Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is
1 : 4
4 : 1
1:2
1:2
Which of the following units denotes the dimensions ML^{2} /Q^{2}, where Q denotes the electric charge?
Weber (Wb)
Henry (H)
Wb/m^{2}
Wb/m^{2}
A charged ball B hangs from a silk thread S which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to
cos θ
cot θ
sin θ
sin θ
Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is
2L
L/4
8L
8L
Two thin wires rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is
zero
zero
A charged particle q is shot towards another charged particle Q which is fixed, with a speed v it approaches Q upto a closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be
r
2r
r/2
r/2
Four charges equal to −Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is
From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:
$\frac{37}{9}M{R}^{2}$
4MR^{2}
$\frac{40}{9}M{R}^{2}$
10 MR^{2}
B.
4MR^{2}
let σ be the mass per unit area.
The total mass of the disc = σ x πR^{2} = 9M
The mass of the circular disc out
$=\sigma x\pi {\left(\frac{R}{3}\right)}^{2}=\sigma x\frac{\pi {R}^{2}}{9}=M$
Let us consider the above system as a complete disc of mass 9M and a negative mass M superimposed on it. Moment of Inertia (I_{1)} of the complete disc (9MR^{2})/2 about an axis passing through O and perpendicular to the plane of the disc.
M.I. of the cut-out portion about an axis passing through O and perpendicular to the plan of disc
$\frac{1}{2}xMx{\left(\frac{R}{3}\right)}^{2}$
Therefore, M.I (I2) of the cut out portion about an axis passing through O and perpendicular to the plane of disc
$=\left[\frac{1}{2}xMx{\left(\frac{R}{3}\right)}^{2}+Mx{\left(\frac{2R}{3}\right)}^{2}\right]$
Using perpendicular axis theorem
Therefore, the total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is
I = I_{1} + I_{2}
$=\frac{1}{2}9M{R}^{2}-\left[\frac{1}{2}xMx{\left(\frac{R}{3}\right)}^{2}+Mx{\left(\frac{2R}{3}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{9M{R}^{2}}{2}-\frac{9M{R}^{2}}{18}=\frac{(9-1)M{R}^{2}}{2}=4M{R}^{2}$