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44101.

The relative permiability of a diamagnetic substance is ........ .. .

Answer»

(A) very large
(B) small but greater than 1
(C) less then 1
(D) negative

Solution :Relative permiability `mu_(r)= (mu)/( mu_0) = 1 + 1 chi_(m)`
For a DIAMAGNETIC substance `chi_(m)` is negative, so `mu_(r)` will be less than 1.
Discussion
44102.

A 1 Omega resistance in series with an ammeter is balanced by 25 cm of potentiometer by 50 cm. The ammeter shows a reding of 1.5 A. Then, the error in ammeter reading is

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`0.03 A`
3A
`1.3 A`
`0.3A`

ANSWER :A
Discussion
44103.

In the above problem, when will be the fourth echo heard

Answer»

4s
1.5 s
5.5 s
3.5 s

Solution :`t_(4) = 2t_(1) + t_(2) = 3 + 2.5 = 5.5s`
Discussion
44104.

A cell of constant emf is first connected a resistance R_1 and then connected to a resistance R_2. If power delivered in both the cases is the same, then internal resistance of the cell is:

Answer»

`SQRT(R_1R_2)`
`sqrt(R_1/R_2)`
`|(R_1+ R_2)/2|`
`(R_1+R_2)^3`

ANSWER :A
Discussion
44105.

Which of the following has negative temperature coefficient of resistance?

Answer»

Copper
ALUMINIUM
Germanium
Iron

Answer :C
Discussion
44106.

Figure shows four plates each of plate area A and separated between plates is d. Two switches S_(w1) and S_(w_2) can activate two batteries in the circuit. Now switch S_(w_2) is opened, what is the charge on plate 4?

Answer»

`-5/3 (epsilon_0AV)/(d)`
`+5/3 (epsilon_0AV)/(d)`
`-1/3 (epsilon_0AV)/(d)`
`+1/3 (epsilon_0AV)/(d)`

Solution :
When both the switches are closed, CIRCUIT is
Charge on plate 2
`C_(EQ)=(2C)/(3)=(2epsilon_(0)A)/(3d)`
`q_(0)=((2C)/(3))(2V)=(4CV)/(3)`
Charge on plate 1
Where switch `Sw_(2)` is opened, circuit is

`[((x-V))/(-2V)]C+(x-2V)C+xC=0`
or `x=(5)/(3)V`
Charge on plate 4 is `-(5)/(3)CV=(-5epsilon_(0)AV)/(2d)`
Alternative method
`(-q_(1))/(C)+V+(q_(2))/(C)=0` or `-q_(1)+q_(2)=-CV`
`(-q_(1))/(C)-((q_(1)+q_(2)))/(C)+2V=0`
or `q_(1)+2q_(2)=2CV`
`q_(1)=(4CV)/(3),q_(2)=(CV)/(3)`
Discussion
44107.

Where it propagates ?

Answer»

SOLUTION :In SPACE
Discussion
44108.

In a metre bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Omega. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

Answer»

SOLUTION :`X =8.2 Omega` to MINIMISE RESISTANCE of the CONNECTIONS which are not accounted for in the bridge formula.
Discussion
44109.

Four transparent slabs having thickness t_(1)= 2cm, t_(2)= 4 cm, t_(3)= 3 cm" and "t_(4)= 5cm are introduced in one of the paths of light emitted by two narrow slits the ascending order of shift of the central fringe

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`t_(1), t_(2), t_(3), t_(4)`
`t_(4), t_(3), t_(2), t_(1)`
`t_(3), t_(2), t_(4), t_(1)`
`t_(1), t_(3), t_(2), t_(4)`

ANSWER :D
Discussion
44110.

Figure shows four plates each of plate area A and separated between plates is d. Two switches S_(w1) and S_(w_2) can activate two batteries in the circuit. What is the charge passing through battery 1?

Answer»

`3/2 (epsilon_0AV)/(D)`
`3/2 (epsilon_0AV)/(d)`
`1/3 (epsilon_0AV)/(d)`
`4/3(epsilon_0AV)/(d)`

Solution :
When both the switches are closed, circuit is
Charge on PLATE 2
`C_(eq)=(2C)/(3)=(2epsilon_(0)A)/(3d)`
`q_(0)=((2C)/(3))(2V)=(4CV)/(3)`
Charge on plate 1
Where switch `Sw_(2)` is OPENED, circuit is

`[((x-V))/(-2V)]C+(x-2V)C+xC=0`
or `x=(5)/(3)V`
Charge on plate 4 is `-(5)/(3)CV=(-5epsilon_(0)AV)/(2d)`
ALTERNATIVE method
`(-q_(1))/(C)+V+(q_(2))/(C)=0` or `-q_(1)+q_(2)=-CV`
`(-q_(1))/(C)-((q_(1)+q_(2)))/(C)+2V=0`
or `q_(1)+2q_(2)=2CV`
`q_(1)=(4CV)/(3),q_(2)=(CV)/(3)`
Discussion
44111.

In n-type semiconductors, majority charge carriers are

Answer»

HOLES
PROTONS
neutrons
electrons

ANSWER :d
Discussion
44112.

Figure-4.50 shows a thermal network of two metal rods of same cross section area. If the heat current from the ends A and B is 130 W, find the heat current through the curved metal rod.

Answer»


ANSWER :`[60W]`
Discussion
44113.

Figure shows four plates each of plate area A and separated between plates is d. Two switches S_(w_1) and S_(w_2) can activate two batteries in the circuit. Switch S_(w_1) and S_(w_2) are closed, What is the charge on plate 2?

Answer»

`3/2 (epsilon_0AV)/(D)`
`3/2 (epsilon_0AV)/(d)`
`1/3 (epsilon_0AV)/(d)`
`4/3(epsilon_0AV)/(d)`

SOLUTION :
When both the switches are closed, CIRCUIT is
Charge on plate 2
`C_(eq)=(2C)/(3)=(2epsilon_(0)A)/(3d)`
`q_(0)=((2C)/(3))(2V)=(4CV)/(3)`
Charge on plate 1
Where switch `Sw_(2)` is opened, circuit is

`[((x-V))/(-2V)]C+(x-2V)C+xC=0`
or `x=(5)/(3)V`
Charge on plate 4 is `-(5)/(3)CV=(-5epsilon_(0)AV)/(2d)`
Alternative method
`(-q_(1))/(C)+V+(q_(2))/(C)=0` or `-q_(1)+q_(2)=-CV`
`(-q_(1))/(C)-((q_(1)+q_(2)))/(C)+2V=0`
or `q_(1)+2q_(2)=2CV`
`q_(1)=(4CV)/(3),q_(2)=(CV)/(3)`
Discussion
44114.

As shown in figure a two slit arrangement with a source (S) which emits unpolarised light. I_(0) is the intensity of principle maxima when no polariseer is present. Now a polarised 'P' is placed as shown with its axis whose direstion is not given. Then

Answer»

the intensity PRINCIPAL MAXIMA is `(5)/(8)I_(0)`
the intensity of principal maxima is `(I_(0))/(2)`
the intensity of firstminima is zero
the intensity of first minima is `(I_(0))/(8)`

Solution :WITHOUT P :
`A=A_(_|_)+A_(11)`
`A_(_|_)=A_(_|_)^(1)+A_(_|_)^(2)=sin(kx-omega t)`
`+A_(_|_)^(@)sin(kx-omega t+phi), A_(11)=A_(11)^((1))+A_(11)^((2))`
`A_(11)=A_(11)^(@)[sin(kx-theta t)+sin(kx-omega t+phi)]`
where `A_(_|_)^(@), A_(11)^(@)` care the amplitudes of either of the beam in `_|_` and 11 polarizations.
`:. "Intensity" =`
`= = {|A_(_|_)^(0)|^(2) + |A_(11)^(@)|^(2)}`
`[sin^(2)(kx-omega t)(1+cos^(2)phi+2sin phi)`
`+sin^(2)(kx-omega)sin^(2)phi]_("average")`
`={|A_(_|_)^(0)|^(2) + |A_(11)^(@)|^(2)} ((1)/(2)).2(1+cos phi)`
`=2|A_(_|_)^(@)|^(2).(1+cos phi)since |A_(_|_)^(@)|_("average") = |A_(11)^(@)|_("average")`
With P:
Assume `A_(_|_)^(2)` is blocked:
Intensity `= (A_(11)^(1)+A_(11)^(2))+(A_(_|_)^(1))^(2)`
`= |A_(_|_)^(@)|^(2) (1+cos phi) + |A_(_|_)^(@)|^(2). (1)/(2)`
Given: `I_(0) = 4|A_(_|_)^(@)|^(2)` = intensity wityhout polariser at principal maxima. intensity principal maxima with polariser `= |A_(_|_)^(@)|^(2) (2+(1)/(2))`
`= (5)/(8)I_(0)`
Intensity at first minima with polariser
`= |A_(_|_)^(@)|^(2) (1-1)+(|A_(_|_)^(@)|^(2))/(2) = (I_(0))/(8)` .
Discussion
44115.

Initially conditions of L-C oscillator circuit ……….

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Q=0 , I=0 and t=0
`Q=Q_0, I=I_0` and t=0
Q=0 , `I=I_0` and t=0
`Q=Q_0 , I=0` and t=0

ANSWER :D
Discussion
44116.

When load is applied to a wire the extension is 3 mm, the extension in the wire of same material, length but half the radius by the same load is :

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0.75 mm
6 mm
1.5 mm
12.0 mm

Solution :`Deltal=(FL)/(Ay)rArr1lprop1/A`
or `Deltalprop1/(R^(2))`
`therefore(Deltal_(2))/(Deltal_(1))=(r_(1)^(2))/(r_(2)^(2))=(2/1)^(2)=4`
`rArrDeltal_(2)=4Deltal_(1)=4xx3=12mm`
So, the correct CHOICE is (d).
Discussion
44117.

The radioactivity of a sample is .X. at time .t_1. . and .Y. at a time .t_2. . If the mean life time of the specimen is .t. , the number of atoms that have disintegrated in the time interval (t_(1) -t_(2))is

Answer»

`Xt_(1) - Yt_(2)`
`X-Y`
`(X-Y)/tau`
`(X-Y) tau `

ANSWER :D
Discussion
44118.

The rate of disintegration of a radioactive sample is R and the number of atoms presentat any time t is N. When R/N is taken along Y-axis and t is taken along X-axis, the correct graphs is

Answer»




Solution :Law of radioactive decay is given as ,

`(dN)/DT = -lambda N `
where,`(dN)/dt=`
rate of disintegration
So,
`R = -lambda N RARR R/N = - lambda =` constant
where, `lambda= ` decay constant
So, the graph `R/N` VERSUS t is as,
So the CORRECT graph shown in option (d).
Discussion
44119.

Tworesistorsalongwitherrorsare kownas R_(1)= (30 +-3) OmegaR_(2) = (60+- 3 ) Omega column- Ilistsfouralagebraticfunctions ofR_(1) R_(2)column-IIlistscorrespondingerrorswhereas column-IIIlisitstheabsoluteerrors pickthe correctoptionparallelcombinationof R1 and R2

Answer»

(IV)(iii)(P)
`(IV)(iii)(Q)
(IV)(II)(R )
`(IV)(iii)(S )

Answer :A
Discussion
44120.

In a car race, car A takes time t less than car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Assuming that the cars start from rest and travel with constant accelerations a_1 and a_2 show that v/t=sqrt(a_1a_2) .

Answer»

Solution :The distance covered by both the cars is the same .THUS ,`s_(1)=s_(2)=s.`
If the cars TAKE time `t_(1)` and `t_(2)` for the race and their velocities at the FINISH be `v_(1)` and `v_(2)` ,then it is given that
`(v_(1))/(2)t_(1)`=s……….(1)
and `(v_(2))/(2)t_(2)=s`.........(2)
`therefore (v_(1))/(t_(2))=(v_(2))/(t_(1))=(v_(1)-v_(2))/(t_(2)-t_(1))=(v)/(t)` ............(3)
`((v)/(t))^(2)=(v_(1)v_(2))/(t_(2)t_(1))=(v_(1))/(t_(1)).(v_(2))/(t_(2))`..........(4)
`therefore(v)/(t)=sqrt(a_(1)a_(2))impliesv=t sqrt(a_(1)a_(2))`
Discussion
44121.

Consider a point charge +q placed at the origin and another point charge -2p placed at a distance of 9 m from the charge +q . Determine the point between the two charges at which electric potential is zero.

Answer»

Solution :According to the superposition PRINCIPLE the total electric potential at a point is equal to the sum of the potentials DUE to each charge at that point. Consider at a distance x from the charge +Q as shown in the figure
The toatal electric potential at P is zero.

`V_("tot")=(1)/(4piepsilon_(0))((q)/(x)-(2q)/((9-x)))=0`
which gives `(q)/(x) =(2q)/((9-x))` or `(1)/(x)=(2)/((9-x))`
Hence x =3 n
Discussion
44122.

Orbital acceleration of an electron is ………..

Answer»

`(pi^(2)N^(2)h^(2))/(2m^(2)r^(3))`
`(n^(2)h^(2))/(4pi^(2)m^(2)r^(3))`
`(4n^(2)h^(2))/(pi^(2)m^(2)r^(3))`
`(n^(2)h^(2))/(pi^(2)m^(2)r^(3))`

Solution :Angular momentum of electron in `n^(th)` orbit,
`MVR=(nh)/(2pi)`
`:.V=(nh)/(2pimr)`
`:.(v^(2))/(r)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(3))`
`:.a_(C)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(3))`
[ `:.(v^(2))/(r)=` centripetal acceleration `a_(c)` ]
Discussion
44123.

Tworesistorsalongwitherrorsare kownas R_(1)= (30 +-3) OmegaR_(2) = (60+- 3 ) Omega column- Ilistsfouralagebraticfunctions ofR_(1) R_(2)column-IIlistscorrespondingerrorswhereas column-IIIlisitstheabsoluteerrors pickthe correctoptionforseriescombinationof R1 and R2 .

Answer»

(III)(ii)(P )
(III)(i)(S )
(III)(I)(Q)
(III)(IV)(S )

ANSWER :B
Discussion
44124.

Tworesistorsalongwitherrorsare kownas R_(1)= (30 +-3) OmegaR_(2) = (60+- 3 ) Omega column- Ilistsfouralagebraticfunctions ofR_(1) R_(2)column-IIlistscorrespondingerrorswhereas column-IIIlisitstheabsoluteerrors pickthe correctoption

Answer»

(IV)(II)(R )
(III)(ii)(S )
(II)(i)(Q)
(I)(iv)(Q)

Answer :D
Discussion
44125.

A mark on the surface of a glass spehre (mu=1.5) is viewed from a diametrically opposite position. It appears to be at a distance 10 cm from its actual position. The radius of the sphere is

Answer»

5 cm
10 cm
15 cm
25 cm

Answer :A
Discussion
44126.

An alternating voltage of frequency f is applied across a series LCR circuit. Let f_r be the resonance frequency for the circuit. Will the current in the circuit lag, lead or remain in phase with the applied voltage when (i) f gt f_r(ii) fltf_r (iii) f=f_r? Explain your answer in each case.

Answer»

Solution :(i) Current will LAG because.
`V_(L) lt V_( C)` HENCE `V_(L) - V_(C) gt O`
(ii) Current will lead, because.
`V_(L) lt V_( C)` Hence `V_(L) - V_( C) lt O`
(III) In phase
Discussion
44127.

In an experiment to find focal length of a concave mirror a graph is drawn. The graph looks like ...........

Answer»




SOLUTION :For concave mirror
OBJECT distance u, image distance v and focal length f
`u=(ALPHA)` when v=f
u=2f(object on centre of CURVATURE)
v=2f
and when u=f then v=`infty` (infinite)
Discussion
44128.

The energy band gap is maximum in

Answer»

metals
SUPERCONDUCTORS
insulators
semiconductors

SOLUTION :The BAND gap is MAXIMUM in insulators.
Discussion
44129.

How manystereoisomers arepossiblefor -

Answer»

`2^(4)`
`2^(5)`
`2^(6)`
`2^(7)`

Solution :
it isunsymetricmoleculeand no. of stereoimers
`=2^(n)=2^(6)`
Discussion
44130.

When a silver foil (Z = 47) was used in an alpharay scattering experiment , the number of alpha-particlesscattered at 30^(@) was found to be 200 per minute. If the silver foil is replaced by alumninium (z = 13) foil of same thickness, the number of alpha-particles scattered per minute at 30^(@) is nearly equal to

Answer»

15
30
10
20

Answer :A
Discussion
44131.

Define electric resistance and write it's SI unit. How does the reisitance of a conductor vary if (a) conductor is stretched to 4 times of it's length (b) Temperature of a conductor is increased.

Answer»

Solution :Electric (R ) : The resistance offered by a flow of electrons in a CONDUCTOR is called electric resistance.
S.I unit of resistance is ohm `(OMEGA)`
The resistance of a conductor
`R = (rho l)/(A) = (rho l^(2))/(V) implies R PROP l^(2)`
(a) In first CASE, `R_(1) = R, l_(1) = l`
(b) In second case, `l_(2) = 4l, R_(2) = ?`
`(R_(2))/(R_(1)) = (l_(2)^(2))/(l_(1)^(2)) implies (R_(2))/(R) = ((4I)/(I))^(2) :. R_(2) = 16 R`
(b) Variation of resistance with temperature is given by `R_(t) = R_(0) (1 + alpha t)`
If temperature increases, resistance also increases.
Discussion
44132.

A galvanometer of100Omega resistance gives full scale deflection when 10mA of current is passed.To convert it into 10 A range ammeter, the resistance of the shunt required will be :

Answer»

`10 OMEGA`
`1 Omega`
`0.1 0mega`
0.01

Answer :C
Discussion
44133.

Is the source of magnetic field analogue to the source of electric field ?

Answer»

SOLUTION :No, the SOURCE of MAGNETIC field is not a magnetic charge, but in CASE of electric field the source of electric field is electric charge
Discussion
44134.

(A) : Dimensions of constant of proportionalities can be derived from dimensional method. (R): Numerical value of constant of proportionality can be found from experiments only.)

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 but (R) is FALSE
Both (A) and (R) are false

ANSWER :D
Discussion
44135.

Atomic hydrogen is in thermodynamics equilibrium with its radiation. Find: (a) the ratio of probabilites of induced and spontaneous radiation of the atoms from the level 2P at a temperature T=3000K, (b) the temperature at which these probabilites become equal.

Answer»

SOLUTION :ATOMIC hydrogen is in thermodynamics equilibrium with its radiation, Find:
(a) the RATIO to probabilites of induced and SPONTANEOUS RADIATIONS of the atom from the level `2P` at a temperature `T=3000K`,
(b) the temperature at which these probabilites become equal.
Discussion
44136.

Figure 33.5 depicts the grid characteristics of a triode plotted at anode voltages of 450 V and 600 V. Find the triode's internal resistance R_(i) in the linear section of the characteristic and its amplification factor mu, i.e. the ration of the change in the anonde voltage to the change in the grid voltage which cauases a given change in the anode current.

Answer»


Solution :Calculations should be done for the linear section of the CHARAC. teristic CURVE. In this case a change in the anode voltage `DeltaU_(a)=150` V, with a constant grid voltage (for instance, with `U_(g)=0` ), causes a change in the anode current of `Deltai_(a)=75mA`. The tube.s internal resistance is `R_(i)=(DeltaU_(a))/(Deltai_(a))` The tube.s AMPLIFICATION factor is `mu=(DeltaU_(a))/(DeltaU_(g))`, i.e. `mu` is the RATIO of the change in the anode voltage to the change in the grid voltage which causes a given change in the anode current. It follows from Fig. that a change of anode current equal to 75 mA may be obtained either by changing the anode voltage by `DeltaU_(a)=150V`, or by changing the grid voltage by `DeltaU_(g)=7.5` V. Therefore the grid is 20 times more effective in controlling the current flowing through the tube on the linear section of its characteristic curve), i.e. amplification factor `mu=20`
Discussion
44137.

In the circuit shown in Fig. 4.68, the cells epsilon_(1) " and " epsilon_(2) have emf's of 4 V and 8 V and internal resistances of 0.5 Omega and 1 Omega respectively. Calculate the current in each resistance.

Answer»


ANSWER :`(1)/(2) A, (1)/(3)A,(1)/(6)A`
Discussion
44138.

What happens to the interference pattern if one of the slits in Young's double-slit experiment is closed ?

Answer»

SOLUTION :The interference PATTERN DISAPPEARS and we observe diffraction pattern DUE to a single-slit.
Discussion
44139.

A pan filled with hot food cools form 94^(@)C to 86^(@)C in 2 minutes . When the roo temperature is 20^(@)C. How long will it cool from 74^(@)C to 66^(@)C ?

Answer»

2 MINUTES
2.8 minutes
2.5 minutes
1.8 minutes

Solution :From Newton.s law of cooling, `(d theta)/(dt)prop(theta-theta_(0))`
In Case (i), `(94-86)/(2)prop((94+86)/(2)-20)`
`IMPLIES (8)/(2)prop(90-20)`
or `4prop70""….(1)`
In Case (ii), `(74-66)/(t)prop((74+66)/(2)-20)`
`(8)/(t)prop(70-20)`
`(8)/(t)prop50""....(2)`
Dividing (1) and (2)
`(t)/(2)=(70)/(50)`
`t=(14)/(5)`
`t=2.8` minutes
Discussion
44140.

A particle of mass 10 g is performing SHM. Its kinetic energies are 4.7 J and 4.6 J when the displacements are 4 cm and 6 cm , respectively. Compute the period of oscillation.

Answer»

SOLUTION :Data : m = 0.01 kg, `KE_(1) = 4.7 J, x_(1) = 4 xx 10^(-2) m, KE_(2) = 4.6 J `
` x_(2) =6 xx 10^(-2) m`
Since the total energy of a PARTICLE is SHM is CONSTANT .
` KE_(1) + PE_(1) = KE_(2) +PE_(2)`
` KE_(1) -KE_(2) = PE_(2) -PE_(1) = 1/2m omega^(2) (x_(2)^(2) -x_(1)^(2))`
` 1/2 m((4pi^(2))/(T^(2)) (x_(2)^(2) -x_(1)^(2)) "" (therefore omega = (2pi)/T)`
` T^(2) = 2mpi^(2) = ((x_(2)^(2)-x_(1)^(2)))/(KE_(1) -KE_(2))`
` = 2(0.01) pi^(2) ([(6^(2)-4^(2)) xx 10^(-4)])/(4.7 - 4.6)`
` = 0.2 pi^(2) xx 20 xx 10^(-4)= 4pi^(2) xx 10^(-4)`
The period of oscillation of the particle.
` T= sqrt(4pi^(2) xx 10^(-4)) = 2pi xx10^(-2) = 6.284 xx 10^(-2)S`
Discussion
44141.

माचिस की डिब्बी दर्शाती है

Answer»

घनीय ज्यामिति
एकनताक्ष ज्यामिति
विषमलम्बाक्ष
द्विसमलम्बाक्ष ज्यामिति

Answer :C
Discussion
44142.

The magnetism of magnet is due to

Answer»

the SPIN MOTION of electrons
Earth
pressure of HUGE MAGNET inside the earth
cosmic rays.

Answer :A::C
Discussion
44143.

Find unit vector in direction of friction force acting on block placed on plank 'p' (Given : bar(V)_(P)=7i - 2j,bar(V)_(B)=3hat(i)+hat(j))

Answer»

`(4)/(5)hat(i)-(3)/(5)hat(j)`
`(4)/(5)hat(i)-(3)/(5)hat(k)`
`(4)/(5)hat(j)-(3)/(5)hat(i)`
`(4)/(5)hat(i)+(3)/(5)hat(j)`

Solution :`v=(DS)/(dt)=12-4t""THEREFORE""u = 12 m//s"and"a=-4m//s^(2),t_(0)=|(u)/(a)|=3s`
Here, `t_(0)` is the time when VELOCITY becomes zero. Since, the given time t = 5s is greater then `t_(0)`. The distance
`d=|(u^(2))/(2a)|+|(1)/(2)a(t-t_(0))^(2)|=(144)/(2xx4)+(1)/(2)xx4xx4=18+8=26m`
Discussion
44144.

In a series combination of two capacitances C and C (Cgt C) (as shown in the circuit),

Answer»

C' STORES more ENERGY than C
C stores more energy than C
potential DIFFERENCE across C is more than that across C'
 potential difference across C is more than that across C

Answer :B::C::D
Discussion
44145.

Potential at a point A is 3 volt and at a point B is 7 volt, an electron is moving towards A from B.

Answer»

It must have some K.E at B to REACH A
It NEED not have any K.E at B to reach A
to reach A it must have more than or equal to 4 eV K.E at B.
when it will reach A, it will have K.E. more than or at least equal to 4 eV if it was relased from REST at B.

Answer :A::C
Discussion
44146.

Answer the questions : In what way is diffraction from each slit related to the interference pattern in a doutler slit experiment ?

Answer»

SOLUTION :The INTENSITY of interference fringes in a double SLIT ARRANGEMENT is modulated by diffraction PATTERN of each slit.
Discussion
44147.

Figure here gives the speed-time graph for a body. The distance travelled between I = 1.0 second and t = 7.0 second is nearest to:

Answer»

1.5 m
2 m
3 m
4 m

Solution :Here the AREAS from t=1 sec. to t =7 sec are
From t=1 to t=2 `x_1=1xx2+(1)/(2)xx1xx2=3 m`
From t=2 to t=3 `x_21xx4=4m`
From t=3 to t=4 `x_3=(1)/(2)xx1xx4=2 m`
From t4 to t=5 `x_4=(1)/(2)xx1xx-4= -2m`
Net distance =3+4+2-2-4=3m.
Discussion
44148.

Mention various (any four) energy losses in a transformer.

Answer»

Solution :Possible causes of energy losses in TRANSFORMERS are as FOLLOWS:
(i) Flux leakage : There is always some leakage of magnetic flux. It can be reduced by WINDING the primary and secondary coils one over the other.
(ii) Resistance of the windings : The wires used for the windings have some resistance and hence some energy is lost due to heat produced in wire. It can be minimised by taking thick insulated copper wire for windings.
(iii) Eddy currents : The alternating magnetic flux INDUCES eddy currents in iron core, which results loss of electrical energy. To minimise it we use a laminated iron core.
(iv) Hysteresis: As the magnetisation cycle of iron core is repeated again and again some loss of energy takes place due to magnetic hysteresis. To minimise it we prefer a SOFT iron core for which hysteresis loss is less.
Discussion
44149.

An alternating current is given by i=(3sin omegat+4cos omegat)A. Find rms current.

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Solution :`i=3sin OMEGA t+4 cos omega t`
`(int_(0)^(t)i^(2).dt)/(int_(0)^(t)dt)=(int_(0)^(T)(3sin OMEGAT+4cos omegat)^(2)dt)/(T)`
`=(1)/(T)int_(0)^(T)(9sin^(2)omegat+16cos^(2)omegat+12sin 2omegat)dt`
`=(25)/(2)""therefore i_("rms")=(5)/(sqrt2)A`
Discussion
44150.

In the circuit diagram shown in figure, potential difference across 3Omega resistance is 20 V. Then, match the following two columns.

Answer»


ANSWER :`(Ararrr;Brarrp;,Crarrp;Drarrs)`
Discussion