Potential of the electrons, V = 30 kV = 3 x 104 V Hence, energy of the electrons, E = 3 x 104 eV Where, e = Charge on an electron = 1.6 x 10⁻19 C Ans (a). Maximum frequency produced by the X-rays = ν The energy of the electrons is given by the relation: E = hν Where, h = Planck’s constant = 6.626 xRead more

Potential of the electrons, V = 30 kV = 3 x 10^{4} V

Hence, energy of the electrons, E = 3 x 10^{4} eV Where,

e = Charge on an electron = 1.6 x 10⁻^{19} C

Ans (a).
Maximum frequency produced by the X-rays = ν

The energy of the electrons is given by the relation: E = hν

Where, h = Planck’s constant = 6.626 x 10⁻^{34} Js

Therefore , ν= E/h = (1.6 x 10⁻^{19}) x (3 x 10^{4}) /(6.626 x 10⁻^{34})

= 7.24 x 10¹⁸ Hz

Hence, the maximum frequency of X-rays produced is 7.24 x 10^{18} Hz.

Ans (b).

The minimum wavelength produced by the X-rays is given as:

λ =c/ν = (3 x 10^{8}) /(7.24 x 10^{18})

=4.14 x 10⁻⁻¹¹ m =0.0414 nm

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

Room temperature, T = 27°C Resistance of the heating element at T, R = 100Ω Let T₁ is the increased temperature of the filament. Resistance of the heating element at T₁, R₁ = 117 Ω Temperature co-efficient of the material of the filament, α= 1.70 x 10⁻⁴ °C-1 α is given by the relation ,α =(R₁-R)/R (Read more

Room temperature, T = 27°C

Resistance of the heating element at T, R = 100Ω

Let T₁ is the increased temperature of the filament.

Resistance of the heating element at T₁, R₁ = 117 Ω

Temperature co-efficient of the material of the filament,

α= 1.70 x 10⁻⁴ °C^{-1 }

α is given by the relation ,α =(R₁-R)/R (T₁-T)

Therefore , T₁-T = (R₁-R )/R α

T₁-27= (117-100 )/100(α 1.70 x 10⁻⁴)

T₁- 27 = 1000

T₁ = 1000+27 = 1027 °C

Therefore, at 1027°C, the resistance of the element is 117 Ω.

Ans (a). There are three resistors of resistances, R₁ = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω They are connected in parallel. Hence, total resistance (R) of the combination is given by, 1/R =1/R₁ + 1/R2 +1/R3 = 1/2 + 1/4 + 1/5 = (10+5+4 )/20 = 19/20 Therefore R = 20/19 Ω Therefore, total resistance of the comRead more

Ans (a).
There are three resistors of resistances,

R₁ = 2 Ω, R_{2} = 4 Ω, and R_{3} = 5 Ω

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

1/R =1/R₁ + 1/R_{2 }+1/R_{3}

= 1/2 + 1/4 + 1/5 = (10+5+4 )/20 = 19/20

Therefore R = 20/19 Ω

Therefore, total resistance of the combination is 20/19 Ω.

Ans (b).

Emf of the battery, V = 20 V

Current (I₁) flowing through resistor R₁ is given by,

I₁ =V/R₁ = 20/2 =10 A

Current (I_{2}) flowing through resistor R_{2} is given by,

I_{2} =V/R_{2}= 20/4 = 5A

Current (I_{3}) flowing through resistor R_{3} is given by,

I_{3}=V/ R_{3 }=20/5 =4 A

Total current, I = I₁ + I2 + I3 = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

Ans (a). Three resistors of resistances 1 Ω, 2 Ω, and 3Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = l + 2 + 3 = 6Ω Ans (b). Current flowing through the circuit = I Emf of the battery, E = 12 V Total resistanRead more

Ans (a).

Three resistors of resistances 1 Ω, 2 Ω, and 3Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = l + 2 + 3 = 6Ω

Ans (b).

Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, R = 6 Ω

The relation for current using Ohm’s law is,

I =E/R

=12/6 =2 A

Potential drop across 1 Ω resistor = V₁

From Ohm’s law, the value of V₁ can be obtained as

V₁ = 2 x 1= 2 V … (i)

Potential drop across 2 Ω resistor = V_{2}

Again, from Ohm’s law, the value of V_{2 }can be obtained as

V_{2} = 2 x 2= 4 V … (ii)

Potential drop across 3 Ω resistor = V_{3}

Again, from Ohm’s law, the value of V_{3 }can be obtained as

V_{3} = 2 x 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

Emf of the battery, E = 10 V Internal resistance of the battery, r = 3fi Current in the circuit, I = 0.5 A Resistance of the resistor = R The relation for current using Ohm's law is, I = E/(R+r) R+r =E/I= 10/0.5 =20 Ω Therefore ,R =20-3 = 17 Ω Terminal voltage of the resistor = V According to Ohm'sRead more

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3fi

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

I = E/(R+r)

R+r =E/I= 10/0.5 =20 Ω

Therefore ,R =20-3 = 17 Ω

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR = 0.5 x 17 = 8.5 V

Therefore, the resistance of the resistor is 17 Q and the terminal voltage is 8.5 V.

Emf of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Maximum current drawn from the battery = I According to Ohm's law, E = Ir Therefore I = E/r =12/0.4 = 30 A The maximum current drawn from the given battery is 30 A.

Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law, E = Ir

Therefore I = E/r =12/0.4

= 30 A

The maximum current drawn from the given battery is 30 A.

## Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Potential of the electrons, V = 30 kV = 3 x 104 V Hence, energy of the electrons, E = 3 x 104 eV Where, e = Charge on an electron = 1.6 x 10⁻19 C Ans (a). Maximum frequency produced by the X-rays = ν The energy of the electrons is given by the relation: E = hν Where, h = Planck’s constant = 6.626 xRead more

Potential of the electrons, V = 30 kV = 3 x 10

^{4}VHence, energy of the electrons, E = 3 x 10

^{4}eV Where,e = Charge on an electron = 1.6 x 10⁻

^{19}CAns (a).Maximum frequency produced by the X-rays = ν

The energy of the electrons is given by the relation: E = hν

Where, h = Planck’s constant = 6.626 x 10⁻

^{34}JsTherefore , ν= E/h = (1.6 x 10⁻

^{19}) x (3 x 10^{4}) /(6.626 x 10⁻^{34})= 7.24 x 10¹⁸ Hz

Hence, the maximum frequency of X-rays produced is 7.24 x 10

^{18}Hz.Ans (b).The minimum wavelength produced by the X-rays is given as:

λ =c/ν = (3 x 10

^{8}) /(7.24 x 10^{18})=4.14 x 10⁻⁻¹¹ m =0.0414 nm

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

## 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.

Room temperature, T = 27°C Resistance of the heating element at T, R = 100Ω Let T₁ is the increased temperature of the filament. Resistance of the heating element at T₁, R₁ = 117 Ω Temperature co-efficient of the material of the filament, α= 1.70 x 10⁻⁴ °C-1 α is given by the relation ,α =(R₁-R)/R (Read more

Room temperature, T = 27°C

Resistance of the heating element at T, R = 100Ω

Let T₁ is the increased temperature of the filament.

Resistance of the heating element at T₁, R₁ = 117 Ω

Temperature co-efficient of the material of the filament,

α= 1.70 x 10⁻⁴ °C^{-1 }α is given by the relation ,

α =(R₁-R)/R (T₁-T)Therefore ,

T₁-T = (R₁-R )/R αT₁-27= (117-100 )/100(α

1.70 x 10⁻⁴)T₁- 27 = 1000

T₁ = 1000+27 = 1027 °C

Therefore, at 1027°C, the resistance of the element is 117 Ω.

See less## (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.

Ans (a). There are three resistors of resistances, R₁ = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω They are connected in parallel. Hence, total resistance (R) of the combination is given by, 1/R =1/R₁ + 1/R2 +1/R3 = 1/2 + 1/4 + 1/5 = (10+5+4 )/20 = 19/20 Therefore R = 20/19 Ω Therefore, total resistance of the comRead more

Ans (a).There are three resistors of resistances,

R₁ = 2 Ω, R

_{2}= 4 Ω, and R_{3}= 5 ΩThey are connected in parallel. Hence, total resistance (R) of the combination is given by,

1/R =1/R₁ + 1/R

_{2 }+1/R_{3}= 1/2 + 1/4 + 1/5 = (10+5+4 )/20 = 19/20

Therefore R = 20/19 Ω

Therefore, total resistance of the combination is 20/19 Ω.

Ans (b).Emf of the battery, V = 20 V

Current (I₁) flowing through resistor R₁ is given by,

I₁ =V/R₁ = 20/2 =10 A

Current (I

_{2}) flowing through resistor R_{2}is given by,I

_{2}=V/R_{2}= 20/4 = 5ACurrent (I

_{3}) flowing through resistor R_{3}is given by,I

_{3}=V/ R_{3 }=20/5 =4 ATotal current, I = I₁ + I2 + I3 = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

See less## (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.

Ans (a). Three resistors of resistances 1 Ω, 2 Ω, and 3Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = l + 2 + 3 = 6Ω Ans (b). Current flowing through the circuit = I Emf of the battery, E = 12 V Total resistanRead more

Ans (a).Three resistors of resistances 1 Ω, 2 Ω, and 3Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = l + 2 + 3 = 6Ω

Ans (b).Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, R = 6 Ω

The relation for current using Ohm’s law is,

I =E/R

=12/6 =2 A

Potential drop across 1 Ω resistor = V₁

From Ohm’s law, the value of V₁ can be obtained as

V₁ = 2 x 1= 2 V … (i)

Potential drop across 2 Ω resistor = V

_{2}Again, from Ohm’s law, the value of V

_{2 }can be obtained asV

_{2}= 2 x 2= 4 V … (ii)Potential drop across 3 Ω resistor = V

_{3}Again, from Ohm’s law, the value of V

_{3 }can be obtained asV

_{3}= 2 x 3= 6 V … (iii)Therefore, the potential drop across 1 Ω, 2 Ω and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

See less## 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?

Emf of the battery, E = 10 V Internal resistance of the battery, r = 3fi Current in the circuit, I = 0.5 A Resistance of the resistor = R The relation for current using Ohm's law is, I = E/(R+r) R+r =E/I= 10/0.5 =20 Ω Therefore ,R =20-3 = 17 Ω Terminal voltage of the resistor = V According to Ohm'sRead more

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3fi

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

I = E/(R+r)

R+r =E/I= 10/0.5 =20 Ω

Therefore ,R =20-3 = 17 Ω

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR = 0.5 x 17 = 8.5 V

Therefore, the resistance of the resistor is 17 Q and the terminal voltage is 8.5 V.

See less## 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?

Emf of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Maximum current drawn from the battery = I According to Ohm's law, E = Ir Therefore I = E/r =12/0.4 = 30 A The maximum current drawn from the given battery is 30 A.

Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law, E = IrTherefore I = E/r =12/0.4

= 30 A

The maximum current drawn from the given battery is 30 A.

See less