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10 months ago

Physics 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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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 * 20 * 1.5 km³ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10m.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Time require to avoid the collision T= l/v where l = mean free path =1/ $\sqrt[]{2} π d^{2} n$

Where n = N/V

n=number of aeroplanes/volume

= $\frac{10}{20 * 20 * 1.5} = 0.0167 k m$ ^-3

T= $\frac{1 V}{\sqrt[]{2} π d^{2} N} * \frac{1}{v}$

T= $\frac{1}{\sqrt[]{2} * 3.14 * 20^{2} * 0.0167 * 10^{- 6} * 150}$ = $\frac{10^{6}}{1776.25 * 2.505}$

$= \frac{10^{6}}{4449.5} = 224.74 h \approx 225 h$

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10 months ago

Physics 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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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Consider an ideal gas with the following distribution of speeds

(a) Calculate V_rms and hence T(m=3 $* 10^{- 26} k g$ )

(b) If all the molecules with speed 1000m/s escape from the system , calculate new V_rms and hence T.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

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10 months ago

Physics 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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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Explain why

(a) There is no atmosphere on moon.

(b) There is fall in temperature with altitude. 13.28 Consider an ideal gas with following distribution of speeds

(i) Calculate Vrms and hence T. 26 ( 3.0 10 kg) m − = *

(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new V_rms and hence T.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) The moon has small gravitational force and hence, the escape velocity is small .

As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds . even though the rms speed of the air molcules is smaller than the escape velocity on the moon, a significant number of molecules have speed grater than the escape velocity.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At g

...more

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10 months ago

Physics 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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10 months ago

Physics 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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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a heat engine as shown in Fig. Q₁and Q₂ are heat added to heat bath T₁ and heat taken from T₂ in one cycle of engine. W is the mechanical work done on the engine. If W > 0, then possibilities are:

(a) Q₁> Q₂> 0

(b) Q₂ > Q₁ > 0

(c) Q₂ < Q₁ < 0

(d) Q₁ < 0, Q₂ > 0

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q₁= W+Q₂

W=Q₁-Q₂>0

Q₁>Q₂>0

We can also write Q₂1

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a cycle followed by an engine 1 to 2 is isothermal 2 to 3 is adiabatic 3 to 1 is adiabatic Such a process does not exist because

(a) Heat is completely converted to mechanical energy in such a process, which is not possible.

(b) Mechanical energy is completely converted to heat in this process, which is not possible.

(c) Curves representing two adiabatic processes don't intersect.

(d) Curves representing an adiabatic process and an isothermal process don't intersect.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

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