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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. |
Suppose the initial charge on the capacitor in Exercise 7 is 6mC. What is the total energy at later time ? |
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Answer» Solution :Given, charge on the capacitor `Q=6mC = 6XX10^(-3)C` `C=30mu F=30xx10^(-6)F` Energy stored in the circuit `E =(Q^(2))/(2C)=((6xx10^(-3))^(2))/(2xx30xx10^(-2))=(36)/(60)=0.6J` After some time, the energy is shared between C and L, but the total energy remains CONSTANT. So, we assume that there is no loss of energy. |
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| 2. |
A proton and an a-particle have the same de-Broglie wavelength. Determine the ratio of their accelerating potentials |
| Answer» Solution :In REGION DE, material offers NEGATIVE RESISTANCE, because slop `(Delta V)/(Delta I) lt 0`. | |
| 3. |
What is the story Deep Water speaking about? |
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Answer» FEAR of WATER and the WAY to OVERCOME it |
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| 4. |
An amplifier is represented by the circuit as shown in Fig. with an input internal resistance r_(i)=100Omega. It is connected to an a.c. voltage source through a series resistance of 300Omega. The no load voltage gain of transistor is 400. What is the apparent gain of the amplifier? |
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Answer» Solution :VOLTAGE gain, `A_(V)=V_(0)/V_(i)=400` or `V_(0)=400V_(i)` If `V_(s)` is the voltage of the source, then voltage ACROSS the INPUT of the amplifier is `V_(i)=(V_(s)xxr_(i))/(r_(i)R_(s))` or `V_(s)=((r_(i)+R_(s))V_(i))/r_(i)` APPARENT voltage gain of amplifier, `A_(V)^(')=V_(0)/V_(s)=(400V_(i)xxr_(i))/((r_(i)+R_(s))V_(i))=(400r_(i))/(r_(i)+R_(s))=(400xx100)/(100+300)=100` |
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| 5. |
In a CE transistor amplifier there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy? |
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Answer» Solution :No, law of CONSERVATION of energy is not violated because here increase in the energy of input AC signal is at the cost of energy STORED in DC SUPPLY. (That is why after USING the amplifier for some time, we have to replace that battery if it is not feedback amplifier). THUS, here energy of output signal = energy of input AC signal + energy of DC supply. Hence, law of conservation of energy is not violated. |
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| 6. |
A thin circular ring of mass m and r is rotating about its axis with a constant angular velocity omega. Two objects, each of mass m are attached gently to the opposite ends of a diameter of the ring wheel now rotates with an angular velocity : |
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Answer» `OMEGAM//(M+m)` `omega_(2)=(MOMEGA)/(M+2m)` |
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| 7. |
Lightof wavelength 5000 A^(0)diffractedby a slit . In diffractionpatternfifth minimum is at a distance of 5 mm from central maximum . If the distance between the screen and the slit is 1 m . The slit width is |
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Answer» 0.5 MM |
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| 8. |
When a coil is joined to a cell, the current through the cell grows with a tiem constant tau. The current will reach 10% of its steady-state value of time |
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Answer» `0.1tau` |
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| 9. |
How much energy is released when a ""^(238)U nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are |
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| 10. |
Which of the following group has the frequency of electromagnetic ray in the ascending form ? |
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Answer» MICROWAVE, ULTRAVIOLET, x - rays. radiowaves `to` microwave `to` infrared radioation `to` visible light `to` ultraviolet rays `to`X - rays `to` gamma rays. |
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| 11. |
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens. |
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Answer» SOLUTION : Here it is GIVEN that `R_1 = 10 cm, R_2 = 15 cm` and `f= 12 cm`. As per sign convention followed for a CONVEX lens, f and `R_1` are +ve but `R_2` is -ve. Hence, applying lens maker.s formula `1/f =(n-1) (1/R_(1)-1/R_(2))`, we have `1/(+12) =(n-1) [1/(+10) -1/(-15)] =(n-1) (1/10 + 1/15) =(n-1) xx 1/6` `rArr (n-1) = 6/12 = 1/2` or `n=1+1/2 = 3/2 = 1.5` |
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| 12. |
Net force acting on an electric dipole placed in a uniform electric field is ___________but net torque acting on the dipole is _____________ |
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Answer» |
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| 13. |
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B . What is the magnitude of the magnetic field? |
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Answer» Solution :we find that there is an upward FORCE F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: MG= IlB `B = (mg)/(Il)` ` = (0.2 xx 9.8)/(2 xx 1.5) = 0.65 T` Note that it would have been sufficient to SPECIFY m/l, the MASS per unit length of the wire. The EARTH’s magnetic field is approximately `4 xx 10^(-5) T`T and we have ignored it. |
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| 14. |
Calculate the energy released by fission from 2 g of ._92^235U in k Wh. Given that the energy released per fission is 200 Mev. |
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Answer» `4.54xx10^4` KWH |
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| 15. |
A charged sphere is connected to a similar uncharged sphere. Then the percentage loss of energy is |
| Answer» ANSWER :A | |
| 16. |
A charged particle gains energy due to _____ . |
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Answer» ELECTRIC field. |
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| 17. |
An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base A and a point B on the wall |
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Answer» `1:2` |
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| 18. |
If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective lens of focal length 5 cm, the focal length of the eye-piece should be close to: |
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Answer» 2 mm If FINAL image is OBTAINED at near point, then magnification, `m=(L)/(f_0)(1-(D)/(f_e))` `375=(150)/(50)(1+(25)/(f_e))` `(375)/(3)=1+(25)/(f_e)` `123-1=(25)/(f_e)` `f_e=(25)/(122)` `thereforef_e=0.2049`mm`therefore~~0.2` mm Case-II : If final image is obtained at infinity, then magnification, `m=(L)/(f_0)((D)/(f_e))` `375=(150)/(50)((25)/(f_e))` `(375)/(3)=(25)/(f_e)` `thereforef_e=(25xx3)/(375)``thereforef_e=0.2` mm |
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| 19. |
What we should use to know the necessity of medical facilities? |
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Answer» Knowledge |
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| 20. |
A current - carrying uniform square frame is suspended from hinged support as shown in the diagram such that it can freely rotate about its upper side. The length and mass of each side of the frame are 2 m and 4 kg respectively. A uniform magnetic field vecB=(3hati+4hatj) is applied. When the wireframe is rotated to 45^(@) from vertical and released it remains in equilibrium. What is the magnitude of current (in A) in the wire frame ? |
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| 21. |
We know that electric and magnetic field are associated with electromagnetic wave travelling in vaccum or some medium. Electric and magnetic fields |
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Answer» remain constant |
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| 22. |
If a stationary charged particle experience an electromagnetic force then _____ . |
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Answer» E = 0, B = 0 `F=F_(e)+F_(m)` INITIALLY (at t = 0) `Fne0` but `F_(m)=0` `(becausev=0)rArrF_(e)ne0rArrEne0` When `tgt0,Fne0` `rArrF_(e)+F_(m)ne0` Here `F_(e)ne0rArrF_(m)=0orF_(m)ne0` `rArrB=0orBne0` `RARR` Option (B) is correct. |
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| 23. |
यदि S={0,1,5,4,7}, तो S के कुल उपसमुच्चयों की संख्या है |
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Answer» 64 |
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| 24. |
An infinite number of charges, each of charge q, are located along the x-axis at x=1, x=2 x=4, x=8 and so on. Find the potential at x=0. |
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Answer» <P> Solution :Potential at x=0 due to other charges is`V=1/(4pi epsi_(0)) [q/1+q/2+q/4+....]=1/(4pi epsi_(0)).q p[1/1 +1/2 +1/4+....]` `=1/(4pi epsi_(0)).q [1/(1-1/2)] ("the term INSIDE the brackets is in GP")=(1)/(4pi epsi_(0)).2q =(q)/(2pi epsi_(0)) "VOLTS"` |
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| 25. |
When m = 1, power carried by side bands is |
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Answer» 11.1% of the TOTAL POWER of AM WAVE |
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| 26. |
The transition from the state n =4 to n =3 is a hydrogen like atom results is ultraviolet radiation. Infrared radiations will be obtained in the transition : |
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Answer» `2 to 1` `E_(4-3)=-Rz^(2) (1/4^(2)-1/3^(2)) =0.049Rz^(2)` `E_(2-1)=-Rz^(2) (1/2^(2)-1/1^(2)) =0.75 Rz^(2)` `E_(3-2)=-Rz^(2) (1/3^(2)-1/2^(2))=0.139Rz^(2)` `E_(4-2)=-Rz^(2) (1/4^(2)-1/2^(2))=0.188 Rz^(2)` `E_(5-4)=-Rz^(2) (1/5^(2)-1/4^(2)) =0.023 Rz^(2)` Since `E_(5-4)`has to be less than `E_(4-3)`, so it is that of Infrared radiation. |
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| 27. |
A charged oil drop remains stationary when situated between two parallel plates 20 mm apart and a p.d. of 500 V is applied to the plates. Find the charge on the drop if it has a mass of 2xx10^(-4) kg . Take g=10 ms^(-2). |
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| 28. |
If the momentum of an electron is required to be same as that of wave of 5200 Å wavelength, its velocity should be …….ms^(-1) |
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Answer» `10^(3)` `h=6.62xx10^(-34)JS` `m=9.1xx10^(-31)Kg` `p=mv=(h)/(lambda)` `therefore v=(h)/(mlambda)` `v=(6.62xx10^(-34))/(9.1xx10^(-31)xx52xx10^(-8))` `v=1.4xx10^(3)m//s` |
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| 29. |
A hollow hemispherical bowl of mass m radius R is placed on a smooth horizontal surface. A particle (also of mass m) strikes the hemi spherical bowl at point P and sticks to it. the velocity of particle just after the collision in vertically upwards direction is (ku)/25, where k is an integer. Find k. |
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Answer» Using CONSERVATION of angular momentum about centre of mass. `"mu" R/4={2/3mR^(2)-(mR^(2))/4+(5mR^(2))/16}omega+(5mR^(2))/16 omega` `=25/24mR^(2)omega` `impliesomega=(6U)/(25R)` `impliesV_(y)=(3u)/25` 
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| 30. |
Two long wires PQR and MNP carry equal current I as shown such that QR and NP are parallel. Find the magnetic field at origin O. |
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Answer» |
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| 31. |
Light of frequency 8.41xx10^(14)Hz is incident on a metal surface. Electrons with their maximum speed of 7.5xx10^(5)ms^(-1) are ejected from the surface. Calculate the threshold frequency for photoemission of electrons. Also find the work function of the metal in electron volt (eV). Given Plank's constant h=6.625xx10^(-34)Js and mass of the electron 9.1xx10^(-31)kg. |
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Answer» Solution :`K_(max)=hv-omega_0` `hv=hv_0+1/2 mv^2` `hv_0=(6.625xx10^(-34)xx8.41xx10^14)-1/2xx9.1xx10^(-31)xx(7.5xx10^5)^2` Calculation of threhold FREQUENTLY `v_0=4.54xx10^14` HZ Work FUNCTION , `omega_0=hv_0` Work function , `omega_0=(6.625xx10^(-34)xx4.54xx10^14)/(1.6xx10^(-19))` Work function , `omega_0`=1.88 eV |
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| 32. |
The nomal temp. of human body is 98 cdot 6^(@) F. What is the corresponding temp. of human body in Celsius scale: |
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Answer» `36^(@)C` `T_(C)=37^(@)C` THUS, correct choice is (b). |
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| 33. |
Why a transistor cannot be used as a rctifier ? |
| Answer» Solution :To use transistor as a RECTIFIER either it’s emitter-.BASE portion or collector, base portion has to be USED. As base is THIN and hightly doped, either of the TWO portions will not work as a p-n junction. So, a transistor cannot be used as a rectifier. | |
| 34. |
Obtain the expressionfor capacitance of a parallel plate capacitor without dielectric medium between the plates. |
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Answer» Solution :A parallel plate capacitor consists of two large plane parallel CONDUCTING plates SEPARATED by a small distance . LET A be the area of each plate and d be the distance between them. The two plates have charges (Q) and -(Q) while the SURFACE charge densities are `(sigma)` and -`(sigma)` respectively.Electric field in the outer region of I and II. `E=sigma/(2epsilon_0)-sigma/(2epsilon_0)=0` In the region between the plates (1) and(2) `E=sigma/(2epsilon_0)+sigma/(2epsilon_0)=sigma/epsilon_0` `E=sigma/epsilon_0=Q/(Aepsilon_0)`...(1) where `sigma=Q/A` As the electric field E is EQUAL to the potentialgradient , for uniform electric field. `E=v/d` v=Ed ....(2) Substituting (2) in equation (1) , `V=Q/(Aepsilon_0)d` The capacitance C of the parallel plate capacitor is `C=Q/V=Q/(Q//Aepsilon_0xxd)=Q/Qxx(Aepsilon_0)/d` `C=(Aepsilon_0)/d` 
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| 35. |
The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V m^(-7). Why then do we not get an electric shock as we step out of ourhouse into the open ? (Assume the house to be a steel cage so there is no field inside) |
| Answer» Solution :Our body and the ground FORM an EQUIPOTENTIAL surface. As we step out into the OPEN, the ORIGINAL equipotential SURFACES of open air change, keeping our head and the ground at the same potential. | |
| 36. |
A block of mass 4 kg is placed on a rough horizontal plane. A time dependent horizontal force F = kt acts on the block (k = 2 N//s). Find the frictional force between the block and the plane at t = 2 second and t = 5 second (mu=0.2) |
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Answer» Solution :When t = 2 sec F = 4N `f_(ms)=mu_(s)MG =0.2xx4xx10=8N` `therefore F lt f_(ms) therefore`FRICTION = 4N When t = 5 sec F = 2 (5) = 10 N `F GT f "" therefore` friction = 8 N |
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| 37. |
Four equal positive charges are fixed at the vertices of a square of side L. Z-axis is perpendicular to the plane of the square. Thepoint z = 0 is the point where the diagonlas of the square intrsect each other. The plot of electric field due to the four charges, as one moves on the z-axis. |
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Answer»
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| 38. |
For minimum dissipation energy in the circuit, the power factor should be |
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Answer» large |
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| 39. |
Represent electric field lines around (i) a positive point charge (ii) a negative point charge and (iiii) an electric dipole. |
Answer» SOLUTION :
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| 40. |
Which of the following is responsible for glittering of a diamond ? |
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Answer» INTERFERENCE |
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| 41. |
A long straight metal rod has a very long hole of radius 'a' drilled parallel to its axis as shown in the figure.If the rod carries current i. find the magnetic field on the axis ofthe hole given C is the centre of the hole and OC = c.x |
Answer» Solution : In the rod current density `j = (i)/(pi(B^2 - a^2))` On the hole AXIS, only the larger rod contributes magnetic field. Imagine an amperean loop of radius C and apply Ampere.s law. `B_("Total") = UNDERSET(underset("solid")"complete")(vec(B_1)) + underset("cavity")(vec(B_2)) = (mu_0 j pi c^2)/(2 pi c) + O` `= (mu_0ic)/(2pi(b^2-a^2)) = (mu_0)/(2pi) (ic)/((b^2 - a^2))` |
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| 42. |
As shown in Fig., a very very thin sheet of aluminium is placed in between the plates of the capacitor. Then the capacitance |
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Answer» will increase. |
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| 43. |
Focal lengthof objective and eyepiece of telescope are 200 cm and 4 cm respectively. What is length of telescope for normal adjustment? |
| Answer» Solution :`L = f_(0) + f_(e) = 200 + 4 = 204 cm` | |
| 44. |
A lamp is hanging along the axis of a circular table of radius r. At what height should the lamp be placed above the table, so that the illuminance at the edge of the table is (1)/(8) of that at its center |
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Answer» `(R)/(2)` |
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| 45. |
If the radius of a nucleus with a mass number 7 is 2 fermi then the radius of the nucleus with mass number 189 is |
| Answer» ANSWER :C | |
| 46. |
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is |
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Answer» 0.4 ln2 |
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| 47. |
Obtain the expression of electric field by thir spherical shell with uniform charge distribution at a point outside it. |
Answer» Solution :Let `sigma`be the UNIFORM surface charge density ofthin spherical shell of radius R.  Consider a point P OUTSIDE the shell with radius vector `vecr`. To calculate `vecE`at P, we take the Gaussian surface to be a SPHERE of radius t and with centre 0 passing through P. The ELECTRIC field at each point of the Gaussian surface has the same magnitude E and is ALONG the radius vector at each point. Thus, `vecE`and `DeltavecS`at every point are parallel and the flux through each element is `EDeltaS`. Summing over all `DeltaS`the flux through the Gaussian surface is `E xx 4pir^(2)`. The charge enclosed is `sigma xx 4piR^(2)` By Gauss.s law, `E xx 4pir^(2) = sigma/epsilon_(0).4piR^(2)` `therefore E = (sigmaR^(2))/(epsilon_(0)r^(2)) =q/(4piepsilon_(0)r^(2)) =(kq)/r^(2)` `therefore vecE = (sigmaR^(2))/(4pi epsilon_(0)r^(2))hatr = (kq)/r^(2).hatr` |
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| 48. |
Draw the V-I characteristics of an LED. State two advantages of LED lamps over conven- tional incandescent lamps. |
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Answer» Solution :V-I characteristics of an LED are drawn here. Main advantages of LED over conventional incandescent lamps are as follows: (1) The operaling vollage is low, power consump- TION is less and no warm up time is required. 2. The lightemitted is NEARLY monochromatic. The bandwidth of emitted light is 100 Åto 500 Å . 3. It is rugged, has a long life and SHOWS fast on-off SWITCHING capability. |
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| 49. |
An atom was earlier assumed to be sphere of radius a having a positively charged point nucleus of charge +Ze at its centre . This nucleus was believed to be surrounded by a uniform desity of negative charge that made the atom neutral as a whole.Use this theorem to find the electric field of this atom at a distance r(rgta) from the centre of the atom. |
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Answer» SOLUTION :Let `rho` be the uniform density of negative charge . Wethen have `(4pia^(3)rho)/(3)=Ze` `thereforerho=((3Ze)/(4pia^(3)))` Taking a sphere of radius r (centre at the nucleus) as the Gaussian SURFACE , we have `E(r)xx4pir^(2)=(q)/(epsilon_(0))`.... (i) where , q is the NET charge enclosed by the Gaussian surface . Now `q=(+Ze)+(-rho.(4pir^(3))/(3))` `=Ze-Ze((r^(3))/(a^(3)))` `=Ze(1-(r^(3))/(a^(3)))` Substituting this value of a , in eq . (i) `E(r)=(Ze)/(4piepsilon_(0)r^(2))(1-(r^(3))/(a^(3)))` `=(Ze)/(4piepsilon_(0))((1)/(r^(2))-(r)/(a^(3)))` |
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| 50. |
When light is polarised by reflection from a transparent surface |
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Answer» reflected and refracted rays are mutually perpendicular |
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