JEE Main 18 March 2021 Shift 1 Solved Paper

Given,
The area of the parallel plate capacitor, A=100m2
The distance between the parallel plate of the capacitor, d=10m
The dielectric constant of the material, K=10
The thickness of the space between the plates,
t=5m
As we know, the expression of the capacitor with dielectric constant
C=‌

Aε0

(d−t+‌ t K )

⇒‌‌C=‌

100×8.85×10−12

(10−5+‌ 5 10 )

⇒‌‌C=161×10−12F=161pF
Hence, the resultant capacitance of the system is 161pF.
So, the value of x to the nearest integer is 161 .