# Electrostatics MCQs – Test 1 6

Electrostatics Questions: MCQs on Electrostatics for JEE Main and NEET, Test Number 01

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## 6 thoughts on “Electrostatics MCQs – Test 1”

• Ambuj tiwari

In q.5 ,Tcos15=mg and Tsin15=F. And when it is immersed in liquid then tan15=F÷K/mg(1-density of liquid÷d.of substance). And tan15 is constant and it is found to divide upper two equation. And when it goes in liquid then Force is changes into F÷K. Where k is dielectric constant which is ask and mg is also changes so easily it is found

• Chirag.K

Hi AMISHA,
As mg’ = mg – Upthrust

= mg – V*sigma*g (since upthrust = v*sigma*g)

mg’ = mg[1- sigma/rho]

T cos Theta=mg –(i)

T sin Theta=F –(ii)

Therefore from equation (i) and (ii)

tan Theta = F/mg —-(iii)

Now for the equilibrium of balls

tan Theta’ = F’/mg’ = F/Kmg[1-(rho/sigma)]

= f/(kmg(1-{sigma/rho }) –(iv)

According to question Theta’=Theta

from equation (iv) & (iii)

K=1/(1-(sigma /rho))

=rho/(rho -sigma ) =1.6/(1.6-0.8)=2

K=2

Hope this helps! 🙂

Chirag K

• king

Coulombic force –

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

– wherein

K – proportionality Constant

Q1 and Q2 are two Point charge

T\cos \Theta=mg –(i)

T\sin \Theta=F_{e} –(ii)

\therefore from equation (i) & (ii)

Tan \Theta=\frac{F_{e}}{mg}—(iii)

if ball is suspended in liquid g’=g[1-\frac{\sigma }{\rho }]

Tan \Theta’=\frac{F’_{e}}{mg’}

=\frac{f}{kmg(1-\frac{\sigma}{\rho })} –(iv)

According to question \Theta’=\Theta

from equation (iv) & (iii)

K=\frac{1}{1-\frac{\sigma }{\rho }}

=\frac{\rho }{(\rho -\sigma )} =\frac{1.6}{(1.6-0.8)}=2
K=2