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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. |
Shown in the figure isa system of three bodies of mass m_(1), m_(2) & m_(3) connected by a single inextensible light string. If all surfaces in contact are smooth & the pulleys P & Q are light, find the tension in the string. |
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Answer» |
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| 2. |
In the circuit shown, the reading of the ammeter is p/q amp. What is the value of p+q? |
Answer»  `2x=4` `x=1/3=p/q` `p+q=4` |
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| 3. |
Three metallic plates out of which the middle is given charge Q as shown in the figure. The outer plates can be earthed with the help of switches S_1 and S_2 . The area of each plates is the same. Answer the following question based on the given passage. The charge that will flow to the earth through S_2 when both switches S_1 and S_2 are grounded simultaneously is |
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Answer» `Q`  `((Q-q))/(epsilon_(0)A)3D=(QD)/(epsilon_(0)A)` or `q=(3Q)/(4)` |
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| 4. |
How did the grandmother die? |
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Answer» during TELLING BEADS LAYING on the bed |
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| 5. |
A potentiometer consists of a wire of length 4 m and resistance 20Omega . If it is connected to a cell ofemf 20 volt, the potential difference per unit length of the wire will be |
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Answer» `0.5 VM^(-1)` |
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| 6. |
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_(1) on a particle placed at P, distance 2R from the centre O of the sphere. A spherical cavity of radius R//2 is now made in the sphere as shown in given figure. The sphere with cavity now applies a gravitational force F_(2) on same particle placed at P. The ratio F_(2)//F_(1) will be |
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Answer» `1//2` `F_(1)=(GMm)/((2R)^(2))`, where M and m are MASS of the solid sphere and particle respectively and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity= force due to solid sphere-force due to sphere creating cavity, ASSUMED to be present above at that position. i.e., `F_(2)=(GMm)/(4R^(2))-(G(M//8)m)/((3R//2)^(2))=(7)/(36)(GM m)/(R^(2))` So, `(F_(2))/(F_(1))=(7 GMm)/(36R^(2))//((GM m)/(4 R^(2)))=(7)/(9)` |
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| 7. |
Three metallic plates out of which the middle is given charge Q as shown in the figure. The outer plates can be earthed with the help of switches S_1 and S_2 . The area of each plates is the same. Answer the following question based on the given passage. The charge that will flow to the earth when only switch S_1 is connected to the earth is |
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Answer» `-(Q//2)`  Field inside the CONDUCTOR is zero. CHARGE distribution is SHOWN in fig. |
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| 8. |
A capacitor connected to a 10V battery collects-a-charge 4muCwith air as dielectric and 100muCwith oil as dielectric. The dielectric constant of oil is |
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Answer» 2 |
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| 9. |
A planet of mass m moves around the sun in an elliptical orbit. The maximum and minimum distance of the planet from the sun is r_(1) and r_(2). The time period is proportional to: |
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Answer» `r_(1)^(3//2)` `=((r_(1)+r_(2))/(2))^(3) or T prop (r_(1)+r_(2))^(3//2)`. Correct choice is (c ). |
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| 10. |
The intensity of the ground waves decrease with increase of distance due to |
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Answer» INTERFERENCE |
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| 11. |
Holes and electronsare ….. |
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Answer» in the same DIRECTION |
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| 12. |
A ray of light from air is incident in air. Then which property of light will not change in water: |
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Answer» velocity |
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| 13. |
यदि द्विघातबहुपद p(x)=x^2 +7x+4 के शून्यक alpha ,betaहो तो alpha+beta-alphabetaका मान होगा - |
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Answer» `-2/3` |
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| 14. |
A beam of light converging towards a point O is intercepted by a convex mirror of focal length 30 cm. The point is 20 cm behind the mirror from its pole. Find the nature and position of the image. What if the point O is at distance 50 cm from the pole? |
| Answer» Solution :(i) Real, at DISTANCE 60 cm from pole in front of REFLECTING surface and erect and magnified 3 TIMES (ii) Virtual, at distance 75cm from pole BEHIND the mirror, inverted and magnified 1.5 times | |
| 15. |
Two plance mirrors are placed at some angle. There are five images formed when an object is placed symmetrically between them. Find the angle between the mirrors. |
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Answer» `60^(@)` |
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| 16. |
Pick out the wrong feature about carbon resistors. |
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Answer» Compact |
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| 17. |
Explain giving reasons for the following: (a) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation. (b) The stopping potential (V_(0)) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces. (c) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation. |
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Answer» Solution :(a) Photoelectric current in a photocell incrases with increase in number of photoelectrons EJECTED from the CATHODE plate which in turn depends on the number of incident radiation photons. On increasing the intensity of incident radiation the number of incident photons INCREASES and hence photoelectric current increases. (b) As per Einstein.s photoelectric equation, we have `h(v-v_(0))=eV_(0) implies V_(0)=(h)/(e)(v-v_(0))` Thus, `V_(0)` varies linearly with frequency v of the incident radiation for a GIVEN photosensitive surface. the slope of `V_(0)-v` curve is `(h)/(e)` which is a constant and independent of the nature of photosensitive surface. thus, the slope remains the same for different surfaces. (c) As per Einstein.s photoelectric equation, the maximum kinetic energy of ejected phtoelectron is given as `h(v-v_(0))=K_(MAX)` Thus, `K_(max)` depends only on the frequency of incident radiation but is independennt of its intensity. |
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| 18. |
Consider a magnetic dipole which on switching ON external magnetic field orient only in two possible was i.e., one along the direction of the magnetic field (parallel to the field ) and another anti-parallel to magnetic field. Compute the energy for the possible orientation . Sketch the graph. |
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Answer» Solution :Let `vec(p_(m)) ` be the DIPOLE and before switching ON the external magnetic field. There is no orientation. Therefore, the energy U = 0 As soon as external magnetic field is switched ON, the magnetic dipole orient parallel `(THETA = 0)^(@)` to the magnetic field with energy . `U_("parallel") = U_("minimum") = - p_(m) B cos 0 ` `U_("parallel") = - p_(m) B "since cos " 0^(@)` = 1 OTHERWISE, the magnetic dipole orients anti-parallel `(theta = 180^(@))` to the magnetic field with energy. `U_("anti-parallel") = U_("minimum") = - p_(m) B cos ` 180 ` rArr U_("anti-parallel") =- p_(m) B "since cos " 180^(@)` = -1 
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| 19. |
A current loop, carrying a current of 5.0 A, is in the shape of a right triangle with sides 30,40, and 50 cm. The loop is in a uniform magnetic field of magnitude 80 mT whose direction is parallel to the current in the 50 cm side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop. |
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| 20. |
Whichof the following is correct ? |
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Answer» The REST mass of a STABLE NUCLEUS is less than the sum of the rest masses of the isolated nucleons |
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| 21. |
Modem is a device which performs |
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Answer» modulation |
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| 22. |
Most of the electron in the base of N-P-N transistor flow |
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Answer» out of the BASE LEAD |
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| 24. |
Why did the author hire the two crewmen? |
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Answer» to take REST from long the voyage |
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| 25. |
State Faraday.s law of electromagneticinduction. Figure shows a reactangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perendicular to the plane of the paper. The field extends from x=0 to x=b and is zero for x gt b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x=0 to x=2b and is then moved backward to x=0 with constant speed v. Obtain the expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 lt x lt 2b. |
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Answer» Solution :The magnitude of the INDUCED EMF in a circuit is equal to the time rate of change of MAGNETIC flux through the circuit. Mathematically, the induced emf is given by ` epsilon = ( - d phi)/(DT)`  First consider the forward motion from `X=0` to `x=2b` The flux `phi_(B)` linked with the section SPQR is `phi_(B) = Blx, 0 le x lt b` `= Blb, b le x lt 2b` The Induced emf is , `epsilon = -(d phi_(B))/(dt)` `= Blv " " 0 le x lt b` `=0 " " b le x lt 2b` 
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| 26. |
The electric field in a certain region is given by E=5 hat(i)-3hat(j) kv//m. The potential difference V_(B)-V_(A) between points a and B having coordinates (4, 0, 3) m and (10, 3, 0) m respectively, is equal to |
| Answer» ANSWER :A | |
| 27. |
A square plate of side 4.0 cm is placed 20cm away from a concave mirror of focal length 15 cm .Calculate the area enclosed by the image of the plate. |
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Answer» `124cm^2` |
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| 28. |
The energy required for electron to transist from n = 1 to n = 3 in Li^(+2) is ...... |
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Answer» 12.1 eV For `LI^(++),Z=3`, `E_(1)=(13.6(3)^2)/((1)^(2))=-122.4eV` `E_(3)=-(13.6xx(3)^(2))/((3)^(2))=-13.6eV` `:.DeltaE=E_(3)-E_(1)=-13.6+122.4=108.8eV` |
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| 29. |
P is power transfer in external circuit and R variable resistance. |
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Answer» |
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| 30. |
If an object is placed at the focus of a concave mirror, where is the image formed ? |
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Answer» Between FOCUS and CENTER of Curvature |
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| 31. |
Pottasium chlorate on heat gives? |
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Answer» POTASSIUM CHLORIDE |
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| 32. |
A motor delivers power which draws 100 liters per minute of water from a pipe. If its power is increased x times it draws 200 liters/min. The value of x is: |
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Answer» 4 |
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| 33. |
If an LED has to emit 662 nm wavelength of light thenwhat should be the band gap energy of its semiconductor? (h=6.62xx10^(-34)Js) |
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Answer» Solution :`LAMBDA=662xx10^(-9)m` `=6.62xx10^(-7)m` `h=6.62xx10^(-34)J` `lambda=(hc)/(E_(g))` `E_(g)=(hc)/(lambda)` `=(6.62xx10^(-34)xx 3xx10^(8))/(6.62xx10^(-7))` `THEREFORE E_(g)=3xx10^(-19)J` `therefore E_(g)=(3xx10^(-19))/(1.6xx10^(-19)) therefore E_(g)=1.875 eV` |
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| 34. |
Magnetic field on t:quatur of Earth is 4 xx 10^(-5) T. Radius of Earth is 6400 km, then magnetic dipole moment of Earth is about ...... Am^(-2). |
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Answer» `10^(23)` `B_(E) = (mu_0 m)/( 4pi R^3)` `THEREFORE m= (4pi B_(E) R^(3) )/( mu_0)` `therefore m= (4 xx 10^(-5) xx (6.4 xx 10^(6) )^(3) )/(10^(-7) ) [ because (mu_0)/( 4pi ) = 10^(-7)]` `therefore m= 1048 . 576 xx 10^(20) ` `therefore m ~~ 1.0xx 10^(23) ""therefore m~~ 10^(23) "Am"^(2)` |
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| 35. |
What is the cause of resistance of a conductor ? |
| Answer» Solution :While drifting, the free ELECTRONS collide with the IONS and atoms of the CONDUCTOR, i.e., motion of the electrons is OPPOSED during COLLISIONS. This is the basic reason of resistance in a conductor. | |
| 36. |
What is the value of electric flux (phi) on a plane of area 1m^2 on which an electric field of 2 V/m crosses with an angle of 30^(@). |
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Answer» 1Vm |
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| 37. |
During interference of light |
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Answer» ENERGY is destroyed at the dark bands |
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| 38. |
Area under speed-time curve gives |
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Answer» distance `tau = (dL)/(dt) rArr dL = tau dt` For constant torque `Delta L = tau Delta I = I Delta OMEGA rArr Delta L = I omega ("if" omega_(1) = 0)` Rotational kinetic ENERGY `= (1)/(2) Iomega^(2) = ((Delta L)^(2))/(2I)` |
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| 39. |
A wire of variable mass per unit length mu = mu_(0)xis hanging from the ceiling as shown in figure. The length of wire is l_(0) . A small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach to the other ends. |
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Answer» `sqrt((6l_(0))/G)` |
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| 40. |
A magnetic system with zero dipole moment |
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Answer» Solenold |
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| 41. |
A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. Draw a ray diagram and find the location of the point at which the beam would now converge. |
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Answer» Solution :As shown in Fig. 9.42, the beam of light now converges at point Q INSTEAD of point P. question for concave lens `u = CP = + 12 cm` and `f= - 16 cm`. Hence, from lens FORMULA `1/v - 1/u = 1/f` We have: `1/v -1/(+12) = 1/(-16)` `rArr 1/v =1/12 -1/16 = 1/48` or `v= +48` cm 
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| 42. |
Explain the use Zener diode as a voltage regulator. |
Answer» Solution : The circuit connections are made as shown in the figure. The circuit contains a zener diode, an unregulated voltage source (`V_"in" > V_Z`), a series resistance `R_S` and a load resistance `R_L`. Any increase/decrease of voltage drop ACROSS `R_S` produce no change in voltage across the zener diode. For (`V_"in" < V_Z`), the zener diode does not function. THUS the zener diode ACTS as voltage REGULATOR. |
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| 43. |
Why are curved railway tracks banked ? |
| Answer» Solution :The curved tracks are banked on the outside to provide the necessary centripetal FORCE because when a fast MOVING train goes round a CURVE, it tends to fly tangentially of the TRACK. | |
| 44. |
An electron enters a magnetic field of 0.01 T with a speed of 10^(7)ms^(-1) and describes a circle of radius 6 mm there. Then specific charge of the electron is given by: |
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Answer» `1.76xx10^(11)Ckg^(-1)` |
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| 45. |
A rod of glass (mu =1.5) and of square cross section is bent into the shape shown in figure. A parallel beam of light falls on the plane flat surface A as shown in figure. If a is the width of a side and R isthe radius of circular arc then for what maximum value of d/Rlight entering the glass slab through surface A emerges from the glass through B |
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Answer» 1.5 |
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| 46. |
Displacement current can be obtained by using the relation I_(D)="_______". |
| Answer» SOLUTION :`epsilon_(0)(dphi_(E))/(DT)` | |
| 47. |
A motor car needs an engine of 3000 watts to keep it moving with a constant velocity of 10 m/sec on horizontal road. The force of friction between the car tyres and the ground is : |
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Answer» 300 dyne `implies F=P/u=(3000)/(10)=300` Newtons. |
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| 48. |
Assertion (A) : When U-235 isotope is bombarded by slow neutrons , nuclear fission takes place with a release of about 200 Me V of energy . Reason (R) : The disintegration energy is fission event first appears as the kinetic energy of fragments and neutrons but eventually it is transferred to the surrounding matter appearing as heat . |
| Answer» Solution :In NUCLEAR FISSION energy is RELEASED on account of higher BINDING energy of fragments than binding energy of URANIUM nucleus . | |
| 49. |
From fig 32-2 approximate the (a) smaller and (b) larger wavelength at which the eye of a standard observer has half the eye's maximum sensitivity . What are the (c ) wavelength (d) frequency and ( e) period of the light at which the eye is the most sensitive ? |
| Answer» Solution :(a) 515 NM , (b) 610 nm (c ) 555 nm , (d) `5.41xx10^(14)` HZ , ( E) `1.85 xx10^(-15)` s | |
| 50. |
A spherical bowl of radius R rotates about the verical diameter with angular velocity omega.The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle theta with the vertical (see figure). Then |
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Answer» `omega=sqrt(g//r COS theta)` |
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