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5151.

Two coherent monochronic light sources are locatedat two vertices of anequilateral triangle .If the intensity due to each of the sourcesare locatedat two vertices of an equilateral triangle. If the intensity due to each of the source independently is1 "Wm"^(2)at the thirdvertex. The resultant intensity due to both the source at the third vertex is ( in "Wm"^(-2))

Answer»


ANSWER :4
Discussion
5152.

A semi-infinite charged rod is arranged so that its one end coincides with the center of a charged ring as shown in Fig. Both of them have a charge per unit length of lambda. The radius of ring is R. The electric field of rof exert force on an element of the ring as shown in Fig. (a) What is the force on the ring due to the rod ?(b) What is the increase in the tension of the ring due to the rod ?

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SOLUTION :(a) `(lambda^(2))/(2 epsilon_(0))`, (B) `(lambda^(2))/(4pi epsilon_(0))`
Discussion
5153.

Two charges q_1and q_2 are placed 30 cm apart as in Fig, A third chargeq_3is moved along the arc of a circle of radius 40 cm from C to D. The change in potential energy of the system is q_3/(4piepsilon_0)k where k is:

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`8q_2`
`8q_1`
`6q_2`
`6q_1`

ANSWER :A
Discussion
5154.

Let f: Rrarr R be defined by f (x) = 3X-4 thenf^(-1)(x)is given by-

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`(x+4)/3`
`(x/3)-4`
`3x+4`
NONE of these

Answer :A
Discussion
5155.

A small square loop of wire of side 'I' is placed inside a large square loop of side L (L > I). If the loops are coplanar and their centres coincide, the mutual inducation of the system is directly proportional to

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Solution :Considering the large loop to be made up of four rods each of length L, the FIELD at the centre, i.e., at a distance `(L//2)` from each rod, will be
`B=4 xx (mu_(0))/(4 PI)(I)/(d)[sin alpha+sin BETA]`
i.e., `B=4 xx (mu_(0))/(4phi)(I)/((L//2)) xx 2sin45`
i.e., `B_(1)=(mu_(0))/(4PI)(8sqrt2)/(L)I`

So the flux linked with smaller loop
`Phi_(2)=B_(1)S_(2)=(mu_(0))/(4pi)(8sqrt2 I)/(LO) l^(2)`
and hence, `M=(phi_(2))/(I) =2sqrt2 (mu_(0))/(pi)(l^(2))/(L) implies M prop (l^(2))/(L)`
Discussion
5156.

Diameter of the objective of a telescope is 200 cm. What is the resolving powerof a telescope ? Take wavelength of light =5000 Å.

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`6.56 XX 10^(6)`
`3.28 xx 10^(5)`
`1 xx 10^(6)`
`3.28 xx 10^(6)`

Solution :`RP = (D)/(1.22 lambda) = (2)/(1.22 xx 0.5 xx 10^(-6)) = (4)/(1.22) xx 10^(6)= 3.28 xx 10^(6)`
Discussion
5157.

One end of a uniform rod of length l and mass m is hinged at A. It is released from rest from horizontal position AB as shown in figure. The force exerted by the rod on the hinge when it becomes vertical is

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`(3)/(2)mg`
`(5)/(2)mg`
3mg
5mg

Solution :`U_(1)=(1)/(2)QE`
`U_(2)=QE`
`U_(1),U_(2)=1 : 2`
NOTE : ENERGY dissipated as HEAT `=U_(2)-U_(1)=(1)/(2)QE`
Discussion
5158.

A 600 W carrier is modulated to a depth of 75% by a 400 Hz sine wave. Find total antenna power.

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769 W
796 W
679 W
637.5 W

Answer :A
Discussion
5159.

Calculatethe wavelenght of H_(alpha)linein Balmerseries of hydrogen . GivenRydberg constant R = 1.097 xx 10^(-7) m^(-1)

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Solution :We knowthatwave numbers and wavelengths of differentspectral LINES of Balmer SERIES aregiven by .
`(1)/(lambda) = bar(v) = R [(1)/(2^(2)) - (1)/(n^(2))]`
`therefore ` Forfirst line of Balmerseries`(1)/(lambda) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5R)/(36)`
`rArr "" lambda = (36)/(5R) = (36)/(5 xx 1.097 xx 10^(7)) = 6.563 xx 10^(-7) m = 656.3 nm`
Discussion
5160.

A long wire carrying current I is connected to a rectangular metallic frame as shown in the adjacent figure. A uniform magnetic field B along y-axis is switched on. The mass linear density of the frame is lambda. The frame lies in y-z plane, The instantaneous angular acceleration of the frame is

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ZERO
`(12lB)/(LAMBDAA)`
`(4lB)/(lambdaa)`
NONE of the above

ANSWER :A
Discussion
5161.

If R and L denote resistance and inductance respectively which of the following has dimension of time ?

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`L/R`
`R/L`
`sqrt(R/L)`
`sqrt(L/R)`

Solution :Unit of `L/R="SECOND X OHM"/"ohm"`
=second
`THEREFORE [L/R]`=[TIME]
Discussion
5162.

An elevator cab of mass m=500 kg is descending with speed v_(i)=4.0m//s when its supporting cable begins to slip, allowing it to fall with constant acceleration vec(a)=vec(g)//5 (Fig. 8-11a). (b) During the 12 m fall, what is the work W_(T) done on the cab by the upward pull vec(T) of hte elevator cable?

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Solution :KEY IDEA
We can CALCULATE work `W_(T)` with Eq. 8-7 `(W=Fd cos phi)` by first writing `F_("net". y)=ma_(y)` for the components in Fig. 8-11b.
Calculations: We get
`T-F_(g)=ma`. (8-18)
Solving for .I., substituting mg for `F_(g)`, and then substituting the result in Eq. 8-7, we obtain
`W_(T)=Td cos phi = m (a+g)d cos phi`. (8-19)

Figure 8-11 An elevator cab, descending with speed `v_(i)`, SUDDENLY begins to accelerate downward. (a) It moves through a displacement `VEC(d)` with constant ACCELERATION `vec(a)=g//5`. (b) A freebody diagram for the cab, displacement included.
Next,substituting `-g//5` for the (downward) acceleration a and then `180^(@)` for the angle `phi` between the direction of forces `vec(T)` and `m vec(g)`, we find
`W_(T)=m(- g/5 + g) d cos phi = 4/5 mgd cos phi`
`=4/5 (500 kg) (9.8 m//s^(2))(12m)cos 180^(@)`
`= -4.70xx10^(4) j~~-47 kJ`.
Caution: Note that `W_(T)` is not simply the negative of `W_(g)` because the cab accelerates during the fall. Thus, Eq. 8-16 (which assumes that the initial and final kinetic energies are equal) does not apply here.
Discussion
5163.

Length of solenoid is 0.4 and it has 240 turns. What is the magnetic induction at the interior point of solenoid when a current of 2.5 A is flowing throght it :

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`1.885xx10^(-3) (WB)//m^2`
`2.89xx10^(-3) (Wb)//m^2`
`4.12xx10^(-3) (Wb)//m^2`
`5XX10^(-3) (Wb)//m^2`

ANSWER :A
Discussion
5164.

Attenuation in optical fibre is mainly due to :

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scattering
absorption and scattering
dispersion
None of the above.

Answer :B
Discussion
5165.

A boat moves perpendicular to the bank with a velocity of 7.2 km/h. The current carries it 150 m downstream, the time required to cross the river. (The width of the river is 0.5 km)

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4.16min
2.15min
3.11min
5.11min

Answer :A
Discussion
5166.

(A): When capacitive reactance is smaller than the inductive reactance in LCR series circuit, e.m.f. leads the current. (R): The phase angle is the angle between the alternating e.m.f. and alternating current of the circuit.

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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 :B
Discussion
5167.

A, B, and C are three large, parallel conducting plates, placed horizontally. A and C are rightly fixed and earthed in figure . B is given some charge . Under the electorstatic and gravitational forces, B may be

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in equilibrium if it is CLOSER to A than to C
in equilibrium midway between A and C
B can never be in stable equilibrium .
in equilibrium if it is closer to C than to A.

Solution :a.,c.
As `A` and `C` are EARTHED, they are connected to each other. Hence, `'A+B'` and `'B+C'` are two capacitors with the same potential difference. If `B` is closer to `A` than to `C`, then the CAPACITANCE `C_(AB)gtC_(BC)`. The upper surface of `B` will have greater charge than the lower surface. As the force of attraction between the plates of a CAPACITOR is proportional to `Q^(2)`, there will be a net upward force on `B`. This can balance its weight.
Discussion
5168.

The multiplication of two irrational numbers is:

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ALWAYS IRRATIONAL
Always rational
Can be irrational or rational
None of these

Answer :C
Discussion
5169.

Six equal point charges are placed at the corners of a regular hexagon of side 'a'. Calculate electric field intensity at the centre of hexagon ?

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Solution :Zero
Similarly electric FIELD DUE to a UNIFORMLY charged ring at the CENTRE of ring :
Discussion
5170.

A palne light wave of intensity I = 0.70 W//cm^(2) illuminates a sphere with ideal mirror surface. The radius of the sphere is R = 5.0cm. From the standpoint of the corpuscular theory find the force that light exerts on the sphere.

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Solution :We consider a strip defined by the angular range `(theta, theta+d theta)`. From the previous problem the normal pressure exerted on this strip is
`(2I)/(c ) cos^(2) theta`
This pressure givens rise to a force whose resultant, by symmetry is in the direction of the INCIDENT light. Thus
`F = (2I)/(c) underset(0)overset(PI//2)int cos^(2).cos theta.2piR^(2)sin thetad theta = piR^(2)(I)/(c )`
Putting in the values
`F = pi xx 25 xx 10^(-4)(0.70xx10^(4))/(3xx10^(8))N = 0.183muN`
Discussion
5171.

In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a

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PURE INDUCTOR
pure CAPACITOR
pure RESISTOR
COMBINATION of a capacitor and an inductor

Answer :A
Discussion
5172.

An elevator cab of mass m=500 kg is descending with speed v_(i)=4.0m//s when its supporting cable begins to slip, allowing it to fall with constant acceleration vec(a)=vec(g)//5 (Fig. 8-11a). (c) What is the net work W done on the cab during the fall?

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Solution :Calculation: The net work is the sum of the WORKS DONE by the forces ACTING on the CAB:
`W=W_(g)+W_(T)=5.88xx10^(4)J-4.70xx10^(4)J`
`=1.18xx10^(4)J~~12kJ`.
Discussion
5173.

A particle is moving along a vertical circle of radius R. At P, what will be the velocity of particle (assume critical condition at C) ?

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`SQRT(GR)`
`sqrt(2gR)`
`sqrt(3gR)`
`sqrt((3)/(2)gR)`

Answer :B
Discussion
5174.

What is the velocity of particle whose mass is double that of rest mass?

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`(2c)/(3)`
`(C )/(2)`
`(sqrt(3)c)/(2)`
`(3c)/(4)`

SOLUTION :`m=(m_(0))/(sqrt(1-(V^(2))/(c^(2))))=2m_(0)`
`therefore sqrt(1-(v^(2))/(c^(2)))=(1)/(2)`
`therefore 1-(v^(2))/(c^(2))=(1)/(4)`
`therefore (v^(2))/(c^(2))=(3)/(4)`
`therefore (v)/(c )=(sqrt(3))/(2)`
`therefore v=(sqrt(3))/(2)c`
Discussion
5175.

The polarizing angle of glass is 57^(@).A ray of light which is incident at this angle will have an angle of refraction as :

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`33^(@)`
`38^(@)`
`25^(@)`
`43^(@)`

Solution :`mu = tan p`
p is POLARISING angle `mu` is REFRACTIVE INDEX
`1.5 = (sin p)/(sin R)(because p = i)`
`sin r = (sin p)/(1.5) = (sin 57^(@))/(1.5)`
` sin r = (0.8387)/(1.5) = 0.5580`
`therefore r = 33^(@)` (angle of refraction).
Discussion
5176.

In a transformer, energy is transferred from the primary to the secondary by

Answer»

the CURRENT in the wires
an ELECTRIC field
magnetic field
P.D. ACROSS the primary

Answer :C
Discussion
5177.

Illustrate Biot-Sevart's law necessary figure.

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SOLUTION :REFER TEXT
Discussion
5178.

If vecA and vecB are non-zero vectors which obey the relation |vecA+vecB|=|vecA-vecB|, then the angle between them is

Answer»

`0^(@)`
`60^(@)`
`90^(@)`
`120^(@)`

Answer :C
Discussion
5179.

A battery of negligible internal resistance is connected with 10m long wire. A standard cell gets balanced on 6m length of this wire. On increasing the length of potentiometer wire by 2m, the null point with the same standard cell in the secondary will be (no series resistance in primary)

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INCREASED by 2m
DECREASED by 2m
increased by 1.2m 
decreased by 1.2m 

ANSWER :C
Discussion
5180.

Find identical capacitor plates are arranged such that they make capacitors each of 2muF . The plates are connected to a source of emf 10V. The charge on plate C is

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`+20muC`
`+40muC`
`+60muC`
`+80muC`

ANSWER :B
Discussion
5181.

For a series L-C-R a.c.circuit at resonance, the statement which is not true ?

Answer»

Wattless current is zero
Power factor is zero
Peak ENERGY stored by a capacitor = peak energy stored by an inductor
Average power = apparent power

Solution :For a SERIES L-C-R, A.C. circuit at resistance, `omegaL=omegaC` So Z = R. So RESULTANT power factor
cos `delta=R/"|Z|"`
`THEREFORE cos delta=R/R=1`
`therefore delta=0^@`
Thus, power factor is 1 and in the circuit only resistance is effective So loss of power is maximum.
Discussion
5182.

The bending of light rays at the edges of corners of an obstacle is called ?

Answer»

SOLUTION :DIFFRACTION
Discussion
5183.

Corrected form of Ampere's is circuital law by Maxwell is ……..

Answer»

`oint vec(B).vec(d)A=0`
`oint vec(B)vec(d)A=mu_(0)I`
`oint vec(B).vec(d)l=(d phi_(B))/(DT)`
`oint vec(B).vec(d)l=mu_(0)i_(C )+mu_(0)in_(0)(d phi_(B))/(dt)`

Answer :D
Discussion
5184.

In an electromagnetic wave vecE and vecB are __________ to each other.

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SOLUTION :MUTUALLY PERPENDICULAR
Discussion
5185.

An isolated capacitor of capacitance C is charged to a potential V. Then a dielectric slab of dielectric constant K is inserted as shown. The net charge on four surfaces 1,2, 3 and 4 would be respectively.

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`0,"CC"-CV,0`
`0,(CV)/(K),(-CV)/K,0`
`CV,0,0,-CV`
`CV,(-CV)/K,(CV)/K,-CV`

ANSWER :B
Discussion
5186.

How threshold frequency and threshold wave length are related ?

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SOLUTION :`v_0 = c/lambda_0`
Discussion
5187.

What's atomic number ? Mass number ?

Answer»

SOLUTION :The number of electrons /protons is called ATOMIC number and TOTAL of number of protons + NEUTRONS is mass number.
Discussion
5188.

The resistance of galvanometer is 999 Omega. A shunt of 1Omega is connected to it. If the main current is 10^(-2)A, what is the current flowing through the galvanometer.

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SOLUTION :`G=999Omega,S=1Omegai=10^(-2)A,i_(g)=?`
`i_(g)=i(S/(G+S))=10^(-2)xx(1/(999+1))=10^(-5)A`
Discussion
5189.

A point charge 50muC is locatedat a point 2 hati+3hatj.Find the electric field vector barE at a point with position vector 8hati-5hatj , when the position vectorsare expressedin metre.

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SOLUTION :`BARE=(1)/(4PI in_(0)) (q)/(r^(3)) barr`, here`q= 50xx10^(-6)C`
`VECR = (vecr_(2) -vecr_(1))= ( 8hati-5hatj) -(2hati +3hatj) = (6hati-8hatj)`
then `barE=450 (6hati-8hatj)NC^(-1)`.
Discussion
5190.

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 Omega as shown in the Fig. Find the value of the current in circuit.

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SOLUTION :VALUE of current in the circuit `I = ((200 -10))/(38 Omega) = 5A`
Discussion
5191.

Current time graph of different source is given which one will have R.M.S. Value V_(0):

Answer»




ANSWER :A
Discussion
5192.

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A= 10 cm^(2) and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (mu_(0) = 4pi xx 10^(-7) T m A^(-1))

Answer»

`2.4pi xx10^(-5)H`
`4.8pi xx10^(-4)H`
`4.8pi xx10^(-5)H`
`2.4pi xx10^(-4)H`

ANSWER :D
Discussion
5193.

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A= 10 cm^(2) and length = 20 cm If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (mu_(0) = 4pi xx 10^(-7) TmA^(-1))

Answer»

`2.4 PI xx 10^(-5)H`
`4.8pi xx 10^(-4)H`
`4.8pi xx 10^(-5)H`
`2.4pi xx 10^(-4)H`

ANSWER :D
Discussion
5194.

The magnetic energy stored in an inductance of 200mH is 1.6J in electric circuit. The inductance will supply the maximum current:

Answer»

0.4A
4A
40A
400A

Answer :B
Discussion
5195.

The energy of photoelectrons emitted from a sensitive plate is 1.56 eV. If its threshold old wavelength is 2500 Å, calculate the wavelength of incident light.

Answer»


ANSWER :`1379 Å`
Discussion
5196.

Five cellseachof emf E adinternalresistancer are connectedin series . Duetooversightone cell is connectedwrongly . The equivalent internall resistanceof the combinationis .

Answer»

3r
2r
5r
4r

Answer :C
Discussion
5197.

12व 15का महत्तम समापवर्तक (HCF) है -

Answer»

3
5
60
20

Answer :A
Discussion
5198.

What happens to conduction band at room temperature ?

Answer»

Solution :Conduction BAND is either empty or partially filled (In SOLIDS like CONDUCTORS and semiconductors.)
Discussion
5199.

A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will depend

Answer»

only on the NUMBER of TIMES radiation is drawn from it
only on the INTENSITY of doses drawn from it
on both (1) and (2)
neither on (1) nor on (2)

Answer :D
Discussion
5200.

A particle slides down on a smooth incline of inclination 30^(@), fixed in an elevator going up with an acceleration 2m//s^(2). The box of incline has width 4m. The time taken by the particle to reach the bottom will be

Answer»

`(8)/(9)sqrt3s`
`(9)/(8)sqrt3s`
`(4)/(3)SQRT((SQRT3)/(2))s`
`(3)/(4)sqrt((sqrt3)/(2))s`

Answer :C
Discussion