answer» NICHROME |
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| 40670. |
The equivalent capacitance between pointsA and B is |
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Answer» `2 MUF` |
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| 40671. |
An infinite line charge produces a field of 9 xx 10^(4)N//C at a distance of 2cm. Calculate the linear charge density. |
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Answer» Solution :(a) `1.45 xx10^(-3) C` (B) `1.6 xx10^(8) NM^(2)//C` |
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| 40672. |
In a compound microscope cross-wires are fixed at the point |
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Answer» Where the image is FORMED by the objective |
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| 40673. |
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? |
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Answer» Solution :Here h = 3.0 cm, u = - 14 cm, f = - 21 cm `therefore 1/v -1/(-14) =1/(-21)` or `1/v =-1/21 - 1/14 =(-2 -3)/42 =-5/42 rArr v=-42/5 = -8.4 cm` NEGATIVE value of v means that the IMAGE is virtual and erect. If SIZE of image be h., then `h. =h(v/u) = (3.0) xx (-8.4)/(-14) = 1.8 cm` If the object is moved farther away from the lens, the virtual image moves towards the principal focus of the lens and size goes on diminishing further. |
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| 40674. |
A spherical air bubble is embedded in a piece of glass. For a light ray passing through it the bubble behaves as a __________. |
| Answer» SOLUTION :CONCAVE LENS | |
| 40675. |
Answer the following questions: (e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter? |
| Answer» Solution :The refractive index of diamond 2.42 is MUCH more greater than the refractive index 1.5 of normal glass. Critical angle of diamond `(24^@)` is much more less than that of normal glass `(41^ @ 48.)` skilled diamond labours develop LARGE incidence series of `24^@` to `90^@`. Hence, the light enters into diamond will undergo TOTAL internal reflection from all surfaces before it emerges out. Hence, diamond gives the sparkling effect. | |
| 40676. |
A conducting coil of resistance R and radius r has its centre at the origin of the coordinate system in a uniform magnetic field of induction B. When it is rotated about y-axis through 90^(@), the change of flux in the coil is directly proportional to |
| Answer» Answer :B::C::D | |
| 40677. |
Explainanalytically how thestationarywaves are formed.Hence show that the distance between node and adjacent antinode is (lambda)/(4). A set of 48 tuning forks is arrangedin a series ofdescending frequencies such thateach fork gives 4 beatsper secondwith preceding one. The frequenciesof first fork is 1.5 timesthe frequency of the last fork, fidn the frequency of the first and 42nd tuning fork. |
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Answer» SOLUTION :AMPLITUDE of antinodesis maximum, `A=+ -2a` `A = 2a cos '(2pix)/(lambda)` `:.+ -2a = 2a cos'(2pix)/(lambda)` `rArr cos'(2pix)/(lambda) = +-1` `:. (2pix)/(lambda) = 0,pi,2pi` or `(2pir)/(lambda) = Ppi` `:. x = (Plambda)/(2) = P(lambda/2)` (where `P = 0,1,2"....."`) For `x = 0, (lambda)/(2), lambda, (3lambda)/(2), "........."` antinodesare produced. Thus,distancebetween any two successive antinodesis `lambda/2`. AMPLITUDEOF nodes is zero , `A = 0` `:. A = 2a cos '(2pix)/(lambda)` `:. 0 = 2a cos'(2pix)/(lambda)` `:. cos '(2pix)/(lambda) = 0` `:. (2pix)/(lambda) = (pi)/(2), (3pi)/(2), (5pi)/(2) "...."` `:.x= (2p - 1) lambda/4` (where ` p = 1,2 "...."`) For `x= lambda/4, (3lambda)/(4), (5lambda)/(4) "...."`nodes areproduced. Thus, distance between any two successive nodes is `lambda/2`. The distance between node and adjacentantinodesis `lambda/4`. Numerical : Given : `n_(1) = 1.5 n_(48)` and 4 beat/second are produced with preceddingone. The set of tuningforks are arrangedin decreasingorderof frequencies . `:. n_(2) = n_(1) - 4` `n_(3)= n_(2) - 4 = n_(1) - 2 xx 4` `n_(48) =n_(47) - 4 = n_(1) - 47 xx 4` `rArr n_(48) =n_(1) - 188` `rArr n_(48) = 1.5n_(48) - 188 ( :' n_(1) = 1.5n_(48))` `rArr 0.5n_(48) = 188` `rArr n_(48) = 376` `rArr n_(1) = 1.5n_(48) = 1.5 xx 376= 564` `n_(42) = n_(1) - 164 = 564 - 164` `:. n_(42) = 400 Hz`. |
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| 40678. |
Two masses of 4 kg and 9 kg are moving with equal kinetic energies. The ratio of magnitudes of their linear momenta is: |
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Answer» `sqrt2:1` |
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| 40679. |
The angle of incident at whch reflected light is totally polarised fro reflection from air to glass (ref. Index n) is : |
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Answer» `sin^(-1)(N)` |
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| 40680. |
Two identical capacitors are placed side by side with a small gap between them as shown in fig. V_(1) and V_2are potential difference across the two capacitors respectively. An electron is projected from the end of upper plate of the left capacitor with velocity v_0=sqrt((3V_(1)1^2e)/(4md^2))towards right as shown. Charge and mass of electron are .e. and .m. respectively. V_2is applied such that the electron just grazes the lower plate of right capacitor horizontally while coming out of it. Assume that there is no gravity and no collision of electron with any plate. What is the ratio V_2/V_1 ? |
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Answer» |
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| 40681. |
A series circuit containing inductance L_1 and capacitance C_1 oscillates at angular frequency omega. A second series circuit, containing inductance L_2 and capacitance C_2 oscillates at the same angular frequency. In terms of omega, what is the angular frequency of oscillation of a series circuit containing all four of these elements? Neglect resistance. |
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Answer» |
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| 40682. |
Find the charge on capacitor in steady state. |
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Answer» Solution :In steady state the equivalent circuit is `5i_(1)-2i_(2)=6` `6i_(2)-2i_(1)=-4` `13i_(2)=14` `i_(1)=(14)/(13)`Amp, `i_(2)=-(4)/(13)` Amp `:. ` Potential difference ACROSS `BC` `V_(A)=6-(14)/(13)=(64)/(13)V` `V_(D)=(64)/(13)-2(-4//13)=(72)/(13)V` `V_(B)=(64)/(13)-1xx{(14)/(13)+(4)/(13)}=(46)/(13)V` `:. V_(DB)=(72-46)/(13)=(26)/(13)=2`volt `:. ` Charge on capacitor `=CV_(BD)=4xx10^(-6)xx2=8muC` 
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| 40683. |
Of the following which pair can be coherent sources |
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Answer» two SODIUM vapour lamps of same power CONNECTED in parallel to the same mains |
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| 40684. |
(A): The induced e.m.f. and current will be same in two identical loops of copper and aluminium, when rotated with same speed in the same magnetic field.(R) : Induced e.m.f. is proportional to rate of change of magnetic field while induced current depends on resistance of wire |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 40685. |
Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 yr. After 1 yr |
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Answer» all the containers will have 5000 atoms of the material. |
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| 40686. |
On what factors does the angle of minimum deviation produced by a prism depend ? |
| Answer» SOLUTION :On the material of the PRISM and the wavelength of LIGHT USED. | |
| 40687. |
सम्बन्ध q=ne में n का मान नहीं हो सकता है - |
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Answer» 3 |
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| 40688. |
The density of ice is x g cm^(-3) and that of water is y g cm^(-3). What is the change in volume when m gram of ice melts ? |
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Answer» `m ( y-x) c.c`. Vol. of m gram WATER FORMED`=m/yc.c` CHANGE in vol. on melting =`(m/y-m/x)` `=m(1/y-1/x)`c.c. Correct choice is (d). |
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| 40689. |
In communication system and RADAR microwave are used, because ……. |
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Answer» its WAVELENGTH is very small. |
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| 40690. |
A conductor with a cavity is charged positively and its surface charge density is sigma. If E and V represent the electric field and potential, then inside the cavity |
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Answer» E = 0 and V = 0 |
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| 40691. |
Magnetic dipole moment is a |
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Answer» Scalar QUANTITY |
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| 40692. |
Give the advantage of AC in long distance power transmission w ith an example. |
Answer» SOLUTION :Advantages of AC in long distance power transmission: Electric power is produced in a large scale at electric power stations with the help of AC generators. These power stations are classified based on the txpe of fuel used as thermal, hydro electric and nuclear pow er stations. Most of these stations are located at remote places. Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually CONSUMED. This process is called power transmission. But there is a difficulty during power transmission. A sizable fraction of electric pow.er is lost due to Joule heating (rR) in the transmission lines which are hundreds of kilometer long. This power loss can be tackled either by reducing current i or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick w.ires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable. Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing pow.er losses to a greater extent. At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically. Illustration: An electric power of 2 MW is transmitted to a place through transmission lines of TOTAL resistance, SAY R = 40 `Omega`, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases. Case (i): P = 2 MW, R = 40 `Omega`, V = 10 kV Power , P=VI `:.` Current, `I=P/V=(2xx10^(6))/(10xx10^(3)=200A` Power loss = Heat produced - `I^(2)R - (200)^(2) xx40 = 1.6xx10^(6)W` % of power loss= `(1.6xx10^(6))/(2xx10^(6)) xx 100%= 0.8xx100% - 80%` Case (ii): P = 2 MW, R= 40 `Omega`, V = 100 kV `:.` Current, `I=P/V=(2xx10^(6))/(100xx10^(3))=20A` Power loss - float produce - `I^(2)R=(20)^(2) xx40=0.016xx10^(6)W` % Power Loss =`(0.016xx10^(6))/(2xx10^(6)) xx100%= 0.8%` |
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| 40693. |
The distance between the object and the real image formed by comvex lens is d. If the magnification is m, then the focal length of the lens is : |
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Answer» `(MD)/((m+1)^(2))` |
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| 40694. |
A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45^(@) with the initial vertical direction is : |
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Answer» `MG(SQRT2+1)` `T sin 45^(@) = F`  `T cos 45^(@)=mg` `:.TAN 45^(@)=(F)/(mg)` `impliesF=mg` Here choice is (b) |
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| 40695. |
Explain the crystalline structure by writing the electronic configuration of elementary semiconductors. |
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Answer» Solution :Elementary semiconductors are Si and Ge. Atomic number of Si is Z = 14. Its ELECTRONIC configuration is `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(2)`. SHELL K and L are filled completely by `1s^(2), 2s^(2)2p^(6)` and the shell for n = 3 is incomplete. It contains `3s^(2)3p^(2)` four valence electron. Atomic number of Ge is Z = 32. Its electronic configuration is `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6)3d^(10), 4s^(2)4p^(2)` shell K and L and are filled completely by `1s^(2), 2p^(2)2s^(2), 2p^(6), 3s^(2)3p^(6)3d^(10)` but for n = 4 shell N is incomplete there are `4s^(2)4p^(2)` valence electrons in it. Hence Si and Ge are tetravalent elements (with four valency) In its crystalline structure , every Si or Ge atom tends to share one of its four valence electrons with each of its four nearest neighbour atoms and also to take share of one electron from each such neighbour. These shared electron pairs forming a covalent BOND or a valence bond. Each bond contains two-two electrons.  Figure depicts one atom of Ge or Si covalent bonding with the four atoms in its neighbour atom in that solid dot are valence electrons. Here figure is in two-dimensional and +4 symbol INDICATES inner CORES of Si or Ge. Each of the bonded electrons bonds tightly to the atom they are attach to. Three-dimensional diamond like crystal structure for carbon, silicon and germanium with respective lattice spacing is shown in the figure.  Here not a single bond is broken so it shows the ideal situation which is at low temperature (means at absolute zero temperature). |
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| 40696. |
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing. |
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Answer» Solution :A simple microscope is a single magnifying (converging) lens of small focal length. The ides is get an erect, MAGNIFIED and virtual image of the obect. For this the object is PLACED between F and P on one side of the lens and viewed from other side of the lens. There are TWO magnificatons to be discussed for two kinds of focusing. Magnificaitons in near Point focusing `m = 1 + (D)/(f)` (ii) Magnification in normal focusing (angular magnification), `m = (D)/(f)` |
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| 40697. |
A trnsistor is a temperature sensitive device. Comment. |
| Answer» Solution :When a TRANSISTOR is properly biased, its ELECTRONS and holes carry current It it’s temperature is INCREASED, many of its covalent bonds may break producing a large number of extra elctron-hole pairs which will set a large current through the transistor. This may produce a large amount of heat resulting in the COMPLETE BREAKDOWN of .the .transistor. | |
| 40698. |
(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working. (b) Answer the following : (i) Why is it necessary to introduce a cylindrical soft iron core inside the coil a galvanometer? (ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain giving reason. |
| Answer» Solution :(b) (i) A CYLINDRICAL soft iron core is introduced inside the coil of a galvanometer so as to MAKE the magnetic field STRONGER. It also HELPS in making the field radial . | |
| 40699. |
The north pole of the earth's magnet is near the geographical |
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Answer» South |
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| 40700. |
यदि X और Y दो ऐसे समुच्चय हैं की X में 40 अवयव, XUY में 60 अवयव और XnnY में 10 अवयव हैं Y में कितने अवयव हैं? |
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Answer» 30 |
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