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Two condensers, one of capacity $C$ and the other of capacity $C/2$ are connected to a $V$ volt battery, as shown in the figure. The energy stored in the capacitors when both condensers are fully charged will be:
Air is pushed carefully into a soap bubble of radius $r$ to double its radius. If the surface tension of the soap solution is $T$, then the work done in the process is:
The capacitance of a parallel plate capacitor with air as a medium is $6 \, \mu\text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30 \, \mu\text{F}$. The permittivity of the medium is: $(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2})$
The wettability of a surface by a liquid depends primarily on:
The net magnetic flux through any closed surface is:
A fluid of density $\rho$ is flowing in a pipe of varying cross-sectional area as shown in the figure. Bernoulli's equation for the motion becomes:
Angular momentum of a body is defined as the product of:
A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B as shown in Figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in Figure (b). The electrostatic energy stored by the system (b) is:
A body weighing $100 \text{ N}$ on the surface of the Earth weighs $x \text{ kg m s}^{-2}$ at a height $\frac{1}{9}R_E$ above the surface of Earth. The value of $x$ is: (take $g=10 \text{ m s}^{-2}$ at the surface of Earth and $R_E$ is the radius of Earth)
The electric potential at a point (x, y, z) is given by V = -x²y - xz³ + 4. The electric field E at that point is
If the plates of a parallel plate capacitor connected to a battery are moved closer to each other, then: (A) The charge stored in it increases. (B) The energy stored in it decreases. (C) Its capacitance increases. (D) The ratio of charge to its potential remains the same. (E) The product of charge and voltage increases. Choose the most appropriate answer from the options given below:
The distance between the two plates of a parallel plate capacitor is doubled, and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to:
Three capacitors, each of capacitance $0.3 \text{ \mu F}$, are connected in parallel. This combination is connected with another capacitor of capacitance $0.1 \text{ \mu F}$ in series. Then the equivalent capacitance of the combination is:
The ratio of escape velocity at the Earth ($v_e$) to the escape velocity at a planet ($v_p$) whose radius and mean density are twice that of the Earth is:
The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is:
A positively charged particle +q is projected with speed v toward a fixed charge +Q, and rebounds after reaching a minimum distance r. What will be the new closest distance of approach if its initial velocity is doubled to 2v?
Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, -σ and σ respectively. If V_A, V_B and V_C denote the potential of the three shells, if c = a + b, we have
The equivalent capacitance of the system shown in the following circuit is:
Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:
The escape velocity of a body on the earth's surface is $11.2 \text{ km/s}$. If the same body is projected upward with a velocity $22.4 \text{ km/s}$, the velocity of this body at an infinite distance from the centre of the earth will be: