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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 42901. |
A bead B of mass 'm' can travel without frictionn on a smooth horizontalwire "xx"^('). The bead is connected to a block of identical mass by an ideal string passing over an ideal pulley. The system, as shown, is in vertical plane. System is initially released from rest in the position shown. Initial acceleration of block A is: |
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Answer» SOLUTION :`mg-T=ma_(1)` `Tcos30^(@)=ma_(2)` and `a_(2) COS 30^(@)=a_(1)` `impliesmg=ma_(1)(1+ 1/(cos^(2)30^(@)))` `implies a_(1)=(3g)/7` |
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| 42902. |
A light wave of frequencynuand wavelength lambda travels from air to glass. Then, |
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Answer» `nu` CHANGES |
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| 42903. |
A ( sinusoidal ) carrier wave C(t) = A_c sin omega_c t is amplitude modulated by a ( sinusoidal ) message signal m(t) = A_m sin omega_m t Write the equation of the (amplitude) modulated signal. Use this equation to obtain the values of the frequencies of all the sinusoidal waves present in the modulated signal. |
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Answer» Solution : The equation of the (amplitude) modulated signal is ` C_m(t) [(A_c + A_m SIN omega_m t)] sin omega_c t` This can be REWRITTEN as `C_m(t) = [A_c ( 1+ MU sin omega_m t)] sin omega_c t` Where `mu=A_m //A_c ` = modulation INDEX `:. C_m (t) = A_c sin omega_ct + ( mu A_c)/(2) 2 sin omega_c t` `=A_c sin omega_c t + ( mu A_c)/(2) [ cos (omega_c - omega_m)t- cos ( omega_c + omega_m) t]` These are the three sinusoidal waves present in the amplitude modulated signal. The frequencies of these three waves are 
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| 42904. |
The wavelengths of given light waves in air and in a medium are 6000 A and 4000 A, respectively. The angle of total internal reflection is . |
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Answer» `SIN^(-1)(2/3)` |
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| 42905. |
A projectile is fired from earth vertically with a velocity Kv_(e ) where v_(e )= escape velocity and K is a constantless than unity. What is the maximum height to which it rises as measured from centre of earth? |
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Answer» `(R )/(K^(2))` `(1)/(2)mk^(2)v_(e )^(2)=(GMm)/(R )-(GMm)/(r )` `(1)/(2) mk^(2)(2GM)/(R )=GMm[(1)/(R )-(1)/(r )]` `(1)/(r )=(1)/(R )-(k^(2))/(R )=(1-k^(2))/(R )` or`r=(R )/(1-k^(2))`. Correct choice is (b). |
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| 42906. |
Two resistances are joined in parallel whose resultant is 6/5 ohms. One of the resistance wire is broken and the effective resistance becomes 2 ohms. The resistance in ohms of the wire that got broken |
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Answer» 2 |
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| 42907. |
Two charges - q and +q are located at points (0, 0, - a) and (0,0, a), respectively. How much work is done in moving a small test charge from the point (5, 0, 0) to (-7,0,0) along the x-axis ? Does the answer change if the path of the test charge between the same points is not along the x-axis ? |
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Answer» Solution :The given charges behave as an electric dipole and the dipole length is 2a, hence dipole moment `p = q.2a`. As the motion of test charge from the POINT (5, 0, 0) to (-7,0,0) is along x-axis and electric potential is only along z-axis, hence work done = 0. The answer does not change because the electric field is a CONSERVATIVE field and the workdone simply depends on the position of initial and FINAL points only. 
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| 42908. |
Direction is a^(to) times^(to)r is |
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Answer» TANGENT to path |
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| 42909. |
A ray of light is incident at 50^(@) onthe middle of one of two mirrors arranged at an angle of 60^(@) between them. They ray then touches the second mirror, get reflected back to the first mirror, making an anlge of incidence of |
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Answer» `50^(@)` |
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| 42910. |
A small spherical ball of density same as that of liquid is released from rest a vessel filled completely with a liquid and accelerating with acceleration 5hati+5hatjm//s as shown in the figure. Find the initial acceleration of ball just after the release with respect to vessel. |
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Answer» |
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| 42911. |
Identify the mismatched pair |
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Answer» a) Microwaves - Aircraft navigation |
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| 42912. |
In the previous example, what is the minimum speed necessary to keep the roller coaster on the track at all times? |
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Answer» Solution :The position where the roller coaster is most in danger of leaving the track is at the top. `F_(N)+F_(w)=(mv^(2))/(r)` The SLOWER the OBJECT moves, the smaller `F_(N)` gets (`F_(W)` is a constant). Therefore, the minimum speed occurs when `F_(N)=0`. (for a MOMENT, the ROLLERCOASTER is touching the track but not pressed against it: it's almost losing contact.) `F_(w)=(mv^(2))/(r)` `mg=(mv^(2))/(r)` `v=sqrt(gr)` |
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| 42913. |
The given circuit shows an arrangement of four capacitors A potential difference 30V is applied across the combination. It is observed that potentials at points .A. and .B. differ by 5V. Also if a conducting wire is connected between .A. and .B. electrons will flow from A to B. Of course, we have not connected any wire actually between A and B, we have described only an .if. situation. Charge on capacitor C_2is |
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Answer» ` 60muC` |
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| 42914. |
The current in a simple series circuit is 5 A. When an additionalresistance of 2Omegais inserted, the current drops to 4 A. The original resistance of the circuit was |
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Answer» `8Omega` |
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| 42915. |
The given circuit shows an arrangement of four capacitors A potential difference 30V is applied across the combination. It is observed that potentials at points .A. and .B. differ by 5V. Also if a conducting wire is connected between .A. and .B. electrons will flow from A to B. Of course, we have not connected any wire actually between A and B, we have described only an .if. situation. Equivalent capacitance between X andY is |
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Answer» ` 2.34muF` |
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| 42916. |
A wall has two layers A and B, each made of defferent material. A has thickness 10 cm, while B has thickness 20 cm, their coefficients of conductivities are in the ratio of 3:1. A constant temperature difference of 35^(@)C exists across the wall. The difference of temperature across the layer A is : |
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Answer» `28^(@)C`  In STEADY state `H_(1)=H_(2)`. `(k_(1)A(DeltaT_(1)))/(l_(1))=(k_(2)A(DeltaT_(2)))/(l_(2))` `(3kDeltaT_(1))/(10)=(kDeltaT_(2))/(20)` `(DeltaT_(1))/(DeltaT_(2))=1/6""rArr""DeltaT_(2)=6DeltaT_(1)` Also `DeltaT_(1)+DeltaT_(2)=35` `DeltaT_(2)=6DeltaT_(1)=35""rArr""DeltaT_(1)=5^(@)C`. Thus correct choice is (c ). |
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| 42917. |
Which of the following statements (s) is//are correct ? |
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Answer» If the electric field due to a point charge varies as `r^(-2.5)` instead of `r^(-2)`, then Gauss's law will still be valid. b. it is almost impossible to calculate the field distribution around an electric dipole using gauss's law because proper symmetry will not be formed c. This is obvious d. This options is not correct because it is not given that charge is moving slowly. |
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| 42918. |
The given circuit shows an arrangement of four capacitors A potential difference 30V is applied across the combination. It is observed that potentials at points .A. and .B. differ by 5V. Also if a conducting wire is connected between .A. and .B. electrons will flow from A to B. Of course, we have not connected any wire actually between A and B, we have described only an .if. situation. Potential difference across C_4 is |
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Answer» 12.5 V |
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| 42919. |
A full wave rectifier is fed with a.c. mains frequency 50 Hz. What is fundamental frequency of the ripple in the output current ? |
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Answer» 25 Hz |
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| 42920. |
Explain the terms 'mass defect' and 'binding energy'. How are they related ? Draw binding energy curve. |
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Answer» Solution :MASS defect : It is the difference between the sum of the masses of the NUCLEONS/constituents of the nucleus and the nuclear mass. Binding energy : The MINIMUM energy REQUIRED to break the nucleus into its constituent nucleons/ particles. Binding energy `=(Deltam)931MeV` Binding energy = (Am) 931 MeV Binding energy is energy EQUIVALENT to mass defect. Binding energy curve: 
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| 42921. |
The base of vessel is a square of side 0.1 m.It's height is 0.3m. It is completely filled with a liquid of density 1200(kg)/m^3. Find The thrust of the liquid on the base |
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Answer» SOLUTION :THRUST on the BASE =`PcdotA`=3528(0.1xx0.1m)` =35.28 NEWTON |
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| 42922. |
The base of vessel is a square of side 0.1 m.It's height is 0.3m. It is completely filled with a liquid of density 1200kg/m^3. FindThe pressure on the vertical walls |
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Answer» SOLUTION :The PRESSURE on the SIDE WALLS =the pressure on the BOTTOM 3528`N/m^2` |
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| 42923. |
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 4 cm at a distance of 40 cm from its mid point? The magnetic moment of the bar magnet is a 0.5Am""^2 |
| Answer» SOLUTION :`B_(E) = 7.8125 XX 10^(-7) T, B_(A) = 15.625 xx 10^(-7) T` | |
| 42924. |
For what distance is ray optics a good approximation when the aperture is 3mm wide and the wavelength is 500 mn? |
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Answer» Solution :For distance `Z le Z_(F)`, RAY optics is the GOOD appropriate Fresnel distance `Z_(F)= (a^2)/(LAMBDA)= ((3xx10^(-3))^2)/(5xx 10^(-7))= 18m`. |
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| 42925. |
The electrostaticforce on a small sphereof charge 0.4 muc dueto another small sphere of charge -0.8 muc in air is 0.2 N (a) What is the distance between the two spheres (b)what is the force on the second sphere due to the first |
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Answer» SOLUTION :(a) 12 CM (B) 0.2 N (ATTRACTIVE) |
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| 42926. |
The base of vessel is a square of side 0.1 m.It's height is 0.3m. It is completely filled with a liquid of density 1200kg/m^3. FindThe pressure at the base |
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Answer» SOLUTION :THEPRESSURE on the base P=h`RHO`G or `rho`=6.3*1200*9.8=3528`N/m^2` |
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| 42927. |
Two ideal solenoids of same dimensions. One is air cored with 600 turns while other is Aluminium is cored with 200 turns (relative permeabililty of Aluminium is 3), are connected in a circuit as shown in the figure. The switch S is closed at t=0. Find the ratio of potential difference across air-cored solenoid to that of Aluminium cored solenoid at any time t |
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Answer» `L_(1)=(mu_(@)N_(1)^(2)A)/1, L_(2)=(mu_(@)N_(2)^(2)Amu_(r))/1` SINCE `V=-L(dl)/(DT)`, So `(V_(1))/(V_(2))=(L_(1))/(L_(2))=((N_(1))/(N_(2)))^(2) 1/(mu_(r))=(600/200)^(2)xx1/3=3` |
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| 42928. |
Aparallel plate capacitor of capacitance 500 pF and plate area 0.05 m^(2) is filled with porcelain . Thefollowing observations are recorded . Mark the one which isnot correct . |
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Answer» The free CHARGE on the plates is `0.1muC`. |
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| 42929. |
You are given the two circuits as shown in fig. show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate . |
Answer» SOLUTION :
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| 42930. |
A : Newton.s rings are formed in the reflected system. When the space between the lens and the glass plate is filled with a liquid of refractive index greater than that of glass, the central spot of the pattern is dark. R : The reflection in Newton.s ring cases will be from a denser to a rarer medium and the two interfering rays are reflected under similar conditions. |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 42931. |
The relation F = ma, cannot be deduced from Newton’s second law, if |
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Answer» Force depends on TIME |
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| 42932. |
Two satellites of masses m_(1) and m_(2)(m_(1) gt m_(2)) are revolving around the earth in circular orbits of radius r_(1) and r_(2)(r_(1) gt r_(2)) respectively, which of the following is true regarding their speeds V_(1) and V_(2) ? |
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Answer» `V_(1)=V_(2)` So the CORRECT CHOICE is (B). |
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| 42933. |
Principle of uncertainty show…… |
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Answer» a RELATION between momentum of particles and wave property |
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| 42934. |
Let n_p, and n_c be the numbers of holes and conduction electrons in an intrinsic semiconductor. |
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Answer» `NP gtn_c` |
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| 42935. |
A wheel having moment of inertia 2 kg m^(2) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be |
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Answer» `(2pi)/15Nm` `thereforealpha=omega_(i)//t,` where `ALPHA` is RETARDATION. The torque on the wheel is GIVEN by `tau=Ialpha=(Iomega_(i))/t=(I.2piv)/t=(2xx2xxpixx60)/(60xx60)` `rArr=pi/15` Nm This is the torque REQUIRED to stop the wheel in 1 min. (or 60 sec.). |
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| 42936. |
How we get greater magnifying power in case of a simple magnifier? |
| Answer» Solution :SMALLER the FOCAL length of simple MAGNIFIER GREATER is its MAGNIFYING power | |
| 42937. |
What physical quantity is the same for X-rays of wavelength 10^(-10)m, red light of wavelength 6800 Å and radiowaves of wavelength 500m ? |
| Answer» Solution :The speed in vacuum is same for all the given WAVELENGTHS, which is 3 `xx 10^(8)` m/s. | |
| 42938. |
What happens when the small bar magnet kept on the glass and iron filings sprinkled on glass ? |
Answer» Solution :When iron filings sprinkled, the arrangement of iron filing is as below :  The pattern suggests that : (i) The patterns of the iron filings suggests that the magnet has two poles similar to the POSITIVE and negative charge of an electric dipole. One pole is DESIGNATED the north pole andthe other the south pole. (ii) When suspended FREELY, these poles POINT towards the GEOGRAPHIC north and south poles respectively. (iii) A similar pattern of iron filings is observed around a current carrying solenoid. |
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| 42939. |
a. The top of the atmosphere is at about 400 kV with respectto the surface of theearth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^(-1). Why then do we not get a electric shock as we step out of our house into the openy (Assume the house to be a steel cage so there is no field inside!) b. A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m^(2). Will he get an electric shock if he touches the metal sheet next morning? c. The discharging current in the atmosphere due to the small conductivity of air is known to be 1800A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? d. What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm^(-1) at its surface in the downward direction, corresponding to a surface charge density =-10^(-9) Cm^(-2) ?. |
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Answer» Solution :a. Our body and the ground form an equipotential surface. As we step out into the open, the original equipotential surfaces of open air change, keeping our head and the ground at the same POTENTIAL. b. Yes. The steady discharging current in the atmosphere charges up the aluminium sheet gradually and raises its VOLTAGE to an extent depending on the capacitance of the capacitor (formed by the sheet, slab and the ground). c. The atmosphere is continually being charged by thunderstorms and lightning all over the globe and DISCHARGED through regions of ordinary weather. The TWO opposing currents are, on an average, in equilibrium. d. Light energy involved in lightning, HEAT and sound energy in the accompanying thunder. |
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| 42940. |
A train of length L move with a constant speed V_(t). A person at the back of the train fires a bullet at time t = 0 towards a target which is at a distance of D ( at time t = 0 ) from the front of the train ( on the same direction of motion ). Another person at the front of the train fires another bullet at time t = T towards the same target. Both bullets reach the target at the same time. Assuming the speed of the bullets V_(b) are same, the length of the train is |
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Answer» `T ( V_(B)XX 2V_(t) )` |
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| 42941. |
For the beta^+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K-shell, is captured by the nucleus and a neutrino is emitted). e^+ + ""_Z^AX to ""_(Z - 1)^(A)Y + upsilon Show that if beta^+ emission is energetically allowed, electron capture is necesarily allowed but not vice - versa. |
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Answer» Solution :Consider the competing processes : `""_(Z)^(A)X to ""_(Z - 1)^(A)Y + E^+ + upsilon_e+ Q_1 ` (positron capture) `e^- + ""_(Z)^(A)X to ""_(Z- 1)^(A)Y + upsilon_(e ) + Q_2` (electron capture) `Q_1 = [m_N(""_Z^AX) - m_N(""_(Z - 1)^(A)Y) - m_e]c^2` `= [m_N(""_Z^AX) - Zm_e - m(""_(Z-1)^(A)Y) - (Z - 1) m_e - m_e] c^2 = [m(""_(Z)^(A)X)- m(""_(Z-1)^(A)Y) - 2m_e]c^2` `Q_2 = [m_N(""_Z^AX) + m_e - m_N(""_(Z-1)^(A)Y)]c^2 = [m(""_Z^AX) - m(""_(Z-1)^(A)Y]c^2` This MEANS `Q_1 > 0` implies `Q_2 > 0` but `Q_2 > 0` does not NECESSARILY mean `Q_1 > 0`. Hence the RESULT. |
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| 42942. |
Making use of Eq.(6.4g), find at T=0: (a) the velocity distribution of free electrons, (b) the ratio of the mean velocity of free electron to their maximum velocity. |
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Answer» Solution :(a)From `dn(E )=(sqrt(2)m^(3//2))/(PI^(2)ħ^(3))E^(1//2)dE` we GET on using `E=(1)/(2)MV^(2), dn(E )=dn(V)` `dn(v)=(sqrt(2)m^(3//2))/(pi^(2)ħ^(3))(1)/(sqrt(2))m^(1//2) vmvdv=(m^(3))/(pi^(2)ħ^(3))v^(2)dv` This holds for `0 lt v lt V_(F)` where `(1)/(2)mV_(F)^(2)=E_(F)` and `dn(v)=0 for v gt v_(F)`. (b)MEAN velocity `lt v gt = int_(0)^(v_(F))v^(3)dv//int_(0)^(v_(F)) v^(2) dv=(3)/(4)V_(F)` `:. (lt v gt)/(V_(F))=(3)/(4)` |
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| 42943. |
A surface charge density of 10^(-2)C//m^2 has been created by friction on a glass disk of h =5 mm thickness and R= 50 mm radius. The disk rotates at 1.6 r.p.s. Find the magnetic field intensity at the centre of the disk. When you have learned to integrate, find the magnetic moment and the ratio of the magnetic moment to the angular momentum. |
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Answer» The total intensity of the field at the centre is `H=triangleH_(1)+triangleH_(2)+.....=1/2 sigma omega (triangler_1+triangler_2+.....)=1/2 sigma omega R` where R is the external radius of the disk. The magnetic moment of the ring is `triangleP_(m)=ipir^2=pir^2 =(triangleq)/T=1/2 triangleq wr^2=pi sigma omegar^2 triangler` To find the total magnetic moment of the rotating disk we must add up all these values. We have `Pm=underset(0)overset(R)INT pi sigma r^3dr=1/4 pi sigma omega R^4` The moment of momentum `L=I omega=1/2 mR^2w=1/2pi wphR^4`, where p is the density of the material. We have `p_m/L=sigma/(2ph)` |
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| 42944. |
Two resistors R_(1) and R_(2) are connected in the left gap and the right gap of a metre bridge, and the null point is obtained at 20 cm from the left. On interchanging the resistors in the two gaps, the null point shifts by |
| Answer» Answer :C | |
| 42945. |
A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle. |
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Answer» Solution :` dq=Q/LRd THETA ` Net electric FIELD have only vertical component ` =intdEcos theta ` ` =1/(4piepsilon_0)Q/Lint_(-pi/2)^(pi/2)(Rd theta)/(R^2)COS theta ` ` =1/(4piepsilon_0)Q/(LR)[SIN theta]_(-pi/2)^(pi/2) ` ` 1/(4piepsilon_0)Q/(LR)[1-0(-1)] ` ` (2Q)/(4piepsilon_0LR)=Q/(2piepsilonL^2) (:.piR=L)`. 
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| 42946. |
A mixture of light, consisting of wavelength 590nm and an unknown wavelength, illuminates Young.s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is |
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Answer» `885.0 NM` |
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| 42947. |
A circular insulated copper wire loop is twisted to form two loops of are A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane ( of the paper). A uniform magnetic vecB point into the plane of the paper . At t=0, the loop starts rotation about the common diameter as axis with a constant angular velocity omega in the magnetic field. Which of the following options is /are correct ? |
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Answer» The NET emf induced due to both the LOOPS is proportional to `cos omega t` |
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| 42948. |
A galvanometer has a current range of 15mA and voltage range of 750mv. To convert this galvanometer into an ammeter of range 25A, the shunt resistance required is nearly |
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Answer» 0.2`OMEGA` |
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| 42949. |
During the process of modulation the RF wave is called |
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Answer» MODULATING wave |
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| 42950. |
Two charges of equal magnitude and at a distance‘r’ exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is |
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Answer» F/8 |
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