Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Two capacitors, each having capacitance $40 \mu \mathrm{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $\mathrm{K}$ such that the equivalence capacitance of the system became $24 \mu \mathrm{F}$ . The value of $\mathrm{K}$ will be :
Option: 1
$1.5$

Option: 2
$2.5$

Option: 3
$1.2$

Option: 4
$3$

Answers (1)

best_answer

$\mathrm{C_{e q}=\frac{c \times c^{\prime}}{c+c^{\prime}}=\frac{k c}{(k+1)} }$

$\mathrm{24 \mu F=\frac{k(40 \mu F)}{k+1} }$

$\mathrm{\frac{3}{5}=\frac{k}{k+1} }$

$\mathrm{3 k+3=5 k }$

$\mathrm{3=2 k }$

$\mathrm{k=\frac{3}{2}=1.5}$

The value of $\mathrm{k}$ is $\mathrm{1.5}$

Hence 1 is correct option.

Posted by

Gautam harsolia

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question