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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.
| 43851. |
The currentI and a coil varies with time as shows inthe figure .Tghe variation of induced emf with time would be |
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| 43852. |
Newton’s law states that the radius vector joining the planet and sun sweeps equal areas in equal intervals of time.Is it true? |
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| 43853. |
Define capacitor reactance. Write its SI units. |
| Answer» Solution :The imaginary/virtual RESISTANCE offered by a capacitor to the flow of an ALTERNATING current is CALLED capacitor REACTANCE, `X_(C )=(1)/(omega C)`. Its SI unit is ohm. | |
| 43854. |
Assuming that the length the of the solenoid is large when compared to its diameter, find the edquation for its inductance. |
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Answer» Solution :Consider a long solenoid of length 1 and cross-sectional area. A Let n be the number of turns per unit length (or TURN density) of the solenoid. ii. When an electric current I is passed through the solenoid, a magnetic field is produced by it which is almost uniform and is DIRECTED along the axis of the solenold as shown in Figure. The magnetic field at any POINT inside the solenoid is given by `B=mu_(0)NI`  III. As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is `Phi_(B)=int_(A)vec(B)*dvec(A)BAcostheta=BA" since " =0^(@)` `(mu_(0)ni)A` iv. The toal magnetic flux linked or flux linkage of the solenoid with N turns (the total number of turns N is given by N = n l) is NPhi_(B)=(nl)(mu_(0)ni)A` NPhi_(B)=(mun^(2)Al)i""(1)` `epsilon=(d(NPhi_(B)))/(dt)=-(d(Li))/(dt)""...(2)` `NPhi_(B)=Li` Comparing equation (1) and (2), we have `L=mun^(2)Al` v. From the above eqution, it isd clear that inductance depends on the geometry of the solenoid (turn density n, cross-sectional area A, length l) and the medium present inside the solenoid. If the solenoid is permeability `mumu_(r)n^(2)Al` |
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| 43855. |
Although ordinary rubber is an insulator, the special rubber tyres of aircraft are made slightly conducting. Explain why. |
| Answer» Solution :The aircraft while in motion gets charged due to atmospheric AIR friction. The CONTINUOUS CHARGING may raise the SHARP CONDUCTING metal parts to discharge state which will cause spark and even explosion. If the tyres are made conducting, charge will be collected to the tyres and at the time of launching it may safely get discharged to the ground. | |
| 43856. |
सबसे बड़ा जीवाणु है : |
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Answer» स्पाइरिलम वॉल्यूटान्स (SPIRILLUM volutans) |
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| 43857. |
A freely falling body is an ideal projectile. It's trajectory is |
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Answer» a STRAIGHT LINE |
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| 43858. |
सबसे छोटा जीवाणु है : |
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Answer» स्पाइरिलम वॉल्युटान्स (SPIRILLUM volutans ) |
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| 43859. |
When electric current is passed through a resis tance wire, it get heated up" What happens to the heat energy developed, if the current through the wire is doubled. |
| Answer» SOLUTION :Heat DEVELOPED in the conductor `H = 1^2R` If current is doubled, Heat developed becomes 4 TIMES | |
| 43860. |
Use i the Ampere's law for H and ii continuity of lines of B, to conclude that inside a bar magnet, (a) lines of H run from the N-pole to S-pole while (b) lies of B must run from the S-pole to N -pole. |
Answer» Solution :Consider amagnetic FIELD line of B through the bar magnet as given in the figure below.  The magnetic field line of B through the bar magnet must be a CLOSED loop. Let C be the amperian loop, then, `int_(Q)^(P)H. di=int_(Q)^(P)(B)/(m_(Q))di` We know that the ANGLE b etween B and Di is less than `90^(@)` inside the bar magnet so , it is positive. i.e., `int_(Q)^(P)H.di=int_(Q)^(R)(B)/(mu_(0))digt0` Hence, the lines of B must run from south POLE (s) to north pole (N) inside the bar magnet. According to Ampere's law, `therefore ointHdi=0` `therefore ointH.di=int_(P)^(Q)H.di+int_(Q)^(P)H.di=0` As `int_(Q)^(P)Hdlgt0,` So `int_(P)^(Q)H.dilt0` It will be so if angle between H and di is more than `90^(@)`, so that `cos theta` negative it means the line of H must run from N-pole inside the b ar magnet. |
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| 43861. |
The ratio of energies of electron in the first excited state to its second excited state in H-atom is ...... |
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Answer» `1:4` `:.Eprop(1)/(2)""` ( `:.` for hydrogen Z=1) For first excited STATE `n_(1)=2` and `n_(2)=3`, `:.(E_(1))/(E_(2))=(n_(2)^(2))/(n_(1)^(2))=(9)/(4)"":.E_(1):E_(2)=9:4` |
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| 43862. |
A man weighing 70 kg, riding a motorbike weighing 230 kg at 54 km hr^(-I), accelerates at Ims^(-2) for 10 s when suddenly a child rushes into the road. The rider manages to apply brakes screeching to bring his vehicle to a halt in 3 s, just in time to save the child. What should have been the average retarding force on the vehicle ? |
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Answer» `1.5N` :. velocity attained in 10 s is given by `upsilon = u + at impliesupsilon = 15 + 1 xx 10 = 25 m//s` For retarded motion `u. = 25 m//s, upsilon = 0,t = 3s` `:.upsilon = u. + a. timplies 0= 25+ a. xx 3` `a. =(-25)/(3)m//s^(2)` :.Retarding force F = (M+ m) RETARDATION `=300xx(25)/(3) =2.5xx10^(3)N` So correct choice is (E) as NONE of the choices given in QUESTION is correct. |
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| 43863. |
What did the man from Udipi pray for, when he was young? |
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Answer» a PAIR of trousers |
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| 43864. |
In a periodic table, the average mass of magnesium is given as 24.312 u. The average value of based on their relative natural abudance on earth. The three isotopes and their masses are ""_(12)^(24)Mg (23.98504 u), ""_(12)^(25)Mg(24.98584 u) and ""_(42)^(26)Mg(25.98259u). The natural abundance of ""_(12)^(24)Mg is 78.99% by mass. Calculate the abundances of ther two isotopes. |
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Answer» Solution :`(""_(12)^(24)Mg) = 24.312 u` Mass of `""_(12)^(24)Mg = 23.985044` `""_(12)^(25)Mg = 24.98584u` `""_(12)^(25)Mg = 24.98584u` `""_(12)^(26)Mg = 25.98259u` Let us assume that there are `n_1` atoms of `I^(ST)` isotope, `n_2` atoms of `2^(nd)` etc. TOTAL `(n_1 + n_2 + n_3)` HENCE `(n_1 + n_2 + n_3)24.312` GIVES the total mass. Relative abundance of first = `(n_1)/((n_1 + n_2 + n_3))` Relative abundance of `2^(nd) = (n^2)/((n_1 + n_2 + n_3))` Relative abdundance of `3^(rd) = (n_3)/((n_1 + n_2 + n_3))` `:. n_1 xx 23.98504 + n_2 xx 24.98584 + n_3 xx 25.98259 = (n_1 + n_2 + n_3)24.3124` so that `(n_1)/((n_1 + n_2 + n_3)) xx 23.98504 + (n_2)/((n_1 + n_2 + n_3)) xx 24.98594 + (n_3)/((n_1 + n_2 + n_3)) xx 25.98259` `= 24.3126` `(n_1)/((n_1 + n_2 + n_3)) = 0.7899, (n_2)/((n_1 + n_2 + n_3)) = x, (n_3)/((n_1 + n_2 + n_3)) = 0.2101 - x` `:. 0.7899 xx 23.98504 + x xx 24.98584 + 0.2101 xx 25.98259 - x xx 25.98259 - x xx 25.98259 = 24.3126` `18.9457515 + 5.458942153 - 24.3126 = (25.98259 - 24.98584)x` `x = (0.092093653)/(0.99675) = 0.09239` i.e, 9.24% Adundance of last isotope = 0.1177 i.e., 11.77 |
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| 43865. |
An airplane flying at a velocity of 900 km h^(-1) loops the loop. If the maximum force pressing the pilot against the seat is five times its weight, the loop radius should be |
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Answer» 1562 m |
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| 43866. |
Find KE of the rod having mess 2kg and length 3m which is moving on smooth horizontal surface such that velocity of it's end points are 3m/s and 6m/s as shown. |
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Answer» 18J `omega(r+3)=6` `omega=1` `K=(1)/(2)Iomega^(2)=(1)/(2)((ML^(2))/(12)+md^(2))xxomega^(2)` `& d=r+(l)/(2)` |
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| 43867. |
(A) : The index of refraction (n) for majority of substances is related to their relative permittivity (in_(r )) as n=sqrt(in_(r )) . (R) : The relative permeability (mu_(r )) is approximately 1 for majority substances. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 43868. |
230V, 50 Hz ac is applied to a resistor. The instantaneous value of voltage is represented as |
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Answer» `230sin100pit` |
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| 43869. |
A 5 cm tall object is placed 10 cm from the concave mirror of focal length 15cm .The motion ,nature and size of the image are |
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Answer» -15CM,real,15cm(ERECT) |
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| 43870. |
The device designed to produce pure spectra of light source is called_____ |
| Answer» SOLUTION :SPECTROMETER | |
| 43871. |
In one series A.C. circuit, maxima of current and voltage occur at the same moment. Then which of following component must have been connected to source ? |
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Answer» Only resistor (`because` For only resistive CIRCUIT , `X_L=X_C=0`) `therefore delta=0` `therefore I_m` and `V_m` are obtained at the same MOMENT. |
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| 43872. |
A moving sphere of mass m suffer a perfect elastic collision (not head on) with equally massive stationary sphere. After collision both fly off at angle theta, value of which is : |
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Answer» 0 |
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| 43873. |
The mass defect for the nucleus of helium is 0.0303 amu. The binding energy per nucleon in MeV is nearly |
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Answer» 28 |
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| 43874. |
In the figure masses m_(1),m_(2) and M are kg. 5 kg and 50 kg respectively. The co-efficient of friction between M and ground is zero. The co-efficient of friction between m_(1) and M and that between m_(2) and ground is 0.3. The pulleys and the string are massless. The string is perfectly horizontal between P_(1) and m_(1) and also between P_(2) and m_(2). The string is perfectly verticle between P_(1) and P_(2). An externam horizontal force F is applied to the mass M. Take g=10m//s^(2). (i) Drew a free-body diagram for mass M, clearly showing all the forces. (ii) Let the magnitude of the force of friction between m_(1) and M be f_(1) and that between m,_(2) and ground be f_(2). For a particular F it is found that f_(1)=2f_(2). Find f_(1) and f_(2). Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. |
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Answer» (II) `a=3//5m//s^(2),T=18N,F=60N` |
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| 43875. |
Which of the following physical quantity does not represent electric field ?(i)V/m, (ii)J/C |
| Answer» SOLUTION :J/C . POTENTIAL , ITIS a SCALAR QUANTITY. | |
| 43876. |
A force factor applied on mass is represented by vec F = 6hati - 8hatj + 10hatk. It accelerates the mass with 1 m / s^2. The mass of the body is |
| Answer» Answer :A | |
| 43877. |
The angle of minimum deviation for a 75^(@) prism of dense glass is found to be 45^(@) when in air and 15^(@) when immersed in certain liquid.The refractive index of the liquid is |
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Answer» `(SQRT(3))/2` |
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| 43878. |
alpha-particles have a high ionizing power. Justify your answer. |
| Answer» Solution :`alpha`-particles have a charge of `+3.2 xx 10^(-19)C` and RESEMBLES helium NUCLUES. | |
| 43879. |
A point source of light B is placed at a distance L in front of the center of a mirror of width 'd' hung vertically on a wall. A man walks in front of the mirror along aline parallel to the mirro at a distance 2L from it as shown in figure. The greatest distance over which he can see the image of the light source in the mirror is |
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Answer» `d//2` In `DELTA`ANM and `Delta`ADC, `angleADC=angleANM=90^(@)``[MN_|_AD]` `angleMAN=angleCAN` (law of REFLECTION) `=DeltaANM` is similar to `DeltaADC` `:. (x)/(2L)=(d//2)/(L) or x=d` So, required distance `=d+d+d=3d`. Therefore, (d) is the correct option. 
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| 43880. |
At a certain place the angle of dip is 30^@ and the horizontal component of earth's magnetic field is 3.2 xx 10^(-5)T. Earth's total magnetic field at the place is |
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Answer» `(3.2)/(sqrt3) xx 10^(-5)` |
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| 43881. |
A long copper wire of diameter 2 mm carries a current of 1.5 A. a. What is the magnetic field (B) at a perpendicular distance 1 m from the middle of the wire? b. What is the field (B) on the surface of the conductor? c. At a distance 1 mm from the axis, what is the value of the magnetic field? |
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Answer» Solution :DATE SUPPLIED, Radius r = 1 mm `= 1 xx 10^(-3)m, 1 = 1.5A` a. `B = (mu_0I)/(2pi r) = (4pi xx 10^(-7)xx 1.5)/(2pixx 1) = 3 xx 10^(-7)T` b. `B_(end) = (mu_0I)/(4pir) = 1.5 xx 10^(-7)T` C. `B = (mu_0I)/(2pia) = (4pi xx 10^(-7)xx 1.5)/(2pi xx 10^(-3)) = 3 xx 10^(-4) T` d. same as (c) |
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| 43882. |
Who is Think- Tank? |
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Answer» CHIEF Commander |
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| 43883. |
A body with heat capacity not depending on the temperature and equal to C = 20.0 J/K is cooled from t_1 = 100^@ C to t_2 = 20^@ C. The heat received by the body is |
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Answer» `+160 J` `=-20xx(100 - 20) = -1600 J = -1.6 kJ ` |
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| 43884. |
A low voltage supply from which one needs high currents must have very low internal resistance. Why ? |
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Answer» SOLUTION :Because `I = (epsilon)/(R + r)`(Where R = EXTERNAL resistance ) When R is constant, we can increase the value of I by DECREASING value of INTERNAL resistance r. |
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| 43885. |
An electric dipole of dipole momentvec(p) consists of point charges + q and - q separated by a distance 2a apart. Deduce the expression for the electricfieldvec(E ) due to the dipole at a distance x from the center of dipole on its axial linein terms of the dipole moment vec(p) . Hence show that in the limit a to 0 vec(E )=(2 vec(p))/((4pi epsilon_(0)x^(3))). |
| Answer» SOLUTION : | |
| 43886. |
The shortest wavelength of X-ray emitted from an X-ray tube depends upon. |
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Answer» the CURRENT in the tube |
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| 43887. |
A hollow cylinder with both sides open generates a frequency fin air. When the cylinder vertically immersed into water by half its length the frequency will be |
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Answer» F `v =v/(2l)` When the cylinder is VERTICALLY immersed into water by half its length then `lambda/4 = l/2`  When the cylinder is vertically immersed into water by half its length, then `(lambda)/4 = l/2`  `rArr lambda =(4l)/2 =2 l` `THEREFORE v=v/(2l)`, the same. So, the frequency will be its ORIGINAL frequency f. |
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| 43888. |
A body of mass m is moving with speed v makesa one-dimensional collision with a stationary body of same mass on a horizontal table . They are in connect for the a very small time interval Deltat.The contact force between them varies variesas shown in thegraph. ( Neglect friction) The coefficient of restitution for the collision will be |
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Answer» `3/4` |
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| 43889. |
Which vector should be added to 2hati + 4hatj -3hatk and 3hati-5hatj+7hatk to get a unit vector along y-axis ? |
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Answer» <P>`5hati+hatj+hatk` Now `2i+4j-3k+3i-5j+7k+vec(P)=vec(j)` or `vec(P)=-5hati+2hatj-4hatk` |
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| 43890. |
A coil of 800 turns and 50 cm^2area makes 10 rps about an axis in its own plane in a magnetic field of 100 gauss perpendicular to this axis. What is the instantaneous induced emf in the coil ? |
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Answer» SOLUTION :`A = 50cm^2 = 50 XX 10^(-4) m^2` n= 10 rps, N= 800 B = 100 gauss = `100 xx 10^(-4) T = 10^(-2) T` Now , `epsi = epsi_0 SIN omega t = NBA omega sin omega t ` ` = 800 xx 10^(-2) xx 50 xx 10^(-4) xx 2pi xx 10 sin (20pi t)` or `epsi = 2.5 sin (20pi t)` VOLT |
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| 43891. |
If length of a given wire increases by 10% due to stretching, the resistance of the wire increasesby |
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Answer» 0.21 `A. = (Al)/(l.) = (Al)/(1.1l) = (A)/(1.1)` ` THEREFORE ` New resistance `R. = (rho l.)/(A.) = (rho (1.1l))/((A//1.1)) = 1.21 (rho l)/(A) = 1.21 R` ` therefore (DELTA R)/(R ) = (R.- R)/(R ) = 0.21 = 21%` |
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| 43892. |
A ring of radius R carries a charge +Q, which is uniformally distributed on the surface of the ring. The electrostatic force acting on a charge +q situated at the centre of ring is zero. |
| Answer» Solution :True-Electrostatic FORCE due to DIFFERENT elements of charged ring on the CENTRAL charge ACT in diffeent directions and net force on the central charge is zero. | |
| 43893. |
The energy band diagram of a Si semiconductor crystal at absolute zero temperature, |
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Answer» has completely emptyvalence band and completelyfilled CONDUCTION band. |
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| 43894. |
The work function of lithium is 2.5 eV. what is the maximum wavelength of light that can cause the photoelectric effect in lithium? |
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Answer» |
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| 43895. |
Two compound pendulums are made of : a disc of radius r and (b) A uniform rod of length L. Find the minimum possible time period and distance between centre and point of suspension in each case. |
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| 43896. |
Electromagnets are made of soft iron because soft iron has .............. . |
| Answer» SOLUTION :HIGH SUSCEPTIBILITY and LOW RETENTIVITY | |
| 43897. |
If n represents the order of a half period zone, the area of this zone is approximately proportional to n^(m)where m is equal to |
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Answer» Zero |
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| 43898. |
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60^@ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.) |
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Answer» SOLUTION :(a) According to formula, we have, `tau=BINAsintheta` = `BIN(PIR^(2))sintheta` = `(1)(6)(30)(3.14)(0.08)^(2)sin60^(@)` = `(180)(3.14)(64xx10^(-4))(0.866)` `thereforetau=3.132Nm` (b) Here, formula `tau=BINAsintheta` is true for any SHAPE of planar coil. Hence, for same area of another planar coil, our answer will not change by changing only shape of planar coil. |
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| 43899. |
The energy gap is max in : |
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Answer» METALS |
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| 43900. |
The energy of an electron revolving in a stationary orbit is always negative because ...... |
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Answer» ELECTRON possesses NEGATIVE charge |
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