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41201.

Assertion: The energy (E) and momentum (p) of a photon are related as p=(E)/(c). Reason: The photon behaves as a particle.

Answer»

If both ASSERTION and reason are TURE and the reason is the correct explanation of the assertion.
If both assertion and reason are TRUE but reason is not the correct explanation of the assertion.
If assertion is true but reason is false
If the assertion is false but reason is true

Solution :For photon particles `p=(H)/(LAMDA) and E=hv=(hc)/(lamda) implies p=(E)/(c)`
Discussion
41202.

The frequency response curve of RC coupled amplifier is shown in figure . The band width of the amplifier will be :

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`f_3 -f_2`
`f_4-f_1`
`(f_4-f_2)/2`
`f_3-f_1`

ANSWER :B
Discussion
41203.

At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m^(3)s^(-1). If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g=9.8ms^(-2)).

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Solution :Hydroelectric power `=h rho g XX A xx v = h rho g beta`
where `beta=Av` is the flow (VOLUME of WATER flowing PER second ACROSS a cross-section).
Electric power available `=0.6xx300xx10^(3)xx9.8xx100W`
`=176MW`
Discussion
41204.

A wire is made by welding two wires (radius r and 2r) at one of its end. On using this combined wire in the sonometer , under tension T the welding point is just in middle of the bridges. On vibrating stationary waves are produced. If welding points is a node then the ratio

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

SOLUTION :As `(p)/(r) ` is a CONSTANT `p_(1) : p_(2) : : r : 2r `
RO `p_(1) : p_(2) `: : 1 : 2 `"" ` (here p = no . Of loops )
correct choice is (d).
Discussion
41205.

A vessel ABCD of 10 cm width has two small slits S_(1) and S_(2) sealed with identical glass plates of equal thickness. The distance between the slits is 0.8mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S_(1). A monochromatic light source is kept at S, 40 cm below P and 2m from the vessel to illuminate the slits as shown in fig. The position of the central bright fringe on the other wall CD w.r.t. the line OQ.

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4 cm
6 cm
5 cm
2 cm

Answer :D
Discussion
41206.

A potential difference of 2V is applied between the opposite faces of a Ge crystal of area1 cm^2 and thickness 0.5 mm . If the concentration of electrons in Ge is 2xx10^(19)//m^3 and mobilities of electrons and holes are 0.36(m^2)/("volt"-s) and 0.14(m^2)/("volt"-s)respectively , then the current flowing through the plate will be

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0.25 A
0.45 A
0.56 A
0.64 A

Solution :`sigma = "NE"=(mu_e+mu_h)=2xx10^(19)xx1.6xx10^(-19)`
`(0.36+0.14)`
`=1.6(Omega -m)^(-1)`
`R=rho (l)/A=l/(sigmaA)=(0.5xx10^(-3))/(1.6xx10^(-4))=25/8Omega`
`:. i=V/R=2/(25//8)=16/25A=0.64A`.
Discussion
41207.

The kinetic energy of a proton and that of an alpha - particle are 4 eV and 1 eV , respectively. The ratio of the de - Broglie wavelengths associated with them , will be

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<P>`2:1`
`1:1`
`1:2`
`4:1`

Solution :[Hint : As per RELATION `lamda_(DE)=h/(sqrt(2mK)),lamda_p/lamda_(ALPHA)=sqrt((m_alphaK_(alpha))/(m_pK_(p)))=sqrt((4m_pxx1eV)/(m_pxx4eV))=1`]
Discussion
41208.

An ionized gas contains both positive and negative ions . If it is subjected simultane-ously to an electric fi eld along the +x direction and a magnetic field along the +z direction, then

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positive IONS DEFLECT TOWARDS + y direction and NEGATIVE ions towards - y direction
all ions deflect towards +y direction
all ions deflect towards - y direction
positive ions deflect towards - y direction and negative ions towards + y direction.

Answer :C
Discussion
41209.

In the previous question, maximum permissible error is resitivity and resistance measurement will be

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`2.14%`, `1.5%`
`1.5%`, `2.45%`
`2.41%`, `1.1%`
`2.45%`, `1.5%`

SOLUTION :`((DR)/(R))_(MAX)=(DELTAI)/(i)+(Deltav)/(v)=`
`(0.1)/(10.0)+(0.1)/(100.0)=1.1% ((dR)/(R))_(max)=2.41%`
Discussion
41210.

Statement-I: The temperature at which centigrade and Fahrenheit thermometers read the same is -40^(@). Stament-II : There is no relation between Fahrenheit and centigrade scale of temperature.

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STATEMENT-I is TRUE, statement-II is true and
statement-II is CORRECT EXPLANATION for statement-I.
Statement-I is true, statement-I is true and
statement-II is not correct explanation of statement-I.
Statement-I is true, statement-II is FALSE.
Statement-I is false, statement-II is false.

Solution :Statement I is true, statement II false. So, correct choice is (c ).
Discussion
41211.

In NOT operation of Y is output and A and B are inputs then truth table in

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0,0,1,0
1,1,1,0
1,1,0,0
1,1,1,1

Answer :A
Discussion
41212.

X rays are produced in an x-ray tube by electrons accelerated through an electric potential difference of 50.0 kV. Let K_(0) be the kinetic energy of an electron at the end of the acceleration. The electron collides with a target nucleus (assume the nucleus remains stationary) and then has kinetic energy K_(1)=0.500K_(0). (a) What wavelength is associated with the photon that is emitted? The electron collides with another target nucleus (assume it, too, remain stationary) and then has kinetic energy K_(2)=0.500K_(1). (b) What wavelength is associated with the photon that is emitted?

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SOLUTION :(a) 49.6 PM, (B) 99.2 pm
Discussion
41213.

Consider the situation shown in the given below. What are the sign ofq_1 and q_2?

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SOLUTION :`q_1 ` is -ve but `q_2`is +ve , because FIELD linesstart from +ve CHARGE and END at -ve charge.
Discussion
41214.

What will happen if the electrodes of a cell are placed closer to each other and if their size is made larger?

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Solution :If the ELECTRODES are placed closer to each other the IONS of the reacting electrolyte can move easily from ONE electrode to another. Again, if the SIZE of the electrodes is large, then ions have the opportunity to move across a large space. So in both the cases internal resistance of the cell decreases. As a result, the cell can send a larger amount of current in the external circuit.
Discussion
41215.

What is dgital communication?

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Solution :The communication system which uses DIGITAL SIGNAL (the signal which has only two values-either HIGH or LOW), is CALLED digital communication system.
Discussion
41216.

Consider sunlight incident on a slit of width 10^(4) Å. The image seen through the slit shall

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be a fine sharp slit white in COLOUR at the CENTRE
a bright slit white at the centre diffusing to zero INTENSITIES at the edges.
a bright slit white at the centre diffusing to regions of different colours
only be diffused slit white in colour.

Solution :Given, width of the slit , `a = 10^4 Å`. The wavelength of sunlight (visible light) varies from `lambda = 4000 Å` to `7500 Å`. As, `a > > lambda`, DIFFRACTION cannot TAKE place.
So, the image seen throught the slit shall be a fine sharp slit, white in colour at the centre.
Discussion
41217.

A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the centre of the coil, such that r > > R, varies as

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`1/R`
`1/(r^(3//2))`
`1/(r^2)`
`1/(r^3)`

ANSWER :D
Discussion
41218.

Discuss the diffraction at a grating and obtain the condition for the m^(th)maximum.

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Solution : (i) A plane transmission grating is represented by AB in Figure. Let a plane wavefront of monochromatic light with wave length `LAMDA` be incident normally on the grating.As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.

(ii) A diffraction pattern is obtained on thescreen when the diffracted waves are focused on a screen using a convex lens. (III) Let US consider a point P at an angle `theta` withthe normal drawn from the center of the grating to the screen. The path difference `delta` between the diffracted waves from one pair of corresponding points is,
` delta = (a+b) sin theta`
This path difference is the same for any pair ofcorresponding points. The point P will be bright,when
`delta= m lamda `where m= 0, 1, 2, 3 .
Combining the above two equations, we get,
`(a+b) sin theta = m lamda `
Here, m is called ORDER of diffraction.
Condition for zero order maximum, m=0 :
For `(a + b) sin theta`= 0, the position, `theta = 0, sin theta = 0 ` and m=0. This is called zero order diffraction or central maximum.
Condition for first order maximum, m=1:
If `(a + b ) sin theta_1 = lamda` ,the diffracted light meet at an angle `theta_1` to the incident direction and the first order maximum is obtained.
Condition for second order maximum, m=2 :
Similarly, `(a + b) sin theta_2 = 2lamda`forms the second order maximum at the angular position `theta_2`
Condition for higher order maximum :
(i) On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.If we take,
` N = (1)/( a+ b)`
(ii)Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is SPECIFIED on the grating itself. Now, the equation becomes
` 1/N sin theta =m lamda " or " sin theta = N m lamda`
Discussion
41219.

{:("(i)Electric charge","(a) Cm"),("(ii) Dipole moment","(b)"Nm^(2)C^(-1)),("(iii) Electric field","(c) coulomb"),((iv)"electric flux","(d)"NC^(-1)):}

Answer»


ANSWER :(i)`RARR` (B) ; (ii)`rarr` (d); (iii) `rarr` (a); (iv)`rarr`(c)
Discussion
41220.

A resistor 30Omega, inductor of reactance 10Omega and capacitor of reactance 10Omega are connected in series to voltage source E = 300sqrt2 sin(omegat). Find the current in the circuit.

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14.14A
10A
30A
20A

Solution :Here, R=30 `Omega`
`X_L =10 Omega`
`X_C=10 Omega`
`E_m =300sqrt2V`
Here, `|Z|=SQRT(R^2+(X_L-X_C)^2)`
`=sqrt((30)^2 +(10-10)^2)`
`therefore |Z|=30Omega`
`E_(rms)=E_m/sqrt2 =(300sqrt2)/2`=300 V
`therefore I_(rms)=E_(rms)/"|Z|"`
`=300/30`
`therefore I_(rms)=10 A`
Discussion
41221.

Which statement is correct for a zone plate and a lens

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ZONE plate has MULTI focii WHEREAS LENS has one
Zone plate has one focus whereas lens has MULTIPLE focii
Both are correct
Zone plate has one focus whereas a lens has infinite

Answer :A
Discussion
41222.

Which places had Evelyn worked for?

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POOR children
prisons and hospitals
schools
none

Answer :B
Discussion
41223.

If each diode forward resistance is 50omega then find current through the 100omega resistor in the given circuit.

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Solution :The DIODE `D_(1)` is forward biased but diode `D_(2)` is reverse biased. So no CURRENT pass through the diode `D_(2)`.
`:.` Current passes through `100OMEGA` resistor `=(6)/(50+150+100)=0.02A`
Discussion
41224.

(a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if lambda 550 nm, d = 0.180 mm, and a = 30.0 mu m? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

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SOLUTION :(a) 11, (B) 0.405
Discussion
41225.

The level of water in a tank is 5 m high. A hole of area 1 cm^(2) is made at the bottom of the tank. The rate of leakage of water from the hole is : (g = 10 m//s^(2))

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`10^(-3)m^(3)//s`
`10^(-4)m^(3)//s`
`10^(-1)m^(3)//s`
`10^(-2)m^(3)//s`

Solution :VELOCITY with which liquid comes out of the hole
`v=sqrt(2gh)`
`THEREFORE` Rate of flow of WATER from the hole
`="area"xx"velocity"=Axxsqrt(2gh)`
`=1xx10^(-4)sqrt(2xx10xx5)=10^(-3)m^(3)s^(-1)`
Thus the correct choice is (a).
Discussion
41226.

An alpha -particle and a proton are fired through the same magnetic fields which is perpendicular to their velocity vectors. The alpha-particle and the proton move such that radius of curvature of their path is same. Find the ratio of their de Broglie wavelengths.

Answer»

<P>

Solution :SINCE, `Bqv=(mv^(2))/(R)ormv=QBR`
The de Broglie WAVELENGTH `lamda=(h)/(mv)=(h)/(qBr)`
`(lamda_(alpha-"particle"))/(lamda_("proton"))=(q_(p)r_(p))/(q_(alpha)r_(alpha))=(1)/(2)(1//2)=(1)/(2)`
Discussion
41227.

Assertion: The ionosphere layer acts as a reflector for all range of frequencies. Reason: Ionosphere does not allow electromagnetic wav e to penetrate and escape.

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Solution :The ionosphere layer acts as a reflector for a certain RANG of frequencies is 3 to 30 MHz. Electromagnetic waves of frequencies higher than 30 MHz PENETRATE the ionosphere and escape.
Discussion
41228.

Consider the RLC circuit shown below connected to an AC source of constant peak voltage V_(0) and variable frequency omega_(0).The value of L is 20 mH.For a certain value omega_(0) = omega_(1), rms voltage across L,C, R are shown in the diagram. At omega_(0) = omega_(2), it is found that rms voltage across resistance is 50 V. Then the value of omega_(2) is

Answer»

`sqrt((3)/(5))omega_(1)`
`sqrt((5)/(3))omega_(1)`
50Hz
cannot be calculated from given data

Solution :If voltage across RESISTOR is 50V then this should be the resonance CONDITION.
At resonance, `X_(L) = X_(C)`
`omega_(2)L = (1)/(omega_(2)C) , omega_(2) = (1)/(sqrtLC)`
Also, At `omega = omega_(1)`
`I = (100)/(X_(L) = (60)/(X_(C)), (100)/(omega_(1)L) = (60)/(1/omega_(1)C)`
`C = (100)/(omega_(1)^(2)L xx 60) = (5)/(3omega_(1)^(2)L)`
`omega_(2) = (1)/(sqrt(L xx (5)/(3omega_(1)^(2)L))) = sqrt((3)/(5))omega_(1)`
Discussion
41229.

What is Galvanometer ? Give application of it.

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Solution :1. A galvanometer is a device to detect current or measure voltage.
2. It detect electric current in circuit.
3. It is generally USED in laboratories as AMMETER, miliammeter or micrometer.
4. By modification in circuit we ALSO can USE it as a voltmeter.
Discussion
41230.

If a p-n junction diode, a square input signal of 10V is applied as shown, then the output signal across R_(L) will be.

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Solution :The junction diode will conduct when it is forward biased. Therefore, the OUTPUT voltage will be OBTAINED during positive half cycle only. So correct option is (3).
Discussion
41231.

Assertion (A) : A bulb connected in series with a solenoid coil is connected to an a.c. source. The bulb glow becomes dimmer when a soft iron core is introduced in the solenoid coil. Reason (R) : On introducing soft iron core in the solenoid coil, its inductance increases.

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If both assertion and reason are true and the reason is the correct explanation of the assertion.
If both assertion and renson are true but reason is not the correct explanation of the assertion.
If assertion is true but reason is FALSE.
If the assertion is false but reason is true.

Solution :On INTRODUCING soft iron core, the INDUCTANCE and hence reactance of solenoid coil increases. So the impedance of L-R CIRCUIT increases and current in the circuit decreases making the BULB dimmer.
Discussion
41232.

When an o-particle of mass m moving with velocity v bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as

Answer»

`(1)/(sqrt(m))`
`(1)/(m_(e))`
m
`(1)/(m)`

Solution :Initial KINETIC ENERGY = Distance from CENTRE -Potential energy
`(1)/(2)MV^(2)=(2Ze^(2))/(4pi epsi_(0)r_(0))`
`:.r_(0)=(2Ze^(2))/(4pi epsi_(0)xx(1)/(2)mv^(2))`
`:. In r_(0) prop (Ze^(2))/(pi epsi_(0)v^(2)m),(Ze^(2))/(pi epsi_(0)v^(2))` constant
`:. r_(0) prop (1)/(m)`
Discussion
41233.

Given below are four logic gate symbol (figure). Those for OR, NOR and NAND are respectively

Answer»

`1,4,3`
`4,1,2`
`1,3,4`
`4,2,1`

ANSWER :C
Discussion
41234.

The charge on the6 muF . Capacitor in the circuit shown is

Answer»

540 `MU`C
`270 mu`C
180` mu`C
90 `mu`C

ANSWER :C
Discussion
41235.

Therectangular box shown is the place of lens. By looking at the ray diagram, answer the following questions. (i) If X is 5cm then what is the focal length of the lens ? (ii) If the point O is 1 cm above the axis then what is the position of the image ? Consider the optical center of the lens to be the origin.

Answer»


ANSWER :10 CM, 10, 2
Discussion
41236.

State the underlying principles of working of a moveing coil galvanometer. Write the two reasons why a galvonmeter cannot be used as such to measure current in a given circuit.Name any two factors on which that current sensitivity of a galvanometer depends.

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Solution :Principle of working of a moveing coil galnometer : The working is bassed on the fact that current CARRYING coil SUSPENDED in a magnetic field experience a torque.
(ii)The galvonmeterf cannot be used to measure the value of the current in a GIVEN circuit due to following two reasons:
(a) Galvanometer is a very DEVICE, its gives a full scale deflection for a current of the order of `muA`.
(b). Fing currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit.
(iii) The current sensitivity of the galvanometer depends upon the number of truns N and cross SECTION area of the coil A.
Discussion
41237.

The echo signal of a RADAR is demodulated by a ______________

Answer»

decoder
transmitter
superhet receiver
rectifier

Answer :C
Discussion
41238.

Which of the following statement about e.m. wave is /are / correct

Answer»

1 and 2
2 and 3
1,2 and 3
2 only

Solution :The ENERGY of a PHOTON, `E=hc//LAMBDA`, since `lambda` for
VISIBLE light is more than that of x-rays, hence
energy of x-rays photon is more than that of light.
Discussion
41239.

Draw the phasor diagram for a series RC circuit connected to an a.c. source.

Answer»

Solution :The PHASOR DIAGRAM for a series RC CIRCUIT is SHOWN in FIG. 7.18.
Discussion
41240.

A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 Me V per nucleon. Calculate the energy Q released per fission in Me V.

Answer»

SOLUTION :`X^(240) to Y^(110) + Z^(130) + Q`
ENERGY released per nucleon `= 8.5 "MeV" -7.6 "MeV"`
`=0.9 "MeV"`
Therefore energy released `=0.9 xx 240`
`= 216 "MeV"`
Alternatively :
Energy released `= [240 xx 8.5 -7.6 (110 + 130) ] :"MeV"`
`=216 "MeV"`
Discussion
41241.

How thermosphere extends from earth's surface?

Answer»

SOLUTION :80 KM to 110 km
Discussion
41242.

A coil has 4000 turns and 500 cm^2 as its area. The plane of the coil is placed at right angles to a magnetic induction field of 4 xx 10^(-5) web/m^2. The coil is rotated through 180^@ in 0.5 seconds. The average emf induced in the coil, inmilli volts, is

Answer»

15
20
16
32

Solution :`E=(NBA)/t(costheta_2-costheta_1)`
`=4xx10^3xx4xx10^-5xx10^-2xx2xx2`
`rArr320xx10^-4=32xx10^-3=32mV`
Discussion
41243.

The pair of equations x + 2y + 5 = 0 and -3x - 6y + 1 = 0 has

Answer»

A UNIQUE solution
Exactly TWO solutions
Infinitely MANY solutions
No solution

Answer :D
Discussion
41244.

An electron with rest mass m_(0) moves with a speed of 0.8c.Its mass when it moves with this speed is…….

Answer»

`m_(0)`
`(m_(0))/(6)`
`(5m_(0))/(3)`
`(3m_(0))/(5)`

Solution :`m=(m_(0))/(SQRT(1-(v^(2))/(C^(2))))=(m_(0))/(sqrt(1-(0.8c)/(c)))^(2)`
`therefore m=(m_(0))/(sqrt(1-0.64))=(m_(0))/(sqrt(0.36))=(m_(0))/(0.6)=(10m_(0))/(6)`
`therefore m=(5m_(0))/(3)`
Discussion
41245.

Prove that total produced in different resistors of the circuit is minimum when the current is divided into a number of braches .

Answer»

SOLUTION :When the current is divided into a number if parallel braches , then`I=I_(1)+I_(2)`
Heat produced in the circuit in time t ,
`H=(I_(1)^(2)r_(1)t)/(J)+((I-I_(1))^(2)r_(2)t)/(J)`

DIFFERENTIATING both SIDE with respect to `I_(1)`,
`(dH)/(dI_(1))=(2I_(1)r_(1)t)/(J)-(2(I-I_(1))r_(2)t)/(J)`
When heat generated in the circuit is minimum ,
`(dH)/(dI_(1))=0" or, "(2I_(1)r_(1)t)/(J)-(2(I-I_(1))r_(2)t)/(J)=0`
or , `(2I_(1)r_(1)t)/(J)=(2(I-I_(1))r_(2)t)/(J)=" or, "I_(1)r_(1)=(I-I_(1))r_(2)=I_(2)r_(2)`
`therefore(I_(1))/(I_(2))=(r_(2))/(r_(1))`
i.e., the heat produed in the circuit will be minimum if the current is divided into the braches such that the current in each branch is INVERSELY proportional to the resistance .
Discussion
41246.

Two masses of 1 kg and 16 kg are moving with equal kinetic energy. The ratio of magnitude of the linear momentum is:

Answer»

`1 : 2`
`1 : 4`
`1 : sqrt(2)`
`sqrt(2) : 1`

SOLUTION :KINETIC energy, `K = 1/2 MV^2 = ((mv)^2)/(2m) = (p^2)/(2m) implies p = sqrt(2 m k) "or" p prop sqrt(m)`
`(P_1)/(P_2) = sqrt(1/16) = 1/4`.
Discussion
41247.

A rectifier is a device which is used to convert a.c. voltage input into d.c. voltage output.

Answer»


ANSWER :TURE
Discussion
41248.

Difference of energy levels goes on______ as we move towards higher energy levels.

Answer»

SOLUTION : DECREASING
Discussion
41249.

Consider the following two statements. a. Linear momentum of a system of particles is zero. b. Kinetic energy of a system of particles is zero. Then

Answer»

a implies B and b implies a.
a does not imply b and b does not imply a
a implies b but b does not imply a
a does not imply b but b implies a

Solution :Here, `E_k=1/2mv^2`
As`E_k=0v=0``(because m ne 0)`
Also, momentum= mv
It v=0 , momentum = 0
Thus (b) implies (a).
But, when there is no RESULTANT momentum, there can NEVER be a net force by second law.This can RESULT in a constant non-zero VELOCITY an hence K.E. Thus (a) does not imply (b).
Discussion
41250.

A spherical Gaussian surface encloses, and is concentric with, a charged spherical conductor. If the TNEI over the Gaussian surface is negative, what can you say about the electric field vecE at a point on the surface?

Answer»

SOLUTION :`VECE`is RADIALLY INWARD.
Discussion