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In a certain region of space with volume $0.2 \text{ m}^3$, the electric potential is found to be $5 \text{ V}$ throughout. The magnitude of the electric field in this region is:
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:
A short electric dipole has a dipole moment of $16 \times 10^{-9} \text{ C m}$. The electric potential due to the dipole at a point at a distance of $0.6 \text{ m}$ from the centre of the dipole situated on a line making an angle of $60^{\circ}$ with the dipole axis is: $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2)$
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?
Dependence of intensity of gravitational field ($E$) of the earth with distance ($r$) from the centre of the earth is correctly represented by: (where $R$ is the radius of the earth)
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
A planet has a mass equal to $(1/10)^{\text{th}}$ of Earth's mass and a diameter equal to half of Earth's diameter. The acceleration due to gravity on this planet is:
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:
If a conducting sphere of radius R is charged. Then the electric field at a distance r (r > R) from the centre of the sphere would be, (V = potential on the surface of the sphere):
A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force:
An increase in the temperature of a gas-filled container would lead to:
A hollow metal sphere of radius R is given +Q charges to its outer surface. The electric potential at a distance R/3 from the centre of the sphere will be:
A particle of mass $m$ is projected with a velocity, $v = kv_e$ ($k < 1$) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where $v_e = $ escape velocity, $R = $ the radius of the earth)
An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at the point P due to the dipole is: (ε₀ = permittivity of free space and 1/4πε₀ = k)
A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\frac{3\pi}{Gd}$ represents: