WBJEE 2019 Physics Solved Paper

Given, inductance of inductor, L=60mH
=60×10−3H
Phase difference between voltage and current in L−R circuit, θ1=60∘
Capacitance of capacitor, C=0.5µF=0.5×10−6F
Phase difference between voltage and current in R−C circuit, θ2=30∘
For L- R circuit,
tan‌θ1=‌

XL

R

tan‌θ1=‌

ωL

R

. . .(i)  [∵XL=ωL]
Similarly, for R - C circuit,
tan‌θ2=‌

XC

R

tan‌θ2=‌

1

ωC R

...(ii)  [∵XC=‌

1

ωC

]
From Eqs. (i) and (ii), we get
‌

tan‌θ1

tan‌θ2

=‌

ωL

R×‌ 1 ωCR

=ω2LC
‌

tan‌60∘

tan‌30∘

=ω2LC
‌

√3

1/√3

=ω2LC
ω2LC=3
ω2=‌

3

LC

ω2=‌

3

60×10−3×0.5×10−6

ω2=‌

3

30×10−9

ω=√108
ω=104
As we know that
ω=2πf
f=‌

ω

2π

=‌

104

2π

Hz=‌

1

2π

×104Hz