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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 light wave in air enters a medium of refractive index 4/3. If the wavelength of light in air is 6000 A, then the wave number of light in the medium is : |
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Answer» `1.11 XX 10^6//m` |
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| 2. |
Name The poet of the poem "ThePanther". |
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Answer» PETER Niblett |
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| 3. |
Two identical simple pendulums P and Q are susupended from a common point. The bobs of both the pendulums are positively charged but the magnitude of charge on the bob of P is double to that on Q, Both the bobs repel and reach equilibrium. If the thread of P makes an angle theta_1 with the vertical and that of Q make an angle theta_2with the vertical, then |
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Answer» `theta_1 GT theta_2` |
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| 4. |
A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when |
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Answer» Frequency of the AC source is decreased |
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| 5. |
The nuclears size of atom depends on its |
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Answer» ATOMIC NUMBER |
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| 6. |
Consider a perfectly conducting uniform disc of mass m and radius 'a' hinged in a vertical plane from its centre and free to rotate with respect to hinge. A resistance R is connected between centre of the disc annd periphery by using two sliding contacts C_(1) & C_(2). A long non -conducting massless string is wrapped around the disk, whose another end is attached with a block of mass m. There exists a uniform horizontal magnetic field B, whose arrangement is shown in figure. Given system is released from rest at t=0. Assume friction between string and disc is sufficient so that there is no slipping between them let at any instant t, velocity of block is v, angular velocity of the disc is omega and current in resistance is i Just after t=0 the acceleration of block is: |
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Answer» `g` `mgv=mv(DV)/(dt)+I(omega d omega)/(dt)+i^(2)R`………..i `v=aomega` `(dv)/(dt)=a(domega)/(dt)` `i=(Bomegaa^(2))/(2R)=(Bva)/(2R)`………ii From equation i and ii `(3M)/2 (dv)/(dt)=mg-(B^(2)a^(2))/(6mR) v` `v=(4mgR)/(B^(2)a^(2)) (1-e^(-((B^(2)a^(2))/(6mr))t))` `(dv)/(dt)=(2g)/3 e^(-(B^(2)a^(2))/(6mR)t)` 
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| 7. |
Consider a perfectly conducting uniform disc of mass m and radius 'a' hinged in a vertical plane from its centre and free to rotate with respect to hinge. A resistance R is connected between centre of the disc annd periphery by using two sliding contacts C_(1) & C_(2). A long non -conducting massless string is wrapped around the disk, whose another end is attached with a block of mass m. There exists a uniform horizontal magnetic field B, whose arrangement is shown in figure. Given system is released from rest at t=0. Assume friction between string and disc is sufficient so that there is no slipping between them let at any instant t, velocity of block is v, angular velocity of the disc is omega and current in resitance is i Find the terminal speed of the block of mass m (Terminal speed = constant speed attained by the block after very long time) |
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Answer» `(mgR)/(B^(2)a^(2))` `mgv=mv(dv)/(dt)+I(OMEGA d omega)/(dt)+i^(2)R`………..i `v=aomega` `(dv)/(dt)=a(domega)/(dt)` `i=(Bomegaa^(2))/(2R)=(BVA)/(2R)`………ii From equation i and ii `(3m)/2 (dv)/(dt)=mg-(B^(2)a^(2))/(6mR) v` `v=(4mgR)/(B^(2)a^(2)) (1-e^(-((B^(2)a^(2))/(6mr))t))` `(dv)/(dt)=(2g)/3 e^(-(B^(2)a^(2))/(6mR)t)` 
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| 8. |
In crystals of the salt cesium, cesium ions Cs^(+) form the eight corners of a cube and a chlorine ion Cl^(-) is at the cube's center Fig. The edge length ofthe cube is 0.40 nm. The Cs^(+) ions are each deficient by one electron (and thus each has a charge of -e), and the Cl^(-) ion has one excess electron (and thus has a charge of -e). (a) What is the magnitude of the net electrostatic force exerted on the Cl^(-) ion by the eight Cs^(+) ions at the corners of the cube? (b) If one of the Cs^(+) ions is missing, the crystal is said to have a defect, what is the magnitude of the net electrostatic force exerted on the Cl^(-) ion by the seven remaining Cl^(+) ions ? |
| Answer» SOLUTION :(a) 0, (B) `1.9 XX 10^(-9)N` | |
| 9. |
निम्नलिखित में से कौन- सा राज्य प्रमुख गन्ना उत्पादक है? |
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Answer» उत्तर प्रदेश |
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| 10. |
Explain the origin of diamagnetism. |
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Answer» Solution :An atomic electron is equivalent to a current loop which has orbital magnetic moment. Further, electron spin givesrise to the spin magnetic moment. The magnetic moment of an atom is equal to the VECTOR sum of the magnetic moments of all its electron. The electronic configuration in an atom of a diamagnetic material is such that the vector sum of the orbital and spin magnetic moments of all the electrons is zero. Thus, the atomic magnatic moment is zero. Hence, a diamagnetic matrial hasno intrinsin magnetic moment assoiated with it. When a diamagnetic matrial is placed in a magnetic field, the oribting electrons are accelerated or decelerated depending on their sense of REVOLUTION. For the electrons that are speeded up, the magnetic moment increases and for the those that are slowed, the magnetic moment dereases. Thus each atom acquires a net magentic moment and the diamagnetic matrial is weakly mgnetized. The induced magnetic moment is opposite in direction to the applied field. Diamagnetic is the WEAKEST magnetic phenomenon. Hence, although DIAMAGNETISM is a universal property, it can be detected only in the4 absence of properties reesulting in PARAMAGNETISM and ferromagnetism. |
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| 11. |
निम्न में से कौन-सा समय का मात्रक नहीं है? |
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Answer» पारसेक |
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| 12. |
ABC is a spherical wavefront centred at 'O' symmetric about BE is incident on slits S_(1) and S_(2) .BS_(1) = 3lambda , S_(1)S_(2) = 4 lambda , BO = 6lambda ,S_(1)E = 128 lambdaand lambdais the wavelength of incident light wave. A mica sheet of refractive index 1.5 is pated on S_(2). Find the minimum value of thickness of mica sheet for whhichcentral fringe forms at E ? |
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Answer» `(31lambda)/(8)` `S_(2)S_(2)+(mu - 1)t+S_(2)E = S_(1)^(1)S_(1)+S_(1)E` `implies(OS_(2)^(1)-OS_(2))+(mu-1)t+sqrt(D^(2)+d^(2)) = (OS_(1)^(1)-OS_(1))+D` `implies(mu-1)t = (OS_(2)-OS_(1))+(sqrt(D^(2)+d^(2))-D)` `implies(mu-1)t = 2lambda - [D(1+(d^(2))/(D^(2)))^(1//2)-D]` `implies2lambda - [D(1+(d^(2))/(2D^(2)))-D] implies 2 lambda - [D+(d^(2))/(2D)-D]` `(mu-1)t=2lambda - (d^(2))/(2D) ((3)/(2) - 1)t = 2xxlambda - (16lambda^(2))/(2xx128lambda)` `(1)/(2)t = 2lambda - (lambda)/(16) (1)/(2)t = (31lambda)/(16) = (31lambda)/(8)` Thickness of mica sheet `t = (31lambda)/(8)` |
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| 13. |
A plane wavefront propagating from a rarer into a denser medium is incident at an angle of incidence i on a refracting surface. Draw a diagram showing incident wavefront and refractedwavefront . Hence verify Snell's laws of refraction. |
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Answer» Solution :Here `v_(1) and v_(2)` are speeds of light in medium 1 and 2 as shown BC= vt to get the shape of WAVE FRONT draw an arc of RADIUS `v_(2)t` form A towards medium 2. If CE is tangent PLANE then `AE=v_(2) t,` CE represents refracted wave front In `triangle ABC and triangle AEC sin i= (BC)/(AC)=(v_(1)t)/(AC) , sin r = (AE)/(AC) = (v_(2)t)/(AC)` So `(sin i)/(sin r)= (v_(1)t)/(cancel (AC))xx (cancel(AC))/(v_(2)t)=v_(1)/v_(2)` If c is speed of light in vaccum then `n_(1)=c/v_(1) and n_(2)= c/v_(2)"" rArr v_(1) =c/n_(1)and v_(2) =c/n_(2)` where `n_(1) and n_(2)` refractive indices of medium 1 and 2 `(sin i) /(sin r)=(c//n_(1))/(c//n_(2)) = n_(2)/n_(1)rArr""n_(1) sin i= n_(2) sin r` `rArr`Snell's law 
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| 14. |
Positive and negative point charges of equal magnitude are kept at (0,0, (a)/(2)) and (0,0, (-a)/(2)) respectively. The work done by the electric field when another positive charge is moved from (-a,0,0) to (0,a,0) is |
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Answer» positive |
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| 15. |
A flat circular coil of 10 cm radius has 200 turns of wire. The coil is connected to a capacitor of 20 muF and placed in a uniform magnetic field whose induction decreases at a constant rate of 10^(-2)T//s. Find the capacitor's charge. The plane of the coil is perpendicular to the lines of induction of the field. |
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Answer» |
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| 16. |
A slit of width d is placed in front of a lens of focal length 0.5 m and is illuminated normally with light of wavelength5.89 xx 10^(-7)m . The first diffraction minima on either side of central diffraction maximum are separated by 2xx10^(-3) m. Find the width of the slit. |
| Answer» SOLUTION :`2.95 XX 10^(-4) m` | |
| 17. |
Caalcualte the capacitance of a parallel-plate capacitor of plate are A and plate separation d. The dielectric consists of two wedges of relative permitlvties espi_(1) and epsi_(2) as shown in the figure (fig.6.31). [Hint: If dC id the capacitance of an elementary strip ata distance x from the left ent then |
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Answer» |
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| 18. |
Electron starting from rest is acclerated between two point having p.d. of 20 V and 40 V. de-Broglie wavelength of electron…… |
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Answer» 0.75 Å `lambda=(H)/(sqrt(2meV))` `=(6.62xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx20))` `0.274xx10^(-9)=2.75Å` |
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| 19. |
A convex lens is placed some where between an object and a screen which are separated by 48 cm. If the numerical value of magnification produced by the lens is 3, what is the focal length oflens? |
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Answer» 6 cm |
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| 20. |
In Question 26, for V_CE=0 the above graph lines will be |
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Answer» STRAIGHT ALONG `I_(B)` |
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| 21. |
Explain how power can be transmitted efficiently to long distance. |
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Answer» Solution :i. The electric power generated in a power station SITUATED in a remote place is transmitted to different regions for domestic and industrial use. ii. For long distance transmission, power lines are MADE of conducting material like aluminium. There is always some power loss associated with these lines. iii. If I is the current through the wire and R the resistance, a considerable amount of electric power `I^(2)R` is dissipated as heat. Hence, the power at the receiving end will be lesser than the actual power generated. iv. However, by transmitting the electrical energy at a HIGHER voltage, the power loss can ve controlled as is evident from the following two cases.  Case (1) A power of 11,000 W is transmitted at 220 V. Power P = VI `I=(P)/(V)=(11,000)/(220)=50A` If R is the resistance of line wires, Power loss `=I^(2)R=50^(2)R=2500(R)" watts. "` Case (2) 11,000 W power is transmitted at 22,000 V. `I=(P)/(V)=(11,000)/(22,000)=0.5A` Power loss `=I^(2)R=(0.5)^(2)=R=0.25(R)` watts. Hence it is evident that if power is transmitted at a higher voltage the loss of energy in the form of heat can be CONSIDERABLY reduced. |
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| 22. |
One plate of a capacitor is connected to a spring as shown. Area of both the plate is A and separation is d when capacitor is uncharged. When capacitor is charged, in steady state separation between the plates is 0.8 d. The force constant of the spring is approximately : |
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Answer» `(epsilon_(0)AE^2)/(2d^3)` |
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| 23. |
An ebonite rod held in hand can be charged by rubbing with flannel but a copper rod cannot be charge like this. Why / |
| Answer» SOLUTION :Both the human body and the copper rod conduct electricity. When it is ATTEMPTED to charge a METAL rod by rubbing, the charge flows from the rod two the earth through hand. However, when ebonite rod is charged by rubbing, the charges, so produced stay on the ebonite RODS, as it is bad conductor of electricity. | |
| 24. |
Arrange six identical cells of emf 2 V and internal resistance 0.5 Omega such that maximum current is passed through the external resistance of (i) 0.001Omega and (ii) 250 Omega . What will be the value of current in each case ? |
| Answer» SOLUTION :`24 A, 4.8 xx10^(-2)A` | |
| 25. |
A charge q is placed at the centre of the line joining two equal charges Q. Show that the system of three charges will be in equilibrium ifq= -(Q)/(4) . |
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Answer» SOLUTION :For EQUILIBRIUM of SYSTEM the net force on any charge MUST be zero. Considering net force on A, we have `oversetto (F_(AC) )+oversetto (F_(AB)) =0 ` ` therefore(1)/(4 pi in _0)(Qq)/(((R)/(2))^(2) )+(1)/(4 piin _0) .(Q^(2))/(r^(2)) =0 ` `rArr ""(1)/(4 pi in _0) .(Q)/(r^(2)) [4q +Q]=0` `or"" 4q +Q =0 ` `or "" q= -(Q)/(4) .` 
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| 26. |
A nuclear reactor is a powerful device, wherein nuclear energy is utilised for peaceful purpose. It is based upon controlled nuclear chain rection. The nuclear reaction is controlled by the use of control rods (of boron or cadmium) and moderators like heavy water, graphite, etc. The whole reactor is protect with concrete walls 2 to 2.5 meter thick, so that radiation emitted during nuclear reactions may not produce harmful effects. Read the above passage and answer the following questions: (i) Give any two merits of nuclear reactors. (ii) What is radiactive waste? (iii) why do people often oppose the location site of a nuclear reactor? What do you suggest? |
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Answer» Solution :(i) Nuclear reactors are used in electric power generation. They are ALSO used to produced radioactiveisotopes- which have APPLICATIONS in medicine, industry and agriculture. ii) Radioactive WASTE consists of fission products and transuranic elements such as plutonium and americium. This waste is extremely hazardous to all forms of life on earth. (iii) People often oppose the location site of a nuclear reactor because any leakage of nuclear radiations can affect adversely miles of area surrounding it. ELABORATE safety maesures are NEEDED not only for reactor operations, but also for handling and disposal of radioactive waste. I would suggest that Government must take stringent safety measures and assure people of safeguards in the event of nuclear accidents. At the same time, people must also be educated accordingly. |
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| 27. |
A ray of light incidents on a refracting surface at 30^@with the surface. If the angle of refraction is 45^@, refractive index of the medium is |
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Answer» `sqrt3` |
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| 28. |
What is the meaning of 'whims and fancies'? |
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Answer» The responsibilities |
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| 29. |
The ballistic pendulum is a block of 3.0 kg mass suspended from a thread 2.5 m long. A bullet with the mass of 9.0 g hits the block and sticks in it, the result being a deflection of the system by an angle of 18^@(Fig). Find the bullet's speed |
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Answer» `V=(M+m)/msqrt(2gh)=(M+m)/msqrt(2gl(1-cosalpha))` |
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| 30. |
What are dimesional formulae for pole strength ? |
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Answer» <P> SOLUTION :P = 2 ML` m = P/(2l) or, [m] = `[M^0L^2T^0A^1]/([L]) = [M^0L^1T^0A^1]` |
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| 31. |
The equation of a travelling wave on a string is y = (0.10 mm ) sin [31.4 m^(-1) x + (314s^(-1)) t] |
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Answer» WAVE is travelling ALONG negative x- axis. |
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| 32. |
A bullet moving with a speed of 150 ms^(-1) strikes a wooden plank. After passing through the plank its speed becomes 125 ms^(-1). Another bullet of the same mass and size strikes the plank with a speed of 90 ms^(-1). Its speed after passing through the plank would be: |
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Answer» `50 ms^(-1)` `:. Fxxx=1/2m[(150)^(2)-(125)^(2)]` Also `Fxxx=1/2m[(90)^(2)-V^(2)]` `:. (150)^(2)-(125)^(2)=(90)^(2)-v^(2)` Solving it,we get v=35 m/s. |
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| 33. |
Frequency of FM radio band is from……. |
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Answer» 88 MHZ to 108 MHz |
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| 34. |
Describe biprism experiment to find the wavelength of monochromatic light. Draw the necessary ray diagram for magnified and diminished images of virtual sources. If the difference in velocities of light in glass and water is 2.7xx10^(7)"m"//"s", find the velocity of light in air. ({:("Refractive index of glass = 1.5"),("Refractive index of water = 1.333"):}) |
Answer» Solution : Numerical : GIVEN : `v_(W)-v_(g)=2.7xx10^(7)"m"//"s",mu_(g)=1.5,mu_(w)=1.333.` `v_(w)=(c)/(mu_(w))` `implies""=(c)/(1.333)` Again, `""v_(g)=(c)/(mu_(g))` `implies""=(c)/(1.5)` `:.""v_(w)-v_(g)=(c)/(mu_(w))-(c)/(mu_(g))` `implies""=c[(1)/(1.333)-(1)/(1.5)]` `implies""2.7xx10^(7)=c[(1.5-1.333)/(1.5xx1.333)]` `implies""c=323272455.1=3.23xx10^(8)"m"//"s".` |
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| 35. |
Quality factor = ? |
| Answer» Solution :The current in the series RLC circuit becomes maximum at resonance. Due to the increase in current, the VOLTAGE across L and C are ALSO increased. This MAGNIFICATION of volatges at series resonance is termed as Q - FACTOR. It gives resanance condition of LCT circuit. | |
| 36. |
In an ideal transformer, the voltage is stepped down from 11 kV to 220 V. If the primary current be 100 A, the current in the secondary should be …… |
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Answer» 5 KA For an ideal transformer `P_1=P_2` `epsilon_1I_1=epsilon_2I_2` `I_2=epsilon_1/epsilon_2I_1` `=(11xx10^3xx100)/220` `=5xx10^3` A `therefore I_2`= 5 kA |
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| 37. |
Do electrostatic field lines form (electrostatic) closed loops? Give reason. |
| Answer» Solution :Electric FIELD lines START from positive charge and END at infinity. Therefore , these lines do not form close loop. | |
| 38. |
{:((i)"Positive and negative","(a) Michael Faraday"),((ii)"Electrostatic force","(b)Gauss"),("(iii)Concept of field","(c) Benjamin Franklin"),("(iv) q in arbitrary closed surface", "(d) coulomb"):} |
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Answer» |
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| 39. |
An A.C. circuit consists of an Indicator of inductance 0.5H and a capacitor of capacitance 8 muF in series. When the current in the circuit is maximum, what is the angular frequency of A.C. source ? |
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Answer» 500 rad/s `=1/sqrt(0.5xx8xx10^(-6))` `=10^3/sqrt4` `=1000/2`= 500 rad/sec |
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| 41. |
An electric dipole of dipole moment oversetto pconsists of point charges +q and -q separated by a distance '2a' apart. Deduce the expression for the electric field oversetto Edue to the dipole on its axial line in term of the dipole moment oversetto p. Hence show that in the limitr gt gt a, oversetto E =(2 oversetto p)/( 4 pi in _0 r^(3)) (b) Given the electric field in the region oversetto E = 2x hatiFind the net electric flux through the cube and the charge enclosed by it. |
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Answer» Solution :As ELECTRIC field is given by ` overset to E = 2x hati ` , it is OBVIOUS that electric flux `phi_in = int oversetto E . overset to (ds) ` is finite for onlytwo surface marked 1 and 2 in which lie in Y-Zplane and for all remaining four surface of cube flux is zero . Here AREA of each surface `s= a^(2) ` For face 1,x =0 and `oversetto E = 0 ` . Therefore electric flux ` phi_1 =0.` For face 2,x = a and `oversetto E =2 a hati` , Therefore electric flux ` phi _2 = overset to E . oversetto s= (2 a hati).( a^(2)hatl )= 2a^(3) ` ` therefore ` Net electric flux through the cube`phi_in =phi_1+phi_2 =0 =2a^(3) =2a^(3) ` and the charge enclosed by the cube `q= in_0 phi_in =in_0 2a^(3) =2 in_0 a^(3) ` ` (##U_LIK_SP_PHY_XII_C01_E11_002_S01.png" width="80%"> |
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| 42. |
The wavelength of the light used in Young's double slit experiment is lambda.The intensity at a point on the screen is I, where the path difference is (lambda)/(6). If I_(0) denotes the maximum intensity, then the ratio of I and I_(0) is : |
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Answer» `0.866` ` =(2pi)/(lambda) xx (lambda)/(6) = (pi)/(3) = 60^(@)` Intensity, `I = I_(0)COS^(2)((phi)/(2))` `(I)/(I_(0)) = cos^(2)(30^(@)) = ((sqrt3)/(2))^(2) = 0.75`. |
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| 43. |
The time period of a freely suspended magnet is 2 sec . If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will b |
| Answer» Answer :D | |
| 44. |
How can we resolve the earth's magnetic field at a place into two rectangular components ? |
| Answer» Solution :If the ANGLE of dip is `THETA`, the HORIZONTAL component of total INTENSITY `B_E`is `H_E=B_Ecostheta` . The other component is `Z_v=B_Esin theta` | |
| 45. |
In Young.s double slit expriment if one of the slits is closed |
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Answer» the contrast between the BRIGHT and dark bandds decreases |
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| 46. |
When the nucleus of an electrically neutral atom undergoes a radioactive decay process, it will remain neutral after the decay if the process is |
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Answer» an `alpha` DECAY |
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| 47. |
Light of wavelength lambda= 500 nm falls on two narrow slits placed a distance d= 50 xx 10^(-4) cm apart, at an angle phi= 30^@ relative to the slits shown in figure. In front of the lower slit, a slab of thickness 0.1 mm and refractie index 3/2 is placed. THe interference pattern is observed on a screen at a distance D= 2m from the slits S_1 O = S_2 O = d//2 If the transparent slab is removed the fringe pattern shifts upward , the number of frings shifted upward are |
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Answer» 50 |
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| 48. |
Light of wavelength lambda= 500 nm falls on two narrow slits placed a distance d= 50 xx 10^(-4) cm apart, at an angle phi= 30^@ relative to the slits shown in figure. In front of the lower slit, a slab of thickness 0.1 mm and refractie index 3/2 is placed. THe interference pattern is observed on a screen at a distance D= 2m from the slits S_1 O = S_2 O = d//2 The interference minima cloest to B is of order |
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Answer» 50th MINIMA |
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| 49. |
Light of wavelength lambda= 500 nm falls on two narrow slits placed a distance d= 50 xx 10^(-4) cm apart, at an angle phi= 30^@ relative to the slits shown in figure. In front of the lower slit, a slab of thickness 0.1 mm and refractie index 3/2 is placed. THe interference pattern is observed on a screen at a distance D= 2m from the slits S_1 O = S_2 O = d//2 The position of central maxima is at an angular position thetawith line OB |
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Answer» `30^(@)` above OB |
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| 50. |
A long straight wire carries a current along the 'z' axis. One can find two points in the plane perpendicular to the conductor such that i) The magnetic fields are equal ii) The direction of the magnetic fields are perpendicular iii) Magnitudes of the magnetic fields are equal iv) The field at one point is opposite to that at the other point. |
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Answer» Only i & II are true |
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