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Syllabus

2. Consider an infinite ladder network shown in figure. A voltage is applied between points A and B. If the voltage is halved after each section, fine the ratio $\frac{{R}_{1}}{{R}_{2}}.$

$\left(a\right)1\left(b\right)\frac{3}{4}\left(c\right)\frac{1}{2}\left(d\right)\frac{1}{4}$

Q.58. For the circuit given below, what is the effective resistance between points a and b.

A. 3R/4.

B. 3R/5.

C. 4R/3

D. 5R/3.

AB and CD are connected in an electric circuit in which an electric current is flowing.No current is flowing in these conductors.AB and CD are connected by conductor EF:-

(1) No current will flow through EF

(2) Current will flow from F to E

(3) Current will flow from E to F and the potential at E will be equal to that at A or B

(4) Current will flow from E to F and the potential at E will be less than at A or B

$1.\u25cb\frac{2ibB}{m}\phantom{\rule{0ex}{0ex}}2.\u25cb\frac{2iaB}{m}\phantom{\rule{0ex}{0ex}}3.\u25cb\frac{ibB}{m}\phantom{\rule{0ex}{0ex}}4.\u25cb\frac{iaB}{m}$

please answer it correctly ?

1. Straight line

2. Circle

3. Helix

4. Cycloid

1.the electric current through DC is zero

2. if we remove anyone of the edges AD, AC, DB or SC, the change in current remain same

3. the current through AB is maximum

4. the change in current will be same if AB or DC is removed

A cylindrical region of uniform magnetic field exists perpendicular to plane of paper which is increasing at a constant rate $\frac{dB}{dt}=\alpha .$ The diameter of cylindrical region is l. A non-contucting rigid rod of length l having two charged particles is kept fixed on the diameter of cylindrical region w.r.t. inertial frame. If two charged particles having charges q each is kept fix at the ends of the non-conducting rod. The net force on any one of the charge is

$1.\u25cb\frac{ql\alpha}{4}\phantom{\rule{0ex}{0ex}}2.\u25cb\frac{ql\alpha}{2}\phantom{\rule{0ex}{0ex}}3.\u25cbZero\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.\u25cbql\alpha .$

please answer it

a 200

b 300

c 400

d 500

a P/Q=R/S1+S2

b P/Q=2R/S1=S2

c P/Q=r(S1+S2)/S1S2

d P/q=R(S1=S2)/2S1S2

_{0}the velocity of other slider after considerably long time will be (neglect the self- induction)1. $\frac{{v}_{0}}{4}$

2. $\frac{{v}_{0}}{2}$

3. v

_{0 4. Zero }explain current distribution in the figure.

ABCDA is a uniform circular wire of resistance 2ohm. AOC and BOD are two wires along two perpendicular diameters of the circle, each having the same resistance 1ohm. A battery of emf E and internal resistance r is connected between points A and D. Calculate the equivalent resistance of the network.

This is the question for the above figure. please explain current distribution in the figure, thus providing a detailed solution.

In figure, infinite conducting rings each having current i in the direction shown are placed concentrically in the same plane as shown in the figure. The radius of rings are r, 2r, 2

^{2}r, 2^{3}r ......∞. The magnetic field at the centre of rings will be$1.\u25cbZero\phantom{\rule{0ex}{0ex}}2.\u25cb\frac{{\mu}_{0}i}{r}\phantom{\rule{0ex}{0ex}}3.\u25cb\frac{{\mu}_{0i}}{2r}\phantom{\rule{0ex}{0ex}}4.\u25cb\frac{{\mu}_{0}i}{3r}.$

please answer correctly

AB and CD are smooth parallel rails, separated by a distance l, and inclined to the horizontal at an angle $\theta $. A uniform magnetic field of magnitude B, directed vertically downwards, exists in the region. EF is a conductor of mass m, carrying a current i. For EF to be in equilibrium,

$1.\u25cbImustflowformEtoF\phantom{\rule{0ex}{0ex}}2.\u25cbBil=mg\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}3.\u25cbBil=mg\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}4.\u25cbNoneofthese.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}pleaseansweritcorrectly$

(1) 60 C

(2) 45 C

(3) 40 C

(4) 75 C

Q7. The current through S when it is closed in the circuit is

(A) 4.8 A

(B) 2.4 A

(C) 1.2 A

(D) 0.6 A

Q8. A voltmeter has a resistance of 1800 $\mathrm{\Omega}$ is used to measure the potential difference across a 200 $\mathrm{\Omega}$ resistor which is connected to the terminals of a DC power supply having an emf of 50 V and an internal resistance of 20 $\mathrm{\Omega}$. Find the percentage decrease in the potential difference across the 200 $\mathrm{\Omega}$ resistor as a result of connected the voltmeter across is?

(A) 1 % (B) 5 % (C) 10 % (D) 25%

should be

1. 4/9

2. 2

3. 8/3

4. 18

1. 122 volt

2. 60 volt

3. 100 volt

4. none

states that :

temperature coefficient of a wire is a PER DEGREE CELSIUS. At temperature T K resistance is 5 ohm.. At what tempearure will the resistance of the wire be 10 ohm.

I got this question incorrect sir.

In the given solution it is mentioned that since the temp is given in kelvin you convert into celsius. Hence T- 273 and the answer having a three digit no.

I got option 3 - which is when u dont convert into celsius. I dont understand why we are converting thhe temp into celsius... when :

I DEGREE CELSIUS = 1 DEGREE KELVIN

I am confused - pls help me out !

Ohm's law may be expressed as

$1.\square Potentialdifference,V\propto I(electriccurrent)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2.\square Electriccurrentdensity,j\propto E(electricfield)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3.\square \frac{V}{I}=\rho \frac{l}{A},symbolshavingusualmeaning\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.\square Resis\mathrm{tan}ce,R=\frac{V}{I}=cons\mathrm{tan}t$