Physics NCERT Exemplar Solutions Class 12th Chapter Three

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R'  also when we decrease the value of R the current will  also very large. According to the relation I=e/R+r

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneVd  , current is directly proportional to drift velocity.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- R= δlA for greater value of R, A should be less and its possible value when connected across 1cm * 1/2cm faces.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation – R/S= (l1/100-l1)= 100 (2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= e2r1+e2r1r1+r2

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as we know that J=E, and current density is directly proportional to electric field, so electric field produced by charges accumulated on the surface of wire.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to ohms law

I= VeffReff+R

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= nVeffnReff+nR= VeffReff+R =

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- R= ρl /A

Resistance of first conductor, RA= ρlπ (10-3*0.5) 2

Resistance of second conductor, RB= ρlπ (10-6- (0.5*0.5*10-6)}

Now ratio RARB=ρlπ (10-3*0.5)2ρlπ (10-6- (0.5*0.5*10-6)} = 3:1

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