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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. |
A force vec(F)=(-y hat(i)+ x hat(j))N acts on a particle as it moves in an anticlockwise circularmotion in x-y plane. The centre of the circle is at the origin. If the work done by the force is 32 pi J in one complete revolution then asSigmaingx, y to be in meters, find the radius of the path. |
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Answer» Solution :The force is perpendicular to the radius vector `VEC(R)= x HAT(i)+ y hat(J) rArr` Force is TANGENTIAL Torque `| tau|=R|vec(R)|=R sqrt(x^(2)+y^(2))=R^(2)` `W= int_(0)^(theta) tau d theta = R^(2)2 pi` `R^(2)2 pi = 32 pi rArr R=4 m` |
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| 2. |
State two characteristic properties distinguishing the behaviour of paramagnetic and diamagnetic materials. |
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Answer» Solution :(i) Diamagnetic substances, when placed in a non-uniform magnetic field, tend to MOVE from stronger region of the field to the weaker region. On the other HAND, PARAMAGNETIC substances move from weaker region of the field to the stronger region. (ii) When a bar of diamagnetic material is placed in an external magnetic field, the field LINES are repelled and the field INSIDE the material is reduced. For a paramagnetic bar, the field lines get concentrated inside the material and, hence, the magnetic field inside is slightly enhanced. |
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| 3. |
A wire having length 2.0 m diameter 1.0 mm and resistivity 1.963 xx 10^(-8) Ω m is connected in series with a battery of emf 3V and internal resistance IΩ. Calculate the resistance of the wire and current in the circuit. |
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Answer» SOLUTION :Given l =2m D =1.0 mm `rho = 1.963 xx10^(-8) omega m EPSILON =3v r = 1 omega R= ? I=?` `R=(rho l)/(A)` `R=5xx10^(-2) omega` `I=2.85 A` |
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| 4. |
Whatis the force between two small charged speres having charges of 2xx10^(-7) c and 3xx10^(-7)c placed 30 cm apartin air |
| Answer» SOLUTION :`6XX10^(-3)` N (REPULSIVE ) | |
| 5. |
Show the magnetic dipole moment in terms of pole strength. |
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Answer» Solution :The strength of magnetic pole is called pole strength. Magnetic dipole MOMENT per unit length of magnet is called pole strength of magnet. Charge `q` is like that in equations of static electricity, likewise pole strength is in magnetism. Pole strength is denoted by `q_(m)` or `P`. Pole strength depend on the type of material of magnet, its MAGNETIZATION position and area of cross section (A) of magnet. It is a scalar quantity. The unit of pole strength is ampere meter (Am). If the length of bar magnet is `(2l)` and pole strength `q_m` then magnetic dipole moment, `overset(to)(m) = q_m ( overset(to) (2l) ) ` The direction of ` overset(to)(m)` is from S to N. From this equation unit of m is `Am^(2)`. (Note : The magnetic force ACTING on a pole of a bar magnet kept in magnetic FIELD of 1 T is known is pole strength). `therefore q_m = (overset(to) (F_m) )/( overset(to)(B) ) therefore q_m = F_m [ because B = 1T] and overset(to) (F_m) = (q_m) (overset(to) (B) )` (This equation is similar to static electricity as per `overset(to) (F ) = q (overset(to)(E) )` ). As electric field by a point charge is`E= (kq)/( r^2)`, similarly magnetic field by a pole of pole strength `q_m` is `B = (mu_0)/( 4pi) (q_m)/( r^2)`. |
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| 6. |
What do you know about INTERNET? Write its few applications? |
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Answer» Solution :Internet is a fast growing technology in the field of communication system with multifaceted tools. It provides new WAYS and means to interact and connect with people Internet is the largest computer NETWORK recognized globally that connects millions of people through computers. It finds extensive applications in all walks of life. Applications: (i)SEARCH engine: The search engine is basically a web-based service tool used to search for information on World Wide Web. (ii) Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services and social networking tools. (iii) E-Commerce: Buying and selling of goods and services, transfer of funds are done over an ELECTRONIC network |
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| 7. |
Consider the following statements (i) for a permanent magnet the area of a hysteresis loop is allowed to be large (ii) Coercivity o the material in permanent magnets should be small. Which of the following statements is/are true? |
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Answer» (i) but not (ii) 
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| 8. |
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence, obtain the conditions for the angular width of secondary maxima and secondary minima. (b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single-slit of aperture 2xx10^(-6)m. the distance between the slit and the screen is 1.5m. calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases. |
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Answer» Solution :(b) Here `lamda_(1)=590nm=590xx10^(-9)m,lamda_(2)=596nm=596xx10^(-9)m`, slit width `a=2xx10^(-6)`m and DISTANCE between the slit and screen D=1.5m. First maxima is obtained at a distannce x from centre point on screen such that `asintheta=(AX)/(D)=(3lamda)/(2)impliesx=(3lamdaD)/(2a)` `therefore`Separation between the positions of first maximas `DELTAX=x_(2)-x_(1)=(3D)/(2a)(lamda_(2)-lamda_(1))` `implies Deltax=(3xx1.5)/(2xx2xx10^(-6))xx[596xx10^(-9)-590xx10^(-9)]=6.75xx10^(-3)m` or 6.75mm. |
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| 9. |
Two moving coil galvanometers 1 and 2 are with identical field magnets and suspension torque constants, but with coil of different number of turns N_(1) and N_(2), area per turn A_(1) and A_(2), and resistance R_(1) and R_(2). When they are connected in series in the same circuit, they show deflections theta_(1) and theta_(2). then theta_(1)//theta_(2) is |
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Answer» `A_(1) N_(1)// A_(2) N_(2)` `:. (K theta_(1))/(N_(1)BA_(1)) = (K theta_(2))/(N_(2)BA_(2))` So `(theta_(1))/(theta_(2)) = (A_(1)N_(1))/(A_(2) N_(2))` |
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| 10. |
A simple pendulum has a metal bob, which is negatively charged. If it is allowed to oscillate above a positively charged metallic plate, then its time period will |
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Answer» INCREASE |
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| 11. |
Assertion : A loosely round helix made of stiff wire is suspended vertically with the lower end just touching a dish of mercury. When a current is passed through the wire, the helical wire executes oscillatory motion with the lower end jumping out of and inside of mercury. Reason : When electric current is passed through helix, amagnetic field is produced both inside and outside the helix. |
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Answer» If both ASSERTION and REASON are TRUE and the reason is the CORRECTEXPLANATION of the assertion. |
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| 12. |
A reflecting surface is represented by the equation Y = (2L)/(pi) sin ((pix)/(L)) , 0 lt= x lt= L.A ray travelling horizontally becomes vertical after reflection. The coordinates of the point (s) where this ray is incident is |
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Answer» `(L/4 , (sqrt2L)/(PI))` |
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| 13. |
Calculate the equivalent capacitance between the points indicated the figures 6.22, 6.23 and 6.24. |
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Answer» |
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| 14. |
In a photo electric experiment, photon of energy 5 eV are incident on a metal surface. They liberate electrons which are just stopped by an electrode at a potential of -3.5V with respect to the metal. The work function of the metal in eV is |
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Answer» 5 |
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| 15. |
Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred. |
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Answer» Solution :`I_(rms) = P/V= (800 XX 1000)/(40000) = 20A` a. Line power loss `= I^2R = 20^2 xx 15 = 6 kW ` b. Power supplied = 800 + 6 = 806 kW C. Voltage DROPPED = IR = 20 X 15 = 300 V d. Step up transformer should be of 440V/40300V Power loss `= 6/806 xx 100 = 0.74 % ` This power loss reduces a lot, if power is transmitted at high voltage. |
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| 16. |
A car approaching a crossing at a speed of 20 m/s sounds a horn of frequency 500Hz. when at 80m from the crossing, Speed of sound in air is 330 m/s. What freqnency is heard by an observer 60 m from the crossing on the straight road which crosses car road at right angles? |
Answer» Solution :The situation is as SHOWN in figure. `cos theta =80/100 = 4/5` `:.` Apparent FREQUENCY is `f_("app") =(V/(v - v_s cos theta)) f = (330/(330 - 20 XX 4/5)) (500)` ` =525.5 HZ` |
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| 17. |
A biconcave lens of power P vertically splits into two identical plano concave parts. The power of each part will be |
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Answer» 2P |
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| 18. |
A particle is projected from origin x-axis with velocity V_0 such that it suffers retardation of magnitude given by Kx^3 (where k is a positive constant).The stopping distance of particle is |
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Answer» `((3V_0^2)/(4K))^(1/3)` |
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| 19. |
Two moving coil meters M_1 and M_2 have the following particulars: R_1 = 10 Omega, N_1 = 30, A_1 = 3.6 xx 10^(-3) m^2, B_1 = 0.25 T R_2 = 14 Omega, N_2 = 42, A_2 = 1.8 xx 10^(-3) m^2 , B_2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity, and (b) voltaage sensitivity of M_2 and M_1. |
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Answer» Solution :(a) We know current sensitivity (C.S.) of a GALVONOMETER `= (phi)/(I) = (NAB)/(k)` `:. ((C.S)_(M_2))/((C.S)_(M_1)) = (N_2 A_2B_2)/(N_1 A_1 B_1) = (42 xx 1.8 xx 10^(-3) xx 0.5)/(30 xx 3.6 xx 10^(-3) xx 0.25) = 1.4 = 1.4 : 1`. (B) VOLTAGE sensitivity (V.S) of galvanometer is `(phi)/(V) = (N A B)/(k) cdot 1/R` `:. ((V.S)_(M_2))/((V.S)_(M_1)) = (N_2 A_2 B_2 R_1)/(N_1 A_1 B_1 R_2) = (42 xx 1.8 xx 10^(-3) xx 0.5 xx 10)/(30 xx 3.6 xx 10^(-3) xx 0.25 xx 14) = 1 = 1 : 1`. |
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| 20. |
Suppose the charge of a proton and an electron differ slightly. One of them is - e, the other is (e + Deltae). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Deltae is of the order of [Given mass of hydrogen m = 1.67 xx 10^(-27)kg] |
| Answer» Answer :C | |
| 21. |
The figure below represents the variation of current with potential for a metal. a. Identify the situation. c. Even when the potential is zero, there is current. Explain. d. Current is zero for a particular potential. How does this potential help in determining the velocity of electrons? |
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Answer» Solution :a. Photoelectric EFFECT C. INCIDENT photon possesses suitable energy in order to PRODUCE photoelectric current eventhough the POTENTIAL is zero. d. Using the relation `eV_(0)=(1)/(2)mv^(2)` `v=sqrt((2eV_(0))/(m))V_(0)` - stopping potential |
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| 22. |
A bus is going due north at a speed of 30 km/h. It makes a 90^@ left turn without changing the speed. The change in the velocity of the bus is |
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Answer» 30 km/h TOWARDS west |
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| 23. |
A light string passes over a pulley. To one of it's ends 6 kg is attached. To the other end a mass of 10 kg attached. The tension in the string will be |
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Answer» 24.5 N |
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| 24. |
The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 5 cm, then the object distance for objective will be (in cm) |
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Answer» 1.8 |
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| 25. |
A wire whose linear density is 5 xx 10^(3 )kg//m is stretched between two points with a tension 450 N The wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 490 Hz. What is the length of the wire? |
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Answer» 1.2m |
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| 26. |
When a moving body is suddenly stopped a long time after coming to rest |
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Answer» FRICTIONAL force increases |
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| 27. |
Radio waves diffract around building although light waves do not. The reason is that radio waves |
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Answer» TRAVEL with SPEED larger than c |
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| 28. |
Light waves can be polarised while sound waves not ? Why ? |
| Answer» Solution :Only TRANSVERSE WAVES can be polarised. Light waves are transverse in nature. So, they cannot be polarised. But, SOUND waves have LONGITUDINAL nature. | |
| 29. |
Answer the following questions: (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain. |
| Answer» SOLUTION :(a) Rays converging to a point ‘behind’ a PLANE or convex mirror are reflected to a point in front of the mirror on a screen. In other words, a plane or convex mirror can produce a real image if the OBJECT is VIRTUAL. Convince yourself by drawing an appropriate ray diagram. | |
| 30. |
A coil of inductance 0.50 H and resistance 100 Omegais connected to a 240 V, 50 Hz a.c. supply. (a) What is the maximum current in the coil ? (b) What is the time lag between the voltage maximum and the current maximum ? |
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Answer» Solution :Here L = 0.5 H, R `=100 Omega, V_(rms) = 240 V` and v=50 Hz `THEREFORE omega = 2pi v= 2 xx 3.14 xx 50 = 314 s^(-1)`. (a) `therefore` Impedance `Z = sqrt(R^(2) + L^(2) omega^(2)) = sqrt((100)^(2) + (0.5)^(2).(3.14)^(2)) = 186 Omega` `therefore I_(m) = V_(m)/Z = (sqrt(2) V_(rms))/Z = (sqrt(2) xx 240)/186 = 1.82 A` (b) If the current lags behind the voltage in phase by `phi`, then `phi = tan^(-1) (L omega)/R = tan^(-1) ((0.5 xx 314)/100) = tan^(-1) (1.5700) = 57.5^(@) = (57 xx pi )/180` rad `therefore` TIME lag between voltage maximum and current maximum `t= pi/(2pi) , T = phi/(2piV) = (57.5 xx pi)/(180 xx 2pi xx 50) = 3.2 xx 10^(-3)` s or 3.2 rms. |
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| 31. |
To pump liquid out of a vessel which is not wetted by it into a vessel wetted by it one may make use of surface tension forces (a capillary pump). What will be the speed of flow of petrol in a capillary of 2 mm diameter and of 10 cm length? The experiment is conducted in conditions of weightlessness. |
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Answer» Solution :TWO FORCES act on a fluid in conditions of weightlessness: the force of surface tension sur rod and the force of hydraulic resistance `F_(res)= lambda(l)/(d)*(rho v^(2))/(2)*(pi d^(2))/(4)` (see 20.17) SINCE the velocity of the fluid is small, for small Reynolds numbers `lambda=(64)/(Re)=(64 eta)/(rho vd)`. EQUATING the forces, we obtain after some simplifications. `v= sigma d//8 eta l` Compute the velocity to find the Reynolds nurober, and make sure that it is much less than 2320. |
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| 32. |
A 0.12kg of impure sugar is dissolved in water and solution is mad eupto 8xx10^(-4)m^(3). A length of 20cm of this solution causes arotation of 15^(@). If the specific rotation of sugar is 60xx10^(@) deg//decimetre kg m^(-3), what is the purity of sugar? |
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Answer» |
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| 34. |
In Rutherford's alpha particle scattering experiment with their gold foil 8100 scintillations per minute are observed at an angle of 60^(@) . The number of seintillations per minute at 120^(@) will be |
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Answer» 100 |
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| 35. |
In an oscillating LC circuit with L = 50mH and C = 4.0 mu F, the current is initially a maximum.How long will it take before the capaciotr is fully discharged for the first time: |
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Answer» `7 xx 10^(-4)s` ` = 28 xx 10^(-4) s` Time taken by capacitor to charge FULLY, `t = (T)/(4) = 7 xx 10^(-4)s`. |
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| 36. |
Safety of nuclear reactors is an important issue. Guess some of the saftey problems that a nuclear engineer must cope within reactor design. |
| Answer» Solution :One of the major SAFETY problem in a REACTION is that the nuclear waste FORM the reactor contains some long lived radio ACTIVE isotopes. Further, accidents due to excessive heating and MELTING of the reactor core have to be prevented by designing appropriate cooling system. | |
| 37. |
When a forward bias is applied to a p-n junction, it |
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Answer» RAISES the POTENTIAL barrier. |
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| 38. |
A ray of light strikes a glass plate at an angle of incidence 57 ^(@) . If the reflected and refracted rays are perpendicular to each other, what is the refractive index of glass. |
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Answer» |
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| 39. |
Discuss three basic postulates of Bohr's model of atom. |
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Answer» In order to EXPLAIN the stability of the atom and its LINE spectra, Bohr gave a set of postulates. 1. An electron in an atom revolves in a circular orbit around the nucleus. The force of attraction between the electron and the nucleus provides the necessary centripetal force for the circular motion. 2. The electron can revolve in those orbits for which its orbital angular momentum is an integralmultiple of `h//2pi`. The allowed orbits are given by `m v r= (nh)/(2 pi)` ...(1) where n is an integer and h is Planck.s constant. The integer n is known as the principalnumber and can take any INTEGRAL value 1, 2, 3, .... The numericalvalue of h is `6.624xx10^(-34)` JS. This POSTULATE is known as quantization of the angular momentum. 3. Electromagnetic radiation is emitted when the electron jumps from a higher orbit having energy `E_(n_(1))` to a lower orbit having `E_(n_(2))`. The frequency v of the emitted radiation is given as `n=(E_(n_(2))-E_(n_(1)))/(h)` ...(2) Eq. (2) represents the Bohr.s frequency condition. |
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| 40. |
A of dipole moment p_(0) is placed at origin. Direction of dipole moment is word x-axis Assuming potential energy of dipole is zero when bar(p)botbar(E) |
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Answer» |
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| 41. |
If the rms current in a 50 Hz a.c. circuit is 5 A, the value of the currents 1/300 s after its value becomes zero is |
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Answer» `5sqrt(2)` A `therefore I = I_(m) sin OMEGAT = sqrt(2) xx 5 sin (100 pi) xx 1/300 = sqrt(2) xx 5 xx sin pi/3 = 5sqrt(3/2)` |
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| 42. |
(A ): Sharper is the curvature of spot on a charged body lesser will be the surface density of charge at that point. (R ): Electric field is zero inside a charged conductor. |
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Answer» Both .A. and .R. are true and .R. is the correct explanation of .A. |
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| 43. |
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity ofa body from this platform is fv, where v is its escape velocity from the surface of the earth. The value of f is |
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Answer» `1//2` Escape velocity of the body from the platform is fv. `:." potential energy "+" kinetic energy "=0` `-(GMM)/(2R)+(1)/(2)MV^(2)=0 rArr v=sqrt((GM)/(R^(2))).R=sqrt(gR)=fv` From the surface of the earth, `v_("escape")=sqrt(2gR)` `:. fv=(v_("escape"))/(sqrt(2)) rArr f=(1)/(sqrt(2))`. |
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| 44. |
Same mass of copper is drawn into 2 wires of 1 mm thick and 3mm thick. Two wires are connected in series and current is passed. Heat produced in the wires is the ratio of |
| Answer» ANSWER :C | |
| 45. |
Choose the wrongstatements (s) |
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Answer» Viscous force alwayssupports the motion |
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| 46. |
A certain region of space is to be shielded from magnetic fields. Suggest a method. |
| Answer» Solution :Surround the region by soft IRON RINGS, Magnetic field LINES will be drawn into the rings, and the enclosed space will be FREE of magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an EXTERNAL electric field. | |
| 47. |
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm. (a) What is the magnetic moment of the magnet ? (b) What is the work done in moving it from its most stable to most unstable position ? (c) The bar magnet is replaced by a solenoid of cross-sectional area 2 xx 10^(-4) m^(2) and 1000 turns, but of the same magneticmoment. Determine the current flowing through the solenoid. |
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Answer» Solution :Here, `B = 600 G = 6000 xx 10^(-4) T 6xx 10^(-2) T` `theta = 30^@` `tau = 0.012 Nm` `N= 1000` `A= 2 xx 10^(-4) m^(2)` (a) `tau = m B sin theta` `therefore m= (tau)/( B sin theta) = (0.012)/( 6 xx 10^(-2) xxsin 30^(@) ) = (0.2)/(½)` `m= 0.4 "Am"^(2)` (b) For most stable position `theta_(1) = 0^(@)` and for most unstable position `theta^(2) = 180^@`. `therefore W= - mB cos theta_(2) - (-mB cos theta_(1) ) ` `=-mB cos 180^(@) + m B cos 0^@` `=-mB (-1) + mB (1)` `=mB+ mB` `= 2mB` `2 xx 0.4 xx 6 xx 10^(-2)` `therefore W=0.048` J (C) Magnetic moment of solenoid, `m_(s) = NIA,"here" m_(s) = m = 0.4 "Am"^2` `I= (m)/( NIA) = (0.4)/( 1000 xx 2 xx 10^(-4) ) = 2A` |
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| 48. |
The empirical formula and molecular mass of compound are CH_2O and 180 g . What will be the molecular formula of compound |
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Answer» `C_9H_18O_9` |
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| 49. |
A cell AB of resistance 500 Omega is connected to a battery of emf 12 V and of negligible internal resistance.The reading of a voltmeter is 5V when it is connected between one end of the cell and the coil centre.What is the resistance of the voltmeter? |
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Answer» |
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| 50. |
A person 1.6 m tall is standing at the center between two walls three meter high. What is the minimum size of a plane mirror fixed on the wall in front of him, if he is to see the full height of the wall behind him? |
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Answer» 0.8 m |
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