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11 months ago

Physics 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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New answer posted

11 months ago

Physics 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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New answer posted

11 months ago

Physics 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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New answer posted

11 months ago

Physics 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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New answer posted

11 months ago

Physics Electricity

3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

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Payal Gupta

Contributor-Level 10

3.4 (a) Let $R_{1}$ = 2 Ω, $R_{2}$ = 4 Ω, $R_{3}$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_{1}}$ + $\frac{1}{R_{2}}$ + $\frac{1}{R_{3}}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10 + 5 + 4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

(b) The EMF of the battery = 20 V

Current through $R_{1,}$ $I_{1}$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_{2}$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_{3}$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_{1}$ + $I_{2}$ + $I_{3}$ = 10 + 5 + 4 = 19 A

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New answer posted

11 months ago

Physics Electricity

3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

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Payal Gupta

Contributor-Level 10

3.3 (a) The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

(b) From Ohm's law, I = $\frac{V}{R}$ we get I = $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I $* R$ = 2 $* 1$ = 2 V

Potential drop across 2 Ω resistor = I $* R$ = 2 $* 2$ = 4 V

Potential drop across 3 Ω resistor = I $* R$ = 2 $* 3$ = 6 V

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New answer posted

11 months ago

Physics Electricity

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

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Payal Gupta

Contributor-Level 10

3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm's law

I = $\frac{E}{(R + r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $* 17 = 8.5 V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

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New answer posted

11 months ago

Physics Electricity

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

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Payal Gupta

Contributor-Level 10

3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM's law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

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New answer posted

11 months ago

Physics Mechanical Properties of Fluids

10.31 (a) It is known that density $ρ$ of air decreases with height y as

where $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ = 1.25 kg m^–3 is the density at sea level, and $ρ_{o}$ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.

(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?

[Take $y_{o}$ = 8000 m and $y_{o}$ = 0.18 kg m–3].

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Vishal Baghel

Contributor-Level 10

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and (ii), we get

$\frac{ρ}{ρ_{o}}$ =&nbs

...more

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and (ii), we get

$\frac{ρ}{ρ_{o}}$ = $e^{- k y}$ , k = $y_{0}$

Substituting the value of k in equation (ii), we get

$\frac{1}{k}$

Density $\frac{1}{y_{o}}$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}} \dots \dots (i i i)$

= $ρ = \frac{M a s s}{V o l u m e} = \frac{M a s s o f t h e p a y l o a d + M a s s o f h e l i u m}{V o l u m e}$ = 0.461 kg/ $\frac{m + V ρ_{H e}}{V}$

From equation (iii), taking log on both sides

$\frac{400 + 1425 * 0.18}{1425}$ = - $m^{3}$

y = -8000 $\log_{e} ? \frac{ρ}{ρ_{o}}$ = 8000 m = 8 km

The balloon will rise to 8 km

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11 months ago

Physics Mechanical Properties of Fluids

10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 * 10^–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 * 103 kg m^–3 (g = 9.8 m s^–2)

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Vishal Baghel

Contributor-Level 10

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $* 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $* 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $* 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $* 10^{- 3}$ m

Surface tension of water, s= 7.3 $* 10^{- 2}$ N/m

Angle of contact between the bore surface and water, $θ = 0$

Density of water, $ρ = 1.0 * 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity. g = 9.8 m/ $s^{2}$

Let $h_{1}$ and $h_{2}$ be the heights to which water rises in 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water in the 2 li

...more

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $* 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $* 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $* 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $* 10^{- 3}$ m

Surface tension of water, s= 7.3 $* 10^{- 2}$ N/m

Angle of contact between the bore surface and water, $θ = 0$

Density of water, $ρ = 1.0 * 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity. g = 9.8 m/ $s^{2}$

Let $h_{1}$ and $h_{2}$ be the heights to which water rises in 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water in the 2 limbs

$h_{1} - h_{2}$ = $\frac{2 s c o s ? θ}{ρ g}$ ( $\frac{1}{r_{1}}$ - $\frac{1}{r_{2}})$

= $\frac{2 * 7.3 * 10^{- 2} * c o s ? 0}{1.0 * 10^{3} * 9.8}$ ( $\frac{1}{1.5 * 10^{- 3}}$ - $\frac{1}{3 * 10^{- 3}})$ = 4.965 $* 10^{- 3}$ m = 4.965 mm

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