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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Electricity

3.14 The earth's surface has a negative surface charge density of 10^–9 C m^–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralize the earth's surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 * 10⁶ m.)

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Payal Gupta

Contributor-Level 10

3.14 Surface charge density of the earth, $σ$ = $10^{- 9}$ C $m^{- 2}$

Current over entire Globe, I = 1800 A

Radius of the earth, r = 6.37 $* 10^{6}$ m

Hence, surface area of the earth, A = 4 $π r^{2}$ = 4 $* π * {(6.37 * 10^{6})}^{2}$ = 5.1 $* 10^{14}$ $m^{2}$

Charge on the earth surface, q = $σ$ $* A$ = 0.51 $* 10^{6}$ C

If t is time taken to neutralize the earth surface, then q = I $* t$

t = $\frac{q}{I}$ = $\frac{0.51 * 10^{6}}{1800}$ = 283.28 seconds

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Physics Ncert Solutions Class 12th Electricity

3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 * 10²⁸ m^–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 * 10^–6 m² and it is carrying a current of 3.0 A.

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Payal Gupta

Contributor-Level 10

3.13 Given, number of free electron in a copper conductor, n = 8.5 $* 10^{28}$ $m^{- 3}$

Length of the copper wire, l = 3.0 m

The area of cross section, A = 2 $* 10^{- 6}$ $m^{2}$

Current carried by the wire, I = 3.0 A

From the relation I = nAe $V_{d}$ where

e = Electric charge = 1.6 $* 10^{- 19}$ C

$V_{d} = D r i f t v e l o c i t y$

We get $V_{d}$ = $\frac{I}{n A e}$

Again, Drift velocity ( $V_{d})$ = $\frac{L e n g t h o f t h e w i r e (l)}{T i m e t a k e n t o c o v e r l (t)}$

t = $\frac{l}{V_{d}}$ = $\frac{l n A e}{I}$ = $\frac{3 * 8.5 * 10^{28} * 2 * 10^{- 6} * 1.6 * 10^{- 19}}{3}$ = 27.2 $* 10^{3}$ seconds

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Physics Ncert Solutions Class 12th Electricity

3.12 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

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Payal Gupta

Contributor-Level 10

3.12 EMF of the cell, $E_{1}$ = 1.25 V

Let the EMF of the replaced cell be $E_{2}$

Existing balance point, $l_{1}$ = 35 cm

New balance point, $l_{2}$ = 63 cm

From the relation of balance condition, we get

$\frac{E_{1}}{E_{2}}$ = $\frac{l_{1}}{l_{2}}$ , we get $E_{2}$ = $\frac{E_{1} * l_{2}}{l_{1}}$ = $\frac{1.25 * 63}{35}$ = 2.25 V

Therefore the emf of the another cell is 2.25V

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Physics Ncert Solutions Class 12th Electricity

3.11 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

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Payal Gupta

Contributor-Level 10

3.11 EMF of the battery, E = 8 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Let the effective voltage in the circuit be $V_{1}$

Then $V_{1}$ = V – E = 120 – 8 = 112 V

If the current flown in the circuit be I then from the relation

I = $\frac{V_{1}}{R + r}$ we get I = $\frac{112}{15.5 + 0.5}$ = 7 A

Voltage across the resistor R = IR = 7 $* 15.5 =$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherw

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3.11 EMF of the battery, E = 8 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Let the effective voltage in the circuit be $V_{1}$

Then $V_{1}$ = V – E = 120 – 8 = 112 V

If the current flown in the circuit be I then from the relation

I = $\frac{V_{1}}{R + r}$ we get I = $\frac{112}{15.5 + 0.5}$ = 7 A

Voltage across the resistor R = IR = 7 $* 15.5 =$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherwise, the current will be extremely high.

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Physics Ncert Solutions Class 12th Electricity

3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if R and S are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

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Payal Gupta

Contributor-Level 10

3.10 (a) Balance point from end A, $l_{1}$ = 39.5 cm

Resistance of the resistor, S = 12.5 Ω

Condition for the balance is given as,

$\frac{R}{S}$ = $\frac{l_{1}}{{100 - l}_{1}}$

$R = S * \frac{l_{1}}{100 - l_{1}}$ = 12.5 $* \frac{39.5}{100 - 39.5}$ = 8.16 Ω

$T h e c o n n e c t i o n b e t w e e n r e s i s t o r s i n a W h e a t s t o n e$ bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

(b) If R & S are interchanged, then $l_{1}$ and (100 - $l_{1}$ ) will also get interchanged. The balance point will be (100- $l_{1})$ from A. So the new balance point is 100 – 39.5 = 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflect

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3.10 (a) Balance point from end A, $l_{1}$ = 39.5 cm

Resistance of the resistor, S = 12.5 Ω

Condition for the balance is given as,

$\frac{R}{S}$ = $\frac{l_{1}}{{100 - l}_{1}}$

$R = S * \frac{l_{1}}{100 - l_{1}}$ = 12.5 $* \frac{39.5}{100 - 39.5}$ = 8.16 Ω

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection, no current will flow through galvanometer.

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Physics Ncert Solutions Class 12th Electricity

3.9 Determine the current in each branch of the network shown in Fig. 3.30:

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Payal Gupta

Contributor-Level 10

3.9

Let us assume

$I_{1}$ = Current flowing through the outer circuit

$I_{2}$ = Current flowing through the branch AB

$I_{3}$ = Current flowing through the branch AD

$I_{4}$ = Current flowing through the branch BD

( $I_{2}$ - $I_{4})$ = Current flowing through the branch BC

( $I_{3}$ + $I_{4})$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10 $I_{2}$ + 5 $I_{4}$ - 5 $I_{3}$ = 0

$I_{3} = I_{4} + 2 I_{2}$ …………(1)

For the close circuit BCDB, potential is zero, i.e.

5( $I_{2}$ - $I_{4})$ - 10( $I_{3}$ + $I_{4})$ - 5 $I_{4}$ = 0

5 $I_{2}$ - 5 $I_{4}$ - 10 $I_{3}$ - 10 $I_{4}$ - 5&nbs

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3.9

Let us assume

$I_{1}$ = Current flowing through the outer circuit

$I_{2}$ = Current flowing through the branch AB

$I_{3}$ = Current flowing through the branch AD

$I_{4}$ = Current flowing through the branch BD

( $I_{2}$ - $I_{4})$ = Current flowing through the branch BC

( $I_{3}$ + $I_{4})$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10 $I_{2}$ + 5 $I_{4}$ - 5 $I_{3}$ = 0

$I_{3} = I_{4} + 2 I_{2}$ …………(1)

For the close circuit BCDB, potential is zero, i.e.

5( $I_{2}$ - $I_{4})$ - 10( $I_{3}$ + $I_{4})$ - 5 $I_{4}$ = 0

5 $I_{2}$ - 5 $I_{4}$ - 10 $I_{3}$ - 10 $I_{4}$ - 5 $I_{4}$ =0

5 $I_{2}$ - 10 $I_{3}$ - 20 $I_{4}$ = 0

$I_{2} - 2 I_{3}$ - 4 $I_{4}$ = 0

$I_{2} = 2 I_{3}$ +4 $I_{4}$ …………….(2)

For the close circuit ABCFEA, potential is zero, i.e.

-10 + 10 $I_{1} +$ 10 $I_{2}$ + 5( $I_{2}$ - $I_{4})$ = 0

10 $I_{1}$ + 15 $I_{2}$ - 5 $I_{4}$ = 10

2 $I_{1}$ + 3 $I_{2}$ - $I_{4}$ = 2 …………(3)

Solving equations (1) and (2) we get

$I_{2} = - 2 I_{4}$

$I_{3} = - 3 I_{4}$

Substituting these values in equation (3), we get

$I_{4} = - \frac{2}{17}$ A

$I_{2} = \frac{4}{17}$ A

$I_{3} = \frac{6}{17}$ A

$I_{1}$ = $I_{2} + I_{3}$ = $\frac{10}{17}$ A

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Physics Ncert Solutions Class 12th Electricity

3.8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 * 10^–4 °C^–1.

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Payal Gupta

Contributor-Level 10

3.8 Given

Supply voltage, V = 230 V

Initial current drawn, I= 3.2 A

Final current drawn, $I_{1}$ = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature, $T_{1}$ = ?

From Ohm's law, we get initial resistance, $R$ = $\frac{V}{I}$ = $\frac{230}{3.2}$ Ω = 71.875 Ω

Final resistance, $R_{1}$ = $\frac{V}{I_{1}}$ = $\frac{230}{2.8}$ Ω = 82.143 Ω

From the relation of α = $\frac{R_{1} - R}{R (T_{1} - T)}$ , where α is the temperature coefficient of resistance, we get

1.7 $* 10^{- 4}$ = $\frac{82.143 - 71.875}{71.875 (T_{1} - 27)}$

$T_{1} - 27 =$ 840.34

$T_{1} = 867.34 ?$

Therefore the steady temperature of heating element required is $867.34 ?$

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Physics Ncert Solutions Class 12th Electricity

3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

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Payal Gupta

Contributor-Level 10

3.7 Let us assume R = 2.1 Ω, T = 27.5 °C, $R_{1}$ = 2.7 Ω, $T_{1}$ = 100 $?$ , α = resistivity of silver

We know the relation of α can be given as

α = $\frac{R_{1} - R}{R (T_{1} - T)}$ = $\frac{2.7 - 2.1}{2.1 (100 - 27.5)}$ = 3.94 $* 10^{- 3}$ $?^{- 1}$

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Physics Ncert Solutions Class 12th Electricity

3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 * 10^–7 m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

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Payal Gupta

Contributor-Level 10

3.6 Given, length of the wire, l = 15 m

Uniform cross section, A = 6.0 $* 10^{- 7}$ $m^{2}$

Resistance measured, R = 5.0 Ω

From the relation of

R = $ρ \frac{l}{A}$ , where $ρ$ is the resistivity of the material, we get

$ρ = \frac{A R}{l}$ = $\frac{6.0 * 10^{- 7} * 5}{15}$ = 2 $* 10^{- 7}$ Ωm.

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Physics Ncert Solutions Class 12th Electricity

3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 * 10^–4 °C^–1.

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Payal Gupta

Contributor-Level 10

3.5 Let T be the room temperature and R be the resistance at room temperature and $T_{1}$ be the required temperature and $R_{1}$ be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27 $?,$ R = 100 Ω, $T_{1}$ = ?, $R_{1}$ = 117 Ω, α = 1.70 $* 10^{- 4}$ ° $C^{- 1}$

We know the relation of α can be given as

α = $\frac{R_{1} - R}{R (T_{1} - T)}$ = $\frac{117 - 100}{100 (T_{1} - 27)}$ = 1.70 $* 10^{- 4}$

or ( $T_{1}$ - 27) = $\frac{117 - 100}{100 * 1.70 * 10^{- 4}}$ = 1000

$T_{1}$ = 1000 +27 = 1027 $?$

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