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4201.

Assertion : The net magnetic force on a current loop in a uniform magnetic field is zero but torque may or may not be zero. Reason : Torque on a current carrying coil in a magnetic field is given by vec(tau)=nI(vec(A)Xxtau(B)).

Answer»

If both ASSERTION and reason are TRUE and reason is the CORRECT EXPLANATION of assertion.
If both assertion and reason are true but reason is the correct explanation of assertion.
If assertion is true but reason is false
If both assertion and reason are false.

Answer :B
Discussion
4202.

O_(2) molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms.

Answer»

is not important because nuclear forces are short-ranged
is as important as ELECTROSTATIC force for binding the two atoms
cancels the repulsive electrostatic force between the nuclei.
is not important because oxygen nucleus have equal number of neutrons and protons.

Solution :Because nuclear force is of short range. It is confined to ONE nucleus only. Where as distance between two nucleus can be LARGER also hence, nuclear force between two nucleus is not important.
Discussion
4203.

In the given circuit values are as follows epsi_1 = 2V, epsi_2 = 4V, R_1 = 1 Omegaand R_2 = R_3 = 1 Omega Calculate the currents through R_1 , R_2 and R_3

Answer»

Solution :Potential difference across BE is EQUAL to the e.m.f of `E_1`.
i) The CURRENT through `R_1 ` is zero because the RESULTANT p.d across AG= 0 . The p.d. across BE due to `E_2 = -2 Omega` VOLTS and `E_1 = +2`volts, hence , p.d. across AG= +2 - 2 =0
ii) p.d. across `R_3 =-2` volts.
`i_3 =(-2)/(R_3) = (-2)/1 = -2A`

iii) p.d. across BE = -2 volts.
`i_2 = (-2)/(R_2) = (-2)/(1) = - 2A`
Note :m You may solve this using Kirchoff.s Law as USUAL.
Discussion
4204.

A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other . The particle will move in a

Answer»

STRAIGHT LINE
circle
helix
cycloid

Answer :A
Discussion
4205.

Consider sunlightincident on a pinhole of width 10^(2)A^(0). The image of the pinhole seen on a screen shall be : a) a sharp white ring b) different from a geometrical image c) a diffused central spot, white in colour d) diffused coloured region around a sharp central white spot

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only ..a.. is true
only ..B.. and ..d.. are true
only ..C.. and ..d.. are true
only ..b.. and ..c.. are true

Answer :B
Discussion
4206.

The branch of electronics which deals with the flow 2. The branch of electronics which deals with the flow of electrons in vacuum, gas or solids is called what ?

Answer»


ANSWER :PHYSICAL ELECTRONICS
Discussion
4207.

The reading of a voltmoter when a cell is connected to it is 2.2 V. When the terminals of the cell are connected to a resistance of 4 Omega, the voltmeter reading drops to 2 V. Find the internal resistance of the cell.

Answer»


ANSWER :`0.4 OMEGA`
Discussion
4208.

The relation between B_(v) , B_(h) and B is ….... .

Answer»

`B= sqrt(B_(H)^(2) + B_(v)^(2))`
`B= B_(h). B_(v)`
`B= (B_v)/( B_h)`
`B= (B_h)/(B_v)`

Solution :KNOWLEDGE based question.
Discussion
4209.

If speed of sound is 330 m/s. The lengths of closed pipe and open pipe are :

Answer»

0.4125 m, 1.65m
3.3m, 1.65m
0.825m, 0.825m
1.65m, 0.825m

Solution :For CLOSED pipe,
`3^(rd)` overtone = `7^(th)` harmonic = `n_(7) = (7v)/(4l_(c))`
`5^(th)` harmonic `n_(5) = (5v)/(4l_(c))`
`n_(7) - n_(5) = (2v)/(4l_(c))` = 400 Hz implies `n_(0) = (v)/(4l_(c)) = 200 Hz` (fundametal frequency of closed pipe) Now `3^(rd)` harmonicof closed pipe is equal to `6^(th)` harmonic of open pipe
`(v)/(4l_(c)) = (6v)/(2l_(0))`
implies`(v)/(l_(0)) = (v)/(4l_(c))` = 200 Hz
FUNDAMENTAL frequency of open pipe = `(v)/(2l_(0))` = 100 Hz
Further `l_(0) = (330)/(4 xx 200)` = 0.4125 m
`l_(0) = (330)/(2 xx 100)` = 1.65 m
Discussion
4210.

(A): If two charges are kept in a conductor medium, then electric force acting between them is zero. R: F=F_(0)/k, For conductors k=infty therefore F=F_(0)/infty=0

Answer»

Both Assertion and REASON are true and the Reason is CORRECT EXPLANATION of the Assertion.
Both Assertion and Reason are true, but Reason is not correct explanation of the Assertion.
Assertion is true, but the Reason is false.
Both Assertion and Reason are false.

Solution :Electric force F = EE, have values are same but directions are opposite for both and as `a = F/m` are different due to different mass. So, statement and reason are correct but not corelated.
Discussion
4211.

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?

Answer»

Solution :We know that the magnetic field INSIDE a solenoid is uniform and directed along its axis. Hence, angle `theta` between the magnetic field and the current carrying wire, placed PERPENDICULAR to the solenoid axis, is `90^@`. Moreover `I = 10 A, l = 3 CM = 3 xx 10^(-2) m and B = 0.27 T`
`:. F = B I l SIN theta = 0.27 xx 10 xx 3 xx 10^(-2) xx sin 90^@ = 8,1 xx 10^(-2) N`.
Discussion
4212.

Assertion In non-uniform magnetic field speed of a charged particle varies. Reason Work done by magnetic force on a charged particle is always zero.

Answer»

If both Assertion and Reason are TRUE and Reason is the correct explanationof Assertion.
If both Assertion and Reason are true but Reason is not correct EXPLANATION of Assertion.
If Assertion is true but Reason is false.
If Assertion is false but Reason is true.

Answer :d
Discussion
4213.

A ball Is projected horizontally from the top of a tower with velocity 4m/s. The velocity of the ball after 0.7s (g = 10m/s^2 ) is

Answer»

11m/s
10m/s
8m/s
3m/s

Solution :VELOCITY of the ball after TIME
t = `sqrt(v^2x +v^2y) =sqrt(4^2+ g t^2)= sqrt(16+(10xx7)^2)` = 8m/s.
Discussion
4214.

An ohm-meter is a device that measures an unknown resistance. A simple ohm-meter can be constructed using a galvanometer as shown in the figure. The cell used in the circuit has emf E = 20 volt. The full scale deflection current and resistance of the galvanometer are 2 mA and 20 Omega respectively. R_(0) is a fixed resistance and R is the unknown resistance whose value is directly given by the galvanometer scale. The galvanometer scale is shown in figure. When an unknown resistance R is placed in the circuit, the pointer deflects by theta = 90^(@). Find R.

Answer»


ANSWER :`3333 OMEGA`
Discussion
4215.

Ina Y oung's double slit experiment, separation between the slits is 2 xx 10^(-3) m and distance of the screen from the slit is 2.5 m. Light in the range of 2000 - 8000 Å is allowed to fall on the slits.Wvelength in the visible region that will be present on the screen at 10^(-3) m from the central maxima will be :

Answer»

`4000 Å`
`5000 Å`
`6000 Å`
`8000 Å`

Solution :`lambda =(xd)/(D) = 8000 Å`
For max. `n_(1)lambda_(1) = n_(2)lambda_(2) = n_(3)lambda_(3)`
If`n_(1) = 2 Rightarrow lambda_(1) = 4000 Å`
So VISIBLE region, SECOND bright line is present.
Discussion
4216.

Is the de Broglie wavelength of a photon of an em radiation equal to the wavelength of the radiation ?

Answer»

SOLUTION :YES, de Broglie wavelength `lambda = h/p`, for a photon `p=(hv)/C
Hence, `lambda = h/p=c/v`.
Hence, the TWO WAVELENGTHS are the same.
Discussion
4217.

In Fig. two cells have equal emf E but internal resistancesare r_1 and r_2. If the reading of the voltmeter is zero, then relation between R, r_1 and r_2 is

Answer»

`R = r_1 - r_2`
`R= r_1+r_2`
`2r_1 - r_2`
`r_1r_2`

SOLUTION :a. `V = E - Ir_1, V = 0`
`:. E = Ir_1`
Total EMF `= Ir_1 = Ir_1 = 2Ir_1`
Total RESISTANCE `= R + r_1 + r_2`
Now, `I = (2Ir_1)/(R+r_1+r_2)`
or `R + r_1 + r_2 = 2r_1`
or `R = 2r_1 - r_1 - r_2 or R= r_1- r_2`
Discussion
4218.

A: Microwave propagation is better than the skywave propagation. R: Microwaves have frequencies 100 to 300 GHz which have very good directional properties.

Answer»


ANSWER :A
Discussion
4219.

A body is projected with velocity 24 ms^(-1) making an angle 30^(@) with the horizontal. The vertical component of its velocity after 2s is (g=10ms^(-1))

Answer»

`8 MS^(-1)` upward
`8 ms^(-1)` down ward
`32 ms^(-1)` upward
`32 ms^(-1)` down word

Answer :B
Discussion
4220.

A plane longitudinal harmonic wave propagates in a medium with density rho. The velocity of the wave propagation si v . Assuming that the density variations of the medium, induced by the propagating wave, Delta rho lt lt rho , demonstrate that ltbr. (a) the pressure incrementin the medium Deltap=- rho v^(2)( delta xi//delta x),where delta xi // delta xis the relative deformation, (b)the wave intensity is defined by Eq. (4.3i)

Answer»

Solution :`(a)` Let us CONSIDER the motion of an element of the medium of thikness `dx` os unit area of cross `-` section. Let `xi=` displacement of the PARTICLES of the medium at location `x`. Then by the equation of motion.
`rho dx dot(xi)=-dp`
where `dp` is the pressure increment over the length `dx`
Recalling the wave equation
`ddot(xi)=v^(2)(delta^(2)xi)/( delta x^(2))`
we can write the foregoing equationas
`rho v^(2)(delta^(2)xi)/(deltax^(2))dx=-dp `
Integrating THIE equation, we get
`Delta p=` surplus pressure `=- rho v^(2) ( deltaxi)/( deltax)+ Const.`
In the absence of a deformation `(` a wave `),` the surplus pressure is `Delta p=0`. So' Const ' `=0` and LTBR. `Deltap=-rho v^(2)( deltaxi)/( deltax)`.
`(b)` We have found earlier that
`w=w_(k)+ w_(p)=` total energy density
`w_(k)=(1)/(2)rho((deltaxi)/( delta))^(2), w_(p) =(1)/(2)E((deltaxi)/( deltax))^(2)=(1)/(2) rho v^(2)((deltaxi)/(deltax))^(2)`
It is easy to see that the space `-` time average of both densities is the same and the space time average of total energy density is then
`lt w gt = lt rho v^(2)((deltaxi)/(deltax))^(2)`
The intensity of the wave is
`I=vlt w GE lt ((Deltap)^(2))/(rhov) gt`
Using `lt (Deltap)^(2) gt =(1)/(2) (Deltap)_(m)^(2)` we get `I=((Deltap)_(m)^(2))/(2 rho v)`
Discussion
4221.

A mass m connected to inextensible string of length l lie on a horizontal smooth ground. Other end of string is fixed. Mass m is imparted a velocity v such that string remains taut & motion occurs in horizontal providely stirng during the time string turns through 90^(@)

Answer»

2mv
mv
`SQRT(2)mv`
`msqrt(2v^(2)-2gl)`

Solution :Impulse `=Deltavecp=vecP_(f)-vecp_(i)=m DELTAVECV`
`=m(vhatj-vhati)`
|impulse| `=sqrt(2)mv`
Discussion
4222.

How does the intensity I of scattered light vary with the wavelength lambda of the incident light ?

Answer»

SOLUTION :CLOUDS have large PARICLES like dust and WATER drplets which scatter light of all colors almost EQUALLY.
Discussion
4223.

What was the turning-point in their relationship?

Answer»

When GRANDMOTHER passed away
When AUTHOR STARTED dancing
When they went to city
None of the above

Answer :C
Discussion
4224.

What will be the inputs of A and B for Boolean expression (bar(A + B))+(bar(A.B)) =0 ?

Answer»

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

Answer :D
Discussion
4225.

Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii?

Answer»

SOLUTION :`R_(1)/R_(2)=(A_(1)/A_(2))^(1/3)=(27/125)^(1/3)=3/5`
Discussion
4226.

Draw a neat diagram of experimental set up for Fraunhoffer diffraction due to a single slit.

Answer»

SOLUTION :
Discussion
4227.

What is positive beta - emission?

Answer»

Solution :When a NEUTRON splits into a proton, neutrino and POSITRON, the emission is called POSITIVE `BETA` emission.
Discussion
4228.

Assertion : Nicol prism is used to produce and analyse plane polarised light. Reason : Nicol prism reduces the intensity of light to zero.

Answer»

Assertion and REASON are CORRECT and Reason is the correct EXPLANATION of Assertion.
Assertion and Reason are TRUE but Reason is the false explanation of the Assertion.
Assertion is true but Reason is false.
Assertion is false but Reason is true.

Answer :C
Discussion
4229.

A block of mass 4kg is placed in contact with the front vertical surface of a lorry. The coefficient of friction between the vertical surface and block is 0.8. The lorry is moving with an acceleration of 15 m/s^2 (g = 10ms^(-2)). The force of friction between lorry and block is

Answer»

48 N
24 N
40 N
Zero

Answer :C
Discussion
4230.

Which of the following assertions are correct ?

Answer»

A neutron can DECAY to a PROTON only INSIDE a nucleus
A proton can CHANGETO a neutron only inside a nucleus
An ISOLATED neutron can charge into a proton
An isolated proton can change into a neutron.

Answer :B::C
Discussion
4231.

Three concentric spherical metal shells A.B.C of radii a,b,c(c gt b gt a) have surfacecharge density +sigma, - sigma and +sigma respectively. The potential of the middle shell is (sigma)/(epsilon_(0)) times

Answer»

`((a^(2))/(B)-b+C)`
(a-b+c)
`((a^(2)-b^(2)+c^(2))/(c))`
`(a-b+(c^(2))/(b))`

ANSWER :A
Discussion
4232.

One kilowatt electric heater is to be used with 220V DC supply. It converts (N xx10)/(3)g of water at 100^(0)C into steam at 100^(0)C in one minute. find the value N.

Answer»


ANSWER :8
Discussion
4233.

Calculate the self-inductance for very long solenoid.

Answer»

SOLUTION :LET us calculate the SELF-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.
The magnetic field DUE to a current I flowing in the solenoid is `B=mu_0nI`(NEGLECTING edge effects, as before)
The total flux linked with the solenoid is `Nphi_B=NBA`
`Nphi_B=(nl)(mu_0nI)(A) ""(because n=N/l)`
`=mu_0n^2 AlI`
where nl is the total number of turns. Thus, the self - inductance is,
`L=(Nphi_B)/I`
`L=mu_0n^2Al`...(1)
If we fill the inside of the solenoid with a material of relative permeability u, (for example soft iron, which has a high value of relative permeability), then,
`L=mu_r mu_0 n^2 Al`...(2) `[because mu_1=mu/mu_0]`
The self-inductance of the coil depends on its geometry and on the permeability of the medium.
Discussion
4234.

Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity v=10m.s^(-1). Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Answer»

Solution :Induced emf,
`E=(mu_(0))/(2pix).(IA^(2)v)/(a+x)=(2xx10^(-7)xx50xx(0.1)^(2)xx10)/(0.2(0.1+0.2))`
`=1.67xx10^(-5)~~1.7xx10^(-5)V`
Discussion
4235.

(a) Figure 9.32 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?

Answer»

Solution :`sini._(c)=1.44/1.68" which GIVES "i._(c)=59^(@)`. Total internal reflection takes place when `i gt 59^(@)` or when `r lt r_("max")=31^(@)`. Now, `(sin i_("max")//sinr_("max"))=1.68`, which gives `i_("max")~=60^(@)`. Thus, all incident rays of ANGLES in the range `0ltilt60^(@)` will suffer total internal reflections in the pipe.(If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.)
(b) If there is no outer coating, `i._(c)=sin^(-1)(1//1.68)=36.5^(@),." Now "i=90^(@)`will have `r = 36.5^(@) and i′_(c) = 53.5^(@)` which is greater than `i′_(c)` . Thus, all incident rays `("in the range "53.5^(@) lt i lt 90^(@))` will suffer total internal reflections.
Discussion
4236.

(a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain the expression for the magnetic energy density. (b) A square loop of sides 5 cm carrying a current of 0.2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown in Fig. 6.66. Calculate (i) the resultant magnetic force, and (ii) the torque, if any, acting on the loop.

Answer»


Answer :(b) (i) `1.33 xx 10^(-8) N, attractive (ii) zero
Discussion
4237.

The intensity of a sound wave gets reduced by 20% of passing through a slab. The reduction in intensity on passage through two such consecutive slabs is :

Answer»

0.4
0.36
0.3
0.5

Answer :B
Discussion
4238.

Separation between the plates of a parallel plate capacitor, connected to a battery (zero resistance) of constant emf. is increased with constant (very slow) speed by external forces. During the process, W is the work done by external forces, DeltaU is the change in potential energy of the capacitor, W_b is the work done by the battery and H is the heat loss in the circuit. Then

Answer»

`W+W_b=DeltaU`
`H NE 0`
`H=DeltaU`
`W=0`

ANSWER :A
Discussion
4239.

Define dew point.

Answer»

Solution :DEW POINT may be defined as the temperature at which GIVEN VOLUME of air becomes just saturated with water vapour actually present in it
Discussion
4240.

Angle which the vector vec A = 2 hat i+3 hat j makes with the y-axis is given by :

Answer»

`TAN^(-1)3/2`
`tan^(-1)2/3`
`sin^(-1)2/3`
`COS^(-1)(3/2)`

ANSWER :B
Discussion
4241.

PQ type incident radiation and RS is the reflected ray. They are both parallel. So which mirror on the right makes this possible? There may be one or more reflections through the mirror.

Answer»

PLANE MIRROR
Convex mirror
Plane and concave mirror
An concave mirror

Solution :
PQ and RS rays parallel to a given concave mirror.
Discussion
4242.

What is the width n cm of depletion layer in p-n junction?

Answer»

`10^(-2)`
`10^(-3)`
`10^(-6)`
`10^(-8)`

Solution :`10^(-6)`
Width of DEPLETION barrier is `0.5mu m =5xx10^(-7)m` MEANS range of `5xx10^(-5)CM`
Discussion
4243.

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from the centre of the magnet ? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

Answer»

Solution :For NULL point on the axis of the magnet at a distance r = 14 CM = 0.14 m from the centre of magnet,the magnetic field of the magnet must NULLIFY the horizontal component of earth.s magnetic field `B_H = 0.36 G = 0.36 XX 10^(-4) T ` (because angle of dip is zero, hence `B_H` = total magnetic field of earth `B_E` ) .
`thereforeB_("axial") =(mu_0)/(4pi) * (2m)/(r^3) = B_H = 0.36 xx 10^(-4)` T
Now magnetic field at the same distance on the normal bisector of the magnetic will be
`B_(eq) =(mu_0)/(4pi) * (m)/(r^3) = (B_("axial"))/(2) = (0.36xx10^(-4))/(2) = 0.18 xx 10^(-4) T` (along the direction of `B_H` ).
`therefore ` Total magnetic field on the normal bisector of the magnetic at a distance of 14 cm from the centre of magnet will be
`B_("total") = B_(eq) + B_H = 0.18 xx10^(-4)`
`=0.54 xx 10^(-4) T = 0.54` G in the direction of earth.s field.
Discussion
4244.

A simple pendulum is made from a 0.65m long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elaspses before it attains its greatest speed?

Answer»

0.40s
0.10s
0.80s
0.20s

Answer :A
Discussion
4245.

What are the methods of producing coherent sources ?

Answer»

Solution :There are TWO general factor of producing coherent sources: Division of WAVEFRONT-In this method the wavefront is divided into two or more PARTS with the help of mirrors,biprism,etc.
II. Division of amplitude - In this method, the division of amplitude of the incoming wave is divided into two or more parts by PARTIAL reflection or refraction.
Discussion
4246.

A photon and electron have got same de Broglie wavelength. Which has greater total energy ?

Answer»

SOLUTION :Energy of a PHOTON = `E_1 = hv=(hc)/LAMBDA`
For electron `lambda = h/(MV)=h/(lambda v)`
Energy of electron `E_2 = mc^2 = (hc^2)/(lambda v)`
`E_1/E_2 = (hc)/lambda XX (hv)/(hc^2)=v/c`
Discussion
4247.

Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?

Answer»

SOLUTION :No! Any slab , HOWSOEVER flat , will have roughness much larger than the inter-atomic crystal spacing (-2 to 3 A) and HENCE continuous contact at the atomic LEVEL will not be possible . The JUNCTION will behave as a discontinuity for the following charge carriers .
Discussion
4248.

In the figure shown, all surfaces are smooth and block. A and wedge B have mass 10 kg and 20 kg respectively. Find normal reaction between block A &B, spring force and normal reaction of ground on block B. (g=10m//s^(2)). .

Answer»


ANSWER :80 N, 48 N, 264 N
Discussion
4249.

A transmitting antenna at the top of tower has a height of the receiving antenna is 32m. The maximum distance between them for satisfactory communication in line of sight mode is

Answer»

15.15 KM
21.25 Km
30.45 Km
35.42 Km

Answer :D
Discussion
4250.

Sixteen copper wires of length I and diameter d are connected in parallel to form a single composite conductor of resistance R. What must the diameter D of a single copper wire of length I be if it is to have the same resistance?

Answer»


ANSWER :4D
Discussion