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42601.

In Young.s double slit experiment the distance between the sources is 7.7 mu""m. If the wavelength of light used is 500 nanometre, the angular position of the third dark ringe from the centre fringe is

Answer»

`10.9^(0)`
`0.15^(0)`
`11.3^(0)`
`9.4^(0)`

ANSWER :D
Discussion
42602.

Compare the radii of the nuclei of mass numbers 27 and 64.

Answer»

Solution :`A_1`=27, `A_2`=125
The radius of a nucleus , `R=R_0A^(1//3)`
`rArr R PROP A^(1//3) rArr R_1/R_2=(A_1/A_2)^(1//3) =(27/125)^(1//3)=3/5`
Discussion
42603.

Of the following forces of faction, the one which is self adjusting is

Answer»

ROLLING friction
sliding friction
static friction
dynamic friction

Answer :C
Discussion
42604.

A wire of length l carries a current I along the Y direction and magnetic field is given byvecB = beta/sqrt3 (hat I + hatj + hatk) T. The magnitude of Lorentz force acting on the wire is

Answer»

`sqrt((2)/(sqrt(3)))BETA I L `
`sqrt((1)/(sqrt(3)))betaI l `
`sqrt(2) beta I l `
`sqrt((1)/(2)) beta I l `

SOLUTION :`sqrt((2)/(sqrt(3)))beta I l `
Discussion
42605.

Orbit of a planet around a star is

Answer»

an ellipse
a circle
a parabola
a byperbola

Solution :PLANETS revolve AROUND SUN in ELLIPTICAL orbit.
Discussion
42606.

The earth's magnetic field at a place with zero declination is 3xx 10^(-4) T. The angle of dip at that place is 30^@. A conducting rod is kept is north south direction and is moved at a constant speed of 1 m/s towards east. The emf induced in the rod is ...... (The length of the rod is 10 cm.)

Answer»

`15muV`
`15mV`
0.15V
1.5V

Solution :
`vecl` and `vecv` are MUTUALLY perpendicular
`therefore veclxxvecv=lv hatn`, where `hatn` is unit VECTOR in the downward direction.
`hatn` makes an angle of `60^@` with `vecB`.
`therefore epsilon=vecB.(veclxxvecn)=vecB.lvhatn`
=Blv cos`theta`
=Blv cos `60^@`
`=3xx10^(-4)xx0.1xx1xx1/2`
`=0.15xx10^(-4)=15xx10^(-6)` V
`therefore epsilon=15muV`
Discussion
42607.

Dimension of resistance is:

Answer»

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

ANSWER :B
Discussion
42608.

Large transformers, when used for some time, become hot and are cooled by circulating oil. The heating of transformer is due to

Answer»

Heating effect of CURRENT alone
HYSTERESIS LOSS alone
Both the hysteresis loss and heating effect of current
NONE of the above

Answer :C
Discussion
42609.

A bubble of 8 moles of helium is submerged at a certain depth in water. The temperature of water increases by 30^(@)C How much heat is added approximately to helium during expansion ?

Answer»

4000 J
3000 J
3500 J
4500 J

Solution :Heat added to HELIUM during expansion ,
`H=nC_(v)DeltaT=8xx(3)/(2)Rxx30(C_(V)" for monoatomic GAS" =(3)/(2)R)`
=360R
360xx8.31 ` (R=8.31J"mol"^(-1)-K^(-1))``H~~3000J`
Discussion
42610.

What is inductive reactance of an inductance of 50 henry tpo the flowof D.C.

Answer»

SOLUTION :`X_L=omegaL=2pivL=0`
`THEREFORE` v=0
Discussion
42611.

Specific heat of a gas undergoing adiabatic change is

Answer»

Zero
infinite
positive
nagative

Answer :A
Discussion
42612.

A solenoid 600 mm long has 50 turns on it and its wound on an iron rod 7.5 mm radius. Find the flux through the solenoid when the current in it is 3A. The relative permeability of iron is 600.

Answer»

1.66 Wb
1.66 nWb
1.66 mWb
1.66 `mu` Wb

Solution :`r=7.5 mm = 7.5xx10^(-3)` m
`l=600xx10^(-3)m , N=50, mu_r=mu/mu_0`=600
`THEREFORE mu=600 mu_0, I=3A`
`PHI=(muN^2AI)/l =(600mu_0N^2(pir^2)I)/l`
`=(600xx4pixx10^(-7)xx2500xxpixx(7.5xx10^(-3))^2XX3)/(600xx10^(-3))`
`=4pi^2xx2500xx(7.5)^2xx3xx10^(-10)`
`=10^4xx(3.14)^2 xx(7.5)^2 xx 3xx10^10`
`=1.66xx10^(-3)` Wb
=1.66 mWb
Discussion
42613.

A charged particle of mass m and charge 'q' is projected with a velocity vec V into uniform magnetic field of induction vec B . The force experienced by the charged particle is

Answer»

`F = Q(VEC VXX vec B)`
`vec F= q(vec V .vec B)`
`F= q /((vec V xx vec B))`
`F= vec

Answer :A
Discussion
42614.

When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

Answer»

1.2
2
4
4.8

Answer :C
Discussion
42615.

Estimate the energy of the exchange interaction of electron spin magnetic moments in iron domains.

Answer»


Solution :The Curie point for iron is `770 ^@C`, the energy or exchange INTERACTION MUST EXCEED the energy of thornial motion at temperatures below the Curie point `phi_(C)`, HENCE `epsi_(ex) gt k phi_(C)`.
Discussion
42616.

A crystal diode having internal resistance 200Omega is used as a half rectifier. If the applied voltage is V=50 sin omegat volt and load resistance is 800Omega, find (i) maximum output current (ii) d.c. output current (iii) d.c. output power and (iv) d.c. output voltage.

Answer»


ANSWER :(i) `61mA` (II) `19.4mA` (III) `0.301W` (IV) `15.52V`
Discussion
42617.

Obtain the answers (a) to(b) in Exercise 15 if the circuit is connected to a 110 V, 12 kHz supply ? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Answer»

Solution :GIVEN, the ems value of voltage,
`V_("rms")=110 V`
The frequency of capacitor
`f=12kHz = 12000 Hz`.
Capacitance of conductor `C = 10^(-4)F`.
Tesistance `R = 40 Omega`
CAPACITIVE Resistance
`X_(C )=(1)/(2pi FC)=(1)/(2xx3.14xx12000xx10^(-4))`
`= 0.133 Omega`
The rms value of current
`I_("rms")=(V_("rms"))/(sqrt(X_(C )^(2)+R^(2)))=(110)/(sqrt((40)^(2)+(0.133)^(2)))`
`= 2.75 A`.
The maximum value of current,
`I_(0)=sqrt(2) I_("rms")`
`= 1.414xx2.75`
`= 3.9 A`
Here, the value of `X_(C )` is very small, so term containing C is negligible.
`tan phi = (1)/(omega CR)`
`= (1)/(2xx3.14xx12000xx10^(-4)xx40)`
`= (1)/(96pi)`
It is very small.
In DC circuits, `omega = 0`
`X_(C )=(1)/(omega C)=oo`
So, it behaves like an open circuit.
Discussion
42618.

The half-life of " "_(38)^(90)Sr is 28 years. What is the disintegration rate of 15 mg of this isotope ?

Answer»

Solution :Here half-life period `T_(1/2)`=28 YEARS` = 28 XX 3.154 xx 1^(7)`s, and mass of sample m=15 mg = 0.015 g
`therefore` Number of NUCLEI in given sample of `" "_(38)^(90)Sr N =(0.015xx6.023xx10^(23))/90`
`therefore` Disintegration rate `abs((dN)/(DT))=lambda N=(0.6931 N)/T_(1/2)=(0.6931xx0.015xx6.023xx10^(23))/(90xx28xx3.154xx10^(7))=7.878xx10^(10)Bq =(7.878xx10^(10))/(3.7xx10^(10))C_(i)=2.13C_(i)`.
Discussion
42619.

String A is streched between two clamps separated by distance L. String B, with the same linear density and under the same tension as string A, is stretched between two clamps separated by distance 3L eight harmonies of string B. For which of these eight har. monics of B (if any) does the frequency match the frequency of (a) A's first harmonic, (b) A's second harmonic, and (C) A sthird hannonic?

Answer»


ANSWER :(a) THIRD harmonic of B; (b) SIXTH harmonic of B; (C) None of the first eight harmonics of B WOULD match A's third harmonic
Discussion
42620.

Ship A is sailing towards north-east with velocity vecr=30hati+50hatJ km//hr where hati points east and hatj, north. Ship B is at a distance of 80km east and 150km norht of Ship A and is sailing towards west at 10km//hr. A will be at minimum distance from B in:

Answer»

4.2 h
3.2 h
2.6 h
2.2 h

Answer :C
Discussion
42621.

A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials ?

Answer»

<P>

Solution :The COMPARISON between the spinning of a proton and an electron can be DONE by comparing their magnetic dipole moment which can be given by.
Magnetic dipole moment of proton,
`M_(p)= (eh)/( 4 pi m_(p) )`
`M_(e) = (eh)/( 4 pi m_(e) ) `
`THEREFORE M prop (1)/( m) ""[because (eh)/( 4 pi) " CONSTANT" ]`
`therefore (M_p)/( M_e) = (m_e)/( m_p) `
`therefore M_p = (M_c xx m_c)/( 1837 m_e) ""[because m_(p) = 1837 m_(e) ]`
`therefore M_(p) lt lt M_(e)`
This shows the effect of magnetic moment of proton is neglected as compared to that of electron.
Discussion
42622.

The distances of the satellites of Mars are 25'' for phobos and 62'' for Deimos at mean opposition when Mars is 0.524AU from earth. Calculate distances of theh two satelife for Mars in astonomical units and in meters.

Answer»


ANSWER :NA
Discussion
42623.

Uniform electric field exists in a region and is given by vecE = E_0hati + E_0hatj. There are four points A(-a,0), B(0,-a), C(a,0), and D(0,a) in the xy plane. Which of the following is the correct relation for the electirc potential?

Answer»

`V_A=V_CgtV_B=V_D`
`V_A=V_BgtV_C=V_D`
`V_AgtV_CgtV_B=V_D`
`V_AltV_CltV_BltV_D`

SOLUTION :B. Since potential decreases along electric FIELD as `E=dVldx, V_(A)gtV_(C)` and `V_(B)gtV_(D)`. Also DUE to symmetry
`V_(A)=V_(B),V_(C)=V_(D)`
`V_(A)=V_(B)gtV_(C)=V_(D)`
Discussion
42624.

Two projectile are thrown at an angle thetaand (pi/2-theta), the ratio of their time of flight are

Answer»

`(1 : 1)`
`TANTHETA: 1`
`1 : tantheta`
`tan^2theta : 1`

ANSWER :B
Discussion
42625.

Surface charge density of a sphere of a radius 10 cm is 8.85 xx 10^(-8)c//m^(2). Potential at the centre of the sphere is

Answer»

1000 V
885 V
`10^(-3)` V
442.5 V

Answer :A
Discussion
42626.

Which word has a silent letter?

Answer»

Nincompoop
Kleptomaniac
Knight
Whisper

Answer :C
Discussion
42627.

How are p-type n-type semiconductors formed ?

Answer»

SOLUTION : p-type : When pure/intrinsic semiconductor (germanium/sillicon) is doped with trivalent impurity (INDIUM, gallium, ALUMINUM, boron).n-type : When pure/intrinsic semiconductor (germanium/sillicon) is doped with pentavalent impurity (phosphorous, arsenic, antimony,BISMUTH).
Discussion
42628.

The displacement y in x direction is given by y = 10^(-4) sin(600t - 2x + pi/3)mwhere x is in m, t in s, then speed of wave motion is x xx 10^(2) m//s, what is the value x ?

Answer»


ANSWER :3
Discussion
42629.

Identify correct statement among the following

Answer»

the greater the mass of a pendulum bob, the shorter is its frequency of oscillation
a simple pendulum with a bob of mass m swings with an anguluar amplitude of `40^(@)` its ANGULAR amplitude is `20^(@)`, the tension in the string earlier is less than the tension in the string later
as the length of a simple pendulum is increased, the maximum VELOCITY of its bob during its oscillations will also increases
the fractional change in the time period of a pendulum on changing the TEMPERATURE is INDEPENDENT of the length of the pendulum

ANSWER :B
Discussion
42630.

Two plane mirrors are inclined at an angle of 60^(@) with each other. An incident ray hits one of the mirror at an angle of 80^(@) with the normal. Its angle of deviation after two reflections is

Answer»

`120^(@)`
`240^(@)`
ZERO
`60^(@)`

ANSWER :A
Discussion
42631.

The signals is affected by noise in communication system

Answer»

At the transmitter
At the MODULATOR
 In the channel
At the RECEIVER

ANSWER :C
Discussion
42632.

Solve the above problem, if 20 V battery is connected with unlike polarities together.

Answer»

`4 XX 10^(-4) J`
`9 xx 10^(-4) J`
`2 xx 10^(-4) J`
`10^(-4) J`

Solution :`q_(1)=20 muC U_(1)=10^(-4) J q_(2)=40 muC`
`U_(2)=4 xx10^(-4) J Delta Q=20+40 =60 muC`
(because, polarity has charged)
W.D by the BATTERY `= 20 xx60 xx10^(-6)=12 xx 10^(-4)J`
Heat `=U_(1)-U_(2)+W.D`
`= 10^(-4)-4 xx10^(-4) +12 xx10^(-4) =9 xx10^(-4)J`.
Discussion
42633.

In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction from the n - side

Answer»

500 MEV
550 meV
500 meV
550 meV

Answer :B
Discussion
42634.

The plane faces of two identical plano convex lenses, each with focal length f are pressed against each other using an optical glue to form a usual convex lens. The distance from the optical centre at which an object must be placed to obtain the image same as the size of object is

Answer»

`F/4`
`f/2`
`f`
`2 f`

ANSWER :C
Discussion
42635.

Obtain lens maker'sformula and medium its signification. Lens maker's formula and lens equation:

Answer»

<P>

Solution :(i) Consider a thin lens made up of a medium of refractive INDEX n, is placed in a medium of refractive index ni:
(ii) Let R, and Ribe the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole as shown in figure.

(iii) Consider a point object on the principal axis. The ray which falls very close to P, after REFRACTION at the surface (1) forms image at I.. Before it does so, it is again refracted by the surface (2). ....(IV) Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given from Equation
`n_2/v - n_1/u =((n_2 - n_1))/R_1`
For the refracting surface (1), the light goes from `n_1` to `n_2`.
`n_1/v-n_1/u=(n_2-n_1)(1/R_1 - 1/R_2)`
Further simplifying and rearranging,
`1/v-1/u =((n_2-n_1)/n_1)(1/R_1-1/R_2)`
`1/v-1/u = (n_2/n_1-1)(1/R_1-1/R_2)"...(3)"`
If the object is at infinity, the image is formed at the focus of the lens. Thus, for uro, v=f. Then the equation becomes.
`1/f - 1/oo=(n_2/n_1 -1)(1/R_1 -1/R_2)`
`1/f =(n_2/n_1 - 1)(1/R_1 -1/R_2)"...(4)"`
(vi) If the refractive, index of the lens is `n_2`, and it is placed in air, then `n_2`, = n and `n_2`, = 1. So . the equation (4) becomes,
`1/f =(n-1)(1/R_1 - 1/R_2)"....(6)"`
(vii) The above equation is called the lens maker.s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired FOCAL length with a material of particular refractive index.
(viii) This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
`1/v + 1/u = 1/f"...(6)"`
(ix) This equation is known as lens equation which relates the object distance u and image distance v with the focal length f of the lens. This formula holds good for a any type of lens.
Discussion
42636.

Wave motion is periodic in

Answer»

space
space and time
time
direction

Answer :B
Discussion
42637.

A flywheel rolls down on an inclined plane. At any instant of time, the ratio of rotational kinetic energy to total kinetic energy is :-

Answer»

`1:2`
`3:1`
`4:3`
`1:3`

ANSWER :D
Discussion
42638.

The energy gap of silicon is 1.14eV What is the maximum wavelength at 'which, silicon will begin absorbing energy ?

Answer»

1086
10860
`10.86A`
`108.6A`

ANSWER :B
Discussion
42639.

How gammarays are prodeuced ?

Answer»

Solution :Due to transition of NUCLIDES from HIGHER QUANTUM state to lower.
Discussion
42640.

A cart is moving with a velocity 20 m/s. Sand is being dropped into the cart at the rate of 50 kg/min. The force required to move the cart with constant velocity will be :

Answer»

50 N
30.33 N
26.45 N
16.66 N.

Answer :D
Discussion
42641.

Find the extrinsic conductivity of germanium doped with indium to a concentration of 2xx10^(22)m^(-3), with antimony to a concentration of 5xx10^(21)m^(-3).

Answer»


SOLUTION :Since indium is a trivalent element, the indium impurity acts as an acceptor and produces hole-type conductivity. Knowing the CONCENTRATION of holes and their MOBILITY, we obtain `gamma=en_(+)b_(+)`. Antimony is a pentavalent element, so the antimony impurity acts as a DONOR. The electric conductivity is `gamma=en_(-)b_(-)`.
Discussion
42642.

Name the type of biasing of a p-n junction diode so that the junction offers very high resistance.

Answer»

SOLUTION :REVERSE BIAS.
Discussion
42643.

Density of nuclear matter is the......... mass of........... and its............ .

Answer»


ANSWER :RATIO of ; NUCLEUS ; VOLUME.
Discussion
42644.

A capacitor has been charged by a d.c. source. What are the magnitudes of conduction and displacement currents, when it is fully charged ?

Answer»

Solution :Magnitudes of both conduction current as well as displacement current are ZERO when a CAPACITOR is fully CHARGED by a d.c. SOURCE.
Discussion
42645.

When viewed normally through the flat surface the apparent thickness of a plano-covex lens appears to be 2mm. If the radius of curvature of the spherical surface is 10cm then find thickness of the lens when viewed from the curved surface. Refractive index of the glass is 1.5.

Answer»


ANSWER :2.22mm
Discussion
42646.

Ozone layer in atmosphere is useful, because it

Answer»

stops ultraviolet radiation
stops GREEN HOUSE effect
stops INCREASE in TEMPERATURE of atmosphere
absorbs pollutant gases

Answer :a
Discussion
42647.

The length of second's hand in a watch is 1 cm. The change in velocity of its tip in a period of 12 s is

Answer»

Zero
`((pi)/(15sqrt2)) CMS^(-1)`
`((pi)/(15)) cms^(-1)`
`((pi SQRT2))/(15)) cms^(-1)`

Answer :B
Discussion
42648.

Binding energy per Nucleon of ""_(3)Li^7and ""_(2)He^4are 3.6MeV and 7.06MeV. Then find the energy of proton in the following reaction , Li^(7)+P rarr 2_(2)He^(4)

Answer»

1.728 MEV
17.28 Mev
172.8 MeV
1728 MeV

ANSWER :B
Discussion
42649.

The repulsive force between two magnetic poles in air is 9 xx 10^(-3) N. if the two poles are equal in strength and are separated by a distance of 10 cm , calculate the pole strength of each pole .

Answer»

Solution :The FORCE between TWO poles are given by `vecF= k (q_(m_(A)) q_(m_(D)))/(r^(2))`
The magnitude of the force is `F=k (q_(m_(A)) q_(m_(B)))/(r^(2))`
Given: `F=9 xx 10^(-3) N, r=10 cm =10 xx 10^(-2) m`
Therefore, `9 xx 10^(-3) =10^(-7) xx (q_(m)^(2))/((10 xx 10^(-2))^(2)) rArr q_(m)=30NT^(-1)`
Discussion
42650.

Due to straight current carrying conductor, a null point occurred at pon east of the conductor. The net magnetic induction at a point Q which is half the distance of 'p on north of the conductor is

Answer»

ZERO
`B_H`
`SQRT(2)B_H`
`sqrt(5) B_H `

ANSWER :D
Discussion