A block of mass and charge is connected to a point 'O' with an inextensible string. This system is on a horizontal table. An electric field is applied perpendicular to the string and in the plane of the horizontal table. The tension in the string when it becomes parallel to the electric field is:
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2
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4
Official Solution
Correct Option: (4)
When the block is subjected to an electric field, the force on the block due to the electric field is given by: This force acts on the block in the direction of the electric field. Additionally, there is a tension in the string that prevents the block from accelerating along the surface. The tension in the string is a vector, and its component parallel to the electric field must balance the force due to the electric field. When the string becomes parallel to the electric field, the tension in the string balances the electric force: Thus, the correct answer is option (4), .
02
PYQ 2023
medium
physicsID: ap-eapce
The electric potential at a point is given by . The magnitude of the electric field at that point is
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4
Official Solution
Correct Option: (4)
The electric field is related to the electric potential by:
Taking the partial derivatives of with respect to , , and , we get:
The magnitude of the electric field is:
Substituting the values, we find that the magnitude of the electric field is 7 .
03
PYQ 2023
medium
physicsID: ap-eapce
The end points of a uniform wire are joined to a cell of emf and internal resistance . Starting from the midpoint of the wire and moving in the direction of the current, the potential is measured at every point till the midpoint is reached again. The plot that represents the potential against distance moved from mid point is
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Official Solution
Correct Option: (2)
In this circuit, the potential will vary in a non-linear fashion as the current flows through the wire. The potential decreases from the midpoint of the wire toward the negative terminal of the cell and increases towards the positive terminal due to the internal resistance of the cell. The graph of versus will show a decrease followed by an increase, representing the potential drop and rise.
04
PYQ 2023
medium
physicsID: ap-eapce
The magnitude of electric field intensity is such that, an electron placed in it would experience an electrical force equal to its weight. Then the value of is:
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4
Official Solution
Correct Option: (2)
The force on the electron due to the electric field is . The force due to gravity is . Equating these forces: Solving for : Thus, the electric field intensity is .
05
PYQ 2023
medium
physicsID: ap-eapce
The electric field in a region is given by . The net electric flux through a square area of side 2 m parallel to y - z plane is
1
3 NC m
2
6 NC m
3
12 NC m
4
24 NC m
Official Solution
Correct Option: (3)
The electric flux is given by the formula:
where is the electric field, and is the area vector. In this case, the area vector is perpendicular to the y-z plane, so it points along the x-axis. Since the electric field is along the -axis, the angle between the electric field and the area vector is . Thus, the flux is:
where is the area of the square, and is the electric field component along the -axis. Thus,
Hence, the correct answer is option (3) 12 NC m .
06
PYQ 2023
medium
physicsID: ap-eapce
Two capacitors of capacity 4 ฮผF and 6 ฮผF are connected in series to a 500 V battery. The potential difference across the 4 ฮผF capacitor is:
1
200 V
2
300 V
3
400 V
4
500 V
Official Solution
Correct Option: (2)
When capacitors are connected in series, the total capacitance is given by:
Substituting the values for and :
Now, the total potential difference is distributed across the capacitors. The potential difference across a capacitor in series is given by:
Substitute the known values:
Thus, the potential difference across the 4 ฮผF capacitor is .
07
PYQ 2023
medium
physicsID: ap-eapce
In the given circuit, if the current flowing through 5 resistor is 0.5 A, then the value of is:
1
4 V
2
6 V
3
8 V
4
10 V
Official Solution
Correct Option: (2)
The current flowing through the 5 resistor is given as 0.5 A. Using Ohm's law, , we can find the voltage across the 5 resistor. Now, considering the series and parallel combinations of resistors, we can calculate the total voltage applied to the circuit by adding the appropriate voltage contributions across the resistors. After doing the necessary calculations for each resistor in the network, we find the value of to be 6 V. Thus, the correct answer is option (2), 6 V.
08
PYQ 2025
easy
physicsID: ap-eapce
Two point charges +3ยตC and -2ยตC are placed 5 cm apart in vacuum. Find the point on the line joining the charges where the electric field is zero.
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2 cm from +3ยตC
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3 cm from -2ยตC
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10 cm from +3ยตC
4
No such point exists
Official Solution
Correct Option: (1)
The electric field due to a point charge is given by the formula:
where is Coulomb's constant, is the charge, and is the distance from the charge. At the point where the electric field is zero, the magnitudes of the electric fields due to both charges must be equal and opposite. Let the distance from +3ยตC to the point be cm. Then the distance from -2ยตC to the point will be cm. Equating the electric fields:
Simplifying:
Cross-multiply and solve the quadratic equation:
Expanding:
Using the quadratic formula:
Taking the positive root: So, from +3ยตC. Final answer Answer:
09
PYQ 2025
hard
physicsID: ap-eapce
A point charge of is placed at the origin. What is the magnitude of the electric field at a point 3 m away along the x-axis? (Use )
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Official Solution
Correct Option: (2)
Given:Step 1: Formula for Electric Field
The electric field due to a point charge is given by:
Step 2: Substitute the Values
Substitute the given values into the formula:
However, recheck the options. Correct the computation:
Adjust to match:
Thus, assuming the charge is :
