Current Electricity - NEET Physics Chapterwise MCQs & PYQs
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NEET Current Electricity MCQs & PYQs
Practice NEET Current Electricity Questions
Question 71:
easy
Two batteries one of emf \(18\text{ V}\) and internal resistance \(3\text{ }\Omega\) while other of emf \(12\text{ V}\) and internal resistance \(2\text{ }\Omega\) are connected in parallel with positive terminals together at one point and negative terminals together to other point. If across these points an ideal voltmeter is connected, the reading of voltmeter will be
The potential difference is the equivalent EMF: \(E_{\text{eq}} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{18/3 + 12/2}{1/3 + 1/2} = \frac{12}{5/6} = 14.4\text{ V}\).
Pick out wrong statement about the Kirchhoff’s laws of electric circuit.
Kirchhoff's first law (the junction rule) is based on the law of conservation of charge, not energy. The second law (loop rule) is based on the conservation of energy.
A wire of a certain material is stretched slowly by 10%. Its new resistance and specific resistance will respectively become
Specific resistance (resistivity) depends only on the material and temperature, so it remains same. Resistance \(R \propto l^2\) when a wire is stretched. If length is increased by 10% (\(l' = 1.1l\)), the new resistance is \(R' = (1.1)^2 R = 1.21 R\).
A \((100 \text{ W}, 200 \text{ V})\) bulb is connected to a \(160 \text{ V}\) supply. The power consumed by the bulb will be
The resistance of the bulb is given by \(R = \frac{V^2}{P} = \frac{200^2}{100} = 400 \, \Omega\). When connected to a \(160 \text{ V}\) supply, the power consumed is \(P' = \frac{V'^2}{R} = \frac{160^2}{400} = 64 \text{ W}\).
A wire of resistance \(12\text{ }\Omega\) is bent to form a circle. The equivalent resistance between two diametrically opposite points will be
Bending the wire into a circle divides it into two halves of equal resistance \(6\text{ }\Omega\) in parallel between diametrically opposite points. The equivalent resistance is \(R_{\text{eq}} = \frac{6 \times 6}{6 + 6} = 3\text{ }\Omega\).
Consider the following statements and choose the correct option
Statement (A): Electromotive force (emf) is not a force, it is the voltage difference between two terminal of a battery in open circuit.
Statement (B): The potential difference across terminals of a battery having some internal resistance can never be greater than the emf of the battery.
Statement (A) is correct as EMF is indeed a potential difference under open circuit. Statement (B) is incorrect because during charging, the terminal potential difference is \(V = E + Ir\), which is greater than the EMF \(E\).
The energy used by a 100 W bulb which is on for 10 hours is:
Energy is calculated as \( E = P \times t = 100 \, \text{W} \times 10 \, \text{h} = 1000 \, \text{Wh} = 1 \, \text{kWh} \). In standard SI units, this is \( 1000 \, \text{W} \times 3600 \, \text{s} = 3.6 \times 10^6 \, \text{J} \). Thus, both Options (1) and (2) are correct.
Assertion (A): When constant current is passing through a conductor of variable area of cross section, electric field inside conductor is inversely proportional to cross sectional area.
Reason (R): Microscopic form of Ohm’s law is \( \vec{E} = \rho \vec{J} \), where \( \vec{E} \) stands for electric field, \( \rho \) stands for resistivity and \( \vec{J} \) stands for current density.
Assertion (A) is true because \( I = JA \) and \( E = \rho J \) imply \( E = \frac{\rho I}{A} \). Reason (R) is true as it's the microscopic form of Ohm's law. Reason correctly explains assertion.
Assertion (A): The current density \( \vec{J} \) at any point in ohmic resistor is in direction of electric field \( \vec{E} \) at that point.
Reason (R): A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.
Assertion (A) is true from \( \vec{J} = \sigma \vec{E} \) where \( \sigma \) is conductivity. Reason (R) is false because a charge released from rest moves along an electric field line only if the field line is straight, which is not always true.
Assertion (A): The rate at which energy is being delivered to a light bulb is lower after it has been on for a few seconds than just after it is turned on.
Reason (R): As the filaments warms up, its resistance rises and the current falls.
Concept: Resistance of metals increases with temperature. Power \(P = V^2/R\).
Formula: \(P = V^2/R\), \(R \propto T\).
Solution: As filament warms, its resistance \(R\) increases. For constant voltage \(V\), current \(I = V/R\) decreases, so power \(P = V^2/R\) delivered to the bulb also decreases. Thus, A and R are true and R explains A.