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Physics Ncert Solutions Class 12th

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

A resistance R is to be measured using a meter bridge, a student chooses the standard resistance S to be 100 Ω He finds the null point at l₁= 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure l₁, more accurately
(b) He should change S to 1000 ? and repeat the experiment
(c) He should change S to 3 ? and repeat the experiment
(d) He should have given up hope of a more accurate measurement with a meter bridge.

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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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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two batteries of emf e1 and e2 (e2>e1) and internal resistance r1 and r2 respectively are connected in parallel as shown in fig

(a) Two equivalent emf e_eq of the cells is between e1 and e2 i.e. e1eq

(b) The equivalent emf e_eq is samller than e₁

(c) The e_eq is given by e_eq=e₁+e₂ always

(d) E_eq is independent of internal resistance r₁ and r₂

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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= $\frac{e_{2} r_{1 +} e_{2} r_{1}}{r_{1} + r_{2}}$

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Consider a current carrying wire (current 7) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
(a) Source of emf
(b) Electric field produced by charges accumulated on the surface of wire
(c) The charges just behind a given segment of wire which push them just the right way by repulsion
(d) The charges ahead.

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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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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all. resistances. Show that currents are unaltered. Do this for circuit of Examples 3, 7 in the NCERT Text Book for Class XII.

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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= $\frac{V_{e f f}}{R_{e f f} + R}$

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= $\frac{n V e f f}{n R e f f + n R} =$ $\frac{V_{e f f}}{R_{e f f} + R}$ =

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB

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alok kumar singh

Contributor-Level 10

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

Explanation- R= $ρ l$ /A

Resistance of first conductor, RA= $\frac{ρ l}{π {(10}^{- 3} * 0.5)}$ 2

Resistance of second conductor, RB= $\frac{ρ l}{π {{(10}^{- 6}) - (0.5 * 0.5 * 10^{- 6})}}$

Now ratio $\frac{R_{A}}{R_{B}} = \frac{\frac{ρ l}{π {(10}^{- 3} * 0.5)} 2}{\frac{ρ l}{π {{(10}^{- 6}) - (0.5 * 0.5 * 10^{- 6})}}}$ = 3:1

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two cells of same emf E but internal resistance r₁ and r₂ are connected in series to an external resistor R (figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero?

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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= $\frac{E + E}{R + r 1 + r 2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2 E}{R + r 1 + r 2}$ r1=0

E= $\frac{2 E r 1}{R + r 1 + r 2}$

1= $\frac{2 r 1}{R + r 1 + r 2}$

R+r1+r2 = 2r1

R= r1-r2

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Let there be n resistors R₁ . . . �� �� with �� max = max ( R₁ . . . . �� ) and R min = min { R₁ . . . R n }. Show that when they are connected in parallel the resultant resistance R p = R min �� �� = �� min and when they are connected in series, the resultant resistance R_s > R_max . Interpret the result physically.

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alok kumar singh

Contributor-Level 10

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

Explanation- when all resistance are in parallel

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \dots \dots \dots \dots . . + \frac{1}{R_{n}}$ by multiplying this equation by R_min we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$

But there exist a term in RHS $\frac{R_{m i n}}{R_{m i n}} = 1$ and other terms are positive so we have

$\frac{R_{m i n}}{R_{p}} = \frac{R_{m i n}}{R_{1}} + \dots \dots \dots \dots . . + \frac{R_{m i n}}{R_{n}}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+….R_n

And R_s> R_max

So equivalent resistance is less than R_max

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

The circuit in figure shows two cells connected in opposition to each other. Cell E₁is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B.

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alok kumar singh

Contributor-Level 10

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

Explanation-

total resistance = 2+8= 10 ohm

I= $\frac{6 - 4}{2 + 8}$ = 0.2A

The direction of flow of current is always from high potential to low potential

If VB $>$ VA

VB-4V-0.2 $*$ 8 = VA

V_B-V_A= 3.6V

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current / is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is V?

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alok kumar singh

Contributor-Level 10

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

Explanation – in series combination of resistors current I =e/R+r

10I= $\frac{e}{R + \frac{r}{n}}$

$\frac{1 + n}{1 + \frac{1}{n}}$ $\Rightarrow$ 10 = ( $\frac{1 + n}{1 + n}$ )n $\Rightarrow$ n=10

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of potential difference across R, versus R.

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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 the relation V=e-Ir and I=e/r+R

The graphical relation between voltage and resistance.

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