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Texts: Calculate the moment of inertia of the shaded area abou

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Texts: Calculate the moment of inertia of the shaded area about the x-axis. 72 mm 72 mm 42 mm^2 Answer: Ix - (10 mm)^4 Save for later.

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Medium

Two equal charges

q

of opposite sign separated by a distance

2 a

constitute an electric dipole of dipole moment

p

. If

P

is a point at a distance

r

from the centre of the dipole and the line joining the centre of the dipole to this point makes and angle

θ

with the axis of the dipole, then the potential at

P

is given by

(r >> 2 a

) (where,

p = 2 q a

)

$V = \frac{p c o s θ}{2 π ε _{0} r ^{2}}$
$V = \frac{p s i n θ}{4 π ε _{0} r}$
$V = \frac{p c o s θ}{4 π ε _{0} r}$
$V = \frac{p c o s θ}{4 π ε _{0} r ^{2}}$

Topic:

Electric Charges and Fields

Book:

Master Resource Book in JEE Main Physics (Arihant)

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Question 2

Easy

A metal of radius

100 cm

is rotated at a constant angular speed of

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in a plane at right angles to an external field of magnetic induction

0.05 Wbm^{- 2}

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$3 V$
$1.5 V$
$6 V$
$9 V$

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The numerical aperture for a human eye is of the order of

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Question Text	Texts: Calculate the moment of inertia of the shaded area about the x-axis. 72 mm 72 mm 42 mm^2 Answer: Ix - (10 mm)^4 Save for later.
Topic	All topics
Subject	Physics
Class	Class 12

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