Q41JEE Main 2024MCQ

A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is $\;\frac{A}{3}$ and the height is ${\text{d}}$ , the capacitance of the arrangement is:

🚀 Solve in Practice Mode

📖 Explanation

All capacitor are in parallel combination.Also effective area is common area only $\;\Rightarrow C_{ {\text{eq}}}=C_1+C_2+C_3$ $\;\Rightarrow C_{ {\text{eq}}}=\;{A \epsilon _0}/{3 {\text{d}}}+\;{ {\text{A}} \epsilon _0}/{3(2 {\text{d}})}+\;{ {\text{A}} \epsilon _0}/{3(3 {\text{d}})}$ $\;\Rightarrow {\text{C}}_{ {\text{eq}}}=\;{ {\text{A}} \epsilon _0}/{3} (\;{11}/{6 {\text{d}}})$ $\;\Rightarrow {\text{C}}_{ {\text{eq}}}=\;{11 {\text{A}} \epsilon _0}/{18 {\text{d}}}$

Q43JEE Main 2024MCQ

Two identical capacitors have same capacitance $C$ . One of them is charged to the potential V and other to the potential $2 {\text{V}}$ . The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is : [1-Feb-2024 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{V}}_{ {\text{C}}}=\;{ {\text{q}}_{\;\text{net }\;}}/{ {\text{C}}_{\;\text{net }\;}}=\;{ {\text{CV}}+2 {\text{CV}}}/{2 {\text{C}}}$ ${\text{V}}_{ {\text{C}}}=\;{3 {\text{V}}}/{2}$ Loss of energy $\;=\;\frac{1}{2} {\text{CV}}^{2+}\;\frac{1}{2} {\text{C}}(2 {\text{V}})^{2-}\;\frac{1}{2} 2 {\text{C}} (\;{3 {\text{V}}}/{2}) ^2$ $\;= (\;{ {\text{CV}}^2}/{4})$

Q44JEE Main 2023NAT

A capacitor of capacitance $900 \mu {\text{F}}$ is charged by a $100 {\text{V}}$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-2} {\text{J}}$ . The value of $x$ is ______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{C}}=900 \mu {\text{F}}$ $\; {\text{Q}}= {\text{CV}}=900 \times 10^{-6} \times 100=9 \times 10^{-2}=90 {\text{MC}}$ NowCommon potential will be developed across both capacitors by ${\text{kVL}}$ Total charge on left plates of capacitors should be conserved. $\therefore \;\; 90 {\text{mc}}+0=2 {\text{cv}}_0$ $\; {\text{cv}}_0=45 {\text{mc}}$ Heat dissipated $= {\text{U}}_{ {\text{i}}}- {\text{U}}_{ {\text{f}}}$ [Change in energy stored in the capacitors] $\;=\;\frac{1}{2} \;{(90 {\text{mc}})^2}/{900 \mu {\text{F}}}-2 \times \;\frac{1}{2} \;{(45 {\text{mc}})^2}/{900 \mu {\text{F}}} [ {\text{U}}=\;{ {\text{Q}}^2}/{2 {\text{c}}}]$ $\;=\;\frac{1}{2 \times 900 \times 10^{-6}}(8100-4050) \times 10^{-6}$ $\;=2.25 {\text{Joule}}$ OR $\;\;\text{ Heat }\;=\;\frac{1}{2} \;{ {\text{C}}_1 {\text{C}}_2}/{ {\text{C}}_1+ {\text{C}}_2} ( {\text{V}}_1- {\text{V}}_2) ^2$ $\;=\;\frac{1}{2} \;{ {\text{C}}^2}/{2 {\text{C}}}(100-0)^2$ $\;=\;\frac{1}{2} \;\frac{900 \times 10^{-6}}{2} \times 10^4=\;\frac{9}{4} \;\text{ Joule }\;=2.25 \;\text{ Joule }\;$

Q45JEE Main 2023MCQ

In this figure the resistance of the coil of galvanometer ${\text{G}}$ is $2 Ω$ . The emf of the cell is $4 {\text{V}}$ . The ratio of potential difference across $C_1$ and $C_2$ is: [8-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

At steady state current will not be in the capacitor branch. $\; {\text{i}}=\;\frac{4}{6+2+8}$ $\; {\text{i}}=\;\frac{1}{4} {\text{A}}$ $\;\Delta {\text{V}}_{ {\text{C}}_1}= {\text{i}}(6+2)$ $\;\Delta {\text{V}}_{ {\text{C}}_2}= {\text{i}}(2+8)$ $\;\;{\Delta {\text{V}}_{ {\text{C}}_1}}/{\Delta {\text{V}}_{ {\text{C}}_2}}=\;\frac{4}{5}$

Q46JEE Main 2023MCQ

In the network shown below, the charge accumulated in the capacitor in steady state will be : [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

In steady state, no current will pass through capacitor hence capacitor will act as open circuit. $\; {\text{i}}_2=0$ $\; {\text{i}}_1=\;\frac{3}{6+4}=\;\frac{3}{10} {\text{A}}$ Potential difference on $6 Ω$ resistor $=6 \times \;\frac{3}{10}=1.8$ volt capacitor will have same potential so charge $= {\text{cv}}=4 \times 1.8=7.2 \mu {\text{c}}$

Q47JEE Main 2023MCQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one. Reason R: Capacitance of metallic spheres depend on the radii of spheres. In the light of the above statements, choose the correct answer from the options given below. [1-Feb-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Potential of a conducting sphere is $\; {\text{V}}=\;{ {\text{KQ}}}/{ {\text{R}}} \;\text{ (Solid as well as hollow) }\;$ $\; {\text{V}}_1= {\text{V}}_2 \;\text{ and }\; {\text{R}}_1= {\text{R}}_2$ $\; \therefore {\text{Q}}_1= {\text{Q}}_2$

Q48JEE Main 2023NAT

A parallel plate capacitor with plate area ${\text{A}}$ and plate separation ${\text{d}}$ is filed with a dielectric material of dielectric constant ${\text{K}}=4$ . The thickness of the dielectric material is ${\text{x}}$ , where ${\text{x}} < {\text{d}}$ .Let $C_1$ and $C_2$ be the capacitance of the system for $x=\;\frac{1}{3} d$ and $x=\;\frac{2 d}{3}$ , respectively. If $C_1=2 \mu {\text{F}}$ the value of ${\text{C}}_2$ is _________ $\mu {\text{F}}$ [6-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$C_1=$ $\;{\;\frac{E_0 A}{\frac{2 d}{3}} \times \frac{4 E_0 A}{\frac{d}{3}}}/{\frac{E_0 A}{2 d / 3}+\frac{4 E_0 A}{d / 3}}$ $=\;\frac{18}{\frac{3}{2}+12} \;\frac{E_0 A}{d}=18 \times \;\frac{2}{27} \;\frac{E_0 A}{d}=\;\frac{4}{3} \;\frac{E_0 A}{d}$ According to ${\text{qn}}, \;\frac{4}{3} \;\frac{E_0 A}{d}=2 \Rightarrow \;\frac{E_0 A}{d}=\;\frac{3}{2}----(i)$ Now, $C_2=$ $\;{\;\frac{E_0 A}{\frac{d}{3}} \times \frac{4 E_0 A}{\frac{2 d}{3}}}/{\frac{E_0 A}{d / 3}+\frac{4 E_0 A}{2 d / 3}}$ $=\;\frac{18}{3+6} \;\frac{E_0 A}{d}=2 \times \;\frac{E_0 A}{d}=2 \times \;\frac{3}{2}=3$

Q49JEE Main 2023NAT

For the given circuit, in the steady state, $| {\text{V}}_{ {\text{B}}}- {\text{V}}_{ {\text{D}}}|=$ [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

In steady state, capacitor behaves as an open circuit. Circuit is : $\Rightarrow i_{A B}=\;\frac{6}{3}=2 {\text{A}}$ $i_{A D}=\;\frac{6}{12}=0.5 {\text{A}}$ $\Rightarrow V_B+2 \times 2-10 \times 0.5=V_D$ $\Rightarrow V_B-V_D=1 \;\text{ volt }\;$

Q50JEE Main 2023NAT

In the given figure the total charge stored in the combination of capacitors is $100 \mu {\text{C}}$ . The value of ' ${\text{x}}$ ' is ________ [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

All the capacitors are in parallel $\; {\text{C}}_{ {\text{eq}}}=(5+ {\text{x}}) \mu {\text{F}}$ $\; {\text{Q}}= {\text{C}}_{ {\text{aq}}} {\text{V}}$ $\;100=(5+ {\text{x}})(10)$ $\; {\text{x}}=5$

Q51JEE Main 2023NAT

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu {\text{F}}$ are individually charged by a $100 {\text{V}}$ D.C. source. Capacitor ${\text{C}}_1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor ${\text{C}}_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor ${\text{C}}_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _______ V. (Assuming Dielectric constant $=10$ ) [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Charge on ${\text{C}}_1= {\text{KCV}}$ And charge on ${\text{C}}_2= {\text{CV}}$ When they are connected in parallel charge will be equally divided so charge on one capacitor is $q=\;\frac{K+1}{2} {\text{CV}}$ So, $V=\;\frac{q}{K C}=\;\frac{K+1}{2 K}=55 {\text{V}}$

Q52JEE Main 2023NAT

A capacitor has capacitance $5 \mu {\text{F}}$ when it's parallel plates are separated by air medium of thickness $d$ . A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\;{ {\text{d}}}/{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be _______ $\mu {\text{F}}.$ [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{C}}_{\;\text{new }\;}=\;{E_0 {\text{A}}}/{{ (\;{ {\text{d}}}/{2}) }/{1.5}+\;{ ({ {\text{d}}}/{2}) }/{1}}$ $\;=\;{E_0 A}/{ (\;{ {\text{d}}}/{3}+\;{ {\text{d}}}/{2}) }=\;{6 E_0 {\text{A}}}/{5 {\text{d}}}$ $\;=\;\frac{6}{5} \times 5 \mu {\text{F}}=6 \mu {\text{F}}$ $\;$

Q53JEE Main 2023NAT

In the circuit shown, the energy stored in the capacitor is $n \mu J$ . The value of $n$ is [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{I}}_1=\;\frac{12}{3+9}=1 {\text{A}}$ $\; {\text{I}}_2=\;\frac{12}{4+2}=2 {\text{A}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{C}}}=3 {\text{I}}_1=3 {\text{V}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{D}}}=2 \times 4=8 {\text{V}}$ $\;\;\text{ So, }\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{D}}}=5 {\text{V}}$ $\; {\text{U}}=\;\frac{1}{2} {\text{CV}}^2=\;\frac{1}{2} \times 6 \times 5^2=75 \mu {\text{J}}$

Q54JEE Main 2023MCQ

A capacitor of capacitance ${\text{C}}$ is charge to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is : [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\varphi =\;{ {\text{q}}}/{\epsilon _0}$ $\;\varphi =\;{ {\text{CV}}}/{\epsilon _0}$

Q55JEE Main 2023MCQ

A parallel plate capacitor has plate area $40 {\text{cm}}^2$ and plates separation $2 {\text{mm}}$ . The space between the plates is filled with a dielectric medium of a thickness $1 {\text{mm}}$ and dielectric constant 5 . The capacitance of the system is : [25-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

This can be seen as two capacitors in series combination so $\;\;{1}/{ {\text{C}}_{ {\text{eq}}}}=\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}$ $\;=\;{1}/{{ {\text{K}} \in _0 {\text{A}}}/{ {\text{t}}}}+\;{1}/{{E_0 {\text{A}}}/{ {\text{d}}- {\text{t}}}}$ $\;=\;{ {\text{t}}}/{ {\text{K}} \in _0 {\text{A}}}+\;{ {\text{d}}- {\text{t}}}/{E_0 {\text{A}}}$ $\;=\;\frac{1 \times 10^{-3}}{5 E_0 \times 40 \times 10^{-4}}+\;\frac{1 \times 10^{-3}}{E_0 40 \times 10^{-4}}$ $\;\;{1}/{ {\text{C}}_{ {\text{eq}}}}=\;\frac{1}{20 E_0}+\;\frac{1}{4 E_0}$ $\; {\text{C}}_{ {\text{eq}}}=\;\frac{20 \times 4 E_0}{24}=\;\frac{10 E_0}{3} {\text{F}}$

Q56JEE Main 2023NAT

As shown in the figure, two parallel plate capacitors having equal plate area of $200 {\text{cm}}^2$ are joined in such a way that $\alpha \ne b$ . The equivalent capacitance of the combination is ${\text{x}} \in _0$ F. The value of ${\text{x}}$ is _______. [6-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

As per the arrangement given, distance between the capacitor plates are ${\text{a}}$ and ${\text{b}}$ and ${\text{a}} \ne {\text{b}}$ using the diagram we can write ${\text{b}}=5- {\text{a}}-1=(4-a)$ in ${\text{mm}}$ as we know capacitance of capacitor $C=\;\frac{\epsilon _0 A}{d}$ and in series arrangement $\;{1}/{ {\text{C}}_{\;\text{eq }\;}}=\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}$ $\;{1}/{ {\text{C}}_{\;\text{eq }\;}}=\;{ {\text{a}}}/{\epsilon _0 {\text{A}}}+\;{4- {\text{a}}}/{\epsilon _0 {\text{A}}}=\;{4(\;\text{ \in mm }\;)}/{\epsilon _0 {\text{A}}}$ or ${\text{C}}_{ {\text{eq}}}=\;{\epsilon _0 {\text{A}}}/{4( {\text{mm}})}$ Given ${\text{A}}=200 {\text{cm}}^2$ ${\text{C}}_{ {\text{eq}}}=\;\frac{\epsilon _0 \times 200 \times 10^{-4}}{4 \times 10^{-3}}$ $=\epsilon _0 50 \times 10^{-1}$ or ${\text{C}}_{ {\text{eq}}}=5 \epsilon _0$ farad therefore ${\text{n}}=5$

Q57JEE Main 2023MCQ

The equivalent capacitance of the combination shown is [10-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Capacitor (3) & (4) are short ckt $\therefore {\text{C}}_1 & {\text{C}}_2$ are in parallel ${\text{C}}_{ {\text{eq}}}= {\text{C}}+ {\text{C}}=2 {\text{C}}$

Q58JEE Main 2023MCQ

A parallel plate capacitor of capacitance $2 {\text{F}}$ is charged to a potential ${\text{V}}$ , The energy stored in the capacitor is ${\text{E}}_1$ . The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is ${\text{E}}_2$ . The ratio ${\text{E}}_2 / {\text{E}}_1$ is : [11-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{C}}=2 {\text{F}}$ ${\text{E}} 1=\;\frac{1}{2} {\text{CV}}^2$ . . . (1)Now $\; {\text{V}}_{ {\text{C}}}=\;{ {\text{C}}_1 {\text{V}}_1+ {\text{C}}_2 {\text{V}}_2}/{ {\text{C}}_1+ {\text{C}}_2}$ $\; {\text{V}}_{ {\text{C}}}=\;{ {\text{CV}}+ {\text{O}}}/{2 {\text{C}}}=\;{ {\text{V}}}/{2}$ $\; {\text{E}}_2= {\text{CV}}_{ {\text{C}}}{ }^2= {\text{C}} \cdot \;{ {\text{V}}^2}/{4}$ . . . (2) $\;\;{ {\text{E}}_2}/{ {\text{E}}_1}=\;{ ({ {\text{CV}}^2}/{4}) }/{ (\;{ {\text{CV}}^2}/{2}) }=\;\frac{2}{1} \;\; \;\text{ option }\; \to (4)$

Q59JEE Main 2023NAT

In the given circuit, ${\text{C}}_1=2 \mu {\text{F}}, {\text{C}}_2=0.2 \mu {\text{F}}, {\text{C}}_3=2 \mu {\text{F}}, {\text{C}}_4=4 \mu {\text{F}}, {\text{C}}_5=2 \mu {\text{F}}, {\text{C}}_6=2 \mu {\text{f}}$ , The charge stored on capacitor ${\text{C}}_4$ is ________ $\mu {\text{C}}$ . [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{C}}_{ {\text{eq}}}=0.5 \mu {\text{F}}$ $\; {\text{Q}}=0.5 \times 10=5 \mu {\text{C}}$ $\; {\text{Q}}^{'}=\;{5 \mu {\text{C}} \times 0.8}/{0.8+0.2}=4 \mu {\text{C}}$

Q60JEE Main 2023NAT

A $600 {\text{pF}}$ capacitor is charged by $200 {\text{V}}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 {\text{pF}}$ capacitor. Electrostatic energy lost in the process is _______ $\mu {\text{J}}$ [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\text{ loss of strength }\;=\;\frac{1}{2} \;{ {\text{c}} \times {\text{c}}}/{ {\text{c}}+ {\text{c}}} ( {\text{v}}_1- {\text{v}}_2) ^2$ $\;=\;\frac{1}{2} \times \$ [;\frac{600 × 10^{-12}}{2}]$ \times (200)^2;=600 \times 10^{-12} \times 10^4=6 \times 10^{-6}=6 \mu {\text{J}}$