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.
| 38551. |
You are to conduct a series of trials. For each trial the inclination of the plane is set to an angle theta, ranging from 0^(@)to 90^(@) and an object is released from rest at the top of the stationary inclined plane. The coefficient of static friction between the object and the inclined plane is mu_(s). In each case below, predict the observed outcome for the trial. In the following equation 'tumble' means 'tip' over and rotate' and 'sliding' means NO tumbling Case1: The object is a sphere and mu_(s)=0 |
|
Answer» The sphere will roll WITHOUT slipping for SMALL `THETA` and slide down only for `theta` greater than a certain non-zero value |
|
| 38552. |
Assertion (A) : In a hydrogen atom the radius of nth orbit is directly proportional to the quantum number n. Reason (R) : The speed of electron in nth orbitis inversely proportional to the quantum number. |
|
Answer» If both ASSERTION and reason are true and the reason is the correct explanation of the assertion. |
|
| 38553. |
Statement A : current is scalar Statement B : current element is vector |
|
Answer» A and B are true |
|
| 38554. |
To avoid bends, tanks used by scuba divers are filled with air diluted with ……………. |
|
Answer» Argon |
|
| 38555. |
Two metal spheres, each of radius 3.0 cm, have a center to center separation of 2.0 m. Sphere 1 has charge 1.0 xx 10^(-8) C, sphere 2 has charge -3.0 xx 10^(-8) C. Assume ,that the separation is large enough for us to assume that the chargeon each sphere is uniformly distributed (the spheres do not affect each other) . With V = 0 at infinity, calculate (a) the potential at the point halfway between the centers and the potential on the surface of (b) sphere 1 and ( c) sphere 2. |
| Answer» SOLUTION :(a) `-1.8 XX 10^(2)` V , (b) `2.9 K V`, (C ) `-8.9 k V` | |
| 38556. |
A uniform magnetic field of 3000G is established along the positive z-direction. A rect angular loop of sides 10 cm and 5 cm carries a current of 12A. What is the torque on the loop in different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium? |
|
Answer» Solution :`tau = 1 ( vec A x vec B)` In fig. a.`1.8 xx 10^(-2)` Nm along -y DIRECTION In figa. B. same as in (a) In fig. C `1.8 xx 10^(-2)` Nm along -=x direction. In fig d.`1.8 xx 10^(-2)` Nm t an angle of `240^@` along the x diarection. In e and f the TORQUE is zero. FORCE is zero in each case. Fig e. corresponds to stable equilibrium . |
|
| 38557. |
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 xx 10^(26) W. The number of photons received by the human eye per second on the average from sunlight is of the order of……………… |
|
Answer» `10^(45)` `therefore""(n)/(t) = (P lambda)/(hc) = (3.8 XX 10^(26) xx 550 xx 10^(-9))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 1 xx 10^(45)` |
|
| 38558. |
A steady current flows in a metallic conductor of non-uniform cross-section. The quantity(s)remaining constant along the length of conductor is/are |
|
Answer» current, electric field and DRIFT VELOCITY. |
|
| 38559. |
Two plane mirrors are inclined at an angle of 60^@.An object is placed between the mirrors. The total number of image formed by the two mirrors is : |
|
Answer» 4 |
|
| 38560. |
If when two equal non parallel forces acting at a point the square of the resultant is 3 times the product of forces, than the angle between the forces is |
| Answer» ANSWER :A | |
| 38561. |
Calculate the ratio of electric and gravitational force between two electrons. |
|
Answer» |
|
| 38562. |
Assertion: A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric field. Reason:When X-ray beam falls on the ball, it emits photoelectrons and metal becomes negatively charged. |
|
Answer» Both Assertion and REASON are TRUE and Reason is the correct EXPLANATION of Assertion |
|
| 38563. |
When a paramagnetic substance is brought near to north or south pole of a bar magnet, then it ....... |
|
Answer» experiences repulsion |
|
| 38564. |
The star recedes from the earth at the speed a 100 kms^(-1) Find the doppler shift of spectrur line 3xx 10^(8)ms^(-1) ms having wavelength 5700Å |
|
Answer» `0.63Å` `(Deltaf)/(FS)=(v_(s))/(c) ....(1) and f=(c)/(lambda)` Taking derivative w.r.t. `lambda` PUTTING `Deltaf=(-c)/(lambda^(2))Delta and f_(s)=(c)/(lambda)` `:.` From equation (1) `(c)/(lambda^(2)xx(c)/(lambda)) delta lambda=(v_(s))/(c)` `:. delta lambda=(v_(s))/(c)xxlambda` `=(100xx10^(3)xx 5.7xx10^(-7))/(3XX10^(8))` `=1.90Å` |
|
| 38565. |
Consider tuning a radio to higher frequency station. The energy of radio photons detected |
|
Answer» Is ZERO because radio WAVES do not carry energy |
|
| 38566. |
A cathode of a photo electric cell is changed such that the work function changes from W_(1) to W_(2)(W_(1)ltW_(2)) if the current before and after changes are I_(1) and I_(2) all other conditions remaining unchanged, then (assuming hVgtW_(2) ) |
|
Answer» `I_(1)=I_(2)` |
|
| 38567. |
Can we apply Coulomb's law to any type of charge distribution. |
| Answer» SOLUTION :No, CHARGE MUST be stationary and point SIZE. | |
| 38568. |
For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of prism is |
|
Answer» `30^(@)` |
|
| 38569. |
A parallel plate capacitor is charged as shown (Q is given). A metal slab with the total charge +Q is placed inside the capacitor as shown. The thickness of the slab is d. The distance between the top plate and the top of the slab is 2d, and the distance between the bottom plate and the botthom of the slab is d. Each plate is grounded through a galvanometer as shown . Find the charge that passes through each galvanometer after both switches are closed simultaneously. |
|
Answer» A total charge of `-Q//3` flows `(through G_1)` from the bottom PLATE to ground. Initial charges on each surface  Chare on OUTER surfaces `q_(a)=q_(b)=(Q+Q-Q)/(2)=(Q)/(2)` and `q_(b)=(Q-(Q-Q))/(2)=(Q)/(2)` `C_(1)=C=(epsilon_(0)A)/(2d)` or `C_(2)=2C=(epsilon_(0)A)/(d)`  When both switches are CLOSED the circuit can be redrawn as SHOWN. Moving from `A` and `B` `Q+(x)/(C)-((Q0x))/(2C)=O` or `x=(Q)/(3)`  Final charges on each plate will be Hence charge moving from top plate to ground is `Q-(-(Q)/(3))` `=(4Q)/(3)` and from bottom plate to gound is `-Q-(-(2Q)/(3))=(Q)/(3)` |
|
| 38570. |
A Solid contains 5 xx10^(21) number of atoms. If an electron is removed from each of 0.01% of number of atoms, find the charge gained by this solid ? |
|
Answer» Solution :The solid gains positive charge hence Q = NE Where `N = (5 XX 10^(21) xx (0.01)/(100)) = 5 xx 10^(17)` `:. Q= 5 xx 10^(17) xx 1.6 xx 10^(-19) :. Q = +0.08 C`. |
|
| 38571. |
In a transistorconnected in the common base configuartion, alpha=0.95,I_E=1mA.Calculate the value of I_C and I_B. |
|
Answer» SOLUTION :`ALPHA=(I_C)/(I_E)` `I_C=alphaI_E=0.95xxI=0.95mA` `I_E=I_B+I_C` `:. I_B=I_C-I_E=1-0.95=0.05mA` |
|
| 38572. |
Kirchhoff's first and second law are based ...... . |
|
Answer» CONSERVATION of momentum and conservation of ELECTRIC CHARGE |
|
| 38573. |
State Lenz's law. Two identical loops, one of copper and the other of aluminium, are rotated with the same speed in a uniform magnetic field acting normal to the plane of the loops. State with reason, for which of the coils (i) induced emf, (ii) induced current will be more. |
|
Answer» Solution :For statement of Lenz.s law, see Point Number 8 under the HEADING "Chapter At A Glance". (ii) When two identical loops (i.e., having same area) are rotated with same speed in a given magnetic field acting normally, the magnitudes of induced emf in both loops will be EXACTLY do same because rate of change of flux is same for both and `varepsilon = - (dphi)/dt.` (ii) Induced current `I = varepsilon/R,` where R is the resistance of the loop. As copper is a BETTER CONDUCTOR than aluminium, resistance offered by the copper loop is less and CONSEQUENTLY, induced current in copper loop will be more than for aluminium loop. |
|
| 38574. |
A frame is formed by nine identical wires of resistances R each as shown in the figure. Choose the incorrect statement. |
|
Answer» the ratio of current through CF and DE is 1 |
|
| 38575. |
A point charge of +10 mu C is placed at a distance of 20 cm from another identical point charge of +10 muC. A point charge of -2mu C is moved from point a to b as shown in the figure . Calculate the change in potential energy of the system ? Interpret your result . |
Answer» Solution : `q_(1)= 10muC= 10xx10^(-6)C` `q_(2) =- 2 MUC = -2 xx10^(-6)C ` DISTANCE ,`r = 5cm = 5xx10^(-2)` m Change in potential energy `DeltaU = (9xx10^(9) xx10^(2)xx10^(6)XX(-2xx10^(-6)))/(5xx10^(-2)) ` `=-36xx10^(9)xx10^(-12)xx10^(2)= -36xx10^(-1)` `DeltaU = 3.6 J` Negative sign implies that to move the charge `-2muC` no external work is required. SYSTEM spends its STORED energy to move the charge from point A to point < `DeltaU=-3.6 J` negtive sign implies that to move the charge `-2muC` no external work is required system spend its stored energy to movethe charge from poitnt a to point b. |
|
| 38576. |
If linear velocity of a body moving on the circumference of a circle be equal to the velocity acquired by a freely falling body by a distance equal to half of the radius of the circle, show that centrepetal acceleration is equal to the acceleration due to gravity. |
|
Answer» SOLUTION :The magnitude of the velocity ACQUIRED by a FREELY falling BODY through distance (r/2) is V= `sqrt(2gxxr/2)=sqrt(gr)` The centripetal acceleration is `v^2/r=gr/r=g` |
|
| 38577. |
In which of the following cases, will the charged particles not experience any electromagnetic force? |
|
Answer» Charge is at rest inside the magnetic field |
|
| 38578. |
Assertion : Binding energy increase with increases atomic mass number. Reason : Density of nucleusincrease with increases in atomicmass number . |
|
Answer» |
|
| 38579. |
If tube length of astronomical telescope is 96 cm and magnification is 15, then focal length of objective is ...... cm. |
|
Answer» 100 `therefore f_e=(f_0)/(m)` and `L=f_0+f_e` `96=f_0+(f_0)/(m)=f_0[1+(1)/(15)]=f_0xx(16)/(15)` `therefore f_0 =(96xx1.5)/(16)=90` cm |
|
| 38580. |
In L-C capacitor oscillator at………. Time energy in capacitor and energy in inductor are equal. |
|
Answer» `T/8` |
|
| 38581. |
A 50 gram bullet moving with a velocity of 10ms^(-1)gets embedded into a 950 g stationary body. The loss in kinetic energy of the system will be |
| Answer» ANSWER :A | |
| 38582. |
A negative charge is placed at the midpoint between two fixed equal positive charges , separated by a distance 2d . If the negative charge is given a small distancement x(xltltd) perpendicular to the line joining the positive charges , how the force (F) developed on it will approximately depend on x ? |
|
Answer» `Fpropx`  Let us consider that , TWO positive CHARGES be `+Q` and negative charge be -q `AC=BC=r=(x^(2)+d^(2))^(1//2)` … (i) Now , electric field at point C due to the charges +Q placed at A and B is `E_("net")=2Ecostheta=2xx(kQ)/(r^(2))xx(x)/(r)=(2kQx)/(r^(3))` `therefore` Force on charge - q will be , `F=-qE_(net)` `=-(2kQqx)/(r^(3))=-(2kQqx)/((x^(2)+d^(2))^(3//2))`(using (i)) Now , for small displacement `(x),xltltd,x^(2)=0` `thereforeF=-(2kQqx)/(d^(3))rArrFpropx` So , the negative charge will OSCILLATE along its mean POSITION and the force acting on the negative charge will be restoring force. |
|
| 38583. |
A wire carrying current I is laid in shape of a curve which is represented in plane polar co- ordinate system as r = b + c/ pitheta for 0 lttheta lt pi/2 Here b and c are positive constants. theta is the angle measured with respect to positive x direction in anticlockwise sense and r is distance from origin (see figure). Calculate the magnetic field at the origin due to the wire. |
|
Answer» |
|
| 38584. |
The needle in the dip circle stands vertical when the plane of the circle is |
|
Answer» Along the earth's MAGNETIC meridian |
|
| 38585. |
What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is 750 nm? Assume that the refractive index for the film is mu = 1.33. |
|
Answer» 282 nm |
|
| 38586. |
A parallel plate capacitor with air between the plates has capacitance of 9 pF. The separation between its plates is .d.. The space between the plates is now filled with the other one dielectric constant K_2=6 and thickness (2d)/3 Capacitance of the capacitor is now . |
|
Answer» 1.8pF |
|
| 38587. |
A person looking through a telescope focuses lens at a point on the edge of the bottom of an empty cylindrical vessel. Next he fills the entire vessel with a liquid of refractive index mu, without disturbing the telescope. Now, he observes the mid point of the vessel. Determine the radius to depth ratio of the vessel |
|
Answer» `1/2sqrt((1-mu^2)/(mu^2 + 1))` |
|
| 38588. |
Discuss the source of electromagnetic waves |
|
Answer» Solution :Sources of electromagnetic waves: Propagation of an Electromagnetic Wave Any stationary source charge produces only electric field। When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field(not time dependent, only space or oscillating Magnetic Electric dependent) around the conductor in which charge flows। If the charged particle accelerates, inaddition to electric field it also produces magnetic field। Both electric and magnetic fields are time varying fields। Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors। Any OSCILLATORY motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves। Suppose the electromagnetic field in free space propagates along z direction, and if the electric field VECTOR points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means` E_y`, = `E_0`, sin(kz – wt) `B_x`, = `B_0`, sin(kz – wt) where, `E_0`, and `B_0`, are amplitude of oscillating electric and magnetic field, k is a wave number, o is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave। Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the PRODUCTION of electromagnetic waves)। In free space or in vacuum, the RATIO between `E_0` and `B_0` is equal to the speed of electromagnetic wave, which is equal to speed of light c। `c=(E_0)/(B_0)` In any medium, the ratio of E and B is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as `v=(E_0)/(B_0)ltc`Further, the energy of electromagnetic waves comes from the energy of the oscillating charge. 
|
|
| 38589. |
The colony of rag-pickers is situated in |
|
Answer» SOUTH of Delhi |
|
| 38590. |
The energy of an excited state of hydrogen is -13.6 eV? |
|
Answer» 1 |
|
| 38591. |
In the circuit below, A and B represent two inputs and C represents the output. The circuit represents |
|
Answer» OR GATE |
|
| 38592. |
How can we increase the capacitance of a conductor? |
|
Answer» |
|
| 38593. |
A magnetic needle of length 10 cm, suspended at its middle point through a thread, stays at an angle of 45^(@) with the horizontal. The horizontal component of the earth's magnetic lield is 18 muT, Which of the following statement(s) is//are correct. |
|
Answer» VERTICAL component of magnetie field, `B_(V) = 18 muT`. `tan45^(@) = B_(v)//B_(H) = 18muT` When the force F is applied figure (II), the needle stays in horizontal position Taking torque about the centre of the magnet, `2mB_(V) xx L = F xx l` or `F =2mB_(V)` `=2 xx (1.6 Am ) xx (18 xx 10^(-6)T) = 5.8 xx 10^(-5)N` 
|
|
| 38594. |
Two identical wires of iron and copper with their Young's modulus in the ratio 3 : 1 are suspended at same level. They are to be loaded so as to have same extension and hence level. Ratio of the weight is : |
|
Answer» `1:3` `THEREFORE(Y_(1))/(Y_(2))=(W_(1))/(W_(2))rArr(W_(1))/(W_(2))=3/1` So the CORRECT choice is (c). |
|
| 38595. |
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Omega,what is the maximum current that can be drawn from the battery ? |
|
Answer» Solution :Here emf `epsi` = 12 V, internal resistance r = 0.4 `Omega ` . The maximum current which can be DRAWN from the BATTERY (in short circuit CONDITION) `I_(max) = epsi/r = (12)/(0.4) = 30 A` |
|
| 38596. |
If light of wavelength 412.5 nm is incident on each of the metals given below, which one will show photoelectric emission and why ? |
|
Answer» Solution :Calculating the energy of the incident phohton Identifying the metals Reason The energy of a photon of incident radiation is GIVEN by ` E= ( h_in ) /( lambda ) ` ` therefore E= ( 6.63xx 10 ^( -34) xx 3xx10 ^(8) ) /( ( 412.5xx 10 ^(-9)) xx (1.6xx10 ^(-19)) eV ` `~= 3.01eV ` Hence ,only Na and K will show photoelectric emission [Note : Award this mark even if the student wirtes the NAME of only one of these metals ] Reason : They energy of the incident photon is more that the work function of only these two metals. |
|
| 38597. |
The formula, theta_("min") = 1.22 lamda//D (in the usual notation), for the limit of resolution of a telescope is due to |
|
Answer» Abbe |
|
| 38598. |
In the circuits (A) and (b) switches S_(1) and S_(2) are closed at t = 0 and are kept closed for a long time.The variation of current in the two circuits for t ge 0 are roughly shown by figure (figures are schematic and not drawn to scale): |
|
Answer»
For inductor circuit, `i = i_(0)(1 - e^(-(Rt)/(L)))` Hence graph (c) CORRECTLY DEPICTS i versus t graph. |
|
| 38599. |
Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of 100 kg and they are 100 m apart. How long will it be before the relative to one another. How long will it be before the gravitational attraction brings them 1 cm closer together ? |
|
Answer» `2.52` DAYS `=(Gm_(1)m_(2))/(r^(2))xx(1)/(m_(1))=(Gm_(2))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))` `=6.67xx10^(-13)m s^(-1)` `a^(2)`=acceleration of second astronaut `=(Gm_(1))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))=6.67xx10^(13)" m "s^(-1)` Net acceleration of approach `a=a_(1)+a_(2)=2xx6.67xx10^(-13)" m s^(-1)` Now, `s=(1)/(2)at^(2)1xx10^(-2)=(1)/(2)xx6.67xx10^(-13)xx2xxt^(2)` Solving, we GET `t=1.41` days. |
|
| 38600. |
During the propagation of electromagnetic waves in a medium: |
|
Answer» electric energy density is double of the MAGNETIC energy density |
|
