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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 copper calorimeter of negligible thermal capacity is filled with a liquid. The mass of the liquid is 250 gm. A heating element found that the temperature of the calorimeter and its contents rise form 25^(@)C to 30^(@)C in 5 minutes when a current of 2.05 A is passed through it at al potential difference of 5 volts. The liquid isthrown off and the heater is switched on again. it is now found that the temperature of the calorimeter alone remains constant at 32^(@)C when the current through the heater is 0.7 A at the potential difference 6 volts. Calculate the specific heat capacity of the liquid. The temperature of the surrounding is 25^(@)C |
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| 2. |
Two identical bodies A and B of mass m each are connected by a spring. The body B is pulled by applying a constant force F. The body A moves with acceleration 'a'. Therefore acceleration of B is given by |
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Answer» `F/m - a` |
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| 3. |
A beam of light falls on a glass plate (mu = 3/2) of thickness 6.0 cm at an angle of 60^@. Find the deflection of the beam on passing through the plate. [in cm]. |
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| 4. |
A metal sphere of radius 10 cm is charged to 200 volt. Electric intensity at a point on its surface is, |
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Answer» `20kV//m` |
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| 5. |
What is exosphere? |
| Answer» SOLUTION :It is TE OUTERMOST ATMOSPHERIC region from the earth.s surface. | |
| 6. |
A parallel plate capacitor is charged by a battery . After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates . How will the energy stored in the capacitor be affected ?Justify your answer |
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Answer» Solution :Let the CAPACITOR has a CAPACITANCE C and charge on it , when charged by a battery , be Q . Hence potential difference between the plates of capacitor `V = (Q)/(C)`. When the battery is disconnected and a dielectric SLAB with its thickness equal to the plate separation is inserted between the plates , the charge Q remains conserved , Now : The energy stored in the capacitor `U. = (Q.^(2))/(2 C.) = (Q^(2))/(2C.) = (Q^(2))/(2 KC) = (U)/(K.)` where U = initial value of the energy stored . |
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| 7. |
(A): Most of good conductors of heat are also good conductors of electricity and vice versa. (R): Mainly electrons are responsible for these conductions. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT explanation of 'A' |
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| 8. |
Light waves from two coherent sources having intensities I and 2I cross each other at a pint with a phase difference of 60^@ what is the resultant intensity at the point? |
| Answer» SOLUTION :`I=I_1+I_2+2sqrt(I_1I_2 ).cosphi=I+2I+2sqrt(Ixx2I) cos60=4.414I` | |
| 9. |
1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is |
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Answer» `55^(@)C` Energy required to CONVERT 1gm of ICE into water at `O^(@)C = 335J` Remaining energy (2345-335)J is used to raise the temperature of water beyond `100^(@)C` |
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| 10. |
(i) Draw a labelled diagram of a step down transformer. State the principle of its working. (ii) Express the turn ratio in terms of voltages. (iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer. (iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to 100 V - 550 W refrigerator ? |
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Answer» Solution :(i) to (III) = N/A (iv) Here, `V_(p) = 220V, V_(s) =110 V` and OUTPUT P = 550 W `therefore` Input POWER = Output power P = 550 W `therefore` Current drawn by the primary of TRANSFORMER `I_(p) = P/V_(p) = 550/220 = 2.5 A` |
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| 11. |
The horizontal component of the earth's magnetic field at a place is B_H and angle of dip is 60^@ What is the value of vertical component of earth's magnetic field at equator? |
| Answer» SOLUTION :At magnetic equator the value of vertical COMPONENT of earth.s magnetic FIELD is zero. | |
| 12. |
Explain with reason, how the resolving power of a compound microscope will change when (i) frequency of the incident light on the objective lens is increased, (ii) focal length of the objective lens is increased, (iii) aperture of objective lens is increased. |
| Answer» SOLUTION : R.P. of a compound Microscope=`(2MU sin THETA)/lamda=2mu sin theta v/c`(i) When frequency v increases, R.P. increases(II) R.P. does not change with change in focal length of objective lens.(iii) When APERTURE increases, `theta` increases`therefore ` R.P increases | |
| 14. |
A galvanometer with a coil of resistance 10.0Omega shows full scale deflection for a current of 50mA. How will you convert the galvanometer into (i). An ammeter of range 0 to 0.5 A and (ii). A voltmeter of range 0 to 5V? |
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| 15. |
A particle moves along a horizontal circle with constant speed. If 'a' is its acceleration and 'E' is its kinetic energy (1) a is constant(B) E is constant (C) a is variable (D) Eis variable |
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Answer» A and B are CORRECT |
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| 16. |
Assume that a neutron break in a proton and an electron. The energy released during this process is: Mass of neutron =1.6725 xx 10^(-27)kg Mass of proton =1.6725 xx 10^(-27)kg Mass of electron =9 xx 10^(-31)kg |
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Answer» 5.4MeV `triangle E=trianglemc^(2)=9 xx 10^(-31) xx (3 xx 10^(8))^(2)` =0.511MeV Which is NEARLY equal to (b) 0.73MeV. |
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| 17. |
….. Can be produced by magnetic field. |
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Answer» Moving ELECTRIC CHARGE. |
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| 18. |
A beam of well collimated cathode rays travelling with a speed of 5 xx 10^(-6) ms^(-1) enters a region of mutually perpendicular electric and magnetic fields and emerge undeviated from the region if |vecB| = 0.02 T. The magnitude of the electric field is |
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Answer» `10^5 V m^(-1)` |
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| 19. |
During propagation of waves, where the electron density is large in ionosphere, the angle of refreaction is __________ |
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Answer» `49^(@)` |
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| 20. |
The instrument which produce high frequency signal called a carrier wave is called? |
| Answer» SOLUTION :OSCILLATOR | |
| 21. |
A radioactive substance emits n beat particles in the first 2 s and 0.5 n beta particles in the next 2 s. The mean life of the sample is |
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Answer» 4 s Mean life, `TAU=(1)/(lamda)=(t_(1//2))/(0.693)=(2)/(2.303log_(2))=(2)/(IN2)s` |
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| 22. |
A cavity of radius R//2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R//2 from the centre of the sphere. The gravitational force on a particle of mass m at a distance R//2 from the centre of the sphere on the line joining both the centre of the cavity) [Here g=(GM)//R^(2), where M is the mass of the sphere |
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Answer» `(MG)/(2)` USING the FIGURE, F=Force due to whole sphere-Force due to cavity `F=(GMm)/(R^(3))-((R )/(2))-(G(M//8)m)/((R^(2)))` `=(GM m)/(R^(2))[(4)/(8)-(1)/(8)]=(3)/(8)(GM m)/(R^(2))=(3)/(8)mg` 
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| 23. |
The frequency range of ultraviolet radiation is |
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Answer» `3XX10^(3)Hz` |
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| 24. |
An ideal heat engine is working between temperature T_(1) adn T_(2) has efficiency eta. If both the temperatures are raised by 100^(@)k each, the new efficiency will be : |
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Answer» `ETA` THUS, correct choice is (b). |
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| 25. |
The potential energy of a particle in a certain fieldhas the form U=a/r^2-b/r,where a and b are positive constants, r is the distance from the center of the field. Then |
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Answer» At `r=(2a)/B`,particle is in STEADY equilibriumn. For `(dU)/(dr)=0, b/r^2=(2a)/r^3 rArr r=(2a)/b` `(d^2U)/(dr^2)=(+6a)/r^4-(2B)/r^3=2/r^3((3a)/r-b)` At `r=(2a)/b, (d^2U)/(dr^2)=2/r^3 ((3axxb)/(2a)-b)=2/r^3 b/2=b/r^3 GT 0` i.e., U is minimum So, it is a positionof stable (steady) equilibrium. `F=-(dU)/(dr)=(2a)/r^3-b/r^2` For maximum force , `(DF)/(dr)=(-d^2U)/(dr^2)=0` `rArr (-2)/r^3 ((3a)/r-b)=0 rArr r=(3a)/b` `F=(2a)/((3a)/b)^3-b/((3a)/b)^2=(2ab^3)/(27a^3)-b^3/(9a^2)=-b^3/(27a^2)` |
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| 26. |
The work functions of two metals are 2.75 eV and 2 eV respectively. If these are irradiated by photons of energy3 eV, the ratio of maximum momenta of the photoelectrons emitted respectively by them is |
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Answer» `1:2` |
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| 27. |
Focal length of plane mirror is ...... |
| Answer» Solution :infinite | |
| 28. |
In one average life |
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Answer» HALF of the NUCLEI decay |
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| 29. |
Compute the work done and power delivered by the Lorentz force on the particle of charge q moving with velocity vec(v). Calculate the angle between Lorentz force and velocity of the charged particle and also interpret the result. |
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Answer» Solution :For a charged particle moving on a magnetic field. `VEC(F) = q(vec(v) XX vec(B))` the work done by the magnetic field is W = `int vec(F).d vec(R) = int vec(F).vec(v)`dt W = q `int (vec(v) xx vec(B) ) .vec(v)`dt = 0 Since `vec(v ) xx vec(B)` is perpendicular to `vec(v)` and hence `(vec(v) xx vec(B)) . vec(v) = vec(0)` This MEANS that Lorentz force do no work on the particle. `(dW)/(dt) = P = 0 ` Since, `vec(F).vec(v) = 0 rArr vec(F) and vec(v)` are perpendicular to each other. the angle between Lorentz force and velocity of the charged prticle is `90^(@)`. thus lorentz force changes the direction of the velocity but not the magnitude of the velocity. hence Lorentz force does no work and also does not alter kinetic ENERGY of the particle. |
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| 30. |
For a wave propagating through a medium , identify the property that is indepent of the others |
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Answer» velocity |
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| 31. |
State Bohr's quantisation condition for defining stationary orbits. |
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Answer» Solution :According to Bohr.s quantum condition, the angular momentum of an electron in its stationary (or stable) orbit should be an integer multiple of `(h)/(2pi)` i.e., `L_(N) = m v_(n)r_(n) = n(h)/(2pi) `, where n is a positiveinterger. |
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| 32. |
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed v_x (like particle 1 in Figure). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is (qEL^2)/(2m v_(x)^2). Compare this motion with motion of a projectile in gravitational field. |
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Answer» Solution :Charge on a particle of mass m = − q Velocity of the particle = `V_(X)` Length of the plates = L Magnitude of the UNIFORM electric field between the plates = E Mechanical force, F = Mass (m) `xx` ACCELERATION (a) `a=(F)/(m)` However, electric force, `F=qE` Therefore, acceleration,`a=(qE)/(m)" "...(1)` Time taken by the particle to cross the FIELDS of length L is given by, `t=("Length of the plate")/("Velocity of the particle")=(L)/(v_(x))" "...(2)` In the vertical direction, INITIAL velocity, u = 0 According to the third equation of motion, vertical deflection s of the particle can be obtained as, `s=ut+(1)/(2)at^(2)` `s=0+(1)/(2)((qE)/(m))((L)/(v_(x)))^(2)` `s=(qEL^(2))/(2mv_(x)^(2))" "...(3)` Hence, vertical deflection of the particle at the far edge of the plate is `qEL^(2)//(2mv_(x)^(2))`. This is similar to the motion of horizontal projectiles under gravity. |
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| 33. |
विस्थापन के लिए क्या सही है |
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Answer» यह कभी शून्य नहीं हो सकता |
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| 34. |
Name an experiment which shows wave nature of the electron. Which phenomenon was observed in this experiment using an electron beam? |
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Answer» Solution :Davisson - GERMER experiment confirmed the WAVE NATURE of electrons. They DEMONSTRATED that electron beams are DIFFRACTED when they fall on crystalline solids. |
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| 35. |
The number of particles given by n=-D(n_(2)-n_(1))/(x_(2)-x_(1)) are crossing a unit area perpendicular to x-axis in unit time, where n_(1) and n_(2) are the number of particles per unit volume for the values x_(1) and x_(2) of x respectively. Then the dimensional formula of diffusion constant D is : |
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Answer» `[M^(0)LT^(2)]` Putting the dimensions `D=(1)/("area" XX "time")xx(L)/(1/(volume))` `=(L^(-2)xxT^(-1)xxL)/(L^(-3))=L^(2)T^(-1)` `D=[M^(0)L^(2)T^(-1)]` `:.(B)` is CORRECT. |
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| 36. |
two concentric conducting thin spherical shells A and B having radii r_(A) and r_(B) (r_(B)gt r_(A) are charged to Q_(A) and Q_(A) and (-Q_(B)//Q_(B)l gt // Q_(A)//) . The electrical field along a line (passing through the centre) is |
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Answer»
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| 37. |
In the figure shown, each capacitor is 1muFand the resistors are indicated. The charges. Q_(1) and Q_(4)on the capacitors C_(1) and C_(4)are respectively under steady condition. Then, |
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Answer» `Q_(1) gt Q_(4)` |
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| 38. |
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be : |
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Answer» `1 : 3^(1//2)` `0=m_(1)v_(1)+m_(2)v_(2)` `m_(1)v_(1)=-m_(2)v_(2)` `m_(1)/m_(2)=-v_(2)/v_(1)=v_(2)/v_(1) ("in MAGNITUDE")` But `m_(1)/m_(2) alpha A_(1)/A_(2) alpha v_(2)/v_(1)` But `v_(2)/v_(1)=1/2 therefore A_(1)/A_(2)=1/2` `R_(1)/R_(2)=(R_(0) A_(1)^(1//3))/(R_(0) A_(2)^(1//3)) = (A_(1))/(A_(2))^(1//3) =(1/2)^(1//3)` `R_(1)/R_(2)=1/2^(1//3)=1:2^(1//3)` |
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| 39. |
Find the value of emf induced in the rod for the following cases. The figures are self explanatory. |
Answer» Solution : Figure SHOWS a closed coil ABCA moving in a uniform magnetic field B with a velocity v. The flux passing through the coil is a constant and therefore the induced emf is zero.  Now consider ROD AB, which is a part of the coil. Emf induced in the rod = B L v. Now SUPPOSE the emf induced in part ACB is E, as shown in figure. Since the emf in the coil is zero, Emf (in ACB) + Emf (in BA) = 0 `-E+vBL=0` E=vBL Thus emf induced in any path joining A and B is same, provided the magnetic field is uniform. ALSO the equivalent emf between A and B is BLv (here the TWO emf’s are in parallel) |
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| 40. |
भारत मे राज्यों की संख्या कितनी है |
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Answer» 28 |
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| 41. |
A bullet of mass 'm' moving with velocity *v' strikes a block of mass 'M' at rest and gets embeded into it. The K.E. of composite block will be: |
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Answer» `1/2mv^2xx(m)/((M+m))` |
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| 42. |
Eddy currents are produced ___ |
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Answer» only within the body of the CONDUCTOR |
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| 43. |
Sound waves with frequency 2500 Hz and speed 343 m // s diffract through the rectangular opening axis of a speaker cabinet and into a large auditorium of length d = 100 m. The opening, which has a horizontal width of 30.0 cm, faces a wall 100 m away. Along that wall, how far from the central axis will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.) |
| Answer» SOLUTION :51.4 m | |
| 44. |
Applying to the system two force couples having torques equal in magnitude and opposite in sign, |
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Answer» Solution :The SOLID arrows in Fig.show the FORCES `F_1`and `F_2`applied to a rigid body at points `A_1 and A_2`Apply the forces `T_1 and T_2`at the same points and the forces `T_3 and T_4`at the point C so that is `T_1= T_3= F_1 and T_2 = T_4 = F_2` .The point C is chosen so that `F_1a_1 = F_ya_y` .Show that the now nystem of six forces is equivalent to the old one, i.e. that `(F_1,F_2,T_1,T_2,T_3,T_4)~(F_1,F_2)` Indeed, the system of these forces reduces to the former forces `F_1 and F_2`and to two force couples `(T_1, T_3) and (T_2, T_4)` whose moments are `M_1=T_1a_1 sin alpha and M_2=-T_2a_2sin alpha` But point was chosen go that `|M_1| = |M_2| ` Therefore these moments are COMPENSATED and do not act on the rigid body. On the other hand, the system of six forces is equivalent to the system of forces `T_3 and T_4` , i.e. `(F_1, F_2, T_1, T_2, T_3, T_4) ~ (T_3, T_4)` Indeed, the forces `F_1`and `T_1` , as WELL as `F_1 and T_2`are compensated, only the forces `T_3 and T_4`remaining uncompensated. Hence, the system of forces `(F_1, F_2)`is equivalent to the system `( T_3 T_4 )` , and this gives the solution to the problem. 
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| 45. |
अण्ड समुच्चय के निर्माण में कौन सहायक |
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Answer» अण्ड, |
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| 46. |
In the circuit shown in fig. net resistance between A and B is : |
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Answer» 2 R |
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| 47. |
The cirtical angle of a certain medium is sin^(-1)((3)/(5)). The polarzing angle of the medium is : |
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Answer» `SIN^(-1)((4)/(5))` `tan I (1)/(sin C)` `THEREFORE "" COT I = sin (sin^(-1) ((3)/(5)))` or "" `tan I = (5)/(3)` or "" `I = tan ^(-1) ((5)/(3))`. |
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| 48. |
In Young's experiment, green light of wavelength 5100 A.U. from a narrow slit is incident on a double slit. Ifthe over all separation of 10 fringes on a screen 2 m away is 2 cm, then the silt separation is: |
| Answer» Answer :A | |
| 49. |
Two constant-volume gas thermometers are assembled, one with nitrogen and the other with hydrogen , Both contain enough gas so that p_3=78kPa . (a) What is the difference between the pressure in the two thermometers if both bulbs are in boiling water ? (Hint : See Fig 18-5) (b) Which gas is at higher pressure ? |
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