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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Strain energy of stretched wire is 18 xx 10^(-3)J and strain energy per unit volume of the same wire under same condition is 6 xx 10^(-3) J/m^3. Find its volume |
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Answer» a)`3 cm^3` |
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| 2. |
A body falls freely for 10 sec Its average velocity during this journey ( take = 10ms^(-2)) |
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Answer» `100MS^(-1)` |
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| 3. |
To increase the hardness of X–rays in coolidge tube we should – |
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Answer» INCREASE FILAMENT current |
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| 4. |
An object with mass 5 kg is acted upon by a force, vecF = (-3hati + 4hatj) N. If its initial velocity at t = 0 isvecv = (6hati – 12hatj) m s^(-1), the time at which it will just have a velocity along y-axis is |
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Answer» 2s ` v_x =U_x +a_x t0=6+(-3)/( 5)timplies t=10s` |
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| 5. |
Two closed surface S_1 and S_2 enclose two charges q_1 and q_2as shown in the figure. |
| Answer» Solution :Remains the same (because electric FLUX is inde pendent of the SIZE and SHAPE of the ENCLOSED surface) | |
| 6. |
The angular velocity , omega of a rotating body or shaft can be measuredby attaching an open cylinder of liquid , as shown in Fig and measuring the change in the fluid level , H- h_(0) , caused by the rotation of the fluid . Determine the relationship between this change in fluid level and the angular velocity . Assume no fluid is spilling out of the vessel. |
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Answer» Solution :The HEIGHT h , of the free surface above the tank bottom can be determined as follows `h - (omega^(2) r^(2))/(2 g) + h_(0)`  The INITIAL volume of fluid in the tank , `V_(i) = pi R^(2) H` The volume of the fluid with the rotating tank can be found by considering a differential ELEMENT of cylindrical shape as shown in Fig . This cylindrical shell is TAKEN at some arbitary radius , r and thickness dr . Thus , its volume is `dV = 2 pi r h dr` The total volume is, therefore , `V. = 2 pi int_(0)^(R) r ((omega^(2) r^(2))/(2 g) + h_(0)) dr = (pi omega^(2) R^(4))/( 4 g) + pi R^(2) h_(0)` Since the volume of the fluid in the tank must remain constant (given that none spills over the top) , it follows that`pi R^(2) H = (pi omega^(2) R^(4))/(4 g) = pi R^(2) h_(0)` or `H - h_(0) = (omega^(2) R^(2))/(4 g)` |
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| 7. |
The mean squqre value of ac is related to the mean of the square of instantaneous current by |
| Answer» SOLUTION :SQUARE ROOT | |
| 8. |
Potential energy of the particle performing S.H.M. is |
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Answer» HARMONIC MOTION and OSCILLATORY |
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| 9. |
If a ray of unpolarized light is made to incident at the polarizing angle of a glass slab then show that the reflected and reflected rays will be along mutually perpendicular directions. |
| Answer» SOLUTION :REFER topic " to show that the reflected RAY is right angle to the refracted ray at polarizing angle". | |
| 10. |
In a chain of identical atoms the vibration frequency omega depends on wave number kas omega= omega_(max) sin (ka//2), where omega_(max) is the maximum vibration frequency omega, a is the distance between neighbouring atoms. Making use of this dispersion relation, find the dependence of the number of longitudinal vibrations per unit. frequency interval on omega i.e., dN//omega, if the length of the chain is l. Having obtained dN//omega, find the total number N of possible longitudinal vibrations of the chain. |
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Answer» SOLUTION :If the CHAIN has `N` atoms, we can assume atom NUMBER `0` and `N+1` held ficed. Then the displacement of the `n^(th)` atom has the form `u_(n)=A("sin"(m pi)/(L),n a) sin omega t` Here `k=(m pi)/(L)`. Allowed frequencies then have the form `omega=omega_(max) "sin"(KA)/(2)` In our from only `+vek` values are allowed. The number of models in a wave number range `dk` is `dN=(Ldk)/(pi)=(L)/(pi)(dK)/(d omega)` But `d omega=(a)/(2) omega_(max)"cos"(ka)/(2)dk` Hence `(d omegaqa)/(dk)=(a)/(2)sqrt(omega_(max)^(2)-omega^(2))` So `dN=(2L)/(pia)(d omega)/(sqrt(omega_(max)^(2)-omega^(2)))` (b) The total number modes is `N= int_(0)^(omega_(max))(2L)/(pi a)(d omega)/(sqrt(omega^(2)_(max)-omega^(2)))=(2L)/(pi a).(pi)/(2)=(L)/(a)` i.e., the number of atoms in the chain. |
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| 11. |
A parallel beam of light falls normally on the first face of a prism of small anlge A. At the second face it is partly transmitted partly reflected. The reflected beam striking the first face again, emerges out through the first face at an angle of 6^(@)30' with the normal to the first face. The refracted beam is found to hae undergone a deviation of 1^(@)15' from the original direction. Calculate the refractive index and angle of the glass prism . |
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Answer» |
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| 12. |
A coil of inductive reactance 31Omega2 has a resistance of 8Omega. It is placed in series with a condenser of capacitive reactance 25Omega. The combination is connected to an ac source of 110 V. The power factor of the circuit is |
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Answer» 0.56 |
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| 13. |
The top of the atmosphere is at about 400 kV ,with respect to the surface or the earth, corresponding to an electric field that decreases with altitude. Near the surracc of the earth, chc field isabout 100 Vm^(-1) . Why then do we not get an electric shock as we step out of our house into the open ? (Assume lhe house to be a steel cage so there is no field inside) . |
| Answer» Solution :Since our body and the SURFACE of earth are conducting they FORM an EQUIPOTENTIAL swface. The equipotential surface of OPEN air are parallel to the surface of earth. As we step out into open from our house, the original equipotential surface of open air get modified keeping our body and ground at the same potential since there is no potential difference between our body and the ground, we do not get any ELECTRIC shock. | |
| 14. |
A ring of radius R is with a uniformly distributed charge on it. A charge q is now placed at the centre of the ring. Find the increment in tension in the ring. |
Answer» Solution :  Consider an ELEMENT of the ring. Its enlarged view is as SHOWN. For equilibrium of this segment, we can write. `F = 2 DeltaT sin ((d theta)/(2))` Here F is the repulsive force between q and elemental CHARGE `dQ [ :. dQ = Q/(2 pi R) (Rd theta)]` The electric outward force on element is `F = (1)/(4 pi epsilon_0) (q dQ)/(R^2)` From the above three EQUATIONS, we can write, `1/(4 pi epsilon_0) q/(R^2) (QRd theta)/(2 pi R) ~~ 2 Delta T ((d theta)/(2)) ( :. sin alpha~~ alpha` for SMALL angle) `Delta T = (Qq)/(8 pi^2 epsilon_0 R^2)` |
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| 15. |
In the previous question films of thickness t_(A) and t_(B) and refractive indices mu_(a) and mu_(b) are placed in front of A ahtf B- respectively,, if mu_(A) t_(A) = mu_(B) t_(B) . the central maximum will- |
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Answer» not shift |
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| 16. |
Light waves of wave length lambda propagate in a medium. If M end N are two points on the wave front and they are separated by a distance lambda"/"4, the phase difference between them will be (in radian) |
| Answer» Answer :D | |
| 17. |
A 1000W transmitter works at a frequency 880 kHz. The number of photons emitted per second is |
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Answer» `1.7xx10^(28)` |
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| 18. |
Calculate the energy released in the given reaction: ""_(6)C^(12) + ""_(6)C^(12) to ""_(10)Ne^(20) + ""_(2)He^(4) Given atomic masses are as follows: M_(He) = 4.002603u , M_(Ne) = 19.992439 u, M_( c)=12.0000u |
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Answer» |
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| 19. |
Product (B) is - |
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Answer» ANTIBIOTIC 
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| 20. |
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour. From which of the photosensitive materials with work functions listed given below and using the results of (i), (ii) and (iü) of (a), can you build a photoelectric device that operates with visible light ? |
| Answer» SOLUTION :For a photoelectric device to OPERATE, we REQUIRE INCIDENT light energy E to be equal to or GREATER than the work function `phi_(0)`, of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with `phi_(0) = 2.75` eV), K (with `phi_(0)= 2.30` eV) and Cs (with `phi_(0) = 2.14` eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (With `phi_(0)= 2.14` eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials | |
| 21. |
The variation of magnetic susceptibility (x) with absolute temperature for antiferromagnetic material is given by |
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Answer»
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| 22. |
A man can swim three kilometre distance in three hours in a still water. A log of wood floating downstream cowers a distance of 1 km during this time. How much distance the man would swim opposite to the log of wood in three hours? |
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Answer» 1 km |
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| 23. |
Out of microwaves, ultraviolet rays and infrared rays , which radiation is most effective for emission of electrons from a metallic surface ? |
| Answer» SOLUTION :ULTRAVIOLET RAYS are most effective for photoelectric emission because they have highest frequency and hence most ENERGETIC. | |
| 24. |
A gas of hydrogen like atoms are in an unknown energy state of principal quantum number n_1 . They can absorb radiations having photons of energy 68eV. Consequently , the emission spectrum of the gas has only three different lines . All the wavelengths are equal or smaller than that of the absorbed radiation . Assuming Bohr's model to be applicable answer the following question . The initial state n_1 of the gas atoms is |
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Answer» 2 |
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| 25. |
A flywheel with moment of inertia 0.86 kg ·m^2and a cylinder of 5 cm radius of negligible mass are fixed to a common shaft (Fig.). A thread is wound around the cylinder, and a weight of 6.0 kg mass is attached to it. What time will the weight take to fall 1 m? What will be its final speed? Assume the initial speed to be zero. |
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Answer» `mgh=1/2mv^2+1/2Iomega^2` The flywheel and the cylinder ROTATE at the same angular VELOCITY `omega = v//r` , therefore `2mgh = v^2 (m + 1//r^2)`. Hence `v=sqrt((2mgh)/(m+I//r^2)` |
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| 26. |
A gas of hydrogen like atoms are in an unknown energy state of principal quantum number n_1 . They can absorb radiations having photons of energy 68eV. Consequently , the emission spectrum of the gas has only three different lines . All the wavelengths are equal or smaller than that of the absorbed radiation . Assuming Bohr's model to be applicable answer the following question . The minimum wavelength in the obtained spectrum will be nearly. |
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Answer» `28Å` |
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| 27. |
A gas of hydrogen like atoms are in an unknown energy state of principal quantum number n_1 . They can absorb radiations having photons of energy 68eV. Consequently , the emission spectrum of the gas has only three different lines . All the wavelengths are equal or smaller than that of the absorbed radiation . Assuming Bohr's model to be applicable answer the following question .The atomic number of the gas is |
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Answer» 2 |
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| 28. |
Two rods of equal length and area are joined in parallel and their conductivities are 4.5 and 3.5, what is the conductivity of the combination ? |
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Answer» 3 Thus CORRECT choice is (d). |
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| 29. |
In the circuit shown, the current flowing from the battery is |
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Answer» 11/50 A |
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| 30. |
A wave of frequency 500 Hz has a velocity 360 m/s. The distance between two nearest point which are 60^(@) out of phase, is : |
| Answer» ANSWER :D | |
| 31. |
Two point charges +2C and +6C repel each other with a force of 12N. If a charge q is given the each of the these charges then they attract with 4N. Then value q is |
| Answer» ANSWER :C | |
| 32. |
Obtain the mirror formula and write the expression for the linear magnification. |
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Answer» Solution :Derivation of mirror FORMULA Expression for inear magification In the avofe FIGURE ` Delta BAP and Delta B.A.P` are similar `rArr (BA)/( B.A.) = ( NF)/( NF.) ` Similarly , `Delta MNF and Delta B.AP.` are similar For the given figure as per the sigh CONVENTION |
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| 33. |
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? ("Take h "= 6.63xx10^(-34)" J s and 1 eV" = 1.6xx10^(-19)J.) |
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Answer» Solution :ENERGY of the INCIDENT photon, `E=hv=(hc)/(lambda)` `E=(6.63xx10^(-34)Js)(3xx10^(8)m//s)//lambda` `=(1.989xx10^(-25)Jm)/(lambda)` (i) For voilet light, `lambda_(1)=390` nm (LOWER wavelength end) Incident photon energy, `E_(1)=(1.989xx10^(-25)Jm)/(390xx10^(-9)m)` `=5.10xx10^(-19)J` `=(5.10xx10^(-19)J)/(1.6xx10^(-19)J//eV)` =3.19eV (ii) For yellow-green light , `lambda_(2)=550nm` (average wavelength) Incident photon energy, `E_(2)=(1.989xx10^(-25)Jm)/(550xx10^(-9)m)` `=3.62xx10^(-19)J` =2.26eV (iii) For red light `lambda_(3)=760nm` (higher wavelength end) Incident photon energy, `E_(3)=(1.989xx10^(-25)Jm)/(760xx10^(-9)m)` `=2.62xx10^(-19)J` =1.64eV |
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| 34. |
As shown in (Fig. 3.90), two large parallel vertical conducting plates separated by distance D are charge so that their potential are +V_0 and - V_0. A small conducting ball of mass m and radius r ("where" R lt lt d) is hung midway between the plates. The thread of length L supporting the ball is a conducting wire connected to ground, so the potential of the ball is fixed at V = 0. The ball hangs straight down in stable equilibrium when V_0 is sufficiently small. Show that the equilibrium of the ball is unstable if V_0 exceeds the critical value [k_e d^2 mg//(4 R L)]^(1//2). |
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Answer» Solution :The plates create uniform electric field to the right in the picture, with magnitude `[V_0 - (-V_0)]// d = 2V_0 //d`. Assume the ball swings a small distance X to the right. It moves to a place where the voltage created by the plates is lower by `- E x = -(2 V_0)/d x` Its ground connection maintains it at `V = 0` by ALLOWING charge q to flow from ground onto the ball, where `- (2V_0 x)/d + (k_e q)/R = 0` `q = (2 V_0 x R)/(k_e d)` Then the ball feels electric force `F = q E = (4 V_0^2 x R)/(k_e d^2)` to the right. for EQUILIBRIUM, this must be balanced by the horizontal component of string tension according to `T cos theta = mg` `T sin theta = (4 V_0^2 x R)/(k_e d^2) tan theta = (4 V_0^2 x R)/(k_e d^2 mg) = (x)/(L)` for small x Then `V_0 = ((k_e d^2 m g)/(4 R L))^(1//2)` If `V_0` is less than this value, the only equilibrium position of the ball is hanging STRAIGHT down. If `V_0` exceeds this value, the ball will saving over to one plate or the other. |
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| 35. |
Four metallic plates, each having area A are as in fig. The distance between the consecutive plates is d. Alternative plates are connected to the points A and. B The equivalent capacitance of the system is : |
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Answer» `(epsilon_0A)/d` |
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| 36. |
A solenoid with large number of turns is in a closed circuit and a short bar magnet is dropped through each with its length along the axis. State the acceleration of the falling magnet when it is : a. Well above A b. At the end A c. At the middle d. At the end B e. Far away, down, from B |
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Answer» Solution :a. Acceleration due to gravity (g) b. Less than g by Lenz.s law. C. Same as in (a) d. Less than g, by Lenz.s law. E. FREE from INDUCTION and hence `a=g` |
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| 37. |
Which law states that the magnitude of pressure within fluid is equal in all parts? |
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Answer» PASCAL's law |
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| 38. |
alpha- particle having 27 MeV has 1.14xx10^(-14) m distance of closest approach from a nucleus of atom. Calculate atomic no. of atom. |
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Answer» 100 `(K(Ze)(2e))/(r_(0))=27MeV` `(kxx2Ze^(2))/(r_(0))=27MeV` `:.Z=(27MeVxxr_(0))/(kxx2e^2)` `=(27xx10^(6)xx1.1xx10^(-14))/(9xx10^(9)xx2xx1.6xx10^(-19))` `=1.03125xx10^(2)=103` |
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| 39. |
Hysteresis cycle of a permanent magnet is ..... and ..... |
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Answer» SHORT and broad |
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| 40. |
The intensity of a traveling wave has units of W//m^(2), give the dimensional formula of intensity : |
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Answer» `[ML^(0)T^(-3)]` `:.` E (intensity)=`("Power")/("area")` `E=(ML^(2)T^(-3))/(L^(2))=ML^(0)T^(-3)` HENCE CHOICE is `(a)`. |
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| 41. |
Two radioactive materials X_1 and X_2 have decay constants 10lambda and lambdarespectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X_1 to that of X_2 will be 1/e after atime |
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Answer» `1/(10lambda)` `therefore (N_0e^(-10lambdat))/(N_0e^(-lambdat))=1/e` or `1/e^(9 lambdat)=1/e` or `9lambdat=1` or `t=1/(9lambda)` |
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| 42. |
The probability that particular nucleus of ""_(38)Cl will undergo beta-decay in time interval 10^(-3) sec (the half life of ""_(38)Cl is 37.2 min) is |
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Answer» `3.1 XX 10^(-7)` |
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| 43. |
A conducting bar of 2m length is allowed to fall freely from a 50m high tower, keeping it aligned along the east-west direction. Find the emf induced in the rod when it is 20 m below the top of the tower. g = 10 "ms"^(-2). Horizontal component of earth's magnetic field is 0.7 xx 10^(-4) T and angle of dip =60^@. |
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Answer» Solution :h=50 m, l=2 m , d=20 m , `g=10 MS^(-2)` `B_h=0.7xx10^(-4)T, theta = 60^@ , N=100`, emf = ? From eqn of free fall BODY In `2ad =v^2-v_0^2, v_0=0` `2gd=v^2"" [because a=g]` `2xx10xx20=v^2` `400=v^2` `THEREFORE` v=20 m/s Now, INDUCED emf according to Faraday.s law `therefore EPSILON=B_h vl` [`because` neglecting negative sign] `therefore epsilon = B vl cos theta [ because B_h =B cos theta]` `therefore epsilon = 0.7xx10^(-4)xx20xx2xxcos 60^@ [ because cos 60^@ =1/2]` =1.4 mV |
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| 44. |
In a series LCR circuit , the potential drop across L, C and R respectively are 40 V , 120V and 60V . Then the source voltage is |
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Answer» 220 V `=sqrt(60^(2)+(120-40)^(2))` `V=100` volt |
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| 45. |
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5) |
| Answer» SOLUTION :`theta_B = 57^@` | |
| 46. |
A rocket, set for vertical launching, has a mass of 50 kg and contains 450 kg of fuel. It can have a maximum exhaust speed of 1 kms^(-1) If g = 10 ms^(-2). What should be the minimum rate of fuel consumption to just lift it off the launching pad ? |
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Answer» `10 kg s^(-1)` to lift the ROCKET `F=Mg` `:.Mg=u(dm)/(dt)implies(dm)/(dt)=(500xx10)/(1000)=5kg s^(-1)` HENCE correct choice is (b) |
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| 47. |
At what angle with the water surface does fish in figure see the setting sun ? |
| Answer» SOLUTION :`C=Sin^-1 0.7518` | |
| 48. |
A cirular coil of wire cansiting of 100 turns ,each of readius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field barB at the centre of the coil ? |
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Answer» Solution :`n= 100 ,I=0.40 A, R= 8.0cm = 8.0 xx 10^(-2) m` `B = (mu_0nl )/(2r) = (4PI xx 10^(-7) xx 100 xx 0.4 )/(2xx8xx10^(-2))= PI xx 10^(-4) T` |
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| 49. |
What are energy bands ? |
| Answer» Solution :The GROUP of DISCRETE but CLOSELY spaced energy levels for the electrons in a particular orbit is CALLED energy band. | |
