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.
| 5801. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vec(E )={(3.1 N//C)cos [(1.8 rad//m)y+(5.4xx10^(6)rad//s)]}hat(i)What is the direction of propagation ? |
| Answer» Solution :Here equation of electric FIELD has the form LIKE `E = E_(0)cos (omega t+ky)` and so given wave must be propagating along Y axis.…..(1) | |
| 5802. |
A potentiometer wire of length 1 m has a resistance of 10 Omega . It is connected to a 6 V battery inseries with a resistance of 5 Omega . Determine the emf of the primary cell which gives a balance point at 40 cm. |
Answer» Solution : Here emf of battery `epsi_0` =6V, length of potentiometer wire L = 1 m, length for balancing the CELL of emf `epsi` l = 40 cm = 0.4 m, resistance of potentiometer `R = 10 Omega` and series resistance `R. =5 Omegs` Potential gradient`k = (epsi_0 R)/((R + R.)L)` andemf of cell `epsi = kl = (epsi_0 RL)/((R + R.)L) = (6 xx 10 xx 0.4)/((10 + 5) xx 1) = 1.6 V` |
|
| 5803. |
A particle of mass mis dropped from a height habove the ground. At the same time another particle of mass 2m is thrown vertically upwards from the ground with a speed of sqrt(gh) . If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of sqrt(h/g)is: |
|
Answer» `(sqrt40 -2)/6`  Particles will collide after time `t_0 h/sqrt(GH)` at collision, `v_A= -g sqrt(h//g), V_B = 0` Time taken by COMBINED MASS to reach the ground Time `= ((sqrt(40)-2)/6)sqrt(h//g)` 
|
|
| 5804. |
The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance 25 cm in front of it 50 cm, the magnitude of its image changes from m_(25) to m_(50). The ratio (m_(25))/(m_(50)) is |
|
Answer» |
|
| 5805. |
A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio (C_(p))/(C_(v)) of the mixture is : |
|
Answer» <P>`1 cdot 54` and `C_(P) =(n_(1)CP_(1)+n_(2)CP_(2))/(n_(1)+n_(2))` `therefore (C_(p))/(C_(v))=(4xx(5)/(2) R+(1)/(2)xx(7)/(2) R)/(4xx(3)/(2)R+(1)/(2)xx(5R)/(2))` `therefore` Correct CHOICE is (b). |
|
| 5806. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vec(E )={(3.1 N//C)cos [(1.8 rad//m)y+(5.4xx10^(6)rad//s)]}hat(i)What is the wavelength lambda ? |
|
Answer» SOLUTION :Comparing given EQUATION, `vec(E )=3.1 COS {(5.4xx10^(6))t+(1.8)y}hat(i)` (With all values in SI UNITS) with standard form `E=E_(0)cos (omega t + ky)` we get, `E_(0)=3.1 (N)/(C ) ""`…(2) `omega = 5.4xx10^(6)(rad)/(s) ""`....(3) `k=1.8(rad)/(m) ""`....(4) `therefore (2pi)/(lambda)=1.8 rArr lambda = (2xx3.14)/(1.8)=3.489 m` `~~3.5 m ""`....(5) |
|
| 5807. |
Which of the following shows greenhouse effect |
|
Answer» Ultraviolet rays |
|
| 5808. |
Find the number of acceptro atoms m^(3)in the silicon specimen on adding one indium atom per 5xx10^(8)the number density of atoms in silicon specimen is 5xx10^(28) atoms / m^(3) |
|
Answer» SOLUTION :Numberdensityof ACCEPTOR atoms `=(5xx 10^(23))/(5 XX 10^(5))` `=10^(20)` ATOM`//ms^(3)` |
|
| 5809. |
The resonance frequency of a certain RLC series circuit is omega_(0) . A source of angular frequency2 omega_(0) is inserted into the circuit. After transients die out, the angular frequency of current oscillation is |
|
Answer» `(omega_(0))/(2)` |
|
| 5810. |
Electrostastic means study of : |
|
Answer» ELECTRIC charges in motion |
|
| 5811. |
The equations of motion of a projectile are given by x = 36t metre and 2y = 96t - 9.8r^(2) metre. The angle of projection is : |
|
Answer» `SIN^(-1)(4/5)` for `t=0v_(y)=48` `:.v=(36^(2)+(48)^(2))^(1//2)=60` Now `sintheta=v_(x)/v=48/60=4/5` `theta=sin^(-1)(4/5)` 
|
|
| 5812. |
Three charges are positioned as indicated in the following figure. What are the horizontal and vertical components of the net force exerted on the +15 mu C charge by the +11 mu C and +13 mu C charges ? |
|
Answer» Horizontal - 95 N, VERTICAL - 310 N |
|
| 5813. |
Deduce the expression for the potential energy of a system of two point charges q_1 and q_2 brought from infinity to the points vecr_1 and vecr_2 respectively in the presence of external electric field vecE. |
|
Answer» Solution :To deduce the potential energy of a system of two point charges `q_(1)` and `q_(2)` being brought from infinity to the points `vecr_(1)` and `vecr_(2)` respectively in an external electric field `vecE_(1)`, first we calculate the WORK DONE in bringing the CHARGE `q_1` from infinity of `vecr_1` . Work done in this step is `W_(1) = q_(1) V (vecr_(1))` , where `V (vecr_(1))` is the electrostatic potential at `r_1` . Now we consider the work done in bringing `q_2` to `vecr_2` . In this step work `W_(2)` is done against the electric field and work `W_(12)` is done against the field DUE to charge `q_(1)` . Obviously `W_(2) = q_(2) V (vecr_(2))` where `V (vecr_(2))` is the electrostatic potential at `vecr_(2)` and `W_(12) = (q_(1) q_(2))/(4 pi in_(0) . r_(12))` , where `r_(12) = |vecr_(1) - vecr_(2)|` = distance between `q_(1)` and `q_(2)` `therefore` Total work done in bringing `q_(2)` to `vecr_(2) = W_(2) + W_(12) = q_(2) V (vecr_(2)) + (q_(1) q_(2))/(4pi in_(0) . r_(12))` `therefore` Potential energy of the system U = The total work done in assembling the configuration `U = W_(1) + W_(2) + W_(12) = q_(1) V (r_(1)) + q_(2) V (r_(2)) + (q_(1) q_(2))/(4 pi in_(0) . r_(12))` |
|
| 5815. |
A parallel -plate capacitor is connected to a cell. A sheet of dielectric and a metal sheet are now introduced between the plates of the capacitor, parallel to these plates. The electric intensity between the positive plate of the capcitor and the metal sheet is E_(2), and between the metal sheet and the negatilve plate is E_(3). Then |
|
Answer» `E_(1)=E_(2)=E_(3)` |
|
| 5816. |
In the problem 1, a coil of 500 turns is wound closely on the toroid. Calculate the induced emf in the coil if the current in toroid is 5 A and it reverses its direction in 0.3 s. |
|
Answer» |
|
| 5817. |
Which of the following has negative magnetic susceptibility ? |
|
Answer» Ferromagnetic substance For paramagnetic substance : `0 lt chi lt in` For ferromagnetic substance : `X gt gt 1` |
|
| 5818. |
vecB : 4.00 m, at + 65.0^(@) vecC = (-4 .00m) hati + (-6.00m)hati vecD : 5.00 m, at - 235^(@) |
| Answer» Solution :`(a) vecR = (-3. 18m) HATI + (4.72m) hatj , (b) 5.69 m, (C ) - 56 .0^(@)` (with-axis) or `( 5.69 angle 124 ^(@))` | |
| 5819. |
A sqare frame withsidea=5 cm and a ,long straight conductor carryingsteady current I=5 A are located in the sameplane. Theinductance andthe resistanceof theframeareL=0.1mH andR=1Omega . Theframe is turned suddenly through180 about the side parallel to the conductor which is ata distance b=10 cm .Findthe chargethrough the frame . |
|
Answer» |
|
| 5820. |
The charge on an electron is extremly small and also drift velocity. Even then we get sufficient amount of current in a wire. Why ? |
| Answer» Solution :Although values of l and `v_d` are small, the value of n (density of electrons) is very large in METALS of the ORDER of `10^28 m^-3`. Thus we get a sizeable CURRENT the WIRE. | |
| 5821. |
A parellel-plate capacity whose electrodes are shaped as round disc is changed slowly. Demonstrate that the flux of the Poynting vector across the capacitor's lateral surface is equal to the increment of the capcitor's enegry per unit time. The dissipation of field at theedge is to be neglected in calculations. |
|
Answer» Solution :If the charge on the capacitor is `Q`, the rate of increase of the capcitor's energy `=(d)/(dt) ((1)/(2)(Q^(2))/(c ))= (Q Q)/(C ) = (d)/(epsilon_(0)PIR^(2))Q Q` Now electric field betweeb the pates (INSIDE it) is, `E = (Q)/(piR^(2)epsilon_(0))`. So discplacement CURRENT `= (delD)/(delt) = (Q)/(piR^(2))` This will lead to a magnetic field, (circuital) insidethe plates . At a radial distance `r` `2pirH_(theta)(r ) =pir^(2)(Q)/(piR^(2))` or `H_(theta) = (Qr)/(2piR^(2))` Hence `H_(theta)(R) = (Q)/(2piR)` at the EDGE. Thus inward polynting vector `=S = (Q)/(2piR) xx (Q)/(piR^(2)epsilon_(0))` Total flow `= 2piR d xx S = (Q Qd)/(piR^(2)epsilon_(0))` Proved |
|
| 5822. |
The velocity - time graph of a body is shown in fig. The ratio of the .....during the intervals OA and AB is _______ |
|
Answer» average velocities: 1:1 |
|
| 5823. |
A current of (2.5 pm 0.05) A flows through a wire and develops a potential difference of (10 pm 0.07) volt. Resistance of the wire in ohm, is |
|
Answer» `4 PM 0.108` |
|
| 5824. |
A short p hat(i) is placed at origin in x - y plane. The locus of equipotential surface around the dipole if a constant electric field Ehat(i) also exists through out the space is (k=(1)/(4piepsilon_(0)))(V_("origin")=0) |
|
Answer» `x^(2)+y^(2)+x^(2)=((kp)/(E))^(2//3)` |
|
| 5825. |
Can three vector not in one plane, give zero resultant ? Can four vectors do ? |
| Answer» SOLUTION :Suppose TWO vetor `vec A` and `vec B` lie in one plane while the third C lies in a different plane. | |
| 5826. |
The eccentricity of the earth's orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is |
Answer» Solution : According to the law of conservation of ANGULAR MOMENTUM `mv_(1)r_(1)=mv_(2)r_(2)` `(v_(1))/(v_(2))=(r_(2))/(r_(1))=(a(1+e))/(a(1-e))` where e is eccentricity of the earth.s ORBIT `=((1+0.0167))/((1-0.0167))=1.034` |
|
| 5827. |
In a common emitter transistor, the base currentI_(b) = 2 mu A, alpha = 0.9 ,then I_(c)is equal to |
| Answer» Solution :N/A | |
| 5828. |
A current of 50mA gives deflection of 40^@ in T.G.For a deflection of 60^@the current will bw : |
|
Answer» `I INFTY THETA` |
|
| 5829. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vec(E )={(3.1 N//C)cos [(1.8 rad//m)y+(5.4xx10^(6)rad//s)]}hat(i)Write an expression for the magnetic field part of the wave. |
|
Answer» Solution :EQUATION for magnetic FIELD, `vec(B)=B_(0)cos (ky + omega t)hat(k)` `therefore vec(B)=1.033xx10^(-8)cos {(1.8 y+5.4xx10^(8)t)}hat(k) ""`….(8) Here if we take direction of `vec(B)` along `hat(k)` then only, direction of `vec(E )xx vec(B)=` direction of `hat(i)xx hat(k)=-hat(j)=` direction of propagation of given electromagnetic wave = direction of - Y axis. |
|
| 5830. |
What is nuclide ? |
| Answer» SOLUTION :When we WANT to point out a particle out a particular nucleus, we use the TERM NUCLIDE (NEUTRON + Proton ). | |
| 5831. |
A metallic rod of 1m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular me tallic ring of radius 1m, about an axis passing through the centre and perpendicular to the plane of the ring (see figure). A constant and uniform magnetic field of 1T parallel to the axis present everywhere. What is the emf between the centre and the metallic ring? |
|
Answer» SOLUTION :Method I As stated in this question 6, during rotation, electrons are shifted to the end Q, creating an emf. Under steady STATE, emf acrosss a length dr is `d varepsilon=Bvdr`. Using equation `varepsilon=Blv`, the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by `therefore varepsilon=intdvarepsilon=underset(0)overset(R)intBvdr=underset(0)overset(R)intBomegardr=(BomegaR^(2))/(2)` But `v=omegar`. This gives `epsilon=(1)/(2)xx1.0xx2pixx50xx(1^(2))=157V` Method II Considering the positions OP and OQ, the emf produced depends on rate of change of flux which itself is proportional to rate of SWEEPING of area of the sector. If `THETA` is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by `piR^(2)xx(theta)/(2pi)=(1)/(2)R^(2)theta`, where R is the radius of the circle. Hence, the induced emf is `epsilon=Bxx(d)/(dt)[(1)/(2)R^(2)theta]=(1)/(2)BR^(2)(d theta)/(dt)=(BomegaR^(2))/(2)` [Note: `(d theta)/(2)=omega=2piv`] The same expression we got earlier. |
|
| 5832. |
Potentiometer can be used a) for measurement of emf of cell b) to determine internal resistance of cell c) for the calibration of Ammeter and Voltmeter d) for measuring thermo emf of a thermo couple |
|
Answer» Only (a), (B) CORRECT |
|
| 5833. |
A small ball suspended from a string is set into oscillation. When the ball passes through the lowest point of the motion, the string is cut. If the ball is then moving with the velocity 0.8 ms^(-1) at a height of 5 m above the ground, the horizontal distance travelled by the ball is : (Given g = 10 ms^(-2) ) |
|
Answer» 0.2 m |
|
| 5834. |
In common base configuration of a transistor, a change of 200 mV in emitter voltage produces a change of 5 mA in emitter current. If base collector voltage V_(CB) remains constant, find the dynamic input resistance of transistor. |
| Answer» SOLUTION :40 `OMEGA | |
| 5835. |
Draw a neat labelled diagram of Cassegrain reflecting telescope. |
Answer» Solution : Note : India.s largest REFLECTING TELESCOPE used at Kavalur in Tamil Nadu has a CONCAVE MIRROR, 2.34 m in diameter and the one at Hawaii is USA has a concave mirror 10 m in diameter. |
|
| 5836. |
A metallic surface when illuminated with light of wavelength 3333 Å emits electrons with energies upto 0.6 eV. Calculate the work function of the metal. Data : lambda=3333 Å, K.E=0.6eV, W=? |
|
Answer» SOLUTION :Work function, W = hv - kinetic energy or `W=(HC)/(lambda)-K.E` `=((6.26 xx 10^(-34) xx 3 xx 10^(8))/(3333xx10^(-19)))-(0.6xx1.6xx10^(-19))` `=(5.96 xx 10^(-19))-(0.96 xx 10^(-19))` `W=5 xx 10^(-19)J` `W=(5 xx 10^(-19))/(1.6 xx 10^(-19))eV.` W = 3.125 eV. |
|
| 5837. |
There are mainly two types of combinations of resistors in a circuit, series and parallel. In the given table, Column I shows voltages of different arrangements of resistors, Column II shows different arrangement of resistors and Column III shows currents of different arrangements of resistors. What are the conditions when resistors are in mixed combination, that is, parallel and series combination respectively? |
|
Answer» (IV) (iv) (K) |
|
| 5838. |
Which of the following source of light is monochromatic source ? |
| Answer» Answer :D | |
| 5839. |
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surfrace tension is T, density of liquid is p and L is its latent heat of vaporization. |
|
Answer» `sqrt(T//PL)` `PR^(2)DeltaRL=T[R^(2)-R^(2)+2RDeltaR-DeltaR^(2)]` `pR^(2)DeltaRL=T2RDeltaR`(`DeltaR` is very small) `R=(2T)/(pL)` So, correct . choice is ( c) |
|
| 5840. |
A 4 muF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 muF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? |
|
Answer» Solution :Here `C_1 = 4 muF= 4 xx 10^(-6)F, V_1 = 200 V, C_2= 2muF= 2 xx10^(-6)F and V_2 = 0` On SHARING of CHARGES loss of electrostatic ENERGY `Delta u = (C_1C_2(V_1-V_2)^2)/(2(C_1 + C_2)) = (4 xx 10^(-6) xx 2 10^(-6) xx (200-0)^2)/(2(4 xx 10^(-6) + 2 xx 10^(-6))` `=(4 xx 10^(-6) xx 2 xx 10^(-6) xx 200 xx 200)/(2 xx 6 xx 10^(-6)) = 2.67 xx 10^(-2)J`. |
|
| 5841. |
A dynamo works on the principle of |
|
Answer» AMPERE's law |
|
| 5842. |
When an electron beam is moving in a magnetic field, then the work done is equal to the - |
|
Answer» CHARGE of electron |
|
| 5843. |
A thin rectangular magnet is suspended freely has a period of ocsillation as 4 sec. If it is broken into two equal halves and similarly suspended, the period of of oscillation of each halves will be equal to : |
|
Answer» 8 sec |
|
| 5844. |
There are mainly two types of combinations of resistors in a circuit, series and parallel. In the given table, Column I shows voltages of different arrangements of resistors, Column II shows different arrangement of resistors and Column III shows currents of different arrangements of resistors. What are the conditions when resistors are in mixed combination, that is series and parallel combination respectively? |
|
Answer» (III) (i) (L) |
|
| 5845. |
The magnetic susceptibility is negative for : |
|
Answer» PARAMAGNETIC material |
|
| 5846. |
A particle executes S.H.M. The acceleration of particle is maximum |
|
Answer» at EXTREME position |
|
| 5847. |
A U^(238) preparation of mass 1 g emits 1.24 xx 10^(4) alpha -particles per second. Find the half life of this nuclide and the activity of the preparation in becquerels: |
|
Answer» `4.5 xx 10^(9) yrs, 1.24 xx 10^(4) Bq` N=No. of ATOMS in 1 g `=(6.023 xx 10^(23))/(238) =2.53 xx 10^(21)` `(d N)/(dt)=lambda N` `1.24 xx 10^(4) =lambda xx 2.53 xx 10^(21)` `or lambda=4.9 xx 10^(-18)` Now `T=(0.6931)/(lambda) =(0.693)/(4.9 xx 10^(-18)) =1.4 xx 10^(17)` `=4.5 xx 10^(9)"yrs"` Acitivity of preparation `=1.24 xx 10^(4) Bq`. |
|
| 5848. |
An inductor coil of inductance L is divided into two equal parts and both parts are connected in parallel. The net inductance is : |
| Answer» ANSWER :D | |
| 5850. |
For a CE-transistor amplifier fig. The audio signal voltage across the collector resistance of 1.0kOmega is 1.0 V. suppose the current amplification factor of the transistor is 100, what should be the value of R_(B) in series with V_(B B) supply of 1.0 V if the de base current has be 10 times the signal current, assuming V_(RE) = 0.6 V. Also calculate the voltage drop across the collector resistance. |
|
Answer» Solution :Here `V_(0)=1.0V`, `I_(c)=(V_(0))/(R_(c))=(1.0)/(1000)=1.0xx10^(-3)A=1.0mA` `I_(c)=(V_(0))/(R_(c))=(1.0)/(1000)=1.0xx10^(-3)A=1.0mA` SIGNAL current through the base, `I_(B)=(I_(c))/(R_(c))=(1.0mA)/(100)=0.010mA` Now the base current has to be increased to be, `I_(B)=10I_(b)=10xx0.01=0.10mA` As, `V_(B B)=I_(B)R_(B)` `therefore R_(B)=(V_(B B)-V_(BE))/(I_(B))=((1.0xx0.6))/(0.10xx10^(-3)A)=4xx10^(3)Omega=4kOmega` dc Collector current, `I_(c)=betaxxI_(B)^(1)=100xx0.10=10mA` VOLTAGE drop across the collector resistance `I_(c)R_(c)=(10xx10^(-3)A)xx(100Omega)=10V` |
|