InterviewSolution
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InterviewSolution
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.
| 4501. |
A circular coil of radius r having number of turns n and carrying a current A produces magnetic induction at its centre of magnitude B. B can be doubled by |
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Answer» keeping the number of TURNS n and CHANGING the CURRENT to A/2 |
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| 4502. |
Three photons coming from excited atomic hydrogen sample are picked up. Their energies are 12.1eV, 10.2 eV and 1.9 eV. These photons must come from |
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Answer» a SINGLE atom |
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| 4503. |
Number of colour band marked on carbon resistors is : |
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Answer» 9 |
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| 4504. |
A man of weight W is standing on a lift which is moving upwards with acceleration 'a'. The apparent weight of the man is : |
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Answer» W `R=mg(g+a)` `R=mg((1+a)/(g))` `R=W((1+a)/(g))` Hence CORRECT CHOICE is (d) 
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| 4505. |
A steel tape measures that length of a copper rod as 90.0 cm when both are at 10^(@)C, the calibration temperature, for the tape. What would the tape read for the length of the rod when both are at 30^(@)C. Given alpha_("steel")=1.2xx10^(-5)" per".^(@)Cand alpha_(Cu)=1.7xx10^(-5)per .^(@)C |
| Answer» Answer :A | |
| 4506. |
A body of mass 2 kg slides down with an acceleration of 3 m/s^2 on a rough inclined plane having a slope of 30^(@). The external force required to take the same body up the plane with the same acceleration will be: (g=10 m/s^(2) ) |
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Answer» 14 N |
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| 4507. |
In an intrinsic semiconductor, the fermi energy level lies |
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Answer» nearer to VALENCE BAND |
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| 4508. |
The temperature of air varies with height linearly from T_(1) at the earth's surface to T_(2) at a height h. Calculate the time t needed for a sound wave produced at a heights to reach the earth's surface. The velocity ofsound near the earth's surface is C. |
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Answer» |
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| 4509. |
The time constant of and R- C cirucit during charging is them time in which the charge on the condenser plates, as compared to maximum charge (q_0) becomes (q//q_0)xx 100, which is equal to …….. . |
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Answer» SOLUTION :`Q = q_(0)(1 - e^(-t/tau))`. PUTTING`t = tau,` we get `(q)/(q_(0)) = 1-e^(-1)` or `(q)/(q_(0))xx100 = (e-1)/(e ) xx100 = 63.2%` |
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| 4510. |
Wave nature of light is verified by |
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Answer» Interference |
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| 4511. |
Charge Q is uniformly distributed on a dielectric rod AB of length 2l. The potential at P shown in the figure is equal to |
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Answer» `Q/(4pi epsi_(0) (2l))` |
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| 4512. |
An infinitely long line charge having a uniform charge per unit length lambda lies a distance d. from center of imaginary sphere of radius R. Then |
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Answer» flux crossing the sphereis zero if `d GT R` |
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| 4513. |
If N number of molecules in a container are enclosed then average number of molecules moving between a pair of wall is |
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Answer» `N//2` |
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| 4514. |
In a Young's double-slit experiment apparatus, two identical slits are separated by 1 mm and distance between slits and screen is 1 m. The wavelength of light used is 6000 Å. What is the minimum distance between two points on the screen having 75% intensity of the maximum intensity? |
| Answer» SOLUTION :0.20 mm | |
| 4515. |
A parallel paraxial beam of light is incident on the arrangement as shown (mu_(A)=3//2, mu_(B)=4//3). The two spherical surfaces are very close and each has a radius of curvature 10 cm. Find the point where the rays are focussed. (w.r.t. point of entry) |
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Answer» |
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| 4516. |
Which of the following statement are incorrect about phenol-formaldehyde resin ? |
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Answer» Novolac or resol is a linear polymer and is used in the MANUFACTURE of adhesive |
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| 4517. |
The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy ? |
| Answer» Solution :CARBON steel piece, because HEAT LOST PER cycle is proportional to the AREA of hysteresis loop and area of hysteresis loop is more for carbon steel. | |
| 4518. |
An anmeter of current range 1A has resistance of 100Omega.What is the required resistance connected in series to convert it into voltmeter of range 300V: |
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Answer» `200OMEGA` |
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| 4519. |
Explain the quantum concept of light. |
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Answer» Solution :Max Planck proposed quantum concept in 1900 in order to explain the thermal radiations emitted by a black body and the shape of its RADIATION curves. According to Plancks, matter is composed of a large number of oscillating particles ( atoms) which vibrate with different frequencies. Each atomic oscillator -which VIBRATES with its characteristic frequency -emits or absorbs electromagnetic radiation of the same frequency. It also says that (i) If an oscillator vibrates with frequency v, its energy can have only certain discrete values, given by the equation. `E_(n) = n hv``n= 1,2,3,"....."` where h is a constant, called Planck's constant. (ii) The osciallators emit or ABSORB energy in small packets or quanta and the energyof each quantum is E = hv. This implies that the energy of the oscillator is quantized-that is, energy is not CONTINUOUS as believed in the wave picture . This is called quantization of energy. |
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| 4520. |
We stand by our creed, proud to be Indians - The tone of this line is |
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Answer» Patriotic |
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| 4521. |
The net charge in a current carrying wire is zero. Them, why does amagnetic field exert a force on it? |
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Answer» STATIONARY CHARGE |
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| 4522. |
What is the nature of field around a moving charge? |
| Answer» SOLUTION : Its natural electric FIELD and a magnetic field are due to its MOTIONAL property. | |
| 4523. |
A gas is filled in container at pressure P_(0). If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be : |
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Answer» <P>`2P_(0)` `thereforePpropmv_("rms")^(2)` As mis halved `v_("rms")` is doubled P BECOMES TWICE= `2P_(0)`. Correct CHOICE is (a) |
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| 4524. |
If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R=0.1 Å and (ii) R=10 Å |
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Answer» Solution :Orbital radius of electron in the ground state in H-atom = BOHR radius `a_(0)=r_(1)=0.53Å=(epsi_(0)h^(2))/(pi me^(2)) ( :.n=1)...(1)` Total energy of electro in above case, `E_(1)=-13.6eV ....(2)` `rArr K_(1)=-E_(1)=13.6eV ....(3)` and `U_(1)=2E_(1)=2(-13.6)=-27.2eV ....(4)` (Because for an electron, making orbital motion around the nucleus in an atom, `E = -K=(U)/(2))` Here as per the statement in the first case, `R=0.1Å rArr R lt a_(0)` ( `:.` From equatio `(1) a_(0)=0.53Å)` `rArr` We can take proton as a point charge. HENCE total energy of H-atom in this case will be -13.6eV as per equation(2). As per statement in the second case, `R=10 Å ""..(5)` `rArr R gt gt a_(0)` `rArr` Electron would make its orbital motion inside proton. In this case, new Bohr radius is suppose `b_(0)`. Now charge e is uniformly distributed in its volume `(4)/(3)piR^(3)`.Hence in volume `(4)/(3)pib_(0)^(3)`. if charge contained is e. then since volume density of electric charge is constant. `(e)/((4)/(3)pi R^(3))=(e.)/((4)/(3)pi b_(0)^(3))` `:. e.=((b_(0)^(3))/(R^(3)))e....(6)` Now writing `b_(0)` in place of `r_(1)` and writing `e^(2)=e.e`, `b_(0)=(epsi_(0)h^(2))/(pi me.e)=(epsi_(0)h^(2))/(pi me)xx(1)/(e.)` `:. b_(0)=(epsi_(0)h^(2))/(pi me)xx(R^(3))/(b_(0)^(3)e)` [ From equation (6)] `:.b_(0)^(4)=((epsi_(0)h^(2))/(pi me^(2)))R^(3)` `=(0.53)(10)^(3)` [From equation (1) and (5)] `=510 Åxx(Å)^(3)` `=510(Å)^(4)` `:.b_(0)=(510)^(1//4)={(510)^(1//2)}^(1//2)=4.75Å....(7)` `rArr b_(0) lt R [ :.` Here `R=10 Å) ....(8)` Now according to formula `r_(n)=(epsi_(0)n^(2)h^(2))/(pi mZe^(2))....(9)` and `v_(n)=(Ze^(2))/(2epsi_(0)nh) ....(10)` Taking multiplication of above two equations, `v_(n)r_(n)=(Ze^(2))/(2epsi_(0)nh)xx(epsi_(0)n^(2)h^(2))/(pi m Ze^(2))` `:.v_(n)r_(n)=(nh)/(2PI m)` `:. v_(1)r_(1)=(h)/(2pi m)`= constant ....(11) `:.v_(1).=v_(1)((a_(0))/(b_(0))) ....(12)` `rArr` Here, new value of kinetic energy of electron, `K_(1).=(1)/(2)mv_(1)^(2)` `:. K_(1).=(1)/(2)mv_(1)^(2)((a_(0))/(b_(0)))^(2)` `=K_(1)((a_(0))/(b_(0)))^(2)` `=13.6((0.53)/(4.75))^(2)` `:. K_(1).=0.1693eV ......(13)` `rArr` Now in the present case, electric potential for `r gt R` is `V=(kQ)/(2R^(3))(3R^(2)r^(2))` ( where k= Coulomb.s constant `=(1)/(4pi epsi_(0))` Taking V=V. and Q=e= charge of proton for `r=b_(0)` `V.=(Ke)/(2R^(3))(3R^(2)-b_(0)^(2))` Now new electrostatic potential energy in ground state, `U_(1).=V_(q).` `:.U_(1).=(ke)/(2R^(3))(3R^(2)-b_(0)^(2))(-e)` ( `:.` Charge of electron q=-e) `=-(ke^(2))/(2R^(3))(3R(-2)-b_(0)^(2))` `=-(9xx10^(9)xx(1.6xx10^(-19))^(2))/(2xx(10xx10^(-10))^(3)) xx{3(10xx10^(-10))^(2)-(4.75xx10^(-10))^(2)` `=-(9xx2.56)/(2)xx10^(9-38-20)(300-22.56)` `=-3.196xx10^(-19)J` `=-(3.196xx10^(-19))/(1.6xx10^(-19))eV` `:.U_(1).=-1.9975eV ...(14)` From equation (13) and (14) `E_(1).=K_(1).+U_(1).` `:.E_(1).=0.1693+(-1.9975)` `:.E_(1).=-1.8282eV` Note : There are mistakes in the answers of NCERT exemplar BOOK. |
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| 4525. |
Three condensers of capacity, C_1,C_2 and C_3 are connected in parallel. The resultant capacity C is given by: |
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Answer» `C=C_1+C_2+C_3` |
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| 4526. |
A pump motor is used to deliver water at a certain rate from a given pipe. To obtain twice as much water from the same pipe in the same time, power of the motor has to be increased to : |
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Answer» <P>16 time `:.` Power `P=W/t=(1/2mv^2)/(t)=(mv^2)/(2t)` Work done in SECOND case when mass of water is DOUBLED `W.=1/2(2m)v^2` Power `P.=(mv^2)/(t)` `:. P/P=(mv^2)/(t)xx(2t)/(mv^2)=2` P.=2P. |
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| 4527. |
A uniform magnetic field, B=B_(0)t (where B_(0) is a positive constant) , fills a cylinderical volume of radius R, then the potential difference in the conducting rod PQ due to electorstatic field is : |
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Answer» `B_(0)lsqrt(R^(2)+l^(2))` |
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| 4528. |
The electric field in a region of space is given by, bar(E) = E_0 hati + 2E_0 hatj, where E_0 = 100N/C. The flux of this field through a circular surface of radius 0.02m parallel to the Y - Z plane is nearly: |
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Answer» `3.14 Nm^2//C` |
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| 4529. |
Electromagents are made of soft iron because soft iron has _________ and __________ . |
| Answer» SOLUTION :HIGH PERMEABILITY , LOW RETENTIVITY | |
| 4530. |
Which of the following connot be stepped up in a transformer ? |
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| 4531. |
A circular coil of 200 mm diameter is made of 100 turns of thin wire and carries a current of 50 mA. Find the magnetic field induction in the centre of the coil and on the coil'axis 100 mm away from its centre. |
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Answer» where to is the NUMBER of turns, a is the radius of a burn, and h is the distance from the centre to the point on the axis of the coil where the field is to be determined. |
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| 4532. |
What is the force between two small charged spheres having charges of 2 xx 10^(-7)C and 3 xx 10^(-7)C placed 30 cm apart in air? |
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Answer» Solution :`q_(1)=2 xx 10^(-7)C =3 xx 10^(-7)C, R=30cm=30 xx 10^(-2)m` `F=9 xx 10^(9) xx (q_(1)q_(2))/(r^(2))=9 xx 10^(9) xx (2 xx 10^(-7) xx 3 xx 10^(-7))/((30 xx 10^(-2))^(2)) =0.06 xx 10^(-1)N =6 xx 10^(-3)N ("repulsive")` |
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| 4533. |
(A) : In a communication system based on amplitude modulation the modulation index is kept less than 1 (R) : If modulation index is less than 1, there is minimum distortion of signal |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 4534. |
What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20 cm? is the system a converging or a diverging lens ? Ignore thickness of the lenses. |
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Answer» Solution :`f_(1) = 30 CM , f_(2) = - 20 cm ` F= `(f_(1)f_(2))/(f_(1) + f_(2))= (30 XX -20)/(30- 20) = (-600)/(10) = - 60 ` cm .f. being negative, the combination is a DIVERGING lens. |
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| 4535. |
A battery of emf 6 V and negligible internal resistance is connected to the terminals of apotentiometer wire of length 4 m. The wire is of uniform cross-section and its resistance is 100 Omega . The difference of potential between two points separated by 40 cm on the wire will be |
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Answer» 0.4 V |
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| 4536. |
What is a rectifier ? Draw circuit diagram for a rectifier in which ouput is obtained continuously. |
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Answer» Solution :Rectification. Rectification is the process of converting alternating VOLTAGE/current into direct voltage/current. A device used for this purpose is called rectifier and this phenomenon is called rectification. Principle. It is based on the principle that a p-n junction diode conducts when it is forward biased and does not conduct when it is reverse biased. Construction. The apparatus for rectification CONSISTS of two diodes `D_(1)` and `D_(2)` connected to two ENDS of the secondary of a step down transformer. Output is taken out from mid point of the secondary and common point N of two diodes. Ouput is taken out from ends of the load R.  Working. During the positive half of input `AC,D_(1)` is forward biased and `D_(2)` reverse biased. Hence the current flows through the upper circuit as shown. During negative half, lower portion `(D_(2))` is forward biased and upper `(D_(1))` is reverse biased. Thus during each half, we get the current either from `D_(1)` or from `D_(2)`. The output voltage is unidirectional having ripple CONTENTS. Ripple factor of a rectifier `=("r.m.s of a.c.component")/("value of d.c. component")` To smoothen the output electric filters are used. The electric filters are combination of INDUCTORS are capacitors. Some of the useful filters are L-filter and `pi`-filter. |
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| 4537. |
Obtain the differential equation for a LC circuit. |
Answer» Solution :L-C circuit is shown in figure. In their circuit capacitor (C )and inductor ( L ) are connected. Let at time t=0, capacitor is charged and charge on it is `q_(m)`.  The moment the circuit is comleted the charge on the capacitor starts decreasing giving rise to currentin the circuit. Let q and I be the chrage and current in the circuit at time t. Since`( dI)/( dt) ` is positive the induced emf in L will have polarity as shown in figure, Means `V_(a) GT V_(b)`. As q decreases I increases, ` :. I = - (dq)/(dt)` Induced emf in inductor at any isntant, `V = epsilon = - L (dI)/(dt)` and p.d. across capacitor `= ( q)/( C )` According to Kirchhoff.s loop rule, `- L (dI)/(dt) + ( q)/( C ) = 0` but `I = - ( dq)/( dt)` `:. (dI)/(dt) = - ( d^(2)q)/( dt^(2))` `:. + L ( d^(2) q)/( dt^(2)) + ( q)/( C ) = 0 ` `:. (d^(2) q)/( dt^(2)) + ( q)/( LC ) = 0 ` `:. (d^(2)q)/( dt^(2)) + omega_(0)^(2) q= 0 ` `[ :. (1)/(LC) = omega_(0)^(2) ]` which is second order DIFFERENTIAL equation of linear charged. This equation has the form of general formula, for a simple harmonic oscillator.The charge therefore, OSCILLATES with a natural frequency. `omega_(0) = (1)/( SQRT(LC))` `:. 2 pi f_(0) = (1)/( sqrt(LC))` `:.` Its frequency is `f_(0) = (1)/(2pi ) sqrt((1)/( LC))` |
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| 4538. |
The deflection magnetometer is most sensitive when the defiectiontheta is |
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Answer» Nearly `0^(@)` |
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| 4539. |
(A) : Repeaters are used to extend the range of communication (R) : Repeater is combination of a receiver and a transmitter. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 4540. |
A bar magnet of magnetic moment 1.5JT^(-1) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii) ? |
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Answer» Solution :Here `m= 1.5 J T^(-1) , B = 0.22 T and theta_1 = 0^2` (a) (i) Work done to turn the MAGNET , so as to ALIGN its magnetic moment normal to the magnetic field i.e. `theta_2 =90^@ ` , is given as : `W= mB[cos theta_1 - cos theta_2] = 1.5xx0.22 [cos 0^@ - cos 90^@]` `=1.5xx0.22xx[1-0] =0.33J`. (ii) Work done to turn the magnet so as to align its magnetic moment opposite to the field direction i.e. `theta_2 = 180^@` `W = 1.5xx0.22xx[cos 0^@ - cos 180^@] =1.5xx0.22xx[1-(-2)] =0.66J` (b) TORQUE in case (i) `tau= m B sin theta_2 = 1.5xx0.22xx SIN90^@ = 1.5xx0.22xx1 = 0.33 ` N m Torque in case (ii) = `tau = m B sin theta_2 = 1.5xx0.22xxsin 180^@ = 1.5xx0.22xx0=0` |
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| 4541. |
The pair of equations 3x - 5y = 7, -6x + 10y = 7, have |
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Answer» UNIQUE solution |
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| 4542. |
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^(-1)Then the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position |
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Answer» Solution :a) We know that at each extreme position, velocity of the bob is zero. If the string is cut at the extreme position, it is only under the action of .G.. Hence the bob will fall vertically downwards. b) At the mean position, velocity of the bob is 1 m/s. ALONG the tangent to the ARC, which is in the HORIZONTAL direction. If the string is cut at mean position, the bob will behave as a horizontal PROJECTILE. |
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| 4543. |
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? |
| Answer» Solution :As the copper plates oscillate in the MAGNETIC between the two poles of the magnet, there is a continuous change of magnetic FLUX linked with the plate. Due to this, eddy currents are SET up in the copper plate which try to OPPOSE the MOTION of the plate according to the Lenz's law and finally bring it to rest. | |
| 4544. |
Consider the circuit shown where C_(1) = 6 mu F, C_(2) = 3 mu F and V=20 V. Capacitor C_(1) is first charged by closing the switch S_(1). Switch S_(1) is then opened, and the charged capacitor is connected to the uncharged capacitor C_(2) by closing S_(2). |
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Answer» Total charge that has FLOWN through the battery is `120 MU C.` |
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| 4545. |
Packing fraction of a nucleus is defined as..............per............. . |
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Answer» |
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| 4546. |
A disc of mass m is connected with an ideal spring of spring constant k inside a lift as shown. The left starts moving with constant acceleration (a hat (i)- a hat (j)). At the same moment disc starts rolling without sliding on the wall of lift (from rest). If spring is in natural length initially then friction force acting on disc as a function of y is given by : (y is the displacement of the disc with respect to lift) |
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Answer» `-ky` `rArr fr = (1)/(3) (mg + ma - ky)` |
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| 4547. |
In the moving coil galvanometer N = number of turs A = cross sectional area: B = magnetic field: C = Torsional constant: I = current R = resistance {:("List-I","List-II"),(P."Torque action on the coil",1. NAB//C),(Q. "Defliction produced",2.NAB//CR),(R. "current sensitivity",3.NIAB),(S."voltage sensitivity",4. NABI//C):} |
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Answer» 3,1,2,4 |
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| 4548. |
Three metals have work functions in the ratio 2 3:4. Graphs are drawn for all connecting stopping potential and incident frequency. The graphs have slopes in the ratio |
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Answer» `2:3:4` |
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| 4549. |
The most commonly used semiconductors are |
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Answer» GERMANIUM and silicon |
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| 4550. |
A ring of radius r=25cm made of lead wire is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. What is the number of rps at which the ring ruptures? |
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Answer» SOLUTION :Let us consider an element of the ring (figure). From NEWTON's law `F_n=mw_n` for this element, we GET, `Tdtheta=((m)/(2pi)dtheta)omega^2r` So, `T=(m)/(2pi)omega^2r` Condition for the problem is: `(T)/(pir^2)lesigma_m` or, `(momega^2r)/(2pi^2r^2)lesigma_m` or, `omega_(max)^2=(2pi^2sigma_mr)/(pir^2(2pirrho))=(sigma_m)/(rhor^2)` Thus SOUGHT number of RPS `n=(omega_(max))/(2pi)=(1)/(2pir)sqrt((sigma_m)/(rho))` Using the table of appendices `n=23rps` 
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