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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.
| 752. |
In the photoelectric phenomenon, if the ratio of the frequency of incident radiation incident on a photosensitive surface is 1:2:3, the ratio of the photoelectric current is |
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Answer» `1:2:3` |
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| 753. |
A gas is confined to a cylindrical container of radius 1 cm and length 1 m. The pressure exerted on an end face, compared with the pressure exerted on the long-curved face, is |
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Answer» SMALLER because its area is smaller. |
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| 754. |
(A): It is necessary to use satellites for long distance T.V transmission (R): The television signals are low frequency signals. |
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Answer» Assertion and REASON are true and reason is the correct EXPLANATION of assertion |
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| 755. |
In hydrogen atom which quantity is integral multiple of |
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Answer» ANGULAR MOMENTUM. |
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| 756. |
What is the meaning of 'stake-by-stake'? |
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Answer» To move from ONE stake to another |
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| 757. |
x^2 - 5x + 2 को x + 1 से भाग देने पर शेषफल बराबर |
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Answer» 6 |
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| 758. |
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23Omegais connected to a 230 V variable frequency supply. a. What is the source frequency for which current amplitude is maximum? Obtain this maximum value. b. What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.c. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? d. What is the Q-factor of the given circuit? |
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Answer» Solution :a. At resonance , `omega= (1)/(sqrtLC) = (1)/(sqrt(0.12 xx 480 xx 10^(-9) ) ) = 4167 rad//sec"" becausev_t = (omega_t)/(2pi) = 66 Hz` `I_(max) = (V_(max) )/(R ) = (230 sqrt2)/(23) = 14.14 A` B. `P_(max) = 1/2 I_(max)^2 R = 1/2 xx (14.14)^2 xx 23= 2300 W` c. ` DELTA omega= (R )/(2L) = (23)/( 2 xx 0.24) = 95.8 rad// sec ` ` Delta v = (Delta omega)/(2pi) = 15.2 Hz ` Power ABSORBED = Half of `v= 663 pm 15 Hz` Current amplitude ` = (I_0)/(sqrt2) = (14.14)/(sqrt2) = 10 A` d. `Q = (X_L)/(R ) = (omega L)/(R ) = (4.67 xx 0.12)/(23) = 21.7 ` |
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| 759. |
(A):Displacement of a body is the signed sum of the area under velocity-time graph. (R ):Displacement is a vector quantity. |
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Answer» |
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| 760. |
Fig. illustrates a segment of a double line carrying direct current whose direction in indicated by the arrows. Taking into account that the potential varphi_(2) gt varphi_(1), and making use of the Poynting vector, establish on which side (left or right) the source of the current is located. |
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Answer» Solution :Given `varphi_(2) gt varphi_(1)` the electric field is as SHOWN by the dashed lines `(……..rarr…….)`. The magnetic field is as shown `(odot)` emerging out of the PAPER. `oversetrarr(S) = oversetrarr(E) xx oversetrarr(H)` is parellel to the wires and towards right. HENCE source must be on the LEFT. 
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| 761. |
The time period of a physical pendulum about some pivot point is T. When we take another pivot point, opposite of the first one such that the centre of mass of the physical pendulum lies on the line joining these two pivot points, we obtain the same time period. If the two points are separated by a distance L, then the time period T is |
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Answer» `2pisqrt((L)/(G))` |
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| 762. |
Show magnetic field lines for current carrying loop. Write rule for finding the direction of a magnetic field due to a circular current loop. |
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Answer» Solution :1. The magnetic field lines due to a circular wire form CLOSED loops are shown in figure. 2. The DIRECTION of the magnetic field is given by RIGHT hand thumb rule stated below. "Curl the PALM of your right hand around the circular wire with the fingers pointing in the direction of the current. The right hand thumb gives the direction of the magnetic field. 
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| 763. |
Radius of 1st Bohr's Hydrogen atom is 0.53A. Radius of 20th Bohr's orbit is: |
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Answer» 10.3A |
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| 764. |
A magnetic field strength (H) 3xx10^(3)Am^(-1) produces a magnetic field of induction (B) of 12piT in an iron rod. Find the relative permeability of iron? |
| Answer» SOLUTION :`10^(4)` | |
| 765. |
A: X-rays can penetrate through the flesh but not through the bones. R: The penetrating power of X-ray depends on voltage. |
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Answer» Both assertion and reason are TRUE and the reason is correct explanation of the assertion. HARDNESS PENETRATING power of X-ray depends upon ACCELERATING voltage applied across X-ray tube characteristic X-ray relate to the material of target X-ray can.t pass through matter of heaviour element like bones (which contain phosphorus and calcium) but can pass through matter of higher element as flesh (which contain oxygen, HYDROGEN and carbon) |
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| 766. |
What fraction of the volume of an iceberg (density 917 kg/m^(3)) would be visible if the iceberg floats (a) in the ocean (salt water, density 1024 kg/m^(3)) and (b) in a river (fresh water, density 1000 kg/m^(3))? (When salt water freezes to form ice, the salt is excluded. So, an iceberg could provide fresh water to a community.) |
| Answer» SOLUTION :(a) FRACTION of 10 % (B) Fraction of 8.3 % | |
| 767. |
A body is projected horizontally from the top of a tall tower with a velocity of 30ms^(-1). At time t_(1), its horizontal and vertical components of the velocity are equal and at time t_(2), its horizontal and vertical displacements are equal. Then t_(2)-t_(1) is (take, g=10ms^(-2)) |
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Answer» 1 s `30=10t_(1)` `t_(1)=3s` As per SECOND condition, `30t_(2)=1/2xx10t_(2)^(2)=t_(2)=6s` `therefore" "t_(2)-t_(1)=3s` |
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| 768. |
यदि L सभी सरल रेखा के समुच्चय को समतल में तथा जो R संबंधपरिभाषित है alphaRbeta hArr a_I_beta,alpha,betaEL, तब R है। |
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Answer» स्वतुल्य |
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| 769. |
A transistor has a current amplification factor of 60. In a CE amplifier, input resistance is 1 kOmega and output voltage is 0.01 V. The transconductance is (in SI units) |
| Answer» Answer :B | |
| 770. |
A conducting circular loop is expanding in a magnetic field of 2T uniformly so that rate of increase of its radius R is 1 cm/s. Find the induced emf in the loop when its radius is 20 cm. The field is perpendicular to the plane of the loop. |
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Answer» `0.2pixx10^(-2)` V `phi=AB=pir^2B` `THEREFORE epsilon=-(dphi)/(DT)` `=-d/(dt)[pir^2B]` `=-piB.(d(r)^2)/(dt)` `=-2piBr. (DR)/(dt)` `therefore |epsilon|=2pixx2xx0.2xx0.01` `=8pixx10^(-3)` V `therefore |epsilon|=0.8pixx10^(-2)` V |
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| 771. |
Explain method of davisson and Germer experiment and its results. |
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Answer» SOLUTION :In davisson and Germer experiment,by moving the detector on circular scale at different position. Angle between incident beam and scattered beam is called angle of scattering `(theta)` For different accelerating voltage (V) and different scattering angle `(theta)` intensity of scattered electron beam is measured. The experiment was performed by VARYING accelerating voltage from 44 to 68 V.  It was noted that a strong peak appeared in intensity (I) of the scattered electron for accelerating voltage of 54 volt and scattering angle at `theta=50^(@)`.This suggest that no.of electron scattered in maximum. Let stationary electrons is accelerated by volt V.Its kinetic energy, K=EV `therefore (1)/(2)mv^(2)=eV` `therefore (1)/(2)(m^(2)v^(2))/(m)=eV` `therefore (p^(2))/(2m)=eV` [`therfore` mv=p momentum] `therefore =sqrt(2meV)` Wavelength of matter WAVE (electron) `lambda=(h)/(p)` `therefore lambda=(h)/(sqrt(2meV))` substituting value of h,m,e `therefore lambda =(1.227)/(sqrt(V))` nm Theoretical value of wavelength of electron is 0.165 nm and experimental value obtained is 0.167 nm which are almost same. Here V =54 volt . Thus in this experiment there is excellent agreement between the theoretical value and experimentally obtained value of de-Broglie wavelength. Recently by double slit experiment,wave nature of light was demonstrated. In 1994 ,interference frings were obtained with the beams of iodine molecules then electron. In electron microscope wave nature of electron is used. |
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| 772. |
A cell of internal resistance r connected across an external resistance R can supply maximum current when |
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Answer» `R=r` |
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| 773. |
In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. on shunting the cell with a resistance of 2 Omegathe balancing length becomes 120 cm. The internal resistance of the cell is |
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Answer» `1 Omega ` |
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| 774. |
A parallel plate capacitor is charged by an exteranl ac source straight the displacement current inside the capacitor is the same as the current charging the capacitor. |
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Answer» Solution :Electric FIELD between the capacitor PLATES `E = (SIGMA)/(epsilon_(0)) = (q)/(epsilon_(0)A)` Where q is the charge accumulated on the positive plate. The electric flux through this plate `phi_(E ) = EA = (q)/(epsilon_(A))cdot A = (q)/(epsilon_(0))` `therefore` Dispacement current `I_(d) = epsilon_(0)cdot (d PHI )/( dt ) = epsilon_(0) (d)/(dt) [(q)/(epsilon_(0))] = (dq)/(dt)` `(dq)/(dt)` is the rate at which charge flows to positive plate through the conducting wire. `I_(d) = I_(C )` i.e., displacement current between the capacitor plates = conduction current through wire. |
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| 775. |
Four moles of carbon monoxide anre mixed with four moles of carbon dioxide. Asuming the gases to be ideal, the ratio of specific heats is : |
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Answer» `15//11` `gamma =(n_(1)C_(p1)+n_(2)C_(P2))/(n_(1)C_(v1)+n_(2)C_(v2))` As `C_(p) and C_(v)` of DIATOMIC gas have values `(7R)/(2) and (5R)/(2)` whereas their values for polyatomic gas are 4R and 3R. So on substituting these values we get `gamma =(15)/(11)`. Correct choice is (a). |
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| 776. |
(a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image. |
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Answer» Solution :(a) Let a parallel beam be incident first on convex lens of focal length l= 30 cm. Naturally, the beam will get focused at a point l 30 cm behind convex lens, in the absence of concave lens. However, the IMAGE Ix behaves as a virtual object for concave lens, of focal length 20 cm, which is placed 8.0 cm from convex lens. Thus, for concave lens U. = 30 - 8 = 22 cm `therefore 1/v. -1/(+22) =1/(-20) rArr 1/v^(.) =-1/20 + 1/22 rArr v^(.) = -220 cm` Thus, an incident parallel beam will appear to diverge from a point I, 220 cm from concave lens (or 216 cm from centre point O of the two lens system). Then let us consider that a parallel beam of light is incident first on the concave lens [Fig. 9.09]. The beam will appear to diverge from a point lv 20 cm behind concave lens, in the absence of convex lens. lx now behaves as a real object for convex lens of focal length 30 cm. Thus, for convex lens u. = - (20 + 8) = - 28 cm `therefore 1/v^(.) -1/(-28) =1/(+30) rArr 1/v^(.) = 1/30 + 1/(-28)` `rArr v. = -420 cm` Thus, the incident parallel beam will appear to diverge from a point I, 420 cm behind convex lens (or 416 cm from centre point O of the two lens system). Thus, it is clear that there is no fixed ANSWER for effective focal length of the combination. (b) Given that h = 1.5 cm and distance of object from convex lens (f = + 30 cm), u = - 40 cm  `therefore 1/v -1/(-40) = 1/30 rArr 1/v = 1/30 -1/40 = 1/120 rArr v= 120 cm` Hence, for concave lens `u. = +(120 -8) = +112 cm` and `f. = -20 cm` `therefore 1/v_(2) - 1/(+112) =1/(-20) rArr 1/v =1/112 -1/20 rArr v_(2) = -24.3` cm Thus, a real image is formed 24.3 cm behind concave lens. Hence, magmfication produced by convex lens`|m_(1)| = v/u = 120/40 = 3` and magnification produced by concave lens `|m_(2)| =(24.3)/112 = 0.22` `therefore` TOTAL magnification of image `|m| = |m_(1)| xx |m_(2)| = 3 xx (0.22) = 0.66` `therefore` Size of image `h. = mh = 0.66 xx 1.5 = 1` cm |
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| 777. |
An atmosphere |
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Answer» a)is a UNIT of pressure |
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| 778. |
The dominant contribution to current comes from holes in case of |
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Answer» metals |
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| 779. |
A uniform charge density of 500 nC//m^(3) is distributed throughout a spherical volume of radius 6.00cm .Consider a cubical Gussian surface with its center at the centre of the sphere. What is the elcetric flux through this cubical surface if its edge length is (a) 4.00cm and (b) 14.0 cm ? |
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Answer» |
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| 780. |
Three concentric spherical metal shells A, B, C of radii a, b, c (c>b> a) have surface charge density + sigma ,- sigma and + sigmarespectively. The potential of the middle shell is( sigma)/(epsi_0)times |
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Answer» `((a^2)/( B )-b +C)` |
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| 781. |
Two wires of same material having lengths and radii in the ratio of 3 : 4 and 3 : 2 respectively are connected in parallel with a potential source of 6V. the ratio of currents flowing through them I_(1) : I_(2)= ..... . |
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Answer» `1: 3 ` `R = (rho l)/(A) = (rho l)/(PI r^(2))` `therefore R prop (l)/(r^(2))` ` therefore (R_(1))/(R_(2)) = (l_(1))/(l_(2)) XX ((r_(2))/(r_(1)))^(2) ` `therefore (R_(2))/(R_(1)) = (l_(2))/(l_(1)) xx ((r_(1))/(r_(2)))^(2)"" `... (1) But V`I_(1) R_(1) = I_(2) R_(2) ` `therefore (I_(1))/(I_(2)) = (R_(2))/(R_(1))`....(2) From equation (1) and (2). `(I_(1))/(I_(2)) = (4)/(3) xx ((3)/(2))^(2) = (4)/(3) xx (9)/(4) = (3)/(1)` |
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| 782. |
In an experiment to determine the speed of sound using a resonance column |
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Answer» prongs of the tuning fork are kept in a verticle plane |
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| 783. |
Consider a transparent hemispher (n=2) in front of which a small object isplaced in air (n=1) as shown in figure. Q.Forwhich value of x, of the following , will the final image of the object O be virtual? |
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Answer» 2R Taking refraction first at curved surface, `(2)/(v_(1))+(1)/(x)=(1)/(R)RARR v_(1)=(2Rx)/(x-R)` For PLANE surface, `v^(')=v_(1)RrArrv^(')=(xR+R^(2))/(x-R)` `rArr (1)/(v)-(2(x-R))/(R(x+R))=0` ` (1)/(v)-(2(x-R))/(R(x+R))` For VIRTUAL image, `(1)/(v)lt0rArr (2(x-R))/(R(x+R))lt0` `x LT R` ii. For `x=2R` `V_(1)=(4R^(2))/(R)= 4RrArru=-2R` `m_(1)=(mu_(1))/(mu_(2)),(v)/(u)=(1)/(2), (4R)/((-2R))=-1` `m_(2)=1rArr m_(1)m_(2)=-1` Image is real, inverted, and of same size. (iii) Hence, correct answer is `90^(@)` 
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| 784. |
What other things the experiment established ? |
| Answer» SOLUTION :Nucleus contains +vely CHARGED protons, and CHARGELESS NEUTRONS. | |
| 785. |
Consider a transparent hemispher (n=2) in front of which a small object isplaced in air (n=1) as shown in figure. Q. Consider a ray starting from O which strikes the spherical surface at grazing incident (i=90^(@)). Takin x=R, what will be the angle (from normal) at which the ray may emerge from the plane surface. |
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Answer» `90^(@)` Taking refraction first at curved surface, `(2)/(v_(1))+(1)/(X)=(1)/(R)rArr v_(1)=(2Rx)/(x-R)` For plane surface, `v^(')=v_(1)RrArrv^(')=(xR+R^(2))/(x-R)` `rArr (1)/(v)-(2(x-R))/(R(x+R))=0` ` (1)/(v)-(2(x-R))/(R(x+R))` For virtual IMAGE, `(1)/(v)lt0rArr (2(x-R))/(R(x+R))lt0` `x lt R` II. For `x=2R` `V_(1)=(4R^(2))/(R)= 4RrArru=-2R` `m_(1)=(mu_(1))/(mu_(2)),(v)/(u)=(1)/(2), (4R)/((-2R))=-1` `m_(2)=1rArr m_(1)m_(2)=-1` Image is real, inverted, and of same size. (iii) Hence, correct answer is `90^(@)` 
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| 786. |
Which of the following graphs correctly represents the variation of beta=-(dV//dP)//V with P for an ideal gas at constant temperature? |
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Answer»
On differentiation, we get `PdV+VdP=0` or `(dV)/(dP)=-(V)/(P)or-(dV)/(VdP)=(1)/(P)orbeta=(1)/(P)orbetaP=1` The EQUATION between `betaandP` is of the form XY = constant which represents a rectangular hyperbola. THUS, the GRAPH between `betaandP` will be a rectangular hyperbola and it is represented by (a). |
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| 787. |
A system of coordinate axes which defines the positive of a paritcle in space is called "_____________". |
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Answer» base |
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| 788. |
State Tangent law. |
| Answer» Solution :If a magnet is ACTED UPON two mutually perpendicular forces B and H, then `B = H tantheta`, where `theta` is the ANGLE between B and H. | |
| 789. |
Consider a transparent hemispher (n=2) in front of which a small object isplaced in air (n=1) as shown in figure. Q. What is the nature of final image of the object when x=2R? |
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Answer» Erect and magnified TAKING refraction first at curved surface, `(2)/(v_(1))+(1)/(x)=(1)/(R)rArr v_(1)=(2Rx)/(x-R)` For plane surface, `v^(')=v_(1)RrArrv^(')=(xR+R^(2))/(x-R)` `rArr (1)/(v)-(2(x-R))/(R(x+R))=0` ` (1)/(v)-(2(x-R))/(R(x+R))` For virtual image, `(1)/(v)lt0rArr (2(x-R))/(R(x+R))lt0` `x LT R` ii. For `x=2R` `V_(1)=(4R^(2))/(R)= 4RrArru=-2R` `m_(1)=(mu_(1))/(mu_(2)),(v)/(u)=(1)/(2), (4R)/((-2R))=-1` `m_(2)=1rArr m_(1)m_(2)=-1` Image is real, inverted, and of same size. (iii) Hence, correct answer is `90^(@)` 
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| 790. |
A double convex lens is placed on a horizontal plane mirror. A pin held horizontally above the lens coincides with its own image when it is 18cm from the lens. The space between the lens and the mirror is filled with glycerine and water, turn by turn, and the positions of coincidence of the pin with the image are 28 cm and 24 cm from the lens, respectively. Calculate the refractive index of glycerine, given that the refractive index of water is 4/3. |
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Answer» |
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| 791. |
Assertion: When two protons are brought closer to each other against their electrostatic repulsion then their combined mass becomes slightly more than their total mass. Reason: If we bring two similar charges closer to each other from infinity then work is needed to be done against electrostatic repulsion. Mechanical work against electrostatic repulsion gets stored in the system as electric energy. We know that E=mc^2, hence due to an increase in energy of the system, equivalent mass of the system increases. |
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Answer» If both ASSERTION and reason are CORRECT and reason is a correct explanation of the assertion . |
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| 792. |
The power is transmitted from a power house on high voltage ac because |
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Answer» Electric current travels faster at higher volts |
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| 793. |
A falvanometer having a coil resistance of 100 ohms gives a full scale deflection when a current of one milli-ampere is passed through it. What is the value of resistance which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 amperes ? A resistance of the required value is available but it will get burnt if the energy dissipated in it is greater then one watt. Can it be used for the above described conversion of the galvanometer ? When this modified galvanometr is connected across the terminals of battery, it shows a current 4 amp. The current drops to 1 amp., When the resistance os 1.5 ohm is conneted in series with modified galvanomter. Find the emf and internal resistance of battery. |
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Answer» |
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| 794. |
An aluminium rod (Y = 7xx10^9 N/m^2) has a breaking strain of 0.2%. What is minimum cross sectional area to support a load of 10^4 N ? |
| Answer» Solution :`Y = (M g L)/(A Delta l)` there for `A = (M g L)/(Y Delta l)` But `(Deltal)/L` = `0.2 XX 10^-2` = Breaking STRAIN. There FORA = `(M g L)/(Y Delta l)` =`(10^4)/ (7 xx 10^9 xx 2 xx 10^-3)` `10^-2/ 14` = `7.1 xx 10^-4m^2` | |
| 795. |
A pointcharge + mu c is a distance 5 cm directly above the centre of a square of side 10 cm aswhat is the magnitudeof the electricflux throughthe square |
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Answer» SOLUTION :(a) 0.07 `MUC` (B) no only thatthe net CHARGES INSIDE is zero |
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| 796. |
A small bulb is placed at the bottom of a tankcontaining water to a depth of 80 cm. What isthe area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) |
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Answer» Solution :`d_(1) = 80 cm = 0.8 m` Refractive index of WATER `,mu = 1.33` The given situation is shown in the figure. Where, I = Angle of INCIDENCE r = Angle of refraction ` = 90^(@)` Since the bulb is a point source, the emergent light can be considered as a CIRCLE of radius `R = (AC)/(2) = AO = OB` USING snell.s LAW, the for the refractive index of water is `mu = (sin r)/(sin i)` `sin i = (sin 90^(@))/(1.33) = 0.75` `i sin^(-1) (0.75) = 48.75^(@)` Using the given figure, we have the relation `tan i = (OC)/(OB) = (R)/(d_(1))` `R = tan i xx d_(1) = tan 48.75^(@) = 0.8` R = 0.91 m Area of the surface of water `= piR^(2)` `=3.14 xx (0.91)^(2) = 2.16m^(2)` Hence, the area of the surface of waterthrough which the light from the bulb can emerge is approximately `2.16 m^(2)` 
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| 797. |
There is a hollow cylinder of radius R and it is rotating with constant angular speed omega. There is a point mass which rotates along with the cylinder and geta carried upward. Friction is enough such that the point mass does not slip with respect to the cylinder as long as the normal force is becomes zero. If the path of the particle after it lost contact with the cylinder is as shown in the figure then, In Order to happen this where most the particle lose contact with the cylinder? |
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Answer» `h=R/2`  `v^(2)=Rgsintheta`……(2) From (1) and (2) `tantheta=1` |
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| 798. |
There is a hollow cylinder of radius R and it is rotating with constant angular speed omega. There is a point mass which rotates along with the cylinder and geta carried upward. Friction is enough such that the point mass does not slip with respect to the cylinder as long as the normal force is becomes zero. If the path of the particle after it lost contact with the cylinder is as shown in the figure then, What is the value of omega (in terms of g and R) |
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Answer» `sqrt(g/(sqrt(3)R))`  `V^(2)=Rgsintheta`……(2) From (1) and (2) `tantheta=1` |
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| 799. |
An 1800 W toaster, a 1.3 kW electric fan and a 100 W lamp are plugged in the same 120 V circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. the sum of current drawn by the three devices) through the circuit ? |
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Answer» 40 A |
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| 800. |
Assume the graph of sepecific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following. |
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Answer» FUSION of two nuclei of mass number LYING in the RANGE of `100 lt A lt 200` will release energy |
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