Saved Bookmarks
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: Simple harmonic motion is not a uniform motion. R: Simple harmonic motion can be regarded as the projection of uniform circular motion. |
|
Answer» If both Assertion & REASON are true and the reason is the CORRECT explanation of the assertion, then mark (1) |
|
| 2. |
Read the following passage and then answer question (a) - (e) on the basis of your understanding of the following passage and the related studied concepts. As per Bohr atom model, in an isolated atom the energy of any of its electrons depends on the orbit in which it revolves and it is characterised by a sharp energy level. However, inside a crystalline solid atoms are close to each other and the outer orbits of electrons from neighbouring atoms would come very close or could even overlap. As a result, each electron will have a different energy level. These different energy levels with continuous energy variation form energy bands. The energy band which includes the energy levels of the valence electrons is called the valence band. All the valence electrons reside in the valence band. The energy band above the valence band is called the conduction band. Normally the conduction band is empty. If the lowest level in the conduction band happens to be lower than the highest level of the valence band, electrons from the valence band may easily move into the conduction band and the solid behaves as a conductor. If there is some gap between the conduction band and the valence band, electrons in the valence band remain confined to it and no free electrons are available in the conduction band. It makes the solid an insulator. If some of the electrons from the valence band may gain external energy to cross the gap between the conduction band and valence band, these electrons will move into the conduction band and simultaneously create vacant energy levels in the valence band. Therefore, there is a possibility of conduction due to electrons in conduction band as well as due to vacancies in the valence band. How are energy bands formed in a crystalline solid ? |
| Answer» SOLUTION :In a crystalline solid large number of atoms are brought close and BEGIN to INFLUENCE each other. CONSEQUENTLY energy levels of their valence electrons are MODIFIED and we get a collection of large number of closely spaced energy levels, thus forming an energy band. | |
| 3. |
An object is placed at 30 cm in front of a concave mirror of radius of curvature 30 cm. Find the position, nature, and magnification of the image in each case. |
| Answer» SOLUTION :M=-1, image is REAL , INVERTED and same SIZE as of object. | |
| 4. |
A TV transmitting antenna is 81 m tall. How much service area it can cover if the receiving antenna is at the ground level? (Radius of earth=6.4xx10^(6)m) |
|
Answer» `3258km^(2)` `A=pi(2HR)=(22)/(7)xx2xx81xx6.4xx10^(6)m^(2)=3258km^(2)` |
|
| 5. |
An alpha- particle and a proton moving with the same kinetic energy enters a region of uniform magnetic field at right angles to the field. The ratio ofthe radii of the paths of alpha-particle to that of the proton is: |
|
Answer» `1:8` `r=(p)/(QB)= (sqrt(2mE))/(qB) :. E=(p^(2))/(2m)` `r_(1):r_(2)=(sqrt(m_(1)))/(q_(1)): (sqrt(m_(2)))/(q_(2))` `=(sqrt(4))/(2):(sqrt(1))/(1)=1:1` |
|
| 6. |
A charged particle of mass m and charge q enters a magnetic field B with a velocity v at an angle theta with the direction of B. The radius of the resulting path is |
|
Answer» `(mvsintheta)/(QB)` |
|
| 7. |
Two cells of emf 2 epsi and epsi and internal resistance 2r and r respectively, are connected in parallel.Obtain the expressions for the equivalent emf and the internal resistance of the combination. |
|
Answer» Solution :Here `epsi_1 = 2 epsi , espi_2 =epsi, r_1 = 2R` and `r_2=r ` and the two cells have been connected in parallel. If equivalent emf and INTERNAL RESISTANCE of the COMBINATION be `epsi_(eq)` and `r_(eq)`respectively, then `(epsi_(eq))/(r_(eq)) = epsi_1/r_1+ epsi_2/r_2 " and " (1)/(r_(eq)) = (1)/(r_1) + (1)/(r_2)` ` therefore r_(eq) = (r_1 r_2)/(r_1 + r_2) = ((2r)(r ) )/(2r + r) = 2/3 r` and `(epsi_(eq))/( (2/3 r) )= (2 epsi)/(2 r) + epsi/r = (2epsi)/(r ) rArr epsi_(eq) = 2/3 r xx (2 epsi)/(r ) = 4/3 epsi` |
|
| 8. |
The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10^(-5) T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west, (b) south to north? |
|
Answer» Solution :`F = Il xx B` `F - IlB SIN theta ` The force per unit length is `f = F//l = l B sin theta ` (a) when the current is flowing from east to west ` theta = 90^@` Hence , `f= IB` ` = 1 xx 3 xx 10^(-5) = 3 xx 10^(-5) Nm^(-1)` This is larger than the value `2 xx 10^(-7) Nm^(-1)`QUOTED in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray FIELDS while standardising the ampere. The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors. (b) When the current is flowing from south to north, `theta = 0^@` ` f = 0` Hence there is no force on the conductor . |
|
| 9. |
Two shortmagnetic dipoles each of magnetic moment 10Am^(2) areplaced in end on positon 0.1 m apart from theircentres the force actingthem is |
|
Answer» `0.6xx10^(7)` N |
|
| 10. |
A circular coil of radius 4 cm and of 20 turns carries a current of 3 Amperes. It is placed in a magnetic field at intensity of 0.5Wb//m^(2). What is the magnetic dipole moment of the coil ? |
|
Answer» `0.15Am^(2)` Magnetic DIPOLE MOMENTS M = NIA `thereforeM=20xx3xx3.14xx10^(-4)xx16=0.3Am^(2)` |
|
| 11. |
भारत का राष्ट्रीय पक्षी है? |
|
Answer» हंस |
|
| 12. |
Compute the binding energy per nucleon of ""_(2)^(4)"He" nucleus. |
|
Answer» Solution :We found that the BE of He = 28 MEV. Binding ENERGY per NUCLEON = `OVERLINE(B.E) = 28 MeV/4 = 7 MeV`. |
|
| 13. |
A railroad train is travelling at 30.0 m/s in still air. The frequency of the note emitted by the train whistle is 262 Hz. Speed of sound in air is 340 m/s. |
|
Answer» Frequency heard by a passenger on another train moving in the OPPOSITE DIRECTION to the FIRST at 18.0 m/s and APPROACHING the first is 302 Hz |
|
| 14. |
How the emf of two cells are compared using potentiometer ? |
|
Answer» Solution :Comparison of emf of two cells with a potentiometer:To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer WIRE CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf `xi_(1)` and `xi_(2)` to be compared are connected to the terminals `M_(1), N_(1) and M_(2), N_(2)` of the DPDT switch. The positive terminals of Bt, `xi_(1) and xi_(2)` should be connected to the same end C. The DPDT switch is pressedtowards `M_(1), N_(1)` so that cell `xi_(1)` is included in the secondarycircuit and the balancing length `l_(1)` is found by adjusting the jockey for zero deflection. Then the second cells `xi_(2)` is included in the circuit and the balancing length `l_(2) ` is determined. Let r be the resistance per UNIT length of the potentiometer wire and I be the current flowing through the wire.  we have `xi_(1)=Irl_(1)""...(1)` `xi_(2)=Irl_(2)""...(2)` By dividing equation (1) by (2) `(xi_(1))/(xi_(2))=(l_(1))/(l_(2)) ""...(3)` By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it. |
|
| 15. |
The maximum possible magnification for a simple microscope is 10. What is the advantage of forming image at infinity? |
| Answer» SOLUTION :Strain for EYE, will be MINIMUM when IMAGE is at infinity | |
| 16. |
When a long capillary tube is immersed in water, mass M of water rises in capillary. What mass of water will rise in another long capillary of half the radius and made of same material? |
|
Answer» `M` `sigma 2 pi r/2 cos theta = M. g` `:. M. = M/2` |
|
| 17. |
What is the basic nuclear process underlying beta^(-) decay ? Write nuclear reaction of this decay for ""_(83)^(210)Bi. |
|
Answer» `""_(Z)^(A)X RARR""_(2+1)^(A)Y+ -1^(""^(@)E)+BAR(v)+E_(beta)` For `""_(83)^(210)Bi` `""_(83)^(210)Bi rarr""_(84)^(210)Po + -1^(""^(@)e)+ bar(v)+E_(beta)` |
|
| 18. |
An object O is placed at the bottom of a container in which water is filled up to a height of 8m. Refractive index of water is 4//3. OAB is a vertical line. Observer is looking at O from point E as shown in figure. For the observer, position of O will appear to be |
|
Answer» SOMEWHERE on line OAB |
|
| 19. |
What will be the change in the radius of a soap bubble if it is given positive charge ? |
|
Answer» The RADIUS will decrease |
|
| 20. |
A: Modulator is an essential component of a transmitter. R: Modulator superimposes a low frequency message signal on a high frequency carrier wave. |
|
Answer» If both Assertion & Reason are true and the reason is the CORRECT EXPLANATION of the assertion, then mark (1) |
|
| 22. |
When target potential in the X-ray tube is increased |
|
Answer» FREQUENCY of X-rays increases |
|
| 23. |
यदि द्विआधारी संक्रिया ∗, प्राकृत संख्याओं के समुच्चय N पर इस प्रकार परिभाषित है कि a∗b=a और b का ल ० स० प० , तो 12∗18 का मान है - |
|
Answer» 6 |
|
| 24. |
Two charges 5 xx 10^(-8) C and -3 xx 10^(-8)C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. |
| Answer» SOLUTION :10 cm, 40 cm AWAY from the POSITIVE charge on the SIDE of the NEGATIVE charge. | |
| 25. |
A wire of length 'l' is bent to form a circular coil of some turns. A current i then established in the coil and it is placed in a uniform magnetic field B. The maximum torque that acts on the coil is |
|
Answer» ZERO |
|
| 26. |
In a resonance column experiment the frequency of tuning fork used is 1000Hz and the length of pipe is 100m. Ignoring end correction find the length (in cm) of air column at which second resonance is observed. [Take speed of sound =330m//s] |
|
Answer» `F=(3V)/(4L)=1000impliesl=24.75cm` |
|
| 27. |
(a) The number of nuclei of a given radioactive sample at time t = 0 and t= T are N_(0) and N_(0)/n respectively. Obtain an expression for the half-life T_(1/2). of the nuclide in terms of n and T. (b) Write the basic nuclear process underlying beta^(-)-decay of a given radioactive nucleus. |
|
Answer» Solution :(a) As per LAW of RADIOACTIVE decay `N_(t)=N_(0)E^(-lambdat) implies 1/n=e^(-lambdaT) or n=e^(lambdaT) or lambda=(log_(e)n)/T` `THEREFORE` Half-life `T_(1/2)=0.693/(lambda)=(0.683T)/(log_(e)n)` (b) |
|
| 28. |
A capacitor of10muF is charged to 100 V is connected in parallel to another capacitor of20muF charged to 25 V Common potential is : |
|
Answer» 50 V |
|
| 29. |
An object is placed in front of a convex mirror of focal length dof f and the maximumand minimum distance of an object from the mirror such that the image formed is real and magnified. |
|
Answer» 2F and c Hint: There is no maximumminimum OBJECTDISTANCE for convex mirror to form real and INVERTED image. |
|
| 30. |
An infinite number of charges each q are placed in the x-axis at distances of 1, 2, 4, 8, ………meter from the origin. If the charges are alternately positive and negative find the intensity of electric field at origin. |
Answer» Solution :The electric field intensities due to positive CHARGES at origin is away from the charges and due to -Ve charges the field intensity is towards the charges  The resultant intensity at the origin `E = E_(1) - E_(2) + E_(3) - E_(4)`...... `E= (Q)/(4pi in_(0))(1 - (1)/(2^(2)) + (1)/(4^(2)) - (1)/(8^(2))+ ......)` Since the expression in the bracket is in GP with a COMMON RATIO `=(-1)/(2^(2)) = (-1)/(4)` `E= (Q)/(4pi in_(0))(1)/([1-((-1)/(4))])= (Q)/(4pi in_(0)).(4)/(5)` `E= (4)/(5).(Q)/(4pi in_(0))` `E= (Q)/(5pi in_(0))` |
|
| 32. |
Which diode converts light waves into electrical signals in the absence of external battery? |
|
Answer» LED |
|
| 33. |
A point charge q_1 = +2muC is placed at the origin of coordinates. A secondcharge q_2 = -3muC is placed on the y-axis at y = 100cm. At what point on the y-axis, potential is zero |
|
Answer» y=-200cm |
|
| 34. |
A black bodyemits maximum radiation of wavelength lambda_1=2000 Å at a certain temperature T_1. On increasing the temperature , the total energy of radiation emitted is increased 16 times at temperature T_2.If lambda_2 is the wavelengthcorresponding to which maximumradiation emitted at temperature T_2. Calculate the value of (lambda_1/lambda_2) |
|
Answer» |
|
| 35. |
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2m/s^2 and the car has acceleration 4m/s2 The car will catch up with the bus after a time of: |
| Answer» Answer :C | |
| 36. |
What should be minimum size of the plane mirror to see full image of one self ? |
| Answer» SOLUTION :HALF the HEIGHT of OBSERVER | |
| 37. |
A ray of light is incident on the surface of a spherical glass paper-weight making an angle alphawith the normal and is refracted in the medium at an angle beta .Calculate the deviation. |
| Answer» SOLUTION :`DELTA =2 ( ALPHA - BETA)` | |
| 38. |
An object is placed at a distance 20cm from the pole of a convex miror of focal length 20cm. The image is produced a |
|
Answer» Solution :(d) `u=-20cm, f=20cm` From mirror FORMULA `1/f=1/v+1/u` `1/20=1/v+1/(-20)` `1/v=1/20+1/20` `1/v=1/20+1/20` `1/v=2/20` `impliesv=10cm` |
|
| 39. |
An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no parallax between the images formed by the two mirrors. What is the radius of curvature of the convex mirror ? |
Answer» Solution : As shown in figure the plane mirror will form erect and virtual image of same size at a distance of 30 CM behind it. So the distance of image formed by plane mirror from convex mirror will be PI = MI - MP But as MI = MO, PI = MO-MP = 30-20 = 10CM Now as this image coincides with the image formed by convex mirror, therefore for convex mirror, u=-50 cm, v = + 10 cm `So(1)/(+10) +(1)/(-50)= (1)/(f)i.e.,f=(50 )/(4) = 12.5cm ` So R=2f = 2 x 12.5 = 25 cm |
|
| 40. |
If C the velocity of light,h Planck's constant and G gravitational constant are taken as fundamental quantities, then the dimensional formula of mass is |
|
Answer» `h^(-1//2)G^(1//2)C^(0)` The dimensional FORMULA of above expression is given as, `[ML^(0)T^(0)] = [LT^(-1)]^(a) [ML^(2)T^(-1)]^(b) [M^(-1)L^(3)T^(-2)]^(c )` `[M]^(1) = M^(b-c) implies b - c = 1`…(1) ` L^(0) = L^(a+2b+3c) implies a+2b+3c = 0 `...(2) `T^(0) = T^(a-b-2c) implies a+b+2c = 0`...(3) On solving EQUATIONS (1),(2)and (3), we get, `a=1/2, b=1/2, c=1/2` `:.M = h^(1//2)C^(1//2)G^(-1//2)` |
|
| 41. |
What will happen when Ge and Si is doped with 5^(th)group element?What type of semiconductor is now formed? |
| Answer» Solution :During doping with `5^(TH)` group element,covalent bounds are formed between the atoms of`5^(th)` group element and the atoms of GE and Si.One ELECTION of each ATOM of the `5^(th)`group element remains free.Thus N-type semiconductor is formed. | |
| 42. |
Five balls numbered 1 to 5 are suspends using separate threads Pairs (1,2) , (2,4) and (4,1) show electrostaic attraction while pairs (2,3) and (4,5) show repulsion Thereforeball 1 must be |
|
Answer» POSITIVELY charged |
|
| 43. |
In Fig., the four particles fixed in palce and have charges q_(1) = q_(2) = +5e, q_(3) = +3e and q_(4) = -1e. Distance d = 8.0 mu m. What is the magnitude of the net electric field at point P due to the particles ? |
| Answer» Solution :`(1)/(4pi epsilon_(0)) ((12Q)/(4d^(2))-(3Q)/(d^(2)))hat(j)` | |
| 44. |
For series L-C-R circuit, L = 10mH and C = 10^(-7) F If resonance frequency is made double without changing inductor, the capacitance must be ……... muF. |
|
Answer» 0.25 `therefore 2omega_0 = 1/sqrt(LC.)` `therefore omega_0/(2omega_0)=sqrt((LC.)/(LC))` `therefore 1/4=(C.)/C` `therefore C.=C/4 =10^(-7)/4` `=0.25xx10^(-7)` `=0.025xx10^(-6)` F `therefore C=0.025 MUF` |
|
| 45. |
Assertion: AM detection is the process of recovering the modulating signal from amplitude modulated waveform wich is carried out using a rectifier and an evelope detector. Reason: Amplitude modulated waves can be produced by application of the message signal and the carrier wave to a non-linear device followed by band pass filter. |
|
Answer» |
|
| 46. |
(a) A student wantsto use tow p-n junction diodes toconvert alternating current intodirectcurent drawthe labelled circuit diagram she would use and explain how it works (b) give the truth table and circuit symbol for NAND gate |
Answer» Solution :The labeled circuit diagram ,for the required circuit is as shown  The working of this circuits is as follows (i) During ONE half cycle (of the input ac) diode ` D_1 ` alone gets forward biased and conducts. During the other half cycle. it is aodel `D_2` (alone ) that conducts. Because of the use of the center tapped TRANSFORMER the current though the lead flows in the same DIRECTION in both the half cycles. Hence we get a unidirectional /direct current through the LOAD , when the input as altermating current [Altermatively : The student may just use the FOLLOWING diagrams to explain the working.] ` (##DBT_SM_PHY_XII_DL_18_E02_026_S02##) ` |
|
| 47. |
Maximum power in a 0.5Omega resistance connected with two batteries of 2V emf and 1Omega internal resistance in parallel, is ........ . |
|
Answer» `(8)/(9) ` W Equivalent resistance of internal resistance `r. = (1 xx 1)/(1 + 1) = (1)/(2) = 0.5 OMEGA` TOTAL resistance of circuit = 0.5 + 0.5 = ` 1.0 Omega`  `therefore` Current flowing through the circuit `I = (E)/(R) = (2)/(1) = 2 `A `therefore ` As 2A current is flowing through the 0.5`Omega` resistor, so P `= I^(2) R = 4 x 0.5 = 2.0 ` W |
|
| 48. |
If degree of dissociation of 2M CH_(3)COOH is10% then degree of dissociation of this acetic 4cid in 3 Molar CH_(3)COONa solution will be - |
|
Answer» `=10%` |
|
| 49. |
Chlorine has two isotopes having masses 34.98u and 36.98u. The relative abundances of these isotopes are 75.4 and 24.6 percent. Then average mass of chlorine atom is….u |
|
Answer» 34.91 |
|