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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.

4601.

A nucleus contains no electrons, yet it ejects them, How?

Answer»

SOLUTION :During `BETA^-` decay. Here a neutron in the nucleus converts into a proton with the EJECTION of an ELECTRON and antineutrino. This is the electron which is ejected from the nucleus.
Discussion
4602.

An ariticial satellite (mass =M) of earth revolves in a circular orbit whose radius is 4 times the radius of the earth. Assume that in the spherical region between the radius 4R_(e ) and 2R_(e ) there is cosmic dust. If mass of the satellite is increasing at the rate of u per sec. Find the time in which satellite will come to orbit of radius 2R_(e ).

Answer»


ANSWER :`(M)/(U)(SQRT(2)-1)`
Discussion
4603.

In column-II, some situations are given, and in column-I, their results are given. Mathc the proper entries from column-2 to column-1 using the codes iven below the columns

Answer»

`{:(p,q,r,s),(3,1,4,2):}`
`{:(p,q,r,s),(2,1,3,4):}`
`{:(p,q,r,s),(3,2,1,4):}`
`{:(p,q,r,s),(2,4,3,1):}`

Solution :(A) When `0^(@)C` water VOLUME decreases slightly, so `W_("system")=-ve`. To melt the ice, some heat has to be given `(Q=mL_(f))` which is almost equal to increase in internal ENERGY
(B)
Since, `P-v` cycle is clockwise, so `W_("net")=+ve`
and `(DeltaU)_("cycle")=0`
(C) By the fan, some work is done on the room air. Done to this, temperature of the gas increaes slightly so internal energy will increase slightly Matematically.
`Q=W+Deltau`
`Q=-ve+DeltaurArr DeltaU=+ve`.
(D) `P-V` diagram for the process is

From the diagram
`W_(ArarrBrarrC)=-ve`
`(PV)_(c)lt(PV)_(A)rArrT_(C)ltT_(A)`
So, internal energy DECREASE.
`(E) dQ=-2dU`
`dQ=2h(5)/(2)RdT`
`C=(dQ//dT)/(n)=5R` .....(i)
`C=C_(V)+(R)/(1-x)=(5)/(2)R+(R)/(1-x)` ....(ii)
From equ. (i) `&` (ii).
get `x=(3)/(5)`
So, process equ. is
`PV^(3//5)=` const.
if `PdownarrowrArrVuparrow rArrW=+ve`
To find relation between `T` and `V`, put `P=(NRT)/(V)`
`((nRT)/(V))(V^(3//5))=` constant
`VuparrowrArrTuparrow rArr` internal energy will increase.
Discussion
4604.

What is threshold frequency ?

Answer»

SOLUTION :The MINIMUM FREQUENCY of incident radiation used for ejection of electrons from metallic surface .
Discussion
4605.

At the momentt = 0 a relativsiticproton files with avelocityv_(0) into theregion where there is a unifromtransverseelectric field of strengthE, withv_(0) _|_ E. Findthe time dependence of (a)theangle theta between the proton's velocity vector v and the initial direction of its motion, (b) the projection v_(x)of the vector v on the initialdirection of motion.

Answer»

Solution :The EQUACTIONS are,
`(d)/(dt) ((m_(0) v_(x))/(sqrt(1 - (v^(2)//C^(2))))) = 0` and `(d)/(dt) ((m_(0) v_(y))/(sqrt(1 - (v^(2)//c^(2))))) = E E`
Hence, `(v_(x))/(sqrt(1 - v^(2)//c^(2)))` = CONSTANT `= (v_(0))/(sqrt(1 - (v_(0)^(2)//c^(2))))`
Also, by energy conservation,
`(m_(0) c^(2))/(sqrt(1 - (v^(2)//c^(2)))) = (m_(0) c^(2))/(sqrt(1 - (v_(0)^(2)//c^(2)))) + e Ey`
Dividing `v_(x) = (v_(0) epsilon_(0))/(epsilon_(0) + e Ey), epsilon_(0) = (m_(0) c^(2))/(sqrt(1 - (v_(0)^(2)//c^(2))))`
Also, `(m_(0))/(sqrt(1 - (v^(2)//c^(2)))) = (epsilon_(0) + e E y)/(c^(2))`
Thus, `(epsilon_(0) + e Ey) v_(y) = c^(2) e E t +` constant
Intergating again,
`epsilon_(0) y + (1)/(2) e E y^(2) = (1)/(2) c^(2) E t^(2) +` CONSTNAT.
"contast" = 0, as `y = 0`, at `t = 0`.
Thus, `(ce E t)^(2) = (e y E)^(2) + 2 epsilon_(0) e E y + epsilon_(0)^(2) - epsilon_(0)^(2)`
or, `ceEt = sqrt((epsilon_(0) + e Ey)^(2) - epsilon_(0)^(2))`
or, `epsilon_(0) + e Ey = sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2))`
Hence, `v_(x) = (v_(0) epsilon_(0))/(sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2)))` also, `v_(y) = (c^(2) e Et)/(sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2)))`
and`tan theta = (v_(y))/(v_(x)) = (e Et)/(m_(0) v_(0)) sqrt(1 - (v_(0)^(2)//c^(2)))`.
Discussion
4606.

A resistor of 200 Omega and a capacitor of 15.0 muF are connected in series to a 220V, 50Hz as source . (a) Calculate the current in the circuit. (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the alebraic sum of these voltages more than the source voltage? If yes resolve the paradox.

Answer»

SOLUTION :Here, for series RC a.c. circuit,
`I_(rms) = ( V_(rms))/(sqrt(R^(2) + X_(C )^(2))) = ( V_(rms))/( sqrt(R^(2) + ( 1)/( 4 pi^(2) f^(2) C^(2))))`
`:. I_(rms)= ( 220)/( sqrt((200)^(2) + (1)/( 4 xx (3.14)^(2) xx ( 50)^(2) xx ( 15 xx 10^(-6))^(2))))`
`:. I_(rms) = 0.755A`
(b) (i) `V_(R )- I_(rms) R `
`= ( 0.77) ( 200)`
= 151 Volt
(ii) `V_(C ) = I_(rms) X_(C )`
`= I_(rms)xx (1)/( 2PI f C )`
`= ( 0.755) xx (1)/( (2) ( 3.14) ( 50) ( 15 xx 10^(-6))) `
= 160.3 Volt
(c ) Here `V_(R )` an `V_(C )` are not in same PHASE. Hence we can not MAKE their algebraic sum to get resultant voltage.
Actually, phase difference between` V_(R )` and `V_(C )` is `( pi )/(2)` rad and so resultant voltage can be found out from following phasor diagram.
`V = sqrt( V_(R )^(2) + V_(C )^(2))`
`= sqrt((151)^(21) + ( 160.3)^(2))`
`:. V = 220 `Volt

Above voltage is equal to rms value of source voltage.
Discussion
4607.

In hydrogen atom, when electron jumps from second to first orbit, then energy emitted is

Answer»

`-13.6 eV`
`-27.2eV`
`-6.8 eV`
Noneof these

Solution :For hydrogen ATOM we know that `E_(1) = - 13.6 eV` and `E_(2) = -3.4 eV`. Hence, energy RADI ated for TRANSITION from n =2 to n = 1 state is `E= -3.4-(-13.6) - 10.2 eV`. Hence, option (d) is the correct answer.
Discussion
4608.

Calculate the mutual inductance between two coils when a current of 2A changes to 6A in 2 seconds and induces an emf of 20mV in the secondary coil.

Answer»

Solution :`E= -M(dI)/(dt)`
`-20 xx 10^(-3)= - M (6-2)/(2)`
or M= 10MH
Discussion
4609.

A particle performing S.H.M. has a velocity of 10 m//s, when it crosses the mean position. If the amplitude of oscillation is 2 m, then what is the velocity at midway between mean position and extreme position ?

Answer»

SOLUTION :`v=omegasqrt(a^2-x^2) = omegasqrt(a^2-(a/2)^2)`
` THEREFORE x=a/2 `
`=omegasqrt((3a^2)/4)=(aomegasqrt3)/2=v_max/2sqrt3=(10sqrt3)/2`
` therefore v=5sqrt3 m/s.`
Discussion
4610.

X-rays are

Answer»

STREAMS of NEGATIVELY charged particles
Streams of protons
Stream of electromagnetic RADIATIONS of Stream of electromagnetic radiations of
Streams of electromagnetic radiations of nuclear origin

Answer :C
Discussion
4611.

A block of wood mass 5kg is placed on a plane making an angle 30^(@) with the horizontal. If the co-efficient of friction between the surface of contact of the body and plane is 0.5. What force is required to keep the body sliding down with uniform velocity

Answer»

1.6 N
4.8 N
6.4 N
0.33 N

Answer :D
Discussion
4612.

A diatomic molecular has moment of inertia I. By Bohr's quantization condition its rotational energy in the n^(th) level (n = 0 is not allowed) is

Answer»

`1/(N^(2))((h^(2))/(8 pi ^(2) I))`
`1/n((h^(2))/(8 pi ^(2)I))`
`n ((h^(2))/(8 pi^(2)I))`
`n^(2)((h^(2))/(8 pi^(2)I))`

Solution :Angular momentum, `L = (nh)/(2pi)`
ROTATION K.E = `(L^(2))/(2I) = (n^(2)h^(2))/(8pi^(2)I)`
Discussion
4613.

What excuse did Beinkensopp give for missing his maths test?

Answer»

He WENT to a dentist
Grandmother’s death
Bad hand
None of the above

Answer :A
Discussion
4614.

If the light is replaced by blue light illuminating the object in a microscope then the resolving power of the microscope

Answer»

decreases
increases
gets halfed
remains unchanged

Answer :B
Discussion
4615.

The SI unit of surface integral of electric field is

Answer»

V
Vm
`NC^(-1)`
`NC^(-1)m`

ANSWER :B
Discussion
4616.

A fish which is at a depth of 12cm in water (mu=4/3) is viewed by an observer on the bank of a lake. Its apparent depth as observed by the observer is …..cm.

Answer»

3
9
12
16

Solution :When image FORMED in denser medium is observed from RARER medium.
imaginary depth= `(REAL depth)/N`
(where n= refractive index of denser medium with respect to rarer medium)
`therefore d_1=d_r/n=12/((4/3))=36/4=9cm`
Discussion
4617.

Horizontal range is same for the direction of projection theta and _____.

Answer»


ANSWER :`[(PI)/2-theta]`
Discussion
4618.

For an electromagnetic wave propagating along x-axis E_x=30 v/m. What is the maximum value of magnetic field ?

Answer»

`10^(-7)T`
`10^(-8)T`
`10^(-9)T`
`10^(-6)T`

Answer :A
Discussion
4619.

Find the magnetic induction due to a straight conductor of length 16cm carrying current of 5A at a distance of 6cm from the midpoint of conductor

Answer»

Solution :`B=(mu_(0))/(4PI)I/r(sintheta+sintheta)`
but `sintheta=8/10=4/5`
`B=10^(-7)XX5/(6xx10^(-2))xx2xx4/5`
`=40/3muT`
Discussion
4620.

A small block is resting on an inclined plane (co-efficient of frictionmu gt tan theta) as shown in the figure. The inclined plane is given a constant horizontal acceleration 'a' towards right. (a) Find the range of 'a' such that the bock does not slide on the plane. (b) Find the value of 'a' such that the firction force between the block and the plane is zero.

Answer»


ANSWER :`0 " to " (mu+TAN THETA)/(1-mu tan theta).G, a=g tan theta`
Discussion
4621.

Suppose that the sound level of a conversation is initially at an angry 75 dB and then drops to a soothing 55 dB. Assuming that the frequency of the sound is 500 Hz, determine the (a) initial and (b) final sound intensities and the( c) initial and (d) final sound wave amplitudes.

Answer»

Solution :`(a) 32mu W//m^2 , (b) 0.32 mu W//m^2 , ( C) 1.24 xx 10^(-7) m ~~ 0.12 MUM , (d) 1.24 xx 10^(-8) m ~~ 0.012 mu m `
Discussion
4622.

What is the advantage of using thick metallic strips to join wiresin a potentiometer ?

Answer»

SOLUTION :RESISTANCE of thick METALLIC strips is extremely small and hence NEGLIGIBLE.
Discussion
4623.

Which wave is used for studying crystal structure?

Answer»

SOLUTION :X - RAYS
Discussion
4624.

When a freely suspended bar magnet is heated, its magnetic dipole moment decreases by 36 %. Then its periodic time would

Answer»

increase by 36 %
increase by 25 %
decrease by 25 %
decrease by 64 %

Solution :Initial periodic time of oscillation,
`T= 2PI sqrt( (I)/( mB_h))`
(Where `B_h=` horizontal component of Earth.s MAGNETIC field)
Here I and `B_h` remain CONSTANT and so,
`T prop (1)/( sqrt(m))`
` therefore (T_1)/( T_2) = sqrt((m_2)/( m_1))`
`therefore (T_1)/( T_2)= sqrt((64)/( 1000)) (because ` If `m_1 =100` unit, then `m_2=100-36 =64` unit)
`therefore (T_1)/( T_2) = (8)/(10)`
`therefore T_2 = (10)/(8) T_(1) = 1.25 T_(1)`
Percentage increase in periodic time,
`(T_2 - T_1)/( T_1) xx 100% = (0.25 T_1)/( T_1)xx 100% = 25%`
Discussion
4625.

प्रक्कथन : किसी वाहन का स्पीडोमीटर वाहन की औसत चाल को मापता है। कारण: औसत वेग, कुल विस्थापन तथा कुल लगे समय के अनुपात के बराबर होता

Answer»

प्रक्कथन और कारण दोनों सही हैं और कारण प्रक्कथन का सही स्पष्टीकरण देता है
प्रक्कथन और कारण दोनों सही हैं किन्तु कारण प्रक्कथन का सही स्पष्टीकरण नही देता है
प्रक्कथन सही है किन्तु कारण गलत है
प्रक्कथन गलतहै किन्तु कारण सही है

Answer :D
Discussion
4626.

What is most important propoerty of gamma rays ?

Answer»

Solution :They can PENETRATE HIGH. They can penetrate through several CENTIMETERS of THICK IRON and lead blocks.
Discussion
4627.

A : The geometrical shape of the wave front when a wavefront passes through a convex lens will be again plane wave front R : Whne a plane wave front is reflected by a concave mirror, it remains as plane wavefront.

Answer»

Both A and R are TRUE and R is the CORRECT explanation of A
Both A and R are true and R is not the correct explanation of A
A is true and R is false
Both A and R are false

Answer :D
Discussion
4628.

A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron (a)in a head-on collision (b) in scatterting at right angles.

Answer»

Solution :In a head on collision
`sqrt(2mT)=p_(d)+p_(n)`
`sqrt(2mT)=p_(d)+p_(n)`
`T=(p_(d)^(2))/(2M)+p_(n^(2))/(2m)`
where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring
`p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT`
`p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT`
or since `p_(d)=0` in a head on collisions
`P_(n)= -(1)/(2)(1-(m)/(M))p_(d)`
Going BACK to energy CON
where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring
`p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT`
`p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT`
or since `p_(d)=0` in a head on collisions
`P_(n)= -(1)/(2)(1-(m)/(M))p_(d)`
Going back to energy conservation
`(p_(d)^(2))/(2M)[1+(M)/(4m)(1-(m)/(M))^(2)]=T`
so `(p_(d)^(2))/(2M)=(4mM)/((m+M)^(2))=(8)/(9)`
(b) In this case neutron is scattered by `90^(@)`. Then
we have from the DIAGRAM
`vec(p)_(d)= p_(n)hat(j)+sqrt(2mT)hat(i)`
Then by energy conservation
`(p_(n)^(2)+2mT)/(2M)+(p_(n)^(2))/(2m)=T`
or `(p_(n)^(2))/(2m)(1+(m)/(M))=T(1-(m)/(M))`
or `(p_(n)^(2))/(2m)=(M-m)/(M+m).T`
The energy LOST by neutron in then
`T-(p_(n)^(2))/(2m)=(2m)/(M+m)T`
or fraction of energy lost is `eta=(2m)/(M+m)=(2)/(3)`
Discussion
4629.

Potenlials of points P and Q are 10Vand -4 V respectively. Work done in taking 100 electrons from P to Q ........

Answer»

`22.4xx10^(-16) J`
`2.24 xx10^(-16) J`
`-9.6xx10^(-17) J`
`9.6 xx10^(-17) ` J

Solution :Let `DeltaV=V_(Q)-V_(P)=-4-10=-14V`
`:. W = q(V_(2)-V_(1))=(V_(Q)-V_(P))q`
`=(-4-10)(-1.6xx10^(-19)xx100) ( because q = `ne )
`=14xx1.6xx10^(-17)=2.24xx10^(-16)J`
Discussion
4630.

Five capacitors each of capacitance value C are connected as shownin the figure . The ratio of capacitance between P and R and the capacitance between P and Q is

Answer»

`3:1`
`5 :2`
`2:3`
`1:1`

Solution :Answer (3)
`(C_(PR))/(C_(PQ))=((C)/(2)+(C)/(3))/(C+(C)/(4))`
Discussion
4631.

Two particles are projected from two points 4m apart on the horizontal ground, simultaneously with speed 20ms^(-1) above 60^(@) and 15ms^(-1) above 30^(@) from horizontal towards each other. Find the shortest distance between them during the flight.

Answer»

3.9m
2.8m
1.7m
1.5m

Solution :DISTANCE `=4 COS (60-53)`
Discussion
4632.

The dimensional formula of electric field intensity is ......

Answer»

`M^(1)L^(1)T^(-1)A^(-1)`
`M^(1)L^(2)T^(-3)A^(-1)`
`M^(1)L^(1)T^(-3)A^(-1)`
`M^(1)L^(0)T^(-3)A^(-1)`

Solution :UNIT of E `=N/C`
`therefore |E| =(N)/(C )`
`=(M^(1)L^(1)T^(-2))/(A^(1)T^(1)) =M^(1)L^(1)T^(-3)A^(-1)`
Discussion
4633.

What are the basic characterstics of Lassers? Mention some of the applications.

Answer»

Solution :Some of the basic characteristics of Lasers are : (i) Laser light is highly monochromatic, (ii) Laser light is highly coherent, (III) Laser light is highly DIRECTIONAL. (iv) Laser light can be sharply ofcussed. Some of the IMPORTANT applications of Laser are in : (i) PERFORMING surgery of many kinds, (ii) voice and data transmission, (iii) surveying , (iv) welding autobodies, (v) manufacturing and reading compact discs and DVDs, (vi) garment industry (CUTTING several hundred layers of cloth at a time), (vii) generting holograms.
Discussion
4634.

The equation of wave on a string of linear mass density 0.04 kg m^(-1)is given by y = 0.02 (m) sin [ 2 pi ( (t)/(0.04(s)) - (x)/(0.50(m)) ) ] The tension in the string is :

Answer»

6.25 N
4.0 N
12.5 N
0.5 N

Solution :` v_(2) sqrt((T)/(mu)) or t = V^(2) mu`
from GIVEN EQUATION v = `(50)/(4) = ((25)/(2)) ms^(-1)`
`therefore "" T = (625)/(4) xx 0.04 ` = 6.25 N
so CORRECT choice is (a).
Discussion
4635.

Accuracy in momentum of electron moving with velocity of 50 ms^(-1) is 0.005 %. Then accuracy in its position is…..

Answer»

`2.32xx10^(-2)m`
`46xx10^(-4)m`
`46xx10^(-3)m`
`46xx10^(-6)m`

SOLUTION :Momentum of electron p=mv
`therefore` uncertainty in momentum
`=mvxx0.005%`
`=9.1xx10^(-31)xx50xx(0.005)/(100)`
`=2.275xx10^(-33)kg ms^(-1)`
Corresponding uncertainty In position ,
`Deltax=(H)/(2piDeltaP)`
`=(6.625xx10^(-34))/(2xx3.14xx2.275xx10^(-33))`
Discussion
4636.

Value of gyromagnetic ratio for an orbital electron is ________

Answer»

`8.8xx10^(10)CKG`
`8.8xx10^(10)C^(-1)KG`
`8.8xx10^(10)Ckg^(-1)`
`8.8xx10^(10)C^(-1)kg^(-1)`

ANSWER :C
Discussion
4637.

An element Delta l = Delta x hatiis placed at the origin and carries a large current I = 10 A . What is the magnetic field on the y-axis at a distance of 0.5 m. Deltax = 1 cm.

Answer»

Solution :`dB=(mu_(0))/(4PI)(I(vec(dl)xxvecr))/(r^(3))`
`=(10^(-7)xx10xx10^(-2))/(25xx10^(-2))HATK`
`4XX10^(-8)T(hatk)`
Discussion
4638.

What wave optic deals ?

Answer»

Solution :It DEALS with the PHENOMENA of INTERFERENCE , diffraction and POLARISATION.
Discussion
4639.

A 100V voltmeter having an internal resistance of 20 kOmega is connected in series with a large resistance R across a 110 V line. What is the magnitude of resistance R if the voltmeter reads 5 V?

Answer»

Solution :
As SHOWN in fig., the voltmeter is in series with R so
`V=V_(1)+V_(R)`, i.e., `110=5+V_(R)`
i.e., `V_(R)=110-5=105V`
And as in series POTENTIAL DIVIDES in proportion to resistance,
`(V_(R))/(V_(1))=R/(R_(1)),` i.e., `105/5=R/(20kOmega)` or
`R=420kOmega`
Discussion
4640.

The refractive index of glass w.r.t. a medium is 4\3. If V_m - V_g = 6.25 xx 10^7 m//s, then the velocity of light in the medium will be :

Answer»

`2.5 XX 10^8m//s`
`2.25 xx 10^8m//s`
`1.875 xx 10^8m//s`
`1.5 xx 10^8m//s`

ANSWER :A
Discussion
4641.

The safe current in an insulated aluminium wire of1mm^(2) cross section is 8 A. Find the average drift velocity of the conduction electrons. Assuming the Density of Al to be 2.7 g/cm^3

Answer»


ANSWER :0.24 mm/s
Discussion
4642.

d/dx (x^6/x^8) =?

Answer»

`(x^2+2x^3 )/x^6`
`-2/x^3`
`2x^3`
-`2x^3`

ANSWER :B
Discussion
4643.

In p-type semiconductor holes move in

Answer»

extra ELECTRON in valence BAND
extra electron in CONDUCTION band
MISSING electron in valence band
missing electron in conduction band

Solution :missing electron in valence band
Discussion
4644.

A train takes t sec to perform a journey , if travel for t/n sec with uniform acceleration then for ((n-3)/(n))t sec with uniform speed v and finally it comes to rest with uniform retardation. Then average speed of train is

Answer»

`(3N - 2)(v)/(2n)`
`(2n - 3)(v)/(2n)`
`(3n- 2)v/(3n)`
`(2n - 3)(v)/(3n)`

ANSWER :B
Discussion
4645.

Aperture of a spherical mirror is 20 cm, then circumference of this spherical mirror will be

Answer»

3.14 CM
6.28 cm
20 cm
62.8 cm

SOLUTION :Diameter of spherical mirror
2R
`therefore 20=2r`
`therefore` r=10 cm
`therefore` Circumference =2r
=`2xx3.14xx10`
=62.8 cm
Discussion
4646.

Why do two magnetic lines of force never intersect each other ?

Answer»

Solution :Becauseif two MAGNETIC lines of FORCE intersect each other, then there will be two DIRECTIONS of magnetic field intensity at the POINT of intersection .
Discussion
4647.

The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a He+ ion in the first excited state will be .....

Answer»

`-13.6eV`
`-27.2eV`
`-54.4eV`
`-6.8eV`

Solution :The ENERGY of a hydrogen like atom in nth orbit is given by
`E_(n)=-Z^(2)xx(13.6)/(n^(2))eV`
For first EXCITED STATE of `He^(+),n=2,Z=2`
`:. E_(He^(+))=-(4)/(2^(2))xx13.6=-13.6eV`
Discussion
4648.

A magnet is moving towards a coil along its axis and the emf induced in the coil is epsilon. If the coil also starts moving towards the magnet with the same speed, the induced emf will be ____

Answer»

`epsilon/2`
`epsilon`
`2epsilon`
`4epsilon`

SOLUTION :INDUCED emf =Bvl (VALUE)
`THEREFORE emf_1 prop v` where v is relative velocity [ `because` B and l are constant]
Here when relative velocity becomes 2V, the induced `emf_2 prop 2v`,
`therefore (emf_2)/(emf_1)=(2v)/v`
`therefore emf_2=2xxemf_1`
`therefore emf_2=2 epsilon "" [ because emf_1=epsilon]`
Discussion
4649.

Figure 8-3 shows four situations in which the same box is pulled by an appiled force vec(F) up a frictionless ramp through (and then past ) the same vertical distance In each situation, The force has a magnitude of 10 N. In situations (b) and (d), the force is directed along the plane: in situations (a) and (c). it is directed at an angle phi=37^(@) to the plane. as shown. Rank the situations according to the work done on the box in the vertical distance by the appiled force. Also, discuss whether answer depends on the initial speed of box or the presence of other forces.

Answer»

Solution :KEY IDEAS
Situations (b) and (d) will involve more work than (a) and (c), respectively, because though displacement in (b) and (a) is same and similarly displacement in (d) and (a) is also same, but in (b) and (d) force is acting ALONG displacement. Thus, scalar product of force and displacement will be larger.
CALCULATIONS: Now, we compare magnitude of displacement in (b) and (d), we can see easily that displacement is larger for (d). Thus, by similar ARGUMENT, we can prove that displacement of (c) is larger than (a).
Hence, `W_(d) gt W_(b)`, and `W_(c) gt W_(a)`
Numerical calculations also reveal that our arguments are correct.We find the relation between`W_(b)` and `W_(c)`.
In order to deriver the reltions for work done, we first need to calculate the displacement, that is length of

Figure 8-3 Boxes being pulled on INCLINES by applying forces in different directions.
the ramp in terms of virtual distance Also, in situations the force (F) needs to be resolved along the direction of displacment to determine the work done.
`W_(d) = Fh cosec 37^(@) = (5Fh)/(3)`
`W_(b) = Fh cosec 53^(@) = (5Fh)/(4)`
`W_(c) = (F cos 37^(@))(h cosec 37^(@))=(4 Fh)/(3)`
`W_(d) = (F cos 37^(@))(h cosec 53^(@))= Fh`
Hence `W_(d) gt W_(c) gt W_(b) gt W_(a)`
Discussion
4650.

Minimum energy required to take out the only one electron from ground state of He is:

Answer»

13.6 eV
54.4 eV
27.2 eV
6.8 eV

Answer :B
Discussion