Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is increased, what happens to the charge?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the distance between the plates is increased?
A.Charge increases
B.Charge decreases
C.Charge remains the same
D.Charge becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the voltage is doubled?
A.Charge doubles
B.Charge halves
C.Charge remains the same
D.Charge quadruples
Solution
The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.
Q. A capacitor of capacitance C is charged to a voltage V and then connected in parallel with another uncharged capacitor of capacitance C. What is the final voltage across the capacitors?
A.V/2
B.V
C.2V
D.0
Solution
When connected in parallel, the total charge is conserved. The final voltage across both capacitors is V/2.
Q. A capacitor of capacitance C is connected to a battery of voltage V. If the battery is removed and the capacitor is connected to another capacitor of capacitance 2C, what is the final voltage across the combination?
A.V/3
B.V/2
C.V
D.2V
Solution
When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.
Q. A capillary tube is dipped in water. What is the shape of the water surface inside the tube?
A.Flat
B.Concave
C.Convex
D.Irregular
Solution
The water surface inside the capillary tube is concave due to the adhesive forces between water and the tube material being stronger than the cohesive forces among water molecules.
Q. A capillary tube is dipped into water. How high will the water rise in the tube if the radius is 1 mm?
A.2.5 cm
B.5 cm
C.10 cm
D.15 cm
Solution
Using the capillary rise formula, h = (2γcosθ)/(ρgr), where γ is surface tension, θ is contact angle, ρ is density, g is acceleration due to gravity, and r is radius.
Q. A capillary tube is dipped into water. The height to which water rises in the tube is determined by:
A.Surface tension and density of the liquid
B.Only surface tension
C.Only density of the liquid
D.Viscosity of the liquid
Solution
The height of the liquid column in a capillary tube is determined by both surface tension and the density of the liquid, as described by the capillary rise formula.
Correct Answer: A — Surface tension and density of the liquid
Q. A capillary tube is dipped into water. The water rises in the tube due to which of the following?
A.Surface tension and adhesion
B.Surface tension and cohesion
C.Only adhesion
D.Only cohesion
Solution
The rise of water in a capillary tube is due to both surface tension (which pulls the liquid up) and adhesion (the attraction between water molecules and the tube's surface).
Q. A capillary tube of radius 0.5 mm is dipped in water. What is the height of the water column raised in the tube? (Surface tension = 0.072 N/m, density of water = 1000 kg/m³)
A.0.5 m
B.0.1 m
C.0.2 m
D.0.3 m
Solution
Using the formula h = 2γ/(ρgr), h = 2 × 0.072 N/m / (1000 kg/m³ × 9.81 m/s² × 0.0005 m) = 0.2 m.
