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.
| 40601. |
Two forces of magnitude F and sqrt3 F act at right angles to each other. Their resultant makes an angle theta with F. The value of theta is : |
|
Answer» `30^(@)` |
|
| 40602. |
Derive an expression for the radius of n^(th) Bohr's orbit of hydrogen atom hence write the expression for the radius of first orbit of hydrogen atom. |
|
Answer» Solution :Consider a ATOM with effective nuclear charge + Ze. LET an electron revolves around the nucleus with speed v in the orbit of radius r as shown in figure. The necessary centripetal force on electron is provided by the electrostatic force between the electron and the nucleus. Therefore we have, centripetal force= electrostatic force `(mv^(2))/(r )= (1)/(4PI epsi_(0))(Z e.e)/(r^(2))` `mv^(2)r= (Ze^(2))/(4pi epsi_(0))` ....(1) From Bohr.s angular momentum quantization rule, `mvr= (nh)/(2pi)` `m^(2)v^(2)r^(2)= (n^(2)h^(2))/(4pi^(2))` ....(2) Dividing equation (2) by equation (1) we have, `(m^(2)v^(2)r^(2))/(mv^(2)r)= (n^(2)h^(2))/(4pi^(2)) XX (4pi epsi_(0))/(Ze^(2))` `mr= (n^(2)h^(2) epsi_(0))/(pi Ze^(2))rArr r= (n^(2)h^(2) epsi_(0))/(pi m Ze^(2))` For `n^(th)` orbit, `r_(n)= (n^(2)h^(2) epsi_(0))/(pi mZe^(2))` For H atom Z=1 and for `n^(th)` orbit, `r_(n)= (n^(2)h^(2)epsi_(0))/(pi m e^(2))` 
|
|
| 40603. |
In the following question a statement of assertion (A) is followed by a statement of reason (R ) A: Absolute value of potential is not defined R : Two equipotential lines cannot intersect each other. |
|
Answer» If both Assertion & REASON are true and the reason is the CORRECT explanation of the assertion , then mark (1). |
|
| 40604. |
What are the constituent particles of neutron and proton? |
| Answer» Solution :PROTONS and neutrons are BARYON which are made up of THREE Quarks . Accordingto quark model , PROTON is made up of two upquarksand one down quark and NEUTRON is made up of one up quark and two down quarks. | |
| 40605. |
The refractive indices for lights of violet, yellow and red colours for a flint glass prism are respectively 1.632, 1.620 and 1.613. Find the dispersive power of the prism material. |
|
Answer» |
|
| 40606. |
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^(-15)m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.) |
|
Answer» <P> Solution :Here `lambda=10^(-15)m,h=6.63xx10^(-34)`Js`c=3xx10^(8)ms^(-1)` `implies` Momentum, `p=(h)/(lambda)=(6.63xx10^(-34))/(10^(-15))` `therefore p=6.63xx10^(-19)kg ms^(-1)` `implies` Rest mass energy of electron, `m_(0)c^(2)=0.511 MeV` `therefore m_(0)c^(2)=0.511xx10^(6)xx1.6xx10^(-19)J` `=-0.8176xx10^(-13)J` `implies` Energy of electron according to EQUATION relativity, `E=sqrt(p^(2)c^(2)+m_(0)^(3)c^(4))` `therefore E^(2)=p^(2)c^(2)+m_(0)^(2)c^(4)=p^(2)c^(2)+(m_(0)c^(2))^(2)` `=(6.63xx10^(-19))^(2)xx(3xx10^(8))^(2)` `+(0.8176xx10^(-13))^(2)` `=395.9xx10^(-28)=0.66846xx10^(-26)` Neglecting rest mass energy, `E^(2)=395.6xx10^(-22)` `therefore E=19.8896xx10^(-11)` `therefore E~~19.89xx10^(-11)J` `therefore E=(19.89xx10^(-11))/(1.6xx10^(-10))BEV` `[because 1 BeV=1.6xx10^(-10)]` Thus energy obtained from accelerator should be of the order of BeV. |
|
| 40607. |
The emf of a battery A is balanced by a length of 80cm on a potentio meter wire. The emf of a standard cell 1v is balanced by 50cm. The emf of A is |
|
Answer» 2v |
|
| 40608. |
In the given figure, the cube is length l on each side. Four straight segments of wire ab, bc, cd and da form a closed loop that carries a current I = i_(0) in the direction shown. A uniform magnetic field of magnitude B = B_(0) is in the positive y direction. Determine the torque on the loop |
Answer» Solution :We will make closed loops in a PLANE by adding WIRES so that calculation is simple.  LOOP `ade`,`vec(M)_(1) = (1)/(2)l^(2)i_(0)(-hat(j))` Loop `cde`, `vec(M)_(2) = (1)/(2)l^(2)i_(0)(-hat(k))` Loop `ABCE`, `vec(M)_(3) = l^(2)hat(i)_(0)(-hat(i))` `vec(M) = vec(M)_(1) + vec(M)_(2) + vec(M)_(3) = - (l^(2)i_(0))/(2)(2 hat(i) + hat(j) + hat(k))` `vec(B) = B_(0) hat(j)` `vec(tau) = vec(M) xx vec(B) = - (l^(2)B_(0)i_(0))/(2)[2(hat(i) xx hat(j)) + (hat(j) xx hat(j)) + (hat(k) xx hat(j))]` `= -(l^(2) B_(0)i_(0))/(2) [2 hat(k) - hat(i)] = (l^(2)B_(0)i_(0))/(2)(hat(i) - 2 hat(k))` `|vec(tau)| = tau = (l^(2) B_(0)i_(0))/(2) sqrt(5)` |
|
| 40609. |
The radius of the innermost electron orbit of a hydrogen atom is 5.3 xx 10^(–11) m. What are the radii of the n = 2 and n =3 orbits? |
|
Answer» SOLUTION :`r_(0)=a_(0)=5.3xx10^(-11)m` `:. In n=2,r ^(2)=?` In `n=3, r^(3)=?` `rArr` Radius of electron in `n^(TH)` orbit in HYDROGEN atom `r_(n)=n^(2)r_(0)` Radius in n=2 orbit, `r_(2)=(2)^(2)xx5.3xx10^(-11)` `=4xx5.3xx10^(-11)` `:.r_(@) =2.12xx10^(-10)m` Radiusin n=3 orbit, `r_(3)=(3)^(2)xx5.3xx10^(-11)` `=9xx5.3xx10^(-11)` `:.r_(3)=4.77xx10^(-10)m` |
|
| 40610. |
When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation. |
| Answer» Solution :When a low flying aircraft passes overhead, we ALSO receive a weak TV signal on account of REFLECTION of radiowaves from the aircraft. This weak signal INTERFERES with the NORMAL TV signal coming through the antenna and it MAY result in a slight shaking of the picture on our TV screen. | |
| 40611. |
What is the cause of induced e.m.f. ? |
| Answer» SOLUTION :The basic CAUSE of INDUCED e.m.f. is the change of magnetic flux LINKED with a circuit. | |
| 40612. |
What is an ideal diode ? |
| Answer» SOLUTION :An ideal diode.is ONE which behaves as a PERFECT conductor when forward fiased and as a perfect insulator when reverse .biased. Under this SITUATION, the forward resistance of the diode is assumed to be ZERO and the potential barrier is neglected. | |
| 40613. |
A metallic pipe with nut and bolt assembly is shown in the figure. as is the coefficient of linear expansion for bolt and prop_p( when prop_Bgtprop_p) is for the material of pipe. The arrangement is heated then (##TRG_PHY_MCQ_XII_C09_E04_001_Q01.png" width="80%"> |
|
Answer» TENSILE STRESS is developed in the bolt |
|
| 40614. |
A long current carrying wire , carrying current I_(1) such that I_(1) is flowing out from the plane of paper is placed at O. A steady state current I_(2) is flowingin the loop ABCD |
|
Answer» the net FORCE is zero Force on `AB` . ` dF xxBxx sin 0@)= 0` Force on CD `: Similarly the magnetic field due to current `I_(1)` is along ` DC` . Because ` theta = 180^(@)` here , therefore force on ` DC` is zero . Force on `BC ` : Consider a small element `dl`. ` dF = I_(2) dl B_(1) sin 90^(@) rArr dF = I_(2) dl B_(1) ` By Fleming's left hand rule , the direction of this force is perpendicular to the plane of the paper directed outwards . Force on ` AD` : ` dF = I_(2) dl B_(1) sin 90^(@) rArr dF = I_(2) dl B_(1) ` By Fleming's left hand rule , the direction of this force is perpendicular to theplane of paper directed inwards. Since the current elements are located symmetrical to current ` I_(1) , therfore force on `BC` will cancel out the effect of force on ` AD`. rArr Net force on loop ` ABCD` is zero. ` NEt Torque on the loop : Theforce on ` BC and AD` will create a torque on ` ABCD` in clockwise direction about ` OO` as seen by the observer at `O`.   
|
|
| 40615. |
Two identical charges of 1muCare kept on 0,2 m) and (0,-2 m), respectively. Calculate the magnitude of the force acting on a 2muC charge placed at (-5m,0). |
| Answer» SOLUTION :`11.5xx10^(-4)` N | |
| 40616. |
In the above problem, when will be the third echo heard. |
|
Answer» 4S |
|
| 40617. |
Angle of contact varies between |
|
Answer» `0^@` to `180^@` |
|
| 40618. |
A plate of crystal quartz is cut with its faces perpendicular to the optic axis. It is found that this plane exactly annuls the rotation of the plane of polarisation ofsodium light produced by a 30 cm length of a 18^(@) solution of lactose. Calculate the thickness of the quartz plate. Given 's' for lactose =5253^(@), 1 mm of quartx rotats the plane of polarisation of sodium light by 21.71^(@). |
|
Answer» |
|
| 40619. |
At time t = 2.3 s, a 4 kg block that initially moves with a constant speed of 6 m/s undergoes an inelastic collision with another block. Any two inertial observers must agree that |
|
Answer» The event took PLACE at t = 2.3 s |
|
| 40620. |
The potential in the depletion layer is due to |
|
Answer» Electrons |
|
| 40621. |
If the momentum of a body increases by 20%, the percentage increases in K.E. will be |
|
Answer» 36 |
|
| 40622. |
Under what condition the directions of sum or difference of two vectors will be equal in magnitude ? |
| Answer» Solution :When the TWO VECTORS are unequal in MAGNITUDE and are in the same direction . | |
| 40623. |
A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the glass is |
|
Answer» 2 cm `because (n_(2))/(v)-(n_(1))/(u)=(n_(2)-n_(1))/(R )` `THEREFORE (1)/(v)-(1.5)/((-6))=(1-1.5)/((-6)) or (1)/(v)=(0.5)/(6)-(1.5)/(6)=-(1.0)/(6)` `rArr""v=-6cm` 
|
|
| 40624. |
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50 cm from its mid-point? The magnetic moment of the bar magnet is 0.40 A m^2, the same as in Example 5.2. |
|
Answer» SOLUTION :From Eq (5.7) `B_(E)=(mu_(0)m)/(4pi r^(2))=(10^(-7) XX 0.4)/((0.5)^(3))=(10^(-7) xx 0.4)/(0.125) =3.2 xx 10^(-7)T` From Eq. (5.8) `B_(A)=(mu_(0)2m)/(4pi r^(3))=6.4 xx 10^(-7)T` |
|
| 40625. |
Who is the father of the modern robotics industry formed the world's first robotic company in 1956? |
|
Answer» Joliot |
|
| 40626. |
A gunis mounted on a gun carriage movable on a smooth planethe gun is elevated at an anglealpha to thehorizontal . A shot, firedleaves the gun in a directioninclined at an angle theta to thehorizontal . If themass of the gum and its carriagebe n times that of short, find tan theta : tan alpha . |
| Answer» SOLUTION :`1+(1)/(N)` | |
| 40627. |
The charge on a parallel plate capacitor varies as q=q_(0)cos 2pi v t. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through th capacitor. |
|
Answer» SOLUTION :Displacement current in capacitor, `I_(d)=I_(c )=(dq)/(dt) ""`….(1) Here, `Q=q_(0)cos 2PI vt` Substituting in EQUATION (1), `I_(d)=I_(c )=(d)/(dt)(q_(0)cos 2pi vt)` `therefore I_(d)=I_(c )=-q_(0)sin 2pi vt xx 2pi V` `therefore I_(d)=I_(c )=-2pi vq_(0)sin 2pi vt` |
|
| 40628. |
If v_(1) and v_(2) are the velocities of light in the two media having angles of incidence and refraction i_(1) and i_(2) respectively, then |
|
Answer» `v_(1) " COSEC " i_(1) = v_(2) " cosec " i_(2)` |
|
| 40629. |
Find the minimum kinetic energy of an alpha - particle to cause the reaction ""^(14)N(alpha.p)""^(17)O. The masses of ""^(14)N,""^(4)He,""^(1)H and ""^(17)O are respectively 14.000307u, 4.00262u,1.00783u, and 16.11913u. |
|
Answer» Solution :Since, the masses are given in ATOMIC mass UNIT , it is easiest to proceed by finding the mass DIFFERENCE between REACTANTS and products in the same unit and then MULTIPLYING by 931. 5 Me V/u Thus , we have `Q=(14.00307u + 4.00260u - 1.00783 u - 16.99913u)` `(931.5 (MeV)/u) =-1.20 MeV` Q value is negative ,it means reaction is endothermic So , the minimum kinetic energy of `alpha` - particle to initiate this reaction would be `K_(min)=|Q|((m_(alpha))/(m_N)+1)=(1.20)((4.00260)/(14.00307)+1)=1.54 MeV` |
|
| 40630. |
Using Huygens principle, show that the angle of incidence is equal to angle of reflection during a plane wave front reflected by a plane surface. |
Answer» Solution :Using Huygen.s wave THEORY of light, SHOW that the angle of incidence is equal to angle of reflection in case of reflection of a PLANE wavefront by a plane surface.  `AE=BC=vt` The TRIANGLES EAC and BAC are congruent and therefore, the angles i and r (as show in Fig. )would be equal. This is the law of reflection. |
|
| 40631. |
The current-carrying wire loop in Fig (a) lies all in one plane and consists of a semicle of radius 10.0 cm, a smaller semicicle is rotated out of thatplkane by angle theta until it is perpendicular to the plane [Fig.(b)]. Figure (c) gives the magnitude of the net magnetic field at the center of curvature versus angle theta . what os the radius of the smaller semicircle ? |
|
Answer» |
|
| 40632. |
A uniform copper wire carries a current i amperes and has p carriers per "metre"^(3). The length of the wire is 1 metro and its cross-section area is s "metre"^(2). If the charge on a carrier is q coulombs, the drift velocity in "ms"^(-1) is given by |
|
Answer» 1/lsq |
|
| 40633. |
In an unbiased n-p junction electrons diffuse from n-region to p-region because: |
|
Answer» ELECTRONS travel across the junction due to potential DIFFERENCE |
|
| 40634. |
The nuclear forces are |
|
Answer» attractive |
|
| 40635. |
What is the rms value of the current for A current I=100 cos (200t + 45^@ ) A ? |
|
Answer» `50sqrt2A` `I=I_m cos (200t+45^@)` , `THEREFORE I_m=100A` `therefore I_(rms)=I_m/sqrt2 = 100/sqrt2=(100xxsqrt2)/2=50sqrt2A` |
|
| 40636. |
Two simple pendulums A and B are made to oscillate simultaneously and it is found that A completes 10 oscillations in 20 sand B completes 8 oscillations in 10 s. The ratio of the lengths of A and B is |
|
Answer» `(8)/(5)` For PENDULUM A, `T_(A) = 2pi sqrt((I_(A))/(g))` `(20)/(10)=2pi sqrt((I_(A))/(g))` For pendulum B, `T_(B)=2pi sqrt((I_(B))/(g))` `(10)/(8)=2pi sqrt((I_(B))/(g))` Dividing Eq. (i) by Eq. (II), we GET `(I_(A))/(I_(B))=((160)/(100))^(2)` `(I_(A))/(I_(B))=(64)/(25)` |
|
| 40637. |
Three charges are placed at the vertices of an equilateral triangle of side 'a' as in fig. The force experienced by the charge pleased at the vertex A in a direction parallel to BC is: |
|
Answer» `Q^2/(4PI in_0 a^2)` |
|
| 40638. |
When dilute acid is added to Zinc granules, gas evolved is |
|
Answer» Hydrogen |
|
| 40639. |
When a certain curcuit consisting of a constant e.m.f. E an inductance L and a resistance R is closed, the current in it increases with time according to curve 1. After one parameter (E, L or R) is changed, the increase in current is closed second time. Which parameter was changed and in what direction? |
| Answer» Answer :A | |
| 40640. |
If resultant of two forces vec F and vec F is F. The angle between two forces is : |
|
Answer» `0^(@)` |
|
| 40641. |
STATEMENT-1: It is necessary to define two molar heat capacitor for a gas. STATEMENT-2: Work is done by a gas when its volume changes. |
|
Answer» Statement-1 is TRUE, Statement-2 is True, Statement-2 is a CORRECT EXPLANATION, for Statement-1. |
|
| 40642. |
What should be the velocity of the electron so that its momentum equals of4000Å wavelength photon. |
|
Answer» Solution :de-Broglie WAVELENGTH of electron `lambda = (h)/(P) = (h)/(mv)` `V = (h)/(m lambda)` `= (6.6 XX 10^(-34))/(9.11 xx 10^(-31)xx 4000 xx 10^(-10)) = (6.6 xx 10^(-34))/(36.44 xx 10^(-38)) = 0.18112 xx 10^(4)` `v = 1811 ms^(-1)` |
|
| 40643. |
Three condensers are connected as shown in series. If the insulated plate of C_1 is at 45V one plate of C_3 is earthed, find the p.d between the plates of C_2 |
| Answer» Answer :A | |
| 40644. |
A spherical metal ball of radius 1 cm is suspended in a room at 300 K temperature. Inside the sphere there is a battery operated heater which maintains the temperature of the ball at 1000 K. Find the power of the heater if emissivity of the metal ball is 0.3. |
|
Answer» |
|
| 40645. |
A loop, made of straight edges has six corners at A(0,0,0), B(L,0,0) ,C(L,L,0), D(0,L,0), E(0,L,L) and F(0,0,L). A magnetic field vecB=B_0 (hati+hatk)T is present in the region. The flux passing through the loop ABCDEFA (in that order) is |
|
Answer» `B_0L^2Wb` Area of loop ABCDA in xy-plane , `vecA_1 = |A| hatk =L^2hatk` Area of loop ADEFA in yz- plane `vecA_2=|A|hati =L^2hati` Total area of LOOPS are `vecA=vecA_1+vecA_2` `THEREFORE vecA=L^2hatk+L^2hati` Now flux linked with these loops is `phi=vecA.vecB=(L^2hati+L^2hatk).B_0 (hati+hatk)` where `vecB=B_0 (hati+hatk)` MAGNETIC flux , `phi=vecB.vecA` `=B_0(hati+hatk).L^2(hati+hatk)` `=B_0L^2(1+1)` `=2B_0L^2` 
|
|
| 40646. |
X-rays and gamma-rays of same energies are distinguished by their : |
|
Answer» frequencies |
|
| 40647. |
How many""^(235)U is consumed in a day in an atomic power house operating at 400 MW , provided the whole of mass ""^(235)Uis converted into energy? |
|
Answer» Solution :Power = 400 MW = `400 xx10^(6)W,` TIME = 1 day = 86 , 400 s. Energy produced , E = power `xx ` time `=400xx10^(6)xx86,400=3.456xx10^(13)J` As the whole of mass is converted into energy , by Einstein.s mass - energy relation `E =Mc^2` `M = E/(c^2) = (3.456xx10^(13))/((3xx10^(8))^2) =3.84xx10^(-4)kg = 0.384 g`. |
|
| 40648. |
Kerala has low Infant Mortaliy Rate because: |
|
Answer» it has good CLIMATIC condition |
|
| 40649. |
The terminal potential difference of a cell is greater than itsemf when it is |
|
Answer» being DISCHARGED |
|
| 40650. |
A metallic rod of length 'I' is rotated with a frequency with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius r , about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere . Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it. |
|
Answer» Solution :Suppose the length of the ROD is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along `+y` direction.Suppose the direction of the magnetic field is along `+z` direction.Then, using Lorentz law, we get `vecF=-e(upsilonhatjxxBhatk) rArr vecF=-eupsilonBhati` THUS, the direction of force on the electrons is along `X`-axis. Thus, the electrons will move towards the center i.e., the fixed end of the rod.This movement of electron will result in current and hence it will produce `emf` in the rod between the fixed end and the point touching the ring Let `theta` be the angle between the rod and radius of the circle at any time `t`. Then, area swept by the rod inside the circle =`1/2pir^(2)theta` Induced `emf` =`Bxxd/(dt)(1/2pir^(2)theta)=1/2pir^(2)B(d theta)/(dt)=1/2pir^(2)omega =1/2pir^(2)B(2piupsilon)=pi^(2)R^(2)Bupsilon`. Note:There will be an induced `emf` between the two ends of the rod also. |
|