Saved Bookmarks
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.
| 38701. |
An arbitrary surface encloses a dipole. What is the electric flux through this surface ? |
|
Answer» Solution :According to Gauss.s law, the FLUX ASSOCIATED with close surface is `phi =(sumq)/epsilon_(0)` where Q is charge enclosed by closed surface. NET charge on dipole `=-q + q=0` `therefore` Flux associated with closed surface enclosing dipole, `phi = (-q+q)/epsilon_(0)=0` |
|
| 38702. |
A prism of refracting angle 60^(@) makes an angular deviation of 40^(@) between two rays having a difference of 11^(@) in their angles of incidence. Find the refractive index of the material of the prism. |
|
Answer» |
|
| 38703. |
Which of the glass and ruber is more elastic and why ? |
| Answer» Solution :GLASS is more elastic than rubber because for given APPLIED force units AERA , the STRAIN produced in glass is MUCH smaller than produced in rubber. | |
| 38704. |
In the previous question, if the second car is overtaking at a relative speed of 15 ms^(-1), how fast will the image be moving ? |
|
Answer» `1.2ms^(-1)` `(1)/(u) + (1)/(v) = (1)/(f) : (-1)/(u^(2))(du)/(dt) - (1)/(v^(2)) (dv)/(dt) = 0` `THEREFORE (dv)/(dt)=(v^(2))/(u^(2))(du)/(dt)=((19.4)/(600))^(2)xx15` |
|
| 38705. |
Assertion (A) : If a proton and an electron are placed in the same uniform electric field ,they experience force of equal magnitude. Reason ® :Electric force on a test charges is independent of its mass. |
|
Answer» |
|
| 38706. |
A ray of light is incident normally on one of the face of a prism of angle 30^@ and refractive index sqrt2. The angle of deviation will be |
| Answer» ANSWER :D | |
| 38707. |
What is the current I shown in the given circuit? |
|
Answer» `(V )/(2R)`  In loop4 weget ` R_1=(2 R xx2R )/(2R+ 2R) = R ` In loop3, weget` R_2= ((R+ R_1)xx 2 R)/(2 R+(R+R_1))=R ` In loop2 we get`R_3= ((R +R_2)xx 2 R)/( 2 R+(R + R_2))=R` In loop1, weget ` R_4= R +R_3= R+ R =2R ` ` thereforeI =(I_1)/( 8 )= (V //2 R) /(8 )= (V )/( 16 R )` |
|
| 38708. |
The following figure shows a block of mass m suspended from a fixed point by means of a vertical spring. The block is oscillatting simple harmonically and carries a charge q. There also exists a uniform electric field in the region. Consider four different cases. The electric field is zero in case 1, mg//q downward in case 2, mg//q upward in case 3, and 2mg//q downward in case 4. The speed at mean position is same in all cases. Select the correct alternative (s). |
|
Answer» Time PERIODS of OSCILLATION are equal in case 1 and case 3. (a) Only equilibrium position changes. Time period remains same `T=2pisqrt(m//K)`. (b) `(1)/(2)kx^(2)=(1)/(2)mv^(2)` So `v` at mean position is same amplitude will be same. (c) In case `4` equilibrium psotion is `x_(0)=3mg//k`. |
|
| 38709. |
In the above question what is the amplitude of the oscillating magnetic field, |
| Answer» Solution :`B_(0)=(E_(0))/(C)=(48)/(3XX10^(10))=1.6xx10^(2)m` | |
| 38710. |
Use Huygens's principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light. When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band ? |
Answer» Solution :Let AB be a slit of width 'a' and a parallel beam of monochromatic light is incident on it. According to Huygen's principle the diffraction pattern is the result of superposition of a large NUMBER of waves, starting from different points of illuminated slit.  At the central point C of the screen, the angle `THETA` is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C. The observation point in now taken at P. Minima : Now we divide the slit into two equal halves AO and OB, each of width `(a)/(2)`. Now for every point, `M_(1)` in AO, there is a corresponding point `M_(2)` in OB, such that `M_(1) M_(2)=(a)/(2)`, Then path difference between waves arriving at P and starting from `M_(1) " and " M_(2)` will be `(a)/(2) sin theta =(lamda)/(2)` `a sin theta = lamda` In general, `a sin theta = n lamda` Secondary Maxima : Similarly, it can be shown that for maxima `a sin theta =(n+(1)/(2))lamda` The intensity pattern on the screen is shown in the GIVEN figure.  When the width of the slit is made double the ORIGINAL width, the size reduces by half according to the RELATION : size `- lamda // d`. Intensity increases four fold. |
|
| 38711. |
__________ waves cause tanning of the skin. |
| Answer» SOLUTION :ULTRAVIOLET | |
| 38712. |
What is presbyopia? |
| Answer» Solution :This DEFECT is SIMILAR to hypermetropia i.e.,a PERSON having this defect cannot see nearby objects DISTINCTLY, but can see distant objects without any difficulty. This defect occurs in elderly persons (AGED persons). | |
| 38713. |
A circular disk of radius 2.0 m rotates, starting from rest, with a constant angular acceleration of 20.0rad//s^(2). What is the tangential acceleration of a point on the edge of the disk at the instant that its angular speed is 1.0 rev/s? |
|
Answer» `40m//s^(2)` |
|
| 38714. |
What does Tommy found? |
|
Answer» A REAL book |
|
| 38715. |
A certain substance has a half life of 28 hours. Calculate its means life (tau). |
|
Answer» |
|
| 38716. |
Explain mobility of conductor and derive equation of mobility. |
|
Answer» Solution :`rArr ` Conductivity of material due to mobile charge carrier insider the conductor. `rArr` In metals electrons are mobile charge carrier. `rArr`In electrolyte positive and negative ION are charge carrier. `rArr`In ionized GAS electron and positive ion are charge carrier. `rArr` In SEMICONDUCTOR electron and hole are charge carrier. `rArr`Mobility : Drift velocity per unit electric field is called mobility . `therefore mu = (|v_(d)|)/(E) ` mks unit `(ms^(-1))/(Vm^(-1)) = m^(2) V^(-1) s^(-1)` CGS unit = `cm^(2) V^(-1) S^(-1)` Dimensional formula = `|M^(-1) L^(0) T^(2) A^(1)|` `(((m)/(s)))/((N)/(C)) = (m)/(s) xx (C)/(N)` `(m)/(s) (C)/([kg ""(m)/(s^(2))]) = (CS)/(kg)` = C.s. `kg^(-1)` |
|
| 38717. |
Whenthe magnetic inclination (dip)was measured at various places on earth, inone of thefollowing countriesit wasfound to be zero. Whichto be zero. Which one was it ? |
|
Answer» |
|
| 38718. |
A square loop of side 10cm and resistance 0.5Omegais placed vertically in east-west plane. A uniform magnetic field of 0.1 tesla is setup across the plane along northeast direction. The magnetic field is decreased to zero in 0.7 seconds. They induced current is |
|
Answer» 1 mA |
|
| 38719. |
Shown in the figure below a meter bridge set up with null deflection in the galvanometer. The value of the unknown resistance R is: |
|
Answer» `55Omega` |
|
| 38720. |
A student performed the experiment of determination of focal length of a convex mirror by u-v method length of the mirror used is 24cm. The maximum using an optical beneath of length 1.5m . The focal error in the location of the image can be 0.2cm. The 5 sets of (u,v) values recorded by the student (in cm) are (42,56), (48,48), (60,40), (66,33), (78,39). The data set (s) that cannot come from experiment and is (are) incorrectly recorded, is (are): |
|
Answer» `(42,56)` |
|
| 38721. |
A plane electromagentic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time , E=6.3 hatj V/m . What is B at the point ? |
|
Answer» Solution :The MAGNITUDE of `hatB` is `B=(E )/(C )=(6.3 V//m)/(3xx10^(8)m//s)=2.1 xx10^(-8)T` To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x-and y-axes. USING vector algebra, E x B should be along x-direction. Since, `(+hatj) XX (+HATK) = hati B` is along the z- direction. Thus, `B = 2.1xx 10^(-8) hatk T` |
|
| 38722. |
In an experiment to determine the focal length (f) of a concave mirror by the u - v method, a student places the object pin A on the principal axis at a distance x from the pole P. The students looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shiftshis/her eye towards left, the image appears to the right of the object pin. Then, |
|
Answer» `x lt F` |
|
| 38723. |
An experimental setup of verification of photoelectric effect is shown in the diagram. The voltage across the electrode is measured with the help of an idealvoltmetar, and which can be varied by moving jockey 'J' on the potentiometer wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is 2omega. The resistance of 100 cm long potentiometer wire is 8 omega. The photo current is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm^(2) at separation 0.5 mm are used in the vacuum tube. Photo current in the circuit is very small so we can treat potentiometer circuit an indepdent circuit. The wavelength of various colours is as follows : |{:("Light",underset("Violet")(1),underset("Blue")(2),underset("Green")(3),underset("Yellow")(4),underset("Orange")(5),underset("Red")(6)),(lambda "in" Årarr,4000-4500,4500-5000,5000-5500,5500-6000,6000-6500,6500-7000):}| When other light falls on the anode plate the ammeter reading remains zero till, jockey till, jockey is moved from the end P to the middle point on the wire PQ. Thereafter the deflection is recorded in the ammeter. The maximum kinetic energy of the emitted electron is : |
|
Answer» Solution :`:'` Stopping POTENTIAL `V_(S)=8V` and `KE=eV_(S)` `:. KE=8eV` |
|
| 38724. |
An experimental setup of verification of photoelectric effect is shown in the diagram. The voltage across the electrode is measured with the help of an idealvoltmetar, and which can be varied by moving jockey 'J' on the potentiometer wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is 2omega. The resistance of 100 cm long potentiometer wire is 8 omega. The photo current is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm^(2) at separation 0.5 mm are used in the vacuum tube. Photo current in the circuit is very small so we can treat potentiometer circuit an indepdent circuit. The wavelength of various colours is as follows : |{:("Light",underset("Violet")(1),underset("Blue")(2),underset("Green")(3),underset("Yellow")(4),underset("Orange")(5),underset("Red")(6)),(lambda "in" Årarr,4000-4500,4500-5000,5000-5500,5500-6000,6000-6500,6500-7000):}| When radiation falls on the cathode plate a current of 2muA is recorded in the ammeter. Assuming that the vecuum tube setup followsohm's law, the equivalent resistance of vacuum tube operating in this case when jockey is at end P. |
|
Answer» `8 xx 10^(8) omega` `R=V/I=(16 V)/(2XX10^(-6) A)=8 xx10^(6) Omega` |
|
| 38725. |
A closely wound solenoidof 2000 turns and area of cross- section 1.6 xx 10^(-4) m^2, carrrying a current 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid ? (b) What is the force and torque on teh solenoid if a uniform horizontal magnetic field . 7.5 xx 10^(-2) T is set up at an angle of 30^@ with the axis of solenoid ? |
| Answer» Solution :(a) 1.28 A `m^2 , `(b) ZERO, `4.8 xx 10^(-2) NM` | |
| 38726. |
In the given circuit, all ammeters and voltmeters are ideal. Find the(a) reading of all meters. . (b) same if (A_(1) and (V_(1) are interchanged and(A_(4) and (V_(2) |
|
Answer» Solution :`A_(1)=A_(2)=2A,A_(3)=A_(4)=0.5A,V_(1)=2V,V_(2)=(20)/(6)` `A_(1)=(7)/(6)A,A_(2)=0V_(1)=(25)/(6)V,A_(3)=0,A_(4)=(7)/(6)A,A_(2)=0,V_(2)=(20)/(6)` |
|
| 38727. |
A narrow slit ofwidth 2 mm is illuminated by monochromatic light of wavelength 500 mm. The distance between the first minima on either side on a screen at a distance of 1 m is |
|
Answer» 5mm |
|
| 38728. |
To small equally charged spheres each of mass m are suspended from the same point by silk threads of length L. The distance between the spheres x lt lt L. Find the rate (dq)/(dt) with which he charges leak off each sphere if their approach velocity varies as vartheta =(a)/(sqrt(x)) where a si a constant. |
|
Answer» Solution :For momentaryequilibrium of the spheres, `T COS THETA=mg and T SIN theta = F` where Tis tensionin the STRING `rArr tan theta =(F)/(mg)` For `x lt lt L`,  we can take `tan theta =(x)/(2L)` So `F=(mg x)/(2L)` Here `F=(1)/(4pi in_(0)) (q^(2))/(x^(2)) and q^(2)=(2pi in_(0) mg x^(3))/(L)` given `(dx)/(dt) =(a) /(sqrt(x))` where a is a constant `rArr (dq)/(dt)=(3q)/(2)sqrt((2pi in_(0) mg)/(L))` |
|
| 38729. |
The time period of revolution of a charge q_1 and of mass m moving in a circular path of radius due to Coulomb force of attraction with another charge q_2, at its centre is |
|
Answer» a) `SQRT((16 pi epsi_(0) MR^(3))/(q_(1)q_(2)))` |
|
| 38730. |
A person throws ball with velocity from the top of a building in vertically upward direction the ball reaches the ground with a speed of 3v then the height of the building is : |
|
Answer» `(4v^(2))/(g)` `x=(v^(2))/(2g)` and `h+x=(3v)^(2)/(2g)` `:. H=(9v^(2))/(2g)-(v^(2))/(2g)=(4v^(2))/(g)` |
|
| 38731. |
3 identical capacitors are joined in parallel and are charged with a battery of 10 V. Now the batteryis removed and they are joined in series with each other in this condition what would be the potential difference between the freed plates in the combination? |
|
Answer» 30 V `:. V = (10 + 10 +10) V` `:. V = 30 V ` |
|
| 38732. |
When an open pipe vibrating for it's third overtone, the no. of nodes formed is |
|
Answer» four |
|
| 38733. |
Answer the question regarding earth's magnetism: If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground? |
| Answer» Solution :(c) Geographically, MELBOURNE is LOCATED near magnetic NORTH pole of EARTH (which lies in the southern hemisphere of globe of Earth) and so field lines of magnetic field of Earth come out of ground. | |
| 38734. |
A convex lens of focal length 10 cm. is used as a magnifying lens. Where should the object be placed if the image is to be 30 cm from the lens? |
Answer» Solution : In CASE of magnifying lens, the lens is CONVERGENT and the image is erect, ENLARGED, virtual, between infinity and object and on the same side of lens as shown in FIGURE So here f = 10 cm and v= -30 cm and hence from lens-formula, `(1)/(v )-(1)/(u ) =(1)/(f) ` we have ` (1)/(-30)- (1)/(u)= (1)/(10)` `i.e., u=-7.5cm` So the object must be placed in front of lens at a DISTANCE of 7.5 cm (which is |
|
| 38735. |
A : The resolving power of a telescope more if the diameter of the objective lens if more. R : Objective lens of large diameter collects more light. |
|
Answer» Both A and R are true and R is the CORRECT EXPLANATION of A |
|
| 38736. |
Which of the following is v-T graph for spectral distribution of radiance in a perfectly black body? (Where v is the frequency corresponding to maximum wavelength.) |
|
Answer»
|
|
| 38737. |
In an ideal refrigerator, heat from inside at 280 K is transferred to a room at 300 K. What is the amount of heat (in joule) which will be transferred to the room for each joule of electrical energy consumed in the process |
|
Answer» 12 J |
|
| 38738. |
Unit of omegaC is ____ |
|
Answer» Solution :We know : `X_C=1/(omegaC)` `therefore omegaC=1/X_C` `rArr` Unit of `omegaC=1/Omega=Omega^(-1)` Unit of `omegaC`=unit of f X unit of C `=S^(-1) xx F = S^(-1) xx C/V=S^(-1) xx (AS)/V` `=A/V` |
|
| 38739. |
The ratio of SI Unit of CGS unit of Planck's constant is |
|
Answer» `10^(7)` |
|
| 38740. |
The radius of curvature of a concave mirror is 2 cm and the real image is magnified by 1.5 times. The object distance is |
|
Answer» 20 cm |
|
| 38741. |
A parallel plate capacitor stores a charge Q at a voltage V . Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change ? |
| Answer» Answer :D | |
| 38742. |
A coil of N turns and radius R carries I . It is unwound and rewound to make a square coil of side 'a' having same number of turns N. Keeping the curretn I same, find the ratio of the magnetic moments of the square coil and the circular coil. |
|
Answer» Solution :Magnetic MOMENT of circular coil of N TURNS and radius R carrying a current I `m_1 = N A_1 I = N pi R^2 I"".........(i) ` When the coil is unwound and is rewound to make another square coil of side .a. and number of turns N, as total LENGTH of wire remains constant , we have `N . 2 pi R = N 4 a implies a= (pi R)/(2)` `:.` Magnetic moment of square coil for SAMECURRENT I `m_2 = NA_2 I = N a^2 I = N ((pi R^2)/(2)) I = N(pi^2 R^2)/(4) I"" .......(ii)` `implies (m_2)/(m_1) = pi/4`. |
|
| 38743. |
Water is boiling in a kettle on an electric hot-plate of 800 W power. Find the steam outflow velocity, if the cross section of the spout is 0.9 cm and the pressure at the output is normal. The officiency of the hot plate is 72%. |
|
Answer» Solution :The outflow VELOCITY is `v=mu//rho S`, where `mu` is the amount of water that evaporates per SECOND. Obviously, `mu=(Q)/(Lt)=(etaP)/(L),` WHEREP is the power of the hot-plate and `eta` is its efficiency. HENCE `v=etaP// rhoLS` |
|
| 38744. |
A thermocouple circuit consists of two thermal junctions and a low-resistance galvanometer , all in series. The galvanmeter has a resistance of 8 Omega and the rest of the circuit has a resistance of 1.6 Omega . The temperature develops an emf of 10 microvolt per degree Celsius difference of temperature between the two junction. When one junction is kept is 0^(@)C and the other in a molten metal, the galvanmeter reads 8 millivolt. What is the temperature of the molten metal? Assume that the emf varies linearly with the temperature difference. |
| Answer» | |
| 38745. |
An audio signal given by e_(s)=15sin 2pi(2000t) modulates a carrier wave given by e_(s)=60sin 2pi (100,000t). If calculate (a) Percentage modulation (b) Frequency spectrum of the modulated wave. |
|
Answer» Solution :(a) Signal AMPLITUDE, `B=15` Carrier amplitude, `A=60""m=B/A=(15)/(60)=0.25` `:.` Percentage modulation `=0.25xx100=25%` (b) By comparing the given equations of signal and carrier with their standard form `e_(s)=E_(s)sin omega_(s)t=E_(s)sin 2pif_(s)t` and `e_(C)=E_(c)sin omega_(c)t=E_(c)sin2pi f_(c)t` we have signal frequency `f_(s)=2000Hz` and carrier frequency `f_(c)=100,000` Hz The frequencies present in modulated wave (i) `f_(c)=100,000` Hz = 100 kHz (ii) `f_(c)-f_(s)=100,000-2000=98` kHz (iii)`f_(c)+f_(s)=100` kHz + 2kHz = 102 kHz THEREFORE, frequency spectrum of modulated wave extends from 98kHz to 102 kHz is called band width. |
|
| 38746. |
A count rate meter is used to measure the activity of a given sample. At one instant meter shows 2400 counts/min. One hour later the count drops to 300 counts per minute. What is half life of the sample ? |
|
Answer» 5 mintues `N/N_(0)=A/A_(0)=(1/2)^(t/T)," A=300/min and"` `A_(0)="2400/min"` `(300)/(2400)=(1/2)^(t/T) rArr 1/8=(1/2)^(t/T)` `rArr (1/2)^(3)=(1/2)^(t/T)` This gives `t/T=3 or T =t/3 =(1HR)/(3)=20"min"` |
|
| 38747. |
Light travels through a glass slab of thickness t and refractive index mu. If c is the velocity of light, then the time taken by light to travel the thickness of glass slab is : |
|
Answer» `(tc)/mu` |
|
| 38748. |
A photon of energy .E. ejects a photoelectron from a metal surface whose work functions is W. If this electron enters into a uniform magnetic field of induction .B. in a direction perpendicular to the field and describes a circular path of radius .r., then the radius .r. is given by, (in the usual notation) |
|
Answer» `(SQRT(2M(E-W_(0))))/(eB)` |
|
| 38749. |
The intermediate product 'X' of following synthesis is identified as: |
|
Answer»
|
|
| 38750. |
In an PNP transistor , 10^(10) holes enter the emitter in 10^(-6) s . If 2 % of holes is lost in the base , then the current amplification factor is |
|
Answer» a. 49 |
|
