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#### Page No 16:

#### Question 1:

*I*) 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.

#### Answer:

Current density is basically electric current per unit area i.e. $j=\frac{I}{A}$.

However, electric field and current density are also inter related by a formula,

$\overrightarrow{j}=\sigma \overrightarrow{E}$

This means that current density changes due to the electric field produced by charges get collected on the surface of wire.

Hence, the correct answer is option (b).

#### Page No 16:

#### Question 2:

_{1}and ε

_{2}(ε

_{2 }> ε

_{1}) and internal resistances

*r*

_{1}and

*r*

_{2}respectively are connected in parallel as shown in the given figure.

(a) The equivalent emf ε

_{eq }of the two cells is between ε

_{1}and ε

_{2}, i.e. ε

_{1}< ε

_{eq }< ε

_{2}.

_{eq }is smaller than ε

_{1 }.

_{eq }= ε

_{1}+ ε

_{2}always.

_{eq }is independent of internal resistances

*r*

_{1 }and

*r*

_{2}.

#### Answer:

For parallel combination, equivalent emf

$\frac{{\epsilon}_{eq}}{{R}_{eq}}=\frac{{\epsilon}_{1}}{{r}_{1}}+\frac{{\epsilon}_{2}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\epsilon}_{eq}=\left(\frac{{\epsilon}_{1}}{{r}_{1}}+\frac{{\epsilon}_{2}}{{r}_{2}}\right)\times {R}_{eq}\phantom{\rule{0ex}{0ex}}\Rightarrow {\epsilon}_{eq}=\left(\frac{{\epsilon}_{1}}{{r}_{1}}+\frac{{\epsilon}_{2}}{{r}_{2}}\right)\times \frac{{r}_{1}{r}_{2}}{{r}_{1}+{r}_{2}}\left(\because {R}_{eq}=\frac{{r}_{1}{r}_{2}}{{r}_{1}+{r}_{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\epsilon}_{eq}=\frac{{\epsilon}_{1}{r}_{2}+{\epsilon}_{2}{r}_{1}}{{r}_{1}+{r}_{2}}$

It is given that,

${\epsilon}_{2}>{\epsilon}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {\epsilon}_{1}<{\epsilon}_{eq}<{\epsilon}_{2}$Hence, the correct answer is option (a).

#### Page No 17:

#### Question 3:

*R*is to be measured using a meter bridge. Student chooses the standard resistance

*S*to be 100â„¦. He finds the null point at

*l*

_{1}= 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?

*l*

_{1}more accurately.

*S*to 1000â„¦ and repeat the experiment.

*S*to 3â„¦ and repeat the experiment.

#### Answer:

Given, standard resistance, S = 100 Ω*l*_{1} = 2.9 cm

When meter bridge is balanced_{,}

$\Rightarrow \frac{2.9}{100\u20132.9}=\frac{R}{100}$

$\Rightarrow R=2.98\mathrm{\Omega}\approx 3\mathrm{\Omega}$â€‹

For improving accuracy, the null point should be in between the meter bridge wire which means resistance for known and unknown resistance should be nearly equal

$\Rightarrow \frac{R}{S}=\frac{1}{1}$

⇒

*R*=

*S*= 3Ω

Thus, the student should change

*S*to 3Ω and repeat the experiment.

Hence, the correct answer is option (c).

#### Page No 17:

#### Question 4:

*R*adjusted so that the potential drop across the wire slightly exceeds 10 V.

#### Answer:

In case of potentiometer experiment, emf of a cell can be measured, if along the potentiometer wire potential drop is more than the emf of the cell. Here, emf of two cells are 5V and 10 V, so potential drop across potentiometer wire should be more than 10 V.

Hence, the correct answer is option (b).

#### Page No 17:

#### Question 5:

(a) maximum when the battery is connected across 1 cm × $\frac{1}{2}$ cm faces.

#### Answer:

Resistance of the wire is given as,*R *= $\rho \frac{l}{A}$

Here, *ρ *is resistivity

*l* is length

*A* is area of cross-section

For greater value of *R*, *l *must be maximum and A should be minimum and this is only possible when battery is connected across

$\left(1\mathrm{cm}\times \frac{1}{2}\mathrm{cm}\right)$ faces.

Hence, the correct answer is option (a).

#### Page No 17:

#### Question 6:

#### Answer:

Current in a conductor is given as*I = n A _{e} v_{d}*

Here,

*I*is current

*n*is number of electrons per unit volume

*A*is area of cross-section

*v*

_{d }is drift velocity

*e*is charge of electron.

⇒

*I $\propto $ v*

_{d }

Hence, the correct answer is option (a).

#### Page No 18:

#### Question 7:

#### Answer:

Kirchhoff's junction law states that the total current into junction will be equal to the total current going out of the junction*I*_{in} – *I*_{out} = 0

Kirchhoff's junction rule is based on the law of conservation of charge and the fact that there is no charge accumulation at a junction.

Hence, the correct options are (b) and (d).

#### Page No 18:

#### Question 8:

*R*′.

*R*′ can vary from

*R*

_{0}to infinity.

*r*is internal resistance of the battery (

*r*<<

*R*<<

*R*

_{0}).

*R*′ is varied.

*R*′ is nearly a constant as

*R*′ is varied.

*I*depends sensitively on

*R*′

(d) $I\ge \frac{V}{r+R}$ always.

#### Answer:

Internal resistance = *r*

Variable resistance = *R*'

Here, *r << R << R'*

Applying Ohm's law, *V = IR*

$\Rightarrow I=\frac{V}{\left(r+\frac{RR\text{'}}{R+R\text{'}}\right)}\phantom{\rule{0ex}{0ex}}$

Since, *r << R << R'*

$\Rightarrow I=\frac{V}{r+{\displaystyle \frac{RR\text{'}}{R+R\text{'}}}}\ge \frac{V}{r+R}$

Potential difference across A and B, *V*_{AB} = $I\times \frac{RR\text{'}}{R+R\text{'}}$

$\Rightarrow {V}_{\mathrm{AB}}=\frac{V}{r+\left({\displaystyle \frac{RR\text{'}}{R+R\text{'}}}\right)}\times \frac{RR\text{'}}{R+R\text{'}}$

$=\frac{V\left(R+R\text{'}\right)}{r\left(R+R\text{'}\right)+RR\text{'}}\times \frac{RR\text{'}}{R+R\text{'}}$

$=\frac{VR}{r+R}$

Therefore,* V _{AB} * is constant if

*R*' is varied.

Hence, the correct options are (a) and (d).

#### Page No 18:

#### Question 9:

*ρ*(

*T*) of semiconductors, insulators and metals is significantly based on the following factors:

*T*.

*T*.

*T*.

*T*.

#### Answer:

Resistivity $\rho \left(T\right)=\frac{m}{n{e}^{2}\tau}$

When number of charge carriers changes with temperature *T*, then the resistivity will also change. Also, here *τ *is the relaxation time which is the time interval between two successive collisions, which decrease with increase in temperature.

Hence, the correct options are (a) and (b).

#### Page No 18:

#### Question 10:

*R*is to be carried out using Wheatstones bridge (see Fig. 3.25 of NCERT Book). Two students perform an experiment in two ways. The first students takes

*R*

_{2}= 10 â„¦ and

*R*

_{1}= 5 â„¦. The other student takes

*R*

_{2}= 1000 â„¦ and

*R*

_{1}= 500 â„¦. In the standard arm, both take

*R*

_{3}= 5 â„¦. Both find $R=\frac{{R}_{2}}{{R}_{1}}{R}_{3}=10\Omega $within errors.

*R*

_{2}and

*R*

_{1}can be measured.

*R*

_{2}and

*R*

_{1}, the currents through the arms will be feeble. This will make determination of null point accurately more difficult.

#### Answer:

Given for first student,*R*_{1} = 5 Ω*R*_{2} = 10 Ω*R*_{3} = 5 Ω

For second student,*R*_{1} = 500 Ω*R*_{2} = 1000 Ω*R*_{3} = 5 Ω

Applying wheat-stone bridge rule,

$\frac{{R}_{2}}{R}=\frac{{R}_{1}}{{R}_{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow R={R}_{3}\times \frac{{R}_{2}}{{R}_{1}}$

For** **student** **1,

$R=\overline{)5}\times \frac{10}{\overline{)5}}=10\mathrm{\Omega}$

For student 2,

$R=\overline{)5}\times \frac{\overline{)1000}}{\overline{)500}}=10\mathrm{\Omega}$ ** **

So, wheat-stone bridge is most sensitive and accurate if resistances are of same value.

When the values of *R*_{1} and *R*_{2} becomes large current through the arms become weak which can make the determination of null point accurately more difficult.

Hence, the correct options are (b) and (c).

#### Page No 19:

#### Question 11:

*R*is increased, the neutral point shifts to left.

#### Answer:

When jockey is at a neutral point D, current will not flow through galvanometer which means potentials at B and D are equal. However, potentials at different point will be different. So, the point D is unique to get null point. Now, when jockey is shifted in right of D on wire, potential in wire towards the right becomes smaller. i.e., *V*_{B} > *V*_{O.}

So, current flows from B to D in the wire. Now, when R is increased, potential drop across R increases which results in in-case in potential at B to get null point.

â€‹Hence, the correct options are (a) and (c).

#### Page No 19:

#### Question 12:

Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?

#### Answer:

In an electric circuit, when an electron approaches a junction, in addition to the uniform electric field *E* that it normally faces (which keep the drift velocity dv fixed), there are accumulation of charges on the surface of wires at the junction. Due to which an electric field is produced Thus these electric fields alter direction of momentum.

Hence, when charge crosses a junction in an electric circuit the momentum is not conserved at the junction.

#### Page No 19:

#### Question 13:

*τ*is nearly independent of applied

*E*field whereas it changes significantly with temperature

*T*. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of

*ρ*with temperature. Elaborate why?

#### Answer:

Relaxation time,

$\tau =\frac{m{v}_{d}}{eE}$

$\Rightarrow \tau \alpha \frac{{v}_{d}}{E}$

This means relaxation time depends upon velocity of electrons and ions and electric field. The applied electric field affects the velocities of electron at order of 1 mm/s which can be neglected and this also supports Ohm's law.

However, change in temperature affects velocities of electrons and ions at order 10^{2} m/s, which is significant.

â€‹

Resistivity, $\rho =\frac{m}{n{e}^{2}\tau}$

Thus, *τ* can be affected to an extent and hence the resistivity.

#### Page No 19:

#### Question 14:

*R*

_{unknown}by any other method?

#### Answer:

The necessary condition for wheatstone bridge is given as $\frac{P}{Q}=\frac{R}{S}$

Here, P and Q are ratio arms, R is known resistance and S is unknown resistance-advantage of the null-point method:-

1. Resistance of galvanometer does not affect the balance point. So, no need for determination of current in resistances and internal resistance of a galvanometer.

2. By calculating least count and readings of ammeter and voltmeter, R unknown resistance can be calculated by applying kirchhoff's law and Ohm's law.

#### Page No 19:

#### Question 15:

#### Answer:

In potentiometer, thick metallic strips are used, as they will introduce least resistance while joining the wires. Being metallic strips of least resistance, they can be ignored while measuring the length of the wire which is used for determination of null point.

#### Page No 19:

#### Question 16:

#### Answer:

For wiring in the home, Cu or Al wires are used the main considerations involved are:

(i) Good conductivity of metal.

(ii) Cost of metal used

#### Page No 19:

#### Question 17:

#### Answer:

Alloys are used for making standard resistance coils, this is because they have:

(i) Low value of temperature coefficient of resistivity.

(ii) High resistivity.

#### Page No 19:

#### Question 18:

*P*is to be delivered to a device via transmission cables having resistance

*R*. If

_{C}*V*is the voltage across

*R*and

*I*the current through it, find the power wasted and how can it be reduced.

#### Answer:

Power delivered,*P* = Voltage × Current = *VI*

$\Rightarrow I=\frac{P}{V}$

If *R*_{c} is the resistance of transmission lines then power wasted:*P*_{c} = *I*^{2}*R _{c}*

$\Rightarrow {p}_{c}=\frac{{p}^{2}}{{V}^{2}}{R}_{c}\phantom{\rule{0ex}{0ex}}\Rightarrow {p}_{c}\alpha \frac{1}{{V}^{2}}$

Thus, in order to reduce

*P*should be increased.

_{c , }VHence, the power wastage can be reduced by transmitting power of high voltage.

#### Page No 19:

#### Question 19:

*R*is increased, in which direction will the balance point J shift?

#### Answer:

When *R* is increased, the current through the wire will decrease and hence the potential gradient will also decrease. Hence, the balance length. will increase. Thus point J will shift towards B to obtain the balance point.

#### Page No 20:

#### Question 20:

*E*

_{1}, is connected at

*X*in case (i) and how is

*E*

_{1}related to

*E*?

*E*

_{1 }is connected at

*X*in case (ii)?

#### Answer:

(i) Current in auxiliary circuit decreases whereas potential difference across A and jockey increases. This is possible when positive terminal of *E*_{1} is connected at *X* and *E*_{1} > *E*.

(ii) Current in auxiliary circuit increases whereas potential difference across A and jockey increases. This is possible when *E*_{1} negative terminal is connected at *X*.

#### Page No 20:

#### Question 21:

*E*and internal resistance

*r*is connected across an external resistance

*R*. Plot a graph showing the variation of P.D. across

*R*, verses

*R*.

#### Answer:

Consider a circuit in which battery of emf *E* is connected to resistance *R* and having internal resistances *r*.

Let *V* be the potential drop across resistance and *I* be the current flowing through the circuit.

Applying, Kirchhoff's voltage law,*E* – *IR* – *Ir* = 0

⇒ *E* = *I*(*R* + *r*)

Potential drop across the battery, *V* = *IR*

⇒ *V* = $\frac{ER}{R+r}$

As *R* becomes large *R* + *r* becomes approximately equal to *R* and *V* ≈ *E*.

#### Page No 20:

#### Question 22:

*n*equal resistors of

*R*each are connected in series to a battery of emf

*E*and internal resistance

*R*.

*A*current

*I*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 ‘

*n*’?

#### Answer:

When *n* resistances each of resistance *R* are connected in series and then in parallel.

For series combination:*Rs* = *R*_{1} + *R*_{2} + . . . . + *R*_{n} = *R* + *R* + . . . . .*n* times

⇒ *R*_{s} = *n *R

For parallel combination

$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+....+\frac{1}{{R}_{n}}=\frac{1}{R}+\frac{1}{R}+....+n\mathrm{time}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{\mathit{p}}\mathit{=}\frac{\mathit{R}}{\mathit{n}}$

In series combination, current, *I *= $\frac{E}{R+nR}$

In parallel combination, current, 10* I* = $\frac{E}{R+{\displaystyle \frac{R}{n}}}$

$\Rightarrow \frac{E}{R+{\displaystyle \frac{R}{n}}}=\frac{10E}{R+nR}$

$\Rightarrow \frac{1}{R\left(1+{\displaystyle \frac{1}{n}}\right)}=\frac{10}{R\left(1+n\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow 10\left(1+\frac{1}{n}\right)=1+n\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\frac{10}{n}-1-n=0\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}-9n-10=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n+1\right)\left(n-10\right)=0$

Here *n* = –1 (not possible)

⇒ *n* = 10

Hence, there are 10 resistance in combination.

#### Page No 20:

#### Question 23:

*n*resistors

*R*

_{1}............

*R*with

_{n}*R*

_{max}_{= max}(

*R*

_{1}.........

*R*) and

_{n}*R*{

_{min = min}*R*

_{1}.....

*R*}. Show that when they are connected in parallel, the resultant resistance

_{n}*R*<

_{P}*R*

_{min}and when they are connected in series, the resultant resistance

*R*>

_{S}*R*. Interpret the result physically.

_{max}#### Answer:

Let *R _{min}* and

*R*be the minimum and maximium resistance among

_{max}*R*

_{1},

*R*

_{2}, . . . . .,

*R*. When resistance are connected in parallel,

_{n}$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+....+\frac{1}{{R}_{n}}$

On multiplying both sides by

*R*, we get

_{min }$\frac{{R}_{min}}{{R}_{p}}=\frac{{R}_{min}}{{R}_{1}}+\frac{{R}_{min}}{{R}_{2}}+....+\frac{{R}_{min}}{{R}_{n}}$

Here,

*R*

_{1},

*R*

_{2}. . . ,

*R*there must be a resistance which is minimum.

_{n}This means there must be a term

*R*in R.H.S. which is equal to 1 and the

_{min}*R*other terms are positive.

_{min}$\Rightarrow \frac{{R}_{min}}{{R}_{p}}=\frac{{R}_{min}}{{R}_{1}}+\frac{{R}_{min}}{{R}_{2}}+....+\frac{{R}_{min}}{{R}_{n}}1\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{min}{R}_{p}$

So, in parallel combination, equivalent resistance

*R*is always less than any smallest resistance in the combination.

_{p}When resistance are connected in series, then equivalent resistance

*R*=

_{s}*R*

_{1}+

*R*

_{2}+ . . . +

*R*

_{n}In R.H.S. there must be a term

*R*which has maximum value among

_{max}*R*

_{1}.

*R*

_{2}. . ,

*R*.

_{n}⇒

*R*=

_{s}*R*

_{1}+

*R*

_{2}+ . . +

*R*+ . . +

_{max}*R*=

_{n}*R*

_{++}

⇒

*R*>

_{s}*R*

_{max}So, in series combination equivalent resistance is always greater than the maximum resistance available in combination of resistance.â€‹

#### Page No 20:

#### Question 24:

*E*

_{1}is of emf 6 V and internal resistance 2 â„¦; the cell

*E*

_{2}is of emf 4 V and internal resistance 8 â„¦. Find the potential difference between the points

*A*and

*B*.

#### Answer:

Given,

emf of cell *E*_{1} = 6 V

Internal resistance, *r*_{1} = 2 Ω

emf of cell *E*_{2} = 4 V

Internal resistance, *r*_{2} = 8 Ω

Effective resistance, *R* = 2 + 8 = 10 Ω

Effective emf, *E* = 6 – 4 = 2 V

Using Ohm's law, *V* = *IR*

$\Rightarrow I=\frac{6-4}{2+8}=\frac{1}{5}\mathrm{A}$

The direction of flow of current is from higher potential to lower potential.

Since, ${V}_{B}>{V}_{A}$

$\Rightarrow {V}_{AB}={V}_{B}-{V}_{A}\phantom{\rule{0ex}{0ex}}=4+\frac{1}{5}\times 8=4+1.6\phantom{\rule{0ex}{0ex}}=5.6\mathrm{V}$

Hence, the potential difference between A and B is 5.6 V.

#### Page No 20:

#### Question 25:

Two cells of same emf *E* but internal resistance *r*_{1} and *r*_{2} are connected in series to an external resistor *R* in the given figure. What should be the value of *R* so that the potential difference across the terminals of the first cell becomes zero.

#### Answer:

Emf of two cells = *E*

â€‹Internal resistance = *r*_{1} and *r*_{2}

External resitance = *R*

Terminal voltage across first cell = 0

Now effective emf,

${E}_{effective}=E-E=0$

Electric current, I = $\frac{E-E}{R+{r}_{1}+{r}_{2}}=0$

Disclaimer: According to the circuit shown in the question, current through the circuit will be zero. So, the question/image seems incorrect.

#### Page No 21:

#### Question 26:

*A*is a solid wire of diameter 1mm. Conductor

*B*is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance

*R*to

_{A}*R*.

_{B}#### Answer:

Diameter of solid wire *A*, *d*_{1} = 1 mm.

Diameter of outer solid hollow tube *D*_{1} = 2 mm

Diameter of inner solid hollow tube, *D*_{2} = 1 mm

$\mathrm{Resistance},R=\frac{\rho l}{A}$

Resistance of first conductor, *R _{A}* = $\frac{\rho l}{\pi {\left({\displaystyle \frac{{d}_{1}}{2}}\right)}^{2}}$

Resistance of second conductor, ${R}_{B}=\frac{\rho l}{\pi \left[{\left({\displaystyle \frac{{D}_{1}}{2}}\right)}^{2}-{\left({\displaystyle \frac{{D}_{2}}{2}}\right)}^{2}\right]}$

Ratio of two resistors, $\frac{{R}_{A}}{{R}_{B}}=\frac{{\left({10}^{-3}\right)}^{2}-{\left(0.5\times {10}^{-3}\right)}^{2}}{{\left(0.5\times {10}^{-3}\right)}^{2}}$

$\Rightarrow \frac{{R}_{A}}{{R}_{B}}=\frac{1-0.25}{0.25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{R}_{A}}{{R}_{B}}=\frac{0.75}{0.25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{R}_{A}}{{R}_{B}}=\frac{3}{1}$

Hence, the ratio of resistance

*R*to

_{A}*R*is 3 : 1.

_{B}#### Page No 21:

#### Question 27:

*n*-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

#### Answer:

Let *n* be the number of batteries in series of the circuit and *R _{eff}* be the effective internal resistance of

*n*resistors âˆŠ

_{eff}be the effective emf of

*n*batteries in series and

*R*be the external resistance in the circuit.

Current through

*R*,

*I*= $\frac{{\epsilon}_{eff}}{{R}_{eff}+R}$

On increasing voltage and resistances

*n*-fold

${\epsilon}_{eff}^{new}=n{\epsilon}_{eff}\phantom{\rule{0ex}{0ex}}{R}_{eff}^{new}=n{R}_{eff}$

${R}^{new}=nR\phantom{\rule{0ex}{0ex}}\Rightarrow {I}^{new}=\frac{{\epsilon}_{eff}^{new}}{{R}_{eff}^{new}+{R}^{new}}\phantom{\rule{0ex}{0ex}}=\frac{n{\epsilon}_{eff}}{n{R}_{eff}+nR}=\frac{n{\epsilon}_{eff}}{n\left({R}_{eff}+R\right)}=\frac{{\epsilon}_{eff}}{{R}_{eff}+R}=I\phantom{\rule{0ex}{0ex}}\Rightarrow {I}^{new}=I$

Thus, the current remains unaltered when all voltage and all resistances are increased to

*n*-folds.

#### Page No 21:

#### Question 28:

#### Answer:

Voltage of cells, *V*_{1} = 10 V and *V*_{2} = 2 V

But,* V* = *IR*

Applying Kirchhoff's junction rule:*I*_{2} =* I*_{1} − *I*

In bigger outer loop having 10 V cell and resistance *R*:

applying Ohm's law,

10 = *IR* + 10*I*_{1 .....}(1)

Now, for outer loop having 2 V cell and resistance *R*:

2 = 5*I*_{2} – *RI*

⇒ 2 = 5(*I*_{1} – *I*) – *RI *.....(2)

Solving equation (1) and (2) we get:

6 = 3*RI* + 10*I*

⇒ 2 = $I\left(R+\frac{10}{3}\right)$ .....(3)

Effective potential difference due to both batteries is,*V*_{eff} = *I* (*R* + *R*_{eff}) .....(4)

On comparing equation (3) and (4), we get:*V*_{eff} = 2 V and *R*_{eff} = $\frac{10}{3}\mathrm{\Omega}$.

#### Page No 21:

#### Question 29:

*ρ*

_{Cu }= 1.7 × ${10}_{}^{-8}\Omega \mathrm{m}$ ,

*ρ*

_{Al }= 2.7 × 10

^{–8 }â„¦m]

#### Answer:

Energy consumed per day, *E* = 10 kWh

Time for using energy per day to 5 h

Power consumed, *P* = $\frac{10\mathrm{kWh}}{5\mathrm{h}}=2\mathrm{kW}=2000\frac{\mathrm{J}}{\mathrm{S}}$

Current supplied,

I = $\frac{P}{V}=\frac{200\overline{)0}}{22\overline{)0}}\approx 9\mathrm{A}$

Power loss in copper wire,*P* = ${I}^{2}R={I}^{2}\times \frac{\rho l}{\pi {r}^{2}}\left(\mathrm{as},R=\rho \frac{l}{A}=\frac{\rho l}{\pi {r}^{2}}\right)$

$\Rightarrow {P}_{\mathrm{cu}}={9}^{2}\times \left[\frac{1.7\times {10}^{-8}\times 10}{{\displaystyle \frac{22}{7}\times {\left({10}^{-3}\right)}^{2}}}\right]\approx 4\mathrm{J}/\mathrm{s}.$

Fractional loss of power in Cu wire,

$=\frac{4}{2000}\times 100=0.2\%$

Similarly power loss in Al wire,

P_{Al} = ^{}${I}^{2}\frac{{\rho}_{Al}\times l}{\pi {r}^{2}}$

$\Rightarrow {P}_{\mathrm{Al}}={9}^{2}\times \frac{\left(2.7\times {10}^{-8}\right)\times 10}{{\displaystyle \frac{22}{7}}\times {\left({10}^{-3}\right)}^{2}}=6.9\mathrm{J}/\mathrm{s}$

Fractional loss of power in Al wire,

= $\frac{6.9}{2000}\times 100=0.34\%.$

#### Page No 21:

#### Question 30:

*V*

_{B}= 10 V. R is adjusted to be 50 â„¦ in the given figure. A student wanting to measure voltage

*E*1 of a battery (approx. 8 V) finds no null point possible. He then diminishes

*R*to 10 â„¦ and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

#### Answer:

Variable resistance, *R* = 50 Ω

Let *R*' be the potentiometer wire resistance.

Current supplying in primary circuit be *I* which have *E*_{B} = 10 V

Here, *I* = $\frac{{V}_{\mathrm{B}}}{R+R\text{'}}\left[\mathrm{Using}\mathrm{Ohm}\text{'}\mathrm{s}\mathrm{law}\right]$

$\Rightarrow I=\frac{10}{50+R}$

Potential difference across wire of potentiometer, *V* = *IR*'

$\Rightarrow V\text{'}=\frac{10R\text{'}}{R+50}$

Null deflection can not be obtained by 8 *V*.

it is possible only when:

$\Rightarrow 10R\text{'}8R+400\phantom{\rule{0ex}{0ex}}\Rightarrow 2R\text{'}400\phantom{\rule{0ex}{0ex}}\Rightarrow R\text{'}200\mathrm{\Omega}$

Similarly, for *R* = 10 Ω, null point is possible when,

$\frac{10R\text{'}}{10+R\text{'}}8$

$\Rightarrow 2R\text{'}80\phantom{\rule{0ex}{0ex}}\Rightarrow R\text{'}40\mathrm{\Omega}$

Null point is obtained on 4th segment or at $\frac{3}{4}$ of the total length.

$\Rightarrow \frac{10\times {\displaystyle \frac{3}{4}}R\text{'}}{10+R\text{'}}<8\left(\mathrm{At}\mathrm{balance}\mathrm{point}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 7.5R\text{'}80+8R\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2013R\text{'}160\phantom{\rule{0ex}{0ex}}\Rightarrow R\text{'}160$

⇒ 160 < *R*' < 200

Any *R*' between 160 Ω and 200 Ω can achieve a null point null pint is on the 4th segment of potentiometer wire.

So, 400 cm of wire > 8 V is the potential drop.

Potential gradient (*k*),*k* × 400 cm > 8 V

⇒ *k* × 4 m > 8 V

⇒ *k* > 2 V/m .....(1)

Similarly, as balance point is at 4th segment of the wire, so, no balance point at 3 m.*k* × 300 cm < 8 V = *k* × 3 m < 8 V

⇒ *k* < $\frac{8}{3}$ V/m .....(2)

By using equation (1) and (2), we get:

â€‹$\frac{8}{3}\mathrm{V}/\mathrm{m}k2\mathrm{V}/\mathrm{m}$

#### Page No 21:

#### Question 31:

(b) Electrons give up energy at the rate of

*RI*

_{2}per second to the thermal energy. What time scale would one associate with energy in problem (a)?

*n*= no of electron/volume = $\frac{{10}^{29}}{{m}^{3}}$10

^{29}/

*m*

^{3}, length of circuit = 10 cm, cross-section =

*A*= (1 mm)

^{2}

#### Answer:

Given.

Area of cross-section, *A* = 1 mm^{2 }= 10^{– 6} m^{2}

Resistance, *R* = 6 Ω

Voltage, V = 6 V*n* = $\frac{{10}^{29}}{{m}^{2}}$

Applying Ohm's law, *V* = *IR*

⇒ 6 = I (6)

⇒ I = 1 A

a. Current, I = Ane V_{d}

⇒ V_{d} = $\frac{I}{\mathrm{Ane}}$

⇒ V_{d = $\frac{1}{{10}^{-6}\times {10}^{29}\times 1.6\times {10}^{-19}}$}

⇒ V_{d = $\frac{{10}^{-4}}{1.6}\mathrm{m}/\mathrm{s}$}

Number off electrons in wire = *n* (volume of the wire)

= *n* × A*l*

Here, energy is absorbed in the form of kinetic energy.

$\Rightarrow E=\frac{1}{2}m{v}_{d}^{2}\mathrm{A}nl$

$=\frac{1}{2}\times 9.1\times {10}^{-31}\times \frac{{10}^{-4}\times {10}^{-4}}{1.6\times 1.6}\times {10}^{29}\times {10}^{-6}\times {10}^{-1}$

$=\frac{9.1}{2\times 2.56}\times {10}^{-17}=1.78\times {10}^{-17}\approx 2\times {10}^{-17}$

b. Power loss, *P* = I^{2}*R* = (1)^{2} × 6 = 6 j/s

Energy, *E* = *P* × *t*

$\Rightarrow t=\frac{E}{P}=\frac{\overline{)2}\times {10}^{-17}}{{\overline{)6}}_{3}}=0.33\times {10}^{-17}\mathrm{sec}$

Hence, the time scale is approximately 3 × 10^{–18} sec.

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