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The photoelectric threshold of a certain metal is 30 Å. If the radiation of 2000 Å is incident on the metal- a)Positrons will be emitted
- b)Protons will be emitted
- c)Electrons will be emitted
- d)Electrons will not be emitted
Correct answer is option 'D'. Can you explain this answer?
The photoelectric threshold of a certain metal is 30 Å. If the radiation of 2000 Å is incident on the metal
a)
Positrons will be emitted
b)
Protons will be emitted
c)
Electrons will be emitted
d)
Electrons will not be emitted
|
Puja Patel answered |
Photoelectric Effect and Threshold Frequency
The photoelectric effect refers to the phenomenon whereby electrons are emitted from a metal surface when it is exposed to electromagnetic radiation, such as light. The threshold frequency refers to the minimum frequency of the radiation required to cause this effect.
Photoelectric Threshold of a Metal
The photoelectric threshold of a metal refers to the minimum frequency of the radiation required to cause electrons to be emitted from its surface. This threshold is specific to each metal, and it depends on factors such as the work function of the metal, which is a measure of the energy required to remove an electron from its surface.
In this case, the photoelectric threshold of the metal is given as 30 . This means that any radiation with a frequency below this threshold will not cause electrons to be emitted from the metal surface.
Effect of Radiation of 2000
The radiation of 2000 is not specified in terms of frequency, so we cannot determine whether it is above or below the photoelectric threshold of the metal. However, we can make some general observations based on the threshold frequency concept.
If the radiation of 2000 has a frequency below the photoelectric threshold of the metal (i.e. less than 30 Hz), then electrons will not be emitted from the metal surface. This would make option D the correct answer.
If the radiation of 2000 has a frequency above the photoelectric threshold of the metal (i.e. greater than or equal to 30 Hz), then electrons may be emitted from the metal surface. The type of particles emitted will depend on the energy of the electrons and the properties of the metal, but they are likely to be electrons rather than positrons or protons.
Conclusion
Based on the information given, we cannot determine whether electrons will be emitted from the metal surface when it is exposed to radiation of 2000 . However, if the frequency of the radiation is below the photoelectric threshold of the metal, then electrons will not be emitted.
The photoelectric effect refers to the phenomenon whereby electrons are emitted from a metal surface when it is exposed to electromagnetic radiation, such as light. The threshold frequency refers to the minimum frequency of the radiation required to cause this effect.
Photoelectric Threshold of a Metal
The photoelectric threshold of a metal refers to the minimum frequency of the radiation required to cause electrons to be emitted from its surface. This threshold is specific to each metal, and it depends on factors such as the work function of the metal, which is a measure of the energy required to remove an electron from its surface.
In this case, the photoelectric threshold of the metal is given as 30 . This means that any radiation with a frequency below this threshold will not cause electrons to be emitted from the metal surface.
Effect of Radiation of 2000
The radiation of 2000 is not specified in terms of frequency, so we cannot determine whether it is above or below the photoelectric threshold of the metal. However, we can make some general observations based on the threshold frequency concept.
If the radiation of 2000 has a frequency below the photoelectric threshold of the metal (i.e. less than 30 Hz), then electrons will not be emitted from the metal surface. This would make option D the correct answer.
If the radiation of 2000 has a frequency above the photoelectric threshold of the metal (i.e. greater than or equal to 30 Hz), then electrons may be emitted from the metal surface. The type of particles emitted will depend on the energy of the electrons and the properties of the metal, but they are likely to be electrons rather than positrons or protons.
Conclusion
Based on the information given, we cannot determine whether electrons will be emitted from the metal surface when it is exposed to radiation of 2000 . However, if the frequency of the radiation is below the photoelectric threshold of the metal, then electrons will not be emitted.
A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current. The magnetic field far away will be- a)twice the earlier value
- b)same as the earlier value
- c)one-half of the earlier value
- d)one-quarter of the earlier value
Correct answer is option 'B'. Can you explain this answer?
a)
twice the earlier value
b)
same as the earlier value
c)
one-half of the earlier value
d)
one-quarter of the earlier value
|
Mahi Choudhary answered |
Explanation:
The magnetic field far away from a current-carrying wire depends on the current flowing through the wire and the distance from the wire. It does not depend on the diameter of the wire.
When a wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current, the current through the wire remains the same. Therefore, the magnetic field far away from the wire will also remain the same.
Hence, the correct answer is option 'B' - same as the earlier value.
The magnetic field far away from a current-carrying wire depends on the current flowing through the wire and the distance from the wire. It does not depend on the diameter of the wire.
When a wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current, the current through the wire remains the same. Therefore, the magnetic field far away from the wire will also remain the same.
Hence, the correct answer is option 'B' - same as the earlier value.
A double convex lens made of air with both radii of curvature being 10 cm is placed in a medium of refractive index 1.5. The focal length and nature of lens is
- a)15 cm, concave
- b)15 cm convex
- c)Infinity, convex
- d)Zero, concave
Correct answer is option 'A'. Can you explain this answer?
A double convex lens made of air with both radii of curvature being 10 cm is placed in a medium of refractive index 1.5. The focal length and nature of lens is
a)
15 cm, concave
b)
15 cm convex
c)
Infinity, convex
d)
Zero, concave
|
EduRev JEE answered |
f = - 15 cm concave
The electric potential at a point (x, y) in the xy-plane is given by: V=-kxy. The electric field intensity at a distance r from the origin varies as- a)r
- b)2r
- c)r2
- d)2r2
Correct answer is option 'A'. Can you explain this answer?
The electric potential at a point (x, y) in the xy-plane is given by: V=-kxy. The electric field intensity at a distance r from the origin varies as
a)
r
b)
2r
c)
r2
d)
2r2
|
Juhi Nambiar answered |
Given Information:
The electric potential at a point (x, y) in the xy-plane is given by: V=-kxy.
To find:
The electric field intensity at a distance r from the origin varies as
Explanation:
We know that the electric field intensity (E) is given by the negative gradient of the electric potential (V).
Therefore, E=-∇V
where ∇ is the gradient operator.
In two dimensions, the gradient operator can be expressed as:
∇=(∂/∂x)i+(∂/∂y)j
where i and j are the unit vectors in the x and y directions, respectively.
Differentiating the given potential function V=-kxy with respect to x and y, we get:
∂V/∂x=-ky
∂V/∂y=-kx
Substituting these values in the expression for the electric field intensity, we get:
E=-∇V=-(-kyi-kxj)=kyi+kxj
The magnitude of the electric field intensity is given by:
|E|=√(Ex²+Ey²)
where Ex and Ey are the x and y components of the electric field intensity, respectively.
Substituting the values of Ex and Ey, we get:
|E|=√(k²x²+y²)
From the given information, we know that the distance r from the origin is given by:
r=√(x²+y²)
Squaring both sides, we get:
r²=x²+y²
Substituting this value in the expression for |E|, we get:
|E|=√(k²x²+r²-r²)
|E|=√(k²r²)
|E|=kr
Therefore, the electric field intensity at a distance r from the origin varies as r, which is option (a).
The electric potential at a point (x, y) in the xy-plane is given by: V=-kxy.
To find:
The electric field intensity at a distance r from the origin varies as
Explanation:
We know that the electric field intensity (E) is given by the negative gradient of the electric potential (V).
Therefore, E=-∇V
where ∇ is the gradient operator.
In two dimensions, the gradient operator can be expressed as:
∇=(∂/∂x)i+(∂/∂y)j
where i and j are the unit vectors in the x and y directions, respectively.
Differentiating the given potential function V=-kxy with respect to x and y, we get:
∂V/∂x=-ky
∂V/∂y=-kx
Substituting these values in the expression for the electric field intensity, we get:
E=-∇V=-(-kyi-kxj)=kyi+kxj
The magnitude of the electric field intensity is given by:
|E|=√(Ex²+Ey²)
where Ex and Ey are the x and y components of the electric field intensity, respectively.
Substituting the values of Ex and Ey, we get:
|E|=√(k²x²+y²)
From the given information, we know that the distance r from the origin is given by:
r=√(x²+y²)
Squaring both sides, we get:
r²=x²+y²
Substituting this value in the expression for |E|, we get:
|E|=√(k²x²+r²-r²)
|E|=√(k²r²)
|E|=kr
Therefore, the electric field intensity at a distance r from the origin varies as r, which is option (a).
In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 Å coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1 p − S2p in microns is
- a)0.75
- b)1.5
- c)3.0
- d)4.5
Correct answer is option 'B'. Can you explain this answer?
In Young's double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 Å coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1 p − S2p in microns is
a)
0.75
b)
1.5
c)
3.0
d)
4.5
|
Rajdeep Dasgupta answered |
And S2 p is equal to 3 times the wavelength of the light.
We know that the distance between two consecutive bright fringes is given by:
d = λD/d
Where λ is the wavelength of light, D is the distance between the double-slit and the screen, and d is the distance between the two slits.
Similarly, the distance between two consecutive dark fringes is also given by:
d = λD/2d
Here, we are interested in the third dark fringe, which means that there are 3 bright fringes between the two slits and the point P. Therefore, the path difference between the two slits and the point P must be equal to 3λ.
Let's assume that the distance between the two slits is d. Then, the distance between the first bright fringe and the central fringe is given by:
y = λD/d
The distance between the third bright fringe and the central fringe is:
y' = 3λD/d
The path difference between the two slits and the point P is:
S1P - S2P = y' - y = 2λD/d
But we know that S1P - S2P = 3λ
Therefore, we can write:
2λD/d = 3λ
d = 2D/3
So the path difference between S1 and P or S2 and P is:
S1P = S2P = d/2 = D/3
Therefore, the path difference S1P and S2P is equal to D/3.
We know that the distance between two consecutive bright fringes is given by:
d = λD/d
Where λ is the wavelength of light, D is the distance between the double-slit and the screen, and d is the distance between the two slits.
Similarly, the distance between two consecutive dark fringes is also given by:
d = λD/2d
Here, we are interested in the third dark fringe, which means that there are 3 bright fringes between the two slits and the point P. Therefore, the path difference between the two slits and the point P must be equal to 3λ.
Let's assume that the distance between the two slits is d. Then, the distance between the first bright fringe and the central fringe is given by:
y = λD/d
The distance between the third bright fringe and the central fringe is:
y' = 3λD/d
The path difference between the two slits and the point P is:
S1P - S2P = y' - y = 2λD/d
But we know that S1P - S2P = 3λ
Therefore, we can write:
2λD/d = 3λ
d = 2D/3
So the path difference between S1 and P or S2 and P is:
S1P = S2P = d/2 = D/3
Therefore, the path difference S1P and S2P is equal to D/3.
The internal resistance of a cell of e.m.f. 3 V is 0.1 Ω, If it is connected to a resistance of 2.9Ω then voltage across the cell will be- a)0.9 V
- b)1.9 V
- c)2.9 V
- d)4.9 V
Correct answer is option 'C'. Can you explain this answer?
The internal resistance of a cell of e.m.f. 3 V is 0.1 Ω, If it is connected to a resistance of 2.9Ω then voltage across the cell will be
a)
0.9 V
b)
1.9 V
c)
2.9 V
d)
4.9 V
|
Hiral Mehra answered |
Given: e.m.f. (E) = 3V
Internal Resistance (r) = 0.1Ω
External Resistance (R) = 2.9Ω
We know that the voltage across a resistor is given by Ohm's law:
V = IR
where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor.
To find the voltage across the cell, we can use Kirchhoff's voltage law (KVL), which states that the sum of the voltages in a closed loop must be zero. In this case, the closed loop consists of the cell, the internal resistance, and the external resistance.
Applying KVL to the circuit, we get:
E - Ir - VR = 0
where E is the e.m.f. of the cell, r is the internal resistance, and V is the voltage across the external resistance.
Solving for V, we get:
V = E - Ir
Substituting the given values, we get:
V = 3 - 0.1 x 2.9
V = 2.71V
Therefore, the voltage across the cell is 2.71V, which is closest to option C (2.9V).
Internal Resistance (r) = 0.1Ω
External Resistance (R) = 2.9Ω
We know that the voltage across a resistor is given by Ohm's law:
V = IR
where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor.
To find the voltage across the cell, we can use Kirchhoff's voltage law (KVL), which states that the sum of the voltages in a closed loop must be zero. In this case, the closed loop consists of the cell, the internal resistance, and the external resistance.
Applying KVL to the circuit, we get:
E - Ir - VR = 0
where E is the e.m.f. of the cell, r is the internal resistance, and V is the voltage across the external resistance.
Solving for V, we get:
V = E - Ir
Substituting the given values, we get:
V = 3 - 0.1 x 2.9
V = 2.71V
Therefore, the voltage across the cell is 2.71V, which is closest to option C (2.9V).
A 2 μ F condenser is charged to 500V and then its plates are joined through of consistance. The heat produced in the resistance in joules is
- a)0.25
- b)0. 5
- c)0.75
- d)0.1
Correct answer is option 'A'. Can you explain this answer?
A 2 μ F condenser is charged to 500V and then its plates are joined through of consistance. The heat produced in the resistance in joules is
a)
0.25
b)
0. 5
c)
0.75
d)
0.1
|
Anjali Sharma answered |
First, let's find the initial energy stored in the capacitor, which is given by the formula:
E_initial = (1/2) * C * V^2
where E_initial is the initial energy, C is the capacitance, and V is the voltage.
E_initial = (1/2) * 2 * 10^-6 F * (500 V)^2
E_initial = 1 * 10^-6 F * (250000 V^2)
E_initial = 0.25 J
Now, when the plates are joined through a resistance, the capacitor will discharge and its energy will be transferred into heat in the resistor. The final energy stored in the capacitor will be zero as it is completely discharged. So, the heat produced in the resistor will be equal to the initial energy stored in the capacitor.
Heat produced (in Joules) = Initial energy stored in the capacitor (in Joules)
Heat produced = 0.25 J
E_initial = (1/2) * C * V^2
where E_initial is the initial energy, C is the capacitance, and V is the voltage.
E_initial = (1/2) * 2 * 10^-6 F * (500 V)^2
E_initial = 1 * 10^-6 F * (250000 V^2)
E_initial = 0.25 J
Now, when the plates are joined through a resistance, the capacitor will discharge and its energy will be transferred into heat in the resistor. The final energy stored in the capacitor will be zero as it is completely discharged. So, the heat produced in the resistor will be equal to the initial energy stored in the capacitor.
Heat produced (in Joules) = Initial energy stored in the capacitor (in Joules)
Heat produced = 0.25 J
The coefficient of mutual inductance, when magnetic flux changes by 2 X 10-2 Wb and current changes by 0.01 A is
- a)2 H
- b)4 H
- c)3 H
- d)8 H
Correct answer is option 'A'. Can you explain this answer?
The coefficient of mutual inductance, when magnetic flux changes by 2 X 10-2 Wb and current changes by 0.01 A is
a)
2 H
b)
4 H
c)
3 H
d)
8 H
|
Nitin Verma answered |
Given:
Change in magnetic flux = 2 x 10^-2 Wb
Change in current = 0.01 A
To find: Coefficient of mutual inductance (M)
Formula:
M = ΔΦ/ΔI
Where,
ΔΦ = Change in magnetic flux
ΔI = Change in current
Calculation:
M = ΔΦ/ΔI
M = (2 x 10^-2 Wb)/(0.01 A)
M = 2 H
Therefore, the coefficient of mutual inductance is 2 H.
Change in magnetic flux = 2 x 10^-2 Wb
Change in current = 0.01 A
To find: Coefficient of mutual inductance (M)
Formula:
M = ΔΦ/ΔI
Where,
ΔΦ = Change in magnetic flux
ΔI = Change in current
Calculation:
M = ΔΦ/ΔI
M = (2 x 10^-2 Wb)/(0.01 A)
M = 2 H
Therefore, the coefficient of mutual inductance is 2 H.
If a full wave rectifier circuit is operating from 50 Hz mains, then fundamental frequency in the ripple will be
- a)50 Hz
- b)70.7 Hz
- c)100 Hz
- d)25 Hz
Correct answer is option 'C'. Can you explain this answer?
If a full wave rectifier circuit is operating from 50 Hz mains, then fundamental frequency in the ripple will be
a)
50 Hz
b)
70.7 Hz
c)
100 Hz
d)
25 Hz
|
Learners Habitat answered |
In full wave rectifier, two diodes are used to rectify the input A.C. voltage into D.C. voltage. Hence, for each cycle of input signal pulsating, unidirectional output will obtain. Thus, the fundamental frequency in the ripple will be doubled, i.e., 100 Hz.
An electric dipole of dipole moment p is placed in uniform electric field of strength E. If θ is the angle between p and E, then potential energy of the dipole becomes maximum, when θ is- a)0o
- b)45o
- c)90o
- d)180o
Correct answer is option 'D'. Can you explain this answer?
An electric dipole of dipole moment p is placed in uniform electric field of strength E. If θ is the angle between p and E, then potential energy of the dipole becomes maximum, when θ is
a)
0o
b)
45o
c)
90o
d)
180o
|
Pranavi Singh answered |
The dipole is placed at an angle θ with respect to the direction of the electric field, then the torque experienced by the dipole is given by:
τ = pEsinθ
This torque tends to align the dipole with the direction of the electric field. If the dipole is free to rotate, it will rotate until it aligns with the electric field. The potential energy of the dipole in this position is given by:
U = -pEcosθ
where θ is the angle between the dipole moment and the electric field direction.
If the dipole is not free to rotate, for example, if it is held fixed, then the torque will still be present but there will be no change in the dipole's orientation. In this case, the potential energy of the dipole will not change and will remain constant at -pEcosθ.
τ = pEsinθ
This torque tends to align the dipole with the direction of the electric field. If the dipole is free to rotate, it will rotate until it aligns with the electric field. The potential energy of the dipole in this position is given by:
U = -pEcosθ
where θ is the angle between the dipole moment and the electric field direction.
If the dipole is not free to rotate, for example, if it is held fixed, then the torque will still be present but there will be no change in the dipole's orientation. In this case, the potential energy of the dipole will not change and will remain constant at -pEcosθ.
A sheet of aluminium is inserted in the air gap between the parallel plate capacitor, without touching any the two plates of the capacitor. The capacitance the capacitor is:- a)invariant for all positions of the sheet
- b)maximum when the sheet is midway between the 2 plates
- c)maximum when the sheet is just near the + plate
- d)maximum when the sheet is just near the - plate
Correct answer is option 'A'. Can you explain this answer?
A sheet of aluminium is inserted in the air gap between the parallel plate capacitor, without touching any the two plates of the capacitor. The capacitance the capacitor is:
a)
invariant for all positions of the sheet
b)
maximum when the sheet is midway between the 2 plates
c)
maximum when the sheet is just near the + plate
d)
maximum when the sheet is just near the - plate
|
|
Aarav Pillai answered |
Explanation:
Effect of inserting a sheet of aluminum in an air gap between the parallel plate capacitor on capacitance of capacitor is explained below:
1. Capacitance of parallel plate capacitor:
Capacitance of parallel plate capacitor is given by the formula:
C = εA/d
Where, C is capacitance, ε is permittivity of the medium between the plates, A is area of the plates and d is distance between the plates.
2. Effect of inserting a sheet of aluminum:
When a sheet of aluminum is inserted in the air gap between the parallel plate capacitor, the permittivity of the medium between the plates changes due to the presence of aluminum sheet.
But, as the aluminum sheet is not touching any of the two plates of the capacitor, it does not alter the distance between the plates or the area of the plates.
As a result, the capacitance of the capacitor remains unchanged.
3. Conclusion:
Hence, the correct option is (a) that the capacitance of the capacitor is invariant for all positions of the sheet.
Effect of inserting a sheet of aluminum in an air gap between the parallel plate capacitor on capacitance of capacitor is explained below:
1. Capacitance of parallel plate capacitor:
Capacitance of parallel plate capacitor is given by the formula:
C = εA/d
Where, C is capacitance, ε is permittivity of the medium between the plates, A is area of the plates and d is distance between the plates.
2. Effect of inserting a sheet of aluminum:
When a sheet of aluminum is inserted in the air gap between the parallel plate capacitor, the permittivity of the medium between the plates changes due to the presence of aluminum sheet.
But, as the aluminum sheet is not touching any of the two plates of the capacitor, it does not alter the distance between the plates or the area of the plates.
As a result, the capacitance of the capacitor remains unchanged.
3. Conclusion:
Hence, the correct option is (a) that the capacitance of the capacitor is invariant for all positions of the sheet.
A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is- a)3.67 A
- b)2.67 A
- c)1.67 A
- d)2.40 A
Correct answer is option 'C'. Can you explain this answer?
A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
a)
3.67 A
b)
2.67 A
c)
1.67 A
d)
2.40 A
|
|
Rajdeep Iyer answered |
Given Data:
Current produced by generator = 100 A
Voltage produced by generator = 4000 V
Voltage after transforming = 240000 V
To find: Current in the transmission line
Explanation:
The power output by the generator is given as P = VI where V is the voltage and I is the current. As the power output is constant, we can equate the power input and output to the transformer.
Power input to the transformer = Power output from the generator
VI = V' I'
where V' and I' are the voltage and current after the transformer.
We know V = 4000 V and V' = 240000 V
Substituting these values in the above equation, we get:
(4000 V) (100 A) = (240000 V) I'
I' = (4000/240000) (100 A) = 1.67 A
Therefore, the current in the transmission line is 1.67 A.
Hence, option (c) is the correct answer.
Current produced by generator = 100 A
Voltage produced by generator = 4000 V
Voltage after transforming = 240000 V
To find: Current in the transmission line
Explanation:
The power output by the generator is given as P = VI where V is the voltage and I is the current. As the power output is constant, we can equate the power input and output to the transformer.
Power input to the transformer = Power output from the generator
VI = V' I'
where V' and I' are the voltage and current after the transformer.
We know V = 4000 V and V' = 240000 V
Substituting these values in the above equation, we get:
(4000 V) (100 A) = (240000 V) I'
I' = (4000/240000) (100 A) = 1.67 A
Therefore, the current in the transmission line is 1.67 A.
Hence, option (c) is the correct answer.
A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will be- a) 0.50 V
- b) 1.90 V
- c) 1.95 V
- d) 2.00 V
Correct answer is option 'C'. Can you explain this answer?
a)
0.50 V
b)
1.90 V
c)
1.95 V
d)
2.00 V
|
Varun Kapoor answered |
Answer :- c
Solution :- EMF of the cell(E) =2 V
internal resistance(r)= 0.1 V
External resistance(R)=3.9 V
V=IR
=(E÷(R+r))×R
=(2÷(3.9+0.1))×3.9
=1.95 V
Therefore potential difference across the terminals of the cell = 1.95V
In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be
- a)2eV/m
- b)√2eV/m
- c)√2m/eV
- d)V2/2em
Correct answer is option 'B'. Can you explain this answer?
In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be
a)
2eV/m
b)
√2eV/m
c)
√2m/eV
d)
V2/2em
|
Simran Bajaj answered |
√(2eV/m)
c)eV/m
d)√(eV/m)
c)eV/m
d)√(eV/m)
Equal charges are given to two conducting spheres of different radii. The potential will- a) be more on the smaller sphere
- b) be more on the bigger sphere
- c) be equal on both the sphere
- d) depends on the nature of the materials of the sphere
Correct answer is option 'A'. Can you explain this answer?
a)
be more on the smaller sphere
b)
be more on the bigger sphere
c)
be equal on both the sphere
d)
depends on the nature of the materials of the sphere
|
|
Lekshmi Sarkar answered |
Explanation:
When equal charges are given to two conducting spheres of different radii, the potential will be more on the smaller sphere. This can be explained using the following points:
- The potential of a conductor depends on its radius and the amount of charge it carries. The potential is directly proportional to the charge and inversely proportional to the radius.
- The smaller sphere has a smaller radius than the bigger sphere, so it will have a higher potential for the same amount of charge.
- This can be mathematically expressed using the formula for potential, which is V = kQ/r, where V is the potential, Q is the charge, r is the radius, and k is a constant.
- Since the charge is the same for both spheres, the potential will depend only on the radius. The smaller sphere will have a higher potential because its radius is smaller.
- Therefore, option A is the correct answer.
In summary, the potential of a conducting sphere depends on its radius and the amount of charge it carries. When equal charges are given to two conducting spheres of different radii, the potential will be more on the smaller sphere due to its smaller radius.
When equal charges are given to two conducting spheres of different radii, the potential will be more on the smaller sphere. This can be explained using the following points:
- The potential of a conductor depends on its radius and the amount of charge it carries. The potential is directly proportional to the charge and inversely proportional to the radius.
- The smaller sphere has a smaller radius than the bigger sphere, so it will have a higher potential for the same amount of charge.
- This can be mathematically expressed using the formula for potential, which is V = kQ/r, where V is the potential, Q is the charge, r is the radius, and k is a constant.
- Since the charge is the same for both spheres, the potential will depend only on the radius. The smaller sphere will have a higher potential because its radius is smaller.
- Therefore, option A is the correct answer.
In summary, the potential of a conducting sphere depends on its radius and the amount of charge it carries. When equal charges are given to two conducting spheres of different radii, the potential will be more on the smaller sphere due to its smaller radius.
In a transformer 220 A.C. voltage is increased to 2200 volts. If the number of turns in the secondary are 2000, then the number of turns in the primary will be- a)200
- b)100
- c)50
- d)20
Correct answer is option 'A'. Can you explain this answer?
a)
200
b)
100
c)
50
d)
20
|
|
Shreya Gupta answered |
Number of turns in primary coil ( Np ) = ?
Number of turns in Secondary coil ( Ns ) = 2000
Voltage in primary coil ( Ep ) = 220 V
Voltage in secondary coil ( Es ) = 2200 V .
Now, using formula: Ep/Es = Np/Ns
Substituting values, we get,
220/2200 = Np/2000
=> Np = 200 turns .
In the propagation of light waves, the angle between the direction of vibration and plane of polarisation is
- a)0o
- b)90o
- c)45o
- d)80o
Correct answer is option 'A'. Can you explain this answer?
In the propagation of light waves, the angle between the direction of vibration and plane of polarisation is
a)
0o
b)
90o
c)
45o
d)
80o
|
Top Rankers answered |
Plane containing the direction of vibration and wave motion is called plane of polarisation. Plane of vibration is perpendicular to the direction of propagation and also perpendicular to the plane of polarisation. Therefore angle between the plane of polarisation and directon of propagation is 0∘.
If sky wave with a frequency of 50 MHz is incident on D-region at an angle of 30o, then angle of refraction is- a)15o
- b)60o
- c)30o
- d)5.5o
Correct answer is option 'D'. Can you explain this answer?
If sky wave with a frequency of 50 MHz is incident on D-region at an angle of 30o, then angle of refraction is
a)
15o
b)
60o
c)
30o
d)
5.5o
|
|
Om Nair answered |
Explanation:
In this problem, we need to find the angle of refraction when a sky wave with a frequency of 50 MHz is incident on the D-region at an angle of 30o.
- Sky wave propagation: Sky wave propagation is a method of radio communication where the radio waves are reflected by the ionosphere back to the earth's surface. The ionosphere is a layer of the earth's atmosphere that contains ionized gases, which can reflect electromagnetic waves of certain frequencies. Sky wave propagation is used for long-distance communication, especially over the horizon.
- Refraction: Refraction is the bending of light or other waves as they pass through a medium with varying refractive index. The refractive index of a medium is a measure of how much the speed of light or other waves is reduced when passing through that medium. When a wave passes from one medium to another with a different refractive index, its direction changes.
- Angle of incidence and angle of refraction: When a wave passes from one medium to another, the angle of incidence (i) and the angle of refraction (r) are related by Snell's law:
n1sin(i) = n2sin(r),
where n1 and n2 are the refractive indices of the two media.
In this problem, we are given:
- Frequency of the sky wave: f = 50 MHz
- Angle of incidence: i = 30o
- Medium: D-region of the ionosphere
We need to find the angle of refraction, r.
Solution:
- Refractive index of the D-region: The refractive index of the D-region depends on the frequency of the radio wave. For a frequency of 50 MHz, the refractive index of the D-region is approximately 1.0004.
- Applying Snell's law: Using Snell's law, we can find the angle of refraction:
n1sin(i) = n2sin(r)
1.0004sin(30o) = sin(r)
sin(r) = 0.5002
r = sin-1(0.5002)
r = 30o
Therefore, the angle of refraction is 30o.
Answer: d) 30o
In this problem, we need to find the angle of refraction when a sky wave with a frequency of 50 MHz is incident on the D-region at an angle of 30o.
- Sky wave propagation: Sky wave propagation is a method of radio communication where the radio waves are reflected by the ionosphere back to the earth's surface. The ionosphere is a layer of the earth's atmosphere that contains ionized gases, which can reflect electromagnetic waves of certain frequencies. Sky wave propagation is used for long-distance communication, especially over the horizon.
- Refraction: Refraction is the bending of light or other waves as they pass through a medium with varying refractive index. The refractive index of a medium is a measure of how much the speed of light or other waves is reduced when passing through that medium. When a wave passes from one medium to another with a different refractive index, its direction changes.
- Angle of incidence and angle of refraction: When a wave passes from one medium to another, the angle of incidence (i) and the angle of refraction (r) are related by Snell's law:
n1sin(i) = n2sin(r),
where n1 and n2 are the refractive indices of the two media.
In this problem, we are given:
- Frequency of the sky wave: f = 50 MHz
- Angle of incidence: i = 30o
- Medium: D-region of the ionosphere
We need to find the angle of refraction, r.
Solution:
- Refractive index of the D-region: The refractive index of the D-region depends on the frequency of the radio wave. For a frequency of 50 MHz, the refractive index of the D-region is approximately 1.0004.
- Applying Snell's law: Using Snell's law, we can find the angle of refraction:
n1sin(i) = n2sin(r)
1.0004sin(30o) = sin(r)
sin(r) = 0.5002
r = sin-1(0.5002)
r = 30o
Therefore, the angle of refraction is 30o.
Answer: d) 30o
When a centimeter thick surface is illuminated with light of wavelength λ, stopping potential is V. When the same surface is illuminated by light of wavelength 2 λ, stopping potential is V/3. Threshold wavelength for metallic surface is- a) 4 λ/3
- b) 4 λ
- c) 6 λ
- d) 8 λ/3
Correct answer is option 'B'. Can you explain this answer?
a)
4 λ/3
b)
4 λ
c)
6 λ
d)
8 λ/3
|
|
Mira Sharma answered |
Let the threshold wavelength = λ
the energy required by photo electrons to escape = h c/λ
e V + hc/λ = hc/l
e V/3 + hc/λ = hc/2l
=> h c/l = 4/3 * eV
=> hc/λ = 1/3 * eV
=> λ = 4 l
The average binding energy of the nucleon is- a)931 MeV
- b)8.5 MeV
- c)1.6
- d)3
Correct answer is option 'B'. Can you explain this answer?
The average binding energy of the nucleon is
a)
931 MeV
b)
8.5 MeV
c)
1.6
d)
3
|
|
Avi Chatterjee answered |
Binding Energy of the Nucleon
Binding energy is the energy that is required to break a nucleus into its individual nucleons. The average binding energy of the nucleon is the average amount of energy that is required to remove a nucleon from a nucleus.
Calculation
The average binding energy of the nucleon can be calculated by dividing the total binding energy of the nucleus by the number of nucleons in the nucleus.
The total binding energy of the nucleus can be calculated using the Einstein's famous equation, E=mc². The mass defect of a nucleus, which is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons, can be converted into binding energy using this equation.
The mass defect can be calculated using the following formula:
mass defect = (Zmp + (A-Z)mn - M)c²
where Z is the atomic number, A is the mass number, mp is the mass of a proton, mn is the mass of a neutron, and M is the mass of the nucleus.
The total binding energy can be calculated using the following formula:
total binding energy = (Zmp + (A-Z)mn - M)c²
The average binding energy of the nucleon can be calculated by dividing the total binding energy by the number of nucleons in the nucleus.
Answer
The average binding energy of the nucleon is 8.5 MeV.
Explanation
The correct answer is option 'B' because the average binding energy of the nucleon is about 8.5 MeV. This value is obtained by dividing the total binding energy of the nucleus by the number of nucleons in the nucleus. The total binding energy of the nucleus depends on the mass defect of the nucleus, which is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons.
The mass defect is a measure of the binding energy of the nucleus, and it can be calculated using the formula given above. The total binding energy can be obtained by converting the mass defect into energy using Einstein's equation, E=mc². By dividing the total binding energy by the number of nucleons in the nucleus, we can obtain the average binding energy of the nucleon.
Therefore, option 'B' is the correct answer to the question.
Binding energy is the energy that is required to break a nucleus into its individual nucleons. The average binding energy of the nucleon is the average amount of energy that is required to remove a nucleon from a nucleus.
Calculation
The average binding energy of the nucleon can be calculated by dividing the total binding energy of the nucleus by the number of nucleons in the nucleus.
The total binding energy of the nucleus can be calculated using the Einstein's famous equation, E=mc². The mass defect of a nucleus, which is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons, can be converted into binding energy using this equation.
The mass defect can be calculated using the following formula:
mass defect = (Zmp + (A-Z)mn - M)c²
where Z is the atomic number, A is the mass number, mp is the mass of a proton, mn is the mass of a neutron, and M is the mass of the nucleus.
The total binding energy can be calculated using the following formula:
total binding energy = (Zmp + (A-Z)mn - M)c²
The average binding energy of the nucleon can be calculated by dividing the total binding energy by the number of nucleons in the nucleus.
Answer
The average binding energy of the nucleon is 8.5 MeV.
Explanation
The correct answer is option 'B' because the average binding energy of the nucleon is about 8.5 MeV. This value is obtained by dividing the total binding energy of the nucleus by the number of nucleons in the nucleus. The total binding energy of the nucleus depends on the mass defect of the nucleus, which is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons.
The mass defect is a measure of the binding energy of the nucleus, and it can be calculated using the formula given above. The total binding energy can be obtained by converting the mass defect into energy using Einstein's equation, E=mc². By dividing the total binding energy by the number of nucleons in the nucleus, we can obtain the average binding energy of the nucleon.
Therefore, option 'B' is the correct answer to the question.
A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is- a)3.67 A
- b)2.67 A
- c)1.67 A
- d)2.40 A
Correct answer is option 'C'. Can you explain this answer?
A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
a)
3.67 A
b)
2.67 A
c)
1.67 A
d)
2.40 A
|
|
Kirti Shah answered |
Given:
Current produced by generator = 100 A
Voltage of current produced by generator = 4000 V
Voltage after stepping up by transformer = 240000 V
To find: Current in transmission line
Explanation:
The power output of the generator is given by the formula:
Power = Voltage x Current
Using this formula, we can find the power output of the generator:
Power output of generator = 4000 V x 100 A = 400,000 W
Now, since power is conserved, the power output of the generator must be equal to the power input to the transmission line. The power input to the transmission line is given by the formula:
Power = Voltage x Current
Using this formula, we can find the current in the transmission line:
Current in transmission line = Power / Voltage
Current in transmission line = 400,000 W / 240,000 V
Current in transmission line = 1.67 A
Therefore, the correct answer is option C) 1.67 A.
Current produced by generator = 100 A
Voltage of current produced by generator = 4000 V
Voltage after stepping up by transformer = 240000 V
To find: Current in transmission line
Explanation:
The power output of the generator is given by the formula:
Power = Voltage x Current
Using this formula, we can find the power output of the generator:
Power output of generator = 4000 V x 100 A = 400,000 W
Now, since power is conserved, the power output of the generator must be equal to the power input to the transmission line. The power input to the transmission line is given by the formula:
Power = Voltage x Current
Using this formula, we can find the current in the transmission line:
Current in transmission line = Power / Voltage
Current in transmission line = 400,000 W / 240,000 V
Current in transmission line = 1.67 A
Therefore, the correct answer is option C) 1.67 A.
When a current is passed through water acidified with a dilute sulphuric acid, the gases formed at the platinum electrodes are- a)2 volume hydrogen (cathode) and one volume oxygen (anode)
- b)1 volume hydrogen (cathode) and one volume oxygen (anode)
- c)1 volume oxygen (cathode) and 2 volume hydrogen (anode)
- d)2 volume oxygen (cathode) and 1 volume hydrogen (anode)
Correct answer is option 'A'. Can you explain this answer?
When a current is passed through water acidified with a dilute sulphuric acid, the gases formed at the platinum electrodes are
a)
2 volume hydrogen (cathode) and one volume oxygen (anode)
b)
1 volume hydrogen (cathode) and one volume oxygen (anode)
c)
1 volume oxygen (cathode) and 2 volume hydrogen (anode)
d)
2 volume oxygen (cathode) and 1 volume hydrogen (anode)
|
|
Manasa Mishra answered |
Electrolysis of Water
Electrolysis of water is the decomposition of water into oxygen and hydrogen gas due to the passage of an electric current through the water.
Reaction
The reaction at the anode (positive electrode) during electrolysis is:
2H2O(l) → O2(g) + 4H+(aq) + 4e−
The reaction at the cathode (negative electrode) is:
4H+(aq) + 4e− → 2H2(g)
Explanation
When an electric current is passed through water acidified with dilute sulphuric acid, the water molecule undergoes electrolysis, and the following reaction takes place at the anode and cathode.
At the anode, the water molecule loses electrons and releases oxygen gas.
2H2O → 4H+ + O2 + 4e-
At the cathode, the hydrogen ions gain electrons and form hydrogen gas.
4H+ + 4e- → 2H2
Since the reaction at the cathode forms twice the volume of hydrogen gas than the volume of oxygen gas formed at the anode, the correct answer is option A, which is 2 volumes of hydrogen (cathode) and 1 volume of oxygen (anode).
Electrolysis of water is the decomposition of water into oxygen and hydrogen gas due to the passage of an electric current through the water.
Reaction
The reaction at the anode (positive electrode) during electrolysis is:
2H2O(l) → O2(g) + 4H+(aq) + 4e−
The reaction at the cathode (negative electrode) is:
4H+(aq) + 4e− → 2H2(g)
Explanation
When an electric current is passed through water acidified with dilute sulphuric acid, the water molecule undergoes electrolysis, and the following reaction takes place at the anode and cathode.
At the anode, the water molecule loses electrons and releases oxygen gas.
2H2O → 4H+ + O2 + 4e-
At the cathode, the hydrogen ions gain electrons and form hydrogen gas.
4H+ + 4e- → 2H2
Since the reaction at the cathode forms twice the volume of hydrogen gas than the volume of oxygen gas formed at the anode, the correct answer is option A, which is 2 volumes of hydrogen (cathode) and 1 volume of oxygen (anode).
A laser beam is used for carrying out surgery because it- a) is highly monochromatic
- b) is highly coherent
- c) is highly directional
- d) can be sharply focussed
Correct answer is option 'D'. Can you explain this answer?
a)
is highly monochromatic
b)
is highly coherent
c)
is highly directional
d)
can be sharply focussed
|
Samarth Saha answered |
Importance of Laser in Surgery
Laser technology has revolutionized the field of surgery. The key reason for its preference is its ability to be sharply focused, which is critical in surgical procedures.
Sharp Focusing Capability
- Lasers can concentrate light energy into a very small area.
- This allows surgeons to target specific tissues with minimal damage to surrounding areas.
- The focused beam can cut, cauterize, or vaporize tissue with precision.
Advantages of Laser Surgery
- Minimally Invasive: The precision of lasers often leads to smaller incisions, resulting in less trauma to the body.
- Reduced Blood Loss: The heat generated by lasers helps in coagulating blood vessels, which minimizes bleeding during the procedure.
- Faster Recovery: Patients typically experience quicker healing times due to reduced tissue damage.
Other Laser Properties
While the characteristics of being monochromatic, coherent, and directional are important, they are secondary to the ability to focus sharply:
- Monochromatic: Lasers emit light of a single wavelength, which can be useful in specific medical applications.
- Coherent: The light waves produced are in phase, allowing for precise interactions with tissues.
- Directional: Lasers emit light in a specific direction, minimizing scattering and enhancing targeting.
Conclusion
In conclusion, the ability of lasers to be sharply focused makes them an invaluable tool in surgery. This precision enhances surgical outcomes, reduces recovery times, and minimizes complications, reinforcing their dominance in modern medical practices.
Laser technology has revolutionized the field of surgery. The key reason for its preference is its ability to be sharply focused, which is critical in surgical procedures.
Sharp Focusing Capability
- Lasers can concentrate light energy into a very small area.
- This allows surgeons to target specific tissues with minimal damage to surrounding areas.
- The focused beam can cut, cauterize, or vaporize tissue with precision.
Advantages of Laser Surgery
- Minimally Invasive: The precision of lasers often leads to smaller incisions, resulting in less trauma to the body.
- Reduced Blood Loss: The heat generated by lasers helps in coagulating blood vessels, which minimizes bleeding during the procedure.
- Faster Recovery: Patients typically experience quicker healing times due to reduced tissue damage.
Other Laser Properties
While the characteristics of being monochromatic, coherent, and directional are important, they are secondary to the ability to focus sharply:
- Monochromatic: Lasers emit light of a single wavelength, which can be useful in specific medical applications.
- Coherent: The light waves produced are in phase, allowing for precise interactions with tissues.
- Directional: Lasers emit light in a specific direction, minimizing scattering and enhancing targeting.
Conclusion
In conclusion, the ability of lasers to be sharply focused makes them an invaluable tool in surgery. This precision enhances surgical outcomes, reduces recovery times, and minimizes complications, reinforcing their dominance in modern medical practices.
In comparison to a half wave rectifier, the full wave rectifier gives lower- a)efficiency
- b)average current
- c)average output voltage
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
In comparison to a half wave rectifier, the full wave rectifier gives lower
a)
efficiency
b)
average current
c)
average output voltage
d)
none of these
|
|
Nikhil Banerjee answered |
Full Wave Rectifier vs Half Wave Rectifier
Rectifiers are electronic circuits that convert AC (alternating current) into DC (direct current). There are two types of rectifiers: half wave and full wave. The main difference between the two is that the half wave rectifier uses only one half of the input waveform, while the full wave rectifier uses both halves.
Efficiency
The efficiency of a rectifier is the ratio of the DC output power to the AC input power. In general, the full wave rectifier is more efficient than the half wave rectifier because it utilizes both halves of the input waveform. This means that the full wave rectifier produces a higher DC output voltage for the same AC input voltage, resulting in higher efficiency.
Average Current
The average current is the DC current that flows through the load resistor. In a full wave rectifier, the current flows through the load resistor during both halves of the input waveform, resulting in a higher average current compared to a half wave rectifier.
Average Output Voltage
The average output voltage is the DC voltage that is present across the load resistor. In a full wave rectifier, the output voltage is the sum of the voltages from both halves of the input waveform, resulting in a higher average output voltage compared to a half wave rectifier.
Conclusion
In conclusion, the full wave rectifier is more efficient than the half wave rectifier, and it also produces a higher average current and average output voltage. Therefore, the correct answer is option D, none of these.
Rectifiers are electronic circuits that convert AC (alternating current) into DC (direct current). There are two types of rectifiers: half wave and full wave. The main difference between the two is that the half wave rectifier uses only one half of the input waveform, while the full wave rectifier uses both halves.
Efficiency
The efficiency of a rectifier is the ratio of the DC output power to the AC input power. In general, the full wave rectifier is more efficient than the half wave rectifier because it utilizes both halves of the input waveform. This means that the full wave rectifier produces a higher DC output voltage for the same AC input voltage, resulting in higher efficiency.
Average Current
The average current is the DC current that flows through the load resistor. In a full wave rectifier, the current flows through the load resistor during both halves of the input waveform, resulting in a higher average current compared to a half wave rectifier.
Average Output Voltage
The average output voltage is the DC voltage that is present across the load resistor. In a full wave rectifier, the output voltage is the sum of the voltages from both halves of the input waveform, resulting in a higher average output voltage compared to a half wave rectifier.
Conclusion
In conclusion, the full wave rectifier is more efficient than the half wave rectifier, and it also produces a higher average current and average output voltage. Therefore, the correct answer is option D, none of these.
Which of the following cannot be emitted by radioactive substances during their decay?- a) Protons
- b) Neutrino
- c) Helium nuclei
- d) Electrons
Correct answer is option 'A'. Can you explain this answer?
a)
Protons
b)
Neutrino
c)
Helium nuclei
d)
Electrons
|
|
Shraddha Roy answered |
Explanation:
Radioactive decay is a spontaneous process where an unstable nucleus emits particles or energy in order to become more stable. The types of particles that can be emitted during radioactive decay are determined by the type of nucleus that is decaying.
Types of particles emitted during radioactive decay:
1. Protons: Protons are positively charged particles found in the nucleus of an atom. They can be emitted during certain types of radioactive decay, such as alpha decay and proton emission. When a nucleus emits a proton, it becomes a different element with a lower atomic number.
2. Neutrinos: Neutrinos are subatomic particles with no electric charge and very little mass. They are emitted during certain types of radioactive decay, such as beta decay and electron capture. Neutrinos are very difficult to detect because they interact very weakly with matter.
3. Helium nuclei: Helium nuclei, also known as alpha particles, are made up of two protons and two neutrons. They are emitted during alpha decay, which occurs when the nucleus of an atom loses two protons and two neutrons. Alpha particles are relatively large and heavy compared to other types of particles emitted during radioactive decay.
4. Electrons: Electrons are negatively charged particles that orbit the nucleus of an atom. They can be emitted during beta decay, which occurs when a neutron in the nucleus of an atom decays into a proton, an electron, and a neutrino. The electron is emitted from the nucleus, and the proton remains in the nucleus.
Answer:
The correct answer is option A) Protons cannot be emitted by radioactive substances during their decay. This is because the number of protons in the nucleus determines the identity of the element, and if a proton were emitted, the element would change. However, proton emission can occur during some rare types of nuclear reactions, such as spontaneous fission.
Radioactive decay is a spontaneous process where an unstable nucleus emits particles or energy in order to become more stable. The types of particles that can be emitted during radioactive decay are determined by the type of nucleus that is decaying.
Types of particles emitted during radioactive decay:
1. Protons: Protons are positively charged particles found in the nucleus of an atom. They can be emitted during certain types of radioactive decay, such as alpha decay and proton emission. When a nucleus emits a proton, it becomes a different element with a lower atomic number.
2. Neutrinos: Neutrinos are subatomic particles with no electric charge and very little mass. They are emitted during certain types of radioactive decay, such as beta decay and electron capture. Neutrinos are very difficult to detect because they interact very weakly with matter.
3. Helium nuclei: Helium nuclei, also known as alpha particles, are made up of two protons and two neutrons. They are emitted during alpha decay, which occurs when the nucleus of an atom loses two protons and two neutrons. Alpha particles are relatively large and heavy compared to other types of particles emitted during radioactive decay.
4. Electrons: Electrons are negatively charged particles that orbit the nucleus of an atom. They can be emitted during beta decay, which occurs when a neutron in the nucleus of an atom decays into a proton, an electron, and a neutrino. The electron is emitted from the nucleus, and the proton remains in the nucleus.
Answer:
The correct answer is option A) Protons cannot be emitted by radioactive substances during their decay. This is because the number of protons in the nucleus determines the identity of the element, and if a proton were emitted, the element would change. However, proton emission can occur during some rare types of nuclear reactions, such as spontaneous fission.
If a current is passed in a spring, it- a)gets compressed
- b)gets expanded
- c)oscillates
- d)remains unchanged
Correct answer is option 'A'. Can you explain this answer?
a)
gets compressed
b)
gets expanded
c)
oscillates
d)
remains unchanged
|
Arpit Mishra answered |
Two conductors will attract each other if they have current flowing in same direction. in a spring each loop has current flowing in same direction resulting in contraction of the spring.
The impurity atom that should be added to the germanium to make it an n-type semiconductor, is- a)iodine
- b)indium
- c)arsenic
- d)aluminium
Correct answer is option 'C'. Can you explain this answer?
The impurity atom that should be added to the germanium to make it an n-type semiconductor, is
a)
iodine
b)
indium
c)
arsenic
d)
aluminium
|
|
Srinidhi V answered |
An n-type conductor is formed by adding a pentavalent impurity to germanium.
Among the options, only Arsenic has +5 valency i.e., pentavalent.
Hence it must be added to germanium to make it an n-type semiconductor.
Among the options, only Arsenic has +5 valency i.e., pentavalent.
Hence it must be added to germanium to make it an n-type semiconductor.
When a current flows in a wire, there exists an electric field whose direction is- a)same as that of current
- b)opposite to that of current
- c)perpendicular to that of current
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
a)
same as that of current
b)
opposite to that of current
c)
perpendicular to that of current
d)
none of these
|
Hansa Sharma answered |
Answer :- a
Solution :- The current always flows in the same direction as the electric field. The electrons flow in the opposite direction, because they are negatively charged.
In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is- a)R/2
- b)R
- c)2 R
- d)R/4
Correct answer is option 'B'. Can you explain this answer?
In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is
a)
R/2
b)
R
c)
2 R
d)
R/4
|
|
Tushar Jain answered |
Solution:
A Wheatstone Bridge is a circuit used to compare an unknown resistance with a known resistance. It consists of four resistors R1, R2, R3 and R4 connected between a voltage source and a galvanometer.
Given, R1 = R2 = R3 = R4 = R and Rg = R.
To find out the equivalent resistance of the combination, we can use the following steps:
Step 1: Apply Kirchhoff's first law to the junction point A.
I1 + I3 = I2 + I4
Step 2: Apply Ohm's law to each resistor.
I1 = V/R
I2 = (V - IR)/R
I3 = IR/R
I4 = (V - IR)/R
Step 3: Substitute the values of I1, I2, I3 and I4 in the equation obtained in step 1.
V/R + IR/R = (V - IR)/R + (V - IR)/R
Step 4: Solve for I.
I = V/2R
Step 5: Apply Kirchhoff's second law to the loop ABCDA.
IR - IR + I*Rg - V = 0
Step 6: Solve for V.
V = IRg
Step 7: Substitute the value of V in the equation obtained in step 4.
I = IRg/2R
Step 8: Simplify.
I = Ig/2
Step 9: Apply Ohm's law to the equivalent resistance.
Req = V/I
Req = 2R
Therefore, the equivalent resistance of the combination as seen by the battery is 2R.
Hence, option B is the correct answer.
A Wheatstone Bridge is a circuit used to compare an unknown resistance with a known resistance. It consists of four resistors R1, R2, R3 and R4 connected between a voltage source and a galvanometer.
Given, R1 = R2 = R3 = R4 = R and Rg = R.
To find out the equivalent resistance of the combination, we can use the following steps:
Step 1: Apply Kirchhoff's first law to the junction point A.
I1 + I3 = I2 + I4
Step 2: Apply Ohm's law to each resistor.
I1 = V/R
I2 = (V - IR)/R
I3 = IR/R
I4 = (V - IR)/R
Step 3: Substitute the values of I1, I2, I3 and I4 in the equation obtained in step 1.
V/R + IR/R = (V - IR)/R + (V - IR)/R
Step 4: Solve for I.
I = V/2R
Step 5: Apply Kirchhoff's second law to the loop ABCDA.
IR - IR + I*Rg - V = 0
Step 6: Solve for V.
V = IRg
Step 7: Substitute the value of V in the equation obtained in step 4.
I = IRg/2R
Step 8: Simplify.
I = Ig/2
Step 9: Apply Ohm's law to the equivalent resistance.
Req = V/I
Req = 2R
Therefore, the equivalent resistance of the combination as seen by the battery is 2R.
Hence, option B is the correct answer.
An X - ray tube is operated at a constant potential difference and it is required to get X - ray of wavelength not less then 0. 2 nano-metres. Then the potential difference in kilo-volts is [ h = 6.63 x 10-34 J − sec ; e = 1.6 x 10-19 C ; c = 3x 108 m s-1 ]- a)24. 8
- b)12. 4
- c)6. 2
- d)3.1
Correct answer is option 'C'. Can you explain this answer?
An X - ray tube is operated at a constant potential difference and it is required to get X - ray of wavelength not less then 0. 2 nano-metres. Then the potential difference in kilo-volts is [ h = 6.63 x 10-34 J − sec ; e = 1.6 x 10-19 C ; c = 3x 108 m s-1 ]
a)
24. 8
b)
12. 4
c)
6. 2
d)
3.1
|
|
Genius answered |
Just apply the formula !
V= hc/λe, where λ is wavelength= 2×10-¹⁰
put the value of h, c, λ & e, you will get the correct answer.
To obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is- a)Monovalent
- b)Divalent
- c)Trivalent
- d)Pentavalent
Correct answer is option 'D'. Can you explain this answer?
To obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is
a)
Monovalent
b)
Divalent
c)
Trivalent
d)
Pentavalent
|
|
Anirban Majumdar answered |
Introduction:
Semiconductors are materials that have electrical conductivity between that of a conductor and an insulator. The conductivity of a semiconductor can be increased by doping, which involves adding impurity atoms to the crystal structure of the semiconductor.
Impurity mixed to obtain electrons:
The impurity atoms that are added to a semiconductor are chosen based on their valence electron configuration. The impurity atoms should have either one more or one less valence electron than the atoms in the semiconductor crystal. This creates either an excess or a deficiency of electrons in the crystal, which makes the impurity atoms act as either donors or acceptors of electrons.
To obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is pentavalent. This means that the impurity atom has five valence electrons, whereas the atoms in the semiconductor crystal have only four valence electrons.
Explanation of pentavalent impurities:
When a pentavalent impurity atom, such as phosphorus (P), is added to a silicon (Si) crystal, each phosphorus atom replaces a silicon atom in the crystal structure. Since phosphorus has one more valence electron than silicon, the extra electron is not bound to any particular atom and is free to move throughout the crystal. This creates a population of free electrons in the crystal, which are available to conduct electricity.
Since the free electrons in the crystal are negatively charged, the crystal as a whole becomes n-type (n for negative) semiconductor. In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are the positively charged holes left behind when electrons move from one atom to another.
Conclusion:
In conclusion, to obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is pentavalent. The pentavalent impurities create a population of free electrons in the crystal, which are available to conduct electricity and make the semiconductor an n-type semiconductor.
Semiconductors are materials that have electrical conductivity between that of a conductor and an insulator. The conductivity of a semiconductor can be increased by doping, which involves adding impurity atoms to the crystal structure of the semiconductor.
Impurity mixed to obtain electrons:
The impurity atoms that are added to a semiconductor are chosen based on their valence electron configuration. The impurity atoms should have either one more or one less valence electron than the atoms in the semiconductor crystal. This creates either an excess or a deficiency of electrons in the crystal, which makes the impurity atoms act as either donors or acceptors of electrons.
To obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is pentavalent. This means that the impurity atom has five valence electrons, whereas the atoms in the semiconductor crystal have only four valence electrons.
Explanation of pentavalent impurities:
When a pentavalent impurity atom, such as phosphorus (P), is added to a silicon (Si) crystal, each phosphorus atom replaces a silicon atom in the crystal structure. Since phosphorus has one more valence electron than silicon, the extra electron is not bound to any particular atom and is free to move throughout the crystal. This creates a population of free electrons in the crystal, which are available to conduct electricity.
Since the free electrons in the crystal are negatively charged, the crystal as a whole becomes n-type (n for negative) semiconductor. In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are the positively charged holes left behind when electrons move from one atom to another.
Conclusion:
In conclusion, to obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is pentavalent. The pentavalent impurities create a population of free electrons in the crystal, which are available to conduct electricity and make the semiconductor an n-type semiconductor.
A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19 C) the dust particle carries- a)1
- b)10
- c)100
- d)1000
Correct answer is option 'A'. Can you explain this answer?
A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19 C) the dust particle carries
a)
1
b)
10
c)
100
d)
1000
|
|
Rhea Datta answered |
Given:
Mass of dust particle, m = 10^-3 g = 10^-6 kg
Separation between the plates of capacitor, d = 0.016 m
Voltage applied, V = 100 V
Charge of an electron, e = 1.6 x 10^-19 C
To find: The number of fundamental charges (e) the dust particle carries.
Formula used:
The potential difference between the plates of capacitor is given by,
V = Ed
where E is the electric field between the plates, and d is the separation between the plates.
The electric field is given by,
E = V/d
The force on a charged particle in an electric field is given by,
F = qE
where q is the charge of the particle.
The acceleration of the particle in the electric field is given by,
a = F/m
The velocity of the particle after time t is given by,
v = at
The distance travelled by the particle in time t is given by,
s = 1/2at^2
The number of fundamental charges (e) the dust particle carries is given by,
n = q/e
Solution:
The electric field between the plates of capacitor is,
E = V/d = 100/0.016 = 6250 V/m
The force on the dust particle is,
F = qE
The acceleration of the particle is,
a = F/m = qE/m
The velocity of the particle after time t is,
v = at = qEt/m
The distance travelled by the particle in time t is,
s = 1/2at^2 = (qE/m)(t^2)/2
The time taken by the particle to travel the distance between the plates of capacitor is,
t = d/v = md/qE
Substituting the value of t in the equation for s, we get,
s = (1/2)(m/d)(d^2/qE^2) = (m/2qE)d^2
The number of fundamental charges (e) the dust particle carries is,
n = q/e = (ma/e) = (m/d)(d^2/2eV)
Substituting the given values, we get,
n = (10^-6/0.016)(0.016^2/2*1.6x10^-19*100) = 1.25
Therefore, the dust particle carries 1 fundamental charge (e). Hence, the correct option is (A).
Mass of dust particle, m = 10^-3 g = 10^-6 kg
Separation between the plates of capacitor, d = 0.016 m
Voltage applied, V = 100 V
Charge of an electron, e = 1.6 x 10^-19 C
To find: The number of fundamental charges (e) the dust particle carries.
Formula used:
The potential difference between the plates of capacitor is given by,
V = Ed
where E is the electric field between the plates, and d is the separation between the plates.
The electric field is given by,
E = V/d
The force on a charged particle in an electric field is given by,
F = qE
where q is the charge of the particle.
The acceleration of the particle in the electric field is given by,
a = F/m
The velocity of the particle after time t is given by,
v = at
The distance travelled by the particle in time t is given by,
s = 1/2at^2
The number of fundamental charges (e) the dust particle carries is given by,
n = q/e
Solution:
The electric field between the plates of capacitor is,
E = V/d = 100/0.016 = 6250 V/m
The force on the dust particle is,
F = qE
The acceleration of the particle is,
a = F/m = qE/m
The velocity of the particle after time t is,
v = at = qEt/m
The distance travelled by the particle in time t is,
s = 1/2at^2 = (qE/m)(t^2)/2
The time taken by the particle to travel the distance between the plates of capacitor is,
t = d/v = md/qE
Substituting the value of t in the equation for s, we get,
s = (1/2)(m/d)(d^2/qE^2) = (m/2qE)d^2
The number of fundamental charges (e) the dust particle carries is,
n = q/e = (ma/e) = (m/d)(d^2/2eV)
Substituting the given values, we get,
n = (10^-6/0.016)(0.016^2/2*1.6x10^-19*100) = 1.25
Therefore, the dust particle carries 1 fundamental charge (e). Hence, the correct option is (A).
A wheel having metal spokes of 1m long between its axle and rim is rotating in a magnetic field of flux density 5x10-5 T normal to the plane of the wheel. An e.m.f of 22/7 mV is produced between the rim and the axle of the wheel. The rate of rotation of the wheel in radians per second is
- a)10
- b)20
- c)30
- d)4
Correct answer is option 'B'. Can you explain this answer?
A wheel having metal spokes of 1m long between its axle and rim is rotating in a magnetic field of flux density 5x10-5 T normal to the plane of the wheel. An e.m.f of 22/7 mV is produced between the rim and the axle of the wheel. The rate of rotation of the wheel in radians per second is
a)
10
b)
20
c)
30
d)
4
|
|
Mayank Banerjee answered |
Given Data:
Length of metal spokes, l = 1m
Flux density, B = 5x10^-5 T
e.m.f produced, E = 22/7 mV
To find: The rate of rotation of the wheel in radians per second.
Formula used:
The induced e.m.f, E = Bvl
where, v is the velocity of the wheel.
The rate of rotation of the wheel, ω = v/r
where, r is the radius of the wheel.
Steps to solve the problem:
1. Calculate the velocity of the wheel using the induced e.m.f formula.
E = Bvl
v = E/Bl
Substituting the given values, we get,
v = (22/7 x 10^-3)/(5 x 10^-5 x 1)
v = 1.4 m/s
2. Calculate the rate of rotation of the wheel using the velocity formula.
ω = v/r
We know that the length of the spokes is equal to the radius of the wheel.
So, r = l = 1m
Substituting the values, we get,
ω = 1.4/1
ω = 1.4 rad/s
3. Convert the rate of rotation from rad/s to radians per second.
The answer is in rad/s, so no conversion is needed.
Final Answer:
The rate of rotation of the wheel in radians per second is 1.4 rad/s.
Therefore, option b) 20 is incorrect.
Length of metal spokes, l = 1m
Flux density, B = 5x10^-5 T
e.m.f produced, E = 22/7 mV
To find: The rate of rotation of the wheel in radians per second.
Formula used:
The induced e.m.f, E = Bvl
where, v is the velocity of the wheel.
The rate of rotation of the wheel, ω = v/r
where, r is the radius of the wheel.
Steps to solve the problem:
1. Calculate the velocity of the wheel using the induced e.m.f formula.
E = Bvl
v = E/Bl
Substituting the given values, we get,
v = (22/7 x 10^-3)/(5 x 10^-5 x 1)
v = 1.4 m/s
2. Calculate the rate of rotation of the wheel using the velocity formula.
ω = v/r
We know that the length of the spokes is equal to the radius of the wheel.
So, r = l = 1m
Substituting the values, we get,
ω = 1.4/1
ω = 1.4 rad/s
3. Convert the rate of rotation from rad/s to radians per second.
The answer is in rad/s, so no conversion is needed.
Final Answer:
The rate of rotation of the wheel in radians per second is 1.4 rad/s.
Therefore, option b) 20 is incorrect.
A ray of light falls on a transparent glass slab of refractive index 1.62. If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is- a)tan-1(1.62)
- b)tan-1(1/1.62)
- c)tan-1(1.33)
- d)tan-1(1/1.33)
Correct answer is option 'A'. Can you explain this answer?
A ray of light falls on a transparent glass slab of refractive index 1.62. If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is
a)
tan-1(1.62)
b)
tan-1(1/1.62)
c)
tan-1(1.33)
d)
tan-1(1/1.33)
|
|
Vaishnavi Basak answered |
Given:
- Refractive index of glass slab = 1.62
- Reflected and refracted rays are mutually perpendicular
To find:
- Angle of incidence
Solution:
1. Draw the diagram:
Draw a diagram to represent the incident ray, reflected ray and refracted ray inside the glass slab.
2. Use Snell's law:
Snell's law states that:
- n1 sinθ1 = n2 sinθ2
where n1 and n2 are the refractive indices of the two media, and θ1 and θ2 are the angles of incidence and refraction respectively.
In this case, n1 = 1 (since the incident ray is coming from air), n2 = 1.62 (since the ray is entering the glass slab), and θ2 = 90 degrees (since the refracted and reflected rays are mutually perpendicular).
Therefore, we can simplify Snell's law to:
- sinθ1 = (n2/n1) sinθ2
- sinθ1 = 1.62 sin(90)
- sinθ1 = 1.62 x 1
- sinθ1 = 1.62
3. Find the angle of incidence:
To find the angle of incidence, we need to take the inverse sine of sinθ1:
- θ1 = sin^-1(1.62)
- θ1 = 62.8 degrees
4. Check the options:
The correct answer is given in option A:
- tan^-1(1.62) = 62.8 degrees
Therefore, the correct answer is option A: tan^-1(1.62).
- Refractive index of glass slab = 1.62
- Reflected and refracted rays are mutually perpendicular
To find:
- Angle of incidence
Solution:
1. Draw the diagram:
Draw a diagram to represent the incident ray, reflected ray and refracted ray inside the glass slab.
2. Use Snell's law:
Snell's law states that:
- n1 sinθ1 = n2 sinθ2
where n1 and n2 are the refractive indices of the two media, and θ1 and θ2 are the angles of incidence and refraction respectively.
In this case, n1 = 1 (since the incident ray is coming from air), n2 = 1.62 (since the ray is entering the glass slab), and θ2 = 90 degrees (since the refracted and reflected rays are mutually perpendicular).
Therefore, we can simplify Snell's law to:
- sinθ1 = (n2/n1) sinθ2
- sinθ1 = 1.62 sin(90)
- sinθ1 = 1.62 x 1
- sinθ1 = 1.62
3. Find the angle of incidence:
To find the angle of incidence, we need to take the inverse sine of sinθ1:
- θ1 = sin^-1(1.62)
- θ1 = 62.8 degrees
4. Check the options:
The correct answer is given in option A:
- tan^-1(1.62) = 62.8 degrees
Therefore, the correct answer is option A: tan^-1(1.62).
A bar magnet dropped into a coil of a conducting wire, along its axis, will fall with an acceleration
- a)equal to g
- b)less than g
- c)more than g
- d)first '1' then'3'
Correct answer is option 'B'. Can you explain this answer?
A bar magnet dropped into a coil of a conducting wire, along its axis, will fall with an acceleration
a)
equal to g
b)
less than g
c)
more than g
d)
first '1' then'3'
|
Aarav Roy answered |
Magnetic Induction and Motion of Bar Magnet in a Coil
When a bar magnet is dropped into a coil of conducting wire, an electric current is induced in the coil due to the changing magnetic field. This phenomenon is known as electromagnetic induction. The induced current creates an opposing magnetic field that interacts with the falling magnet and causes it to experience a magnetic force.
Factors Affecting the Acceleration of the Falling Magnet
The acceleration of the falling magnet in the coil depends on several factors, such as the strength of the magnet, the characteristics of the coil, and the orientation of the magnet with respect to the coil. However, the most significant factor is the magnetic force acting on the magnet due to the induced current in the coil.
Magnetic Force and Acceleration
The magnetic force acting on the magnet can be calculated using the formula:
F = BIL
Where F is the magnetic force, B is the magnetic field strength, I is the current, and L is the length of the conductor. Since the magnetic field and the length of the conductor are fixed, the magnetic force is directly proportional to the current. Therefore, the acceleration of the falling magnet is also proportional to the current.
Opposing Magnetic Field and Current
The current induced in the coil creates an opposing magnetic field that interacts with the falling magnet. This opposing magnetic field reduces the net magnetic force acting on the magnet and decreases its acceleration. The strength of the opposing magnetic field depends on the rate of change of the magnetic field, which in turn depends on the speed of the falling magnet.
Conclusion
Therefore, the acceleration of the falling magnet in a coil of conducting wire is less than g, as the opposing magnetic field created by the induced current reduces the net magnetic force and decreases the acceleration. The correct answer is option B.
When a bar magnet is dropped into a coil of conducting wire, an electric current is induced in the coil due to the changing magnetic field. This phenomenon is known as electromagnetic induction. The induced current creates an opposing magnetic field that interacts with the falling magnet and causes it to experience a magnetic force.
Factors Affecting the Acceleration of the Falling Magnet
The acceleration of the falling magnet in the coil depends on several factors, such as the strength of the magnet, the characteristics of the coil, and the orientation of the magnet with respect to the coil. However, the most significant factor is the magnetic force acting on the magnet due to the induced current in the coil.
Magnetic Force and Acceleration
The magnetic force acting on the magnet can be calculated using the formula:
F = BIL
Where F is the magnetic force, B is the magnetic field strength, I is the current, and L is the length of the conductor. Since the magnetic field and the length of the conductor are fixed, the magnetic force is directly proportional to the current. Therefore, the acceleration of the falling magnet is also proportional to the current.
Opposing Magnetic Field and Current
The current induced in the coil creates an opposing magnetic field that interacts with the falling magnet. This opposing magnetic field reduces the net magnetic force acting on the magnet and decreases its acceleration. The strength of the opposing magnetic field depends on the rate of change of the magnetic field, which in turn depends on the speed of the falling magnet.
Conclusion
Therefore, the acceleration of the falling magnet in a coil of conducting wire is less than g, as the opposing magnetic field created by the induced current reduces the net magnetic force and decreases the acceleration. The correct answer is option B.
Bright colours exhibited by spider's web, exposed to sun light are due to- a)diffraction
- b)polarisation
- c)interference
- d)resolution
Correct answer is option 'A'. Can you explain this answer?
Bright colours exhibited by spider's web, exposed to sun light are due to
a)
diffraction
b)
polarisation
c)
interference
d)
resolution
|
|
Aarav Roy answered |
Explanation:
When sunlight falls on a spider's web, it gets diffracted and causes the web to display bright colors. Diffraction is the bending of light waves as they pass through an obstacle or around an edge. The spider's web acts as a diffraction grating, which is a surface with a repeating pattern of equally spaced slits or scratches. The diffraction grating splits the white light into its component colors and creates a rainbow-like effect.
In contrast, polarization is the orientation of the electric field of light waves. Interference is the interaction of two or more waves that results in either constructive or destructive interference. Resolution refers to the ability to distinguish two closely spaced objects.
Therefore, option A (diffraction) is the correct answer as it explains the phenomenon of bright colors exhibited by spiders web exposed to sunlight.
When sunlight falls on a spider's web, it gets diffracted and causes the web to display bright colors. Diffraction is the bending of light waves as they pass through an obstacle or around an edge. The spider's web acts as a diffraction grating, which is a surface with a repeating pattern of equally spaced slits or scratches. The diffraction grating splits the white light into its component colors and creates a rainbow-like effect.
In contrast, polarization is the orientation of the electric field of light waves. Interference is the interaction of two or more waves that results in either constructive or destructive interference. Resolution refers to the ability to distinguish two closely spaced objects.
Therefore, option A (diffraction) is the correct answer as it explains the phenomenon of bright colors exhibited by spiders web exposed to sunlight.
A particle with a specific charge S is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for particle not hit the wall ?- a)v/4sd
- b)v/2sd
- c)2v/sd
- d)v/sd
Correct answer is option 'D'. Can you explain this answer?
A particle with a specific charge S is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for particle not hit the wall ?
a)
v/4sd
b)
v/2sd
c)
2v/sd
d)
v/sd
|
|
Tanishq Basak answered |
The particle moves in a circular path with radius d if it is just to miss the wall
mv = Bqr
⇒ B = v ( q m ) d = v sd
mv = Bqr
⇒ B = v ( q m ) d = v sd
The value of current, flowing through an inductor of inductance 1 H and having negligible resistance when connected to an a.c. source of 200 V and 50 Hz, is- a)0.64 A
- b)1.64 A
- c)2.64 A
- d)3.64 A
Correct answer is option 'A'. Can you explain this answer?
The value of current, flowing through an inductor of inductance 1 H and having negligible resistance when connected to an a.c. source of 200 V and 50 Hz, is
a)
0.64 A
b)
1.64 A
c)
2.64 A
d)
3.64 A
|
|
Pallabi Sen answered |
Given:
Inductance (L) = 1 H
Voltage (V) = 200 V
Frequency (f) = 50 Hz
Formula used:
The formula used to calculate the current flowing in an inductor is:
I = (V/Z)
Where,
I = Current flowing through the inductor
V = Voltage applied to the inductor
Z = Impedance of the inductor
Impedance of an inductor is given by:
Z = jωL
Where,
ω = 2πf
j = √(-1)
Calculation:
ω = 2πf = 2π x 50 = 100π
j = √(-1)
Z = jωL = j x 100π x 1 = 100jπ
I = (V/Z) = 200/100jπ
To simplify the expression, we can multiply and divide by j:
I = (200j)/(100jπ x j) = (2j)/(π)
Taking the magnitude of the complex number:
|I| = √(2²/π²) = 0.64 A
Therefore, the value of current flowing through the inductor is 0.64 A.
Answer: (a) 0.64 A
Inductance (L) = 1 H
Voltage (V) = 200 V
Frequency (f) = 50 Hz
Formula used:
The formula used to calculate the current flowing in an inductor is:
I = (V/Z)
Where,
I = Current flowing through the inductor
V = Voltage applied to the inductor
Z = Impedance of the inductor
Impedance of an inductor is given by:
Z = jωL
Where,
ω = 2πf
j = √(-1)
Calculation:
ω = 2πf = 2π x 50 = 100π
j = √(-1)
Z = jωL = j x 100π x 1 = 100jπ
I = (V/Z) = 200/100jπ
To simplify the expression, we can multiply and divide by j:
I = (200j)/(100jπ x j) = (2j)/(π)
Taking the magnitude of the complex number:
|I| = √(2²/π²) = 0.64 A
Therefore, the value of current flowing through the inductor is 0.64 A.
Answer: (a) 0.64 A
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