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 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 x 10⁻34 Js
Therefore , ν= E/h = (1.6 x 10⁻19) x (3 x 104) /(6.626 x 10⁻34)
= 7.24 x 10¹⁸ Hz
Hence, the maximum frequency of X-rays produced is 7.24 x 1018 Hz.
Ans (b).
The minimum wavelength produced by the X-rays is given as:
λ =c/ν = (3 x 108) /(7.24 x 1018)
=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, 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 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 (I2) flowing through resistor R2 is given by,
I2 =V/R2= 20/4 = 5A
Current (I3) flowing through resistor R3 is given by,
I3=V/ R3 =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 = V2
Again, from Ohm’s law, the value of V2 can be obtained as
V2 = 2 x 2= 4 V … (ii)
Potential drop across 3 Ω resistor = V3
Again, from Ohm’s law, the value of V3 can be obtained as
V3 = 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 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 x 10⁻34 Js
Therefore , ν= E/h = (1.6 x 10⁻19) x (3 x 104) /(6.626 x 10⁻34)
= 7.24 x 10¹⁸ Hz
Hence, the maximum frequency of X-rays produced is 7.24 x 1018 Hz.
Ans (b).
The minimum wavelength produced by the X-rays is given as:
λ =c/ν = (3 x 108) /(7.24 x 1018)
=4.14 x 10⁻⁻¹¹ m =0.0414 nm
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
See lessAt 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 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.
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, 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 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 (I2) flowing through resistor R2 is given by,
I2 =V/R2= 20/4 = 5A
Current (I3) flowing through resistor R3 is given by,
I3=V/ R3 =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.
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 = V2
Again, from Ohm’s law, the value of V2 can be obtained as
V2 = 2 x 2= 4 V … (ii)
Potential drop across 3 Ω resistor = V3
Again, from Ohm’s law, the value of V3 can be obtained as
V3 = 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 lessA 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 lessThe 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 = Ir
Therefore I = E/r =12/0.4
= 30 A
The maximum current drawn from the given battery is 30 A.
See less