Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
An electrical technician requires a capacitance of 2 F in a circuit across a potential dif- ference of 1KV. A large number of 1 F capacitors are available to him, each of which can withstand a potential difference of not than 400 V. suggest a possible arrangement that requires a minimum number of capacitors.
Two spherical conductors of radii $r_1$ and $r_2$ are at potentials $V_1$ and $V_2$ respectively, then what will be the common potential when the conductors are brought in contant ?
When two spherical conductors are brought into contact, charge is redistributed until they reach the same potential. The common potential $V$ is given by the formula: $$ V = rac{r_1 V_1 + r_2 V_2}{r_1 + r_2} $$ Here, $r_1$ and $r_2$ are the radii of the conductors, and $V_1$ and $V_2$ are their respective potentials.
Capacitance of a parallel plate capacitor becomes 4 /3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is
The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ inserted can be calculated using the formula: $$ C' = rac{ ext{original capacitance} imes ext{dielectric constant}}{1 + rac{t}{d}(K - 1)} $$ Given that $C' = rac{4}{3}C$ and $t = rac{d}{2}$, solving this equation yields a dielectric constant $K$ of 2.5.
The plates of a parallel capacitor are charged up to 100 V. If 2 mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6mm the dielectric constant of the plate is
To maintain the same potential difference, the effective distance $d_{eff}$ between the plates of the capacitor is given by: $$ d_{eff} = d - t + rac{t}{K} $$ Given that $d - t + rac{t}{K} = d + 1.6 ext{ mm}$ and $t = 2 ext{ mm}$, solving this equation yields a dielectric constant $K$ of 5.
A parallel plate air capacitor has a capacitance 18 F . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes 72 F. The dielec- tric constant of the medium is
The capacitors of capacitance 4 F, 6 F and 12 F are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases ?
When capacitors are connected in series, the equivalent capacitance \( C_s \) is given by: $$ rac{1}{C_s} = rac{1}{4} + rac{1}{6} + rac{1}{12} = rac{1}{2} \rightarrow C_s = 2 ext{ F} $$ When capacitors are connected in parallel, the equivalent capacitance \( C_p \) is given by: $$ C_p = 4 + 6 + 12 = 22 ext{ F} $$ The ratio of the equivalent capacitance in series to that in parallel is: $$ rac{C_s}{C_p} = rac{2}{22} = rac{1}{11} ). Therefore, the ratio is 1:11.
Large number of capactors of rating 10 F/200V V are available. The minimum number of capacitors required to design a 10 F/700V capacitor is
To design a 10 F/700V capacitor using capacitors of rating 10 F/200V, we need to ensure the voltage rating is met by arranging capacitors in series and the capacitance requirement is met by arranging these series combinations in parallel. For the voltage rating: $$ rac{700V}{200V} = 3.5 ightarrow 4 ext{ capacitors in series} $$ Each series combination will have a total capacitance of: $$ rac{10F}{4} = 2.5F $$ To achieve the total capacitance of 10 F, we need: $$ rac{10F}{2.5F} = 4 ext{ such series combinations in parallel} $$ Therefore, the minimum number of capacitors required is: $$ 4 imes 4 = 16 $$ Thus, the answer is 16.
A variable condenser is permanently connected to a 100 V battery. If capacitor is changed from 2 F to 10 F . then energy changes is equal to
The energy stored in a capacitor is given by the formula $U = rac{1}{2}CV^2$. Initially, the energy stored in the capacitor is $U_1 = rac{1}{2} imes 2 imes 100^2 = 10^4$ J. After changing the capacitance to 10 F, the energy stored is $U_2 = rac{1}{2} imes 10 imes 100^2 = 5 imes 10^4$ J. The change in energy is $ riangle U = U_2 - U_1 = 5 imes 10^4 - 10^4 = 4 imes 10^4$ J.
1000 similar electrified rain drops merge together into one drop so that their total charge remains unchanged. How is the electric energy affected ?
There are 10 condensers each of capacity 5 F . The ratio between maximum and mini- mum capacities obtained from these condensers will be
The maximum capacitance is obtained when all capacitors are connected in parallel. For 10 capacitors each of capacity 5F, the maximum capacitance is $C_{max} = 10 imes 5 = 50$ F. The minimum capacitance is obtained when all capacitors are connected in series. For 10 capacitors each of capacity 5F, the minimum capacitance is $C_{min} = rac{5}{10} = 0.5$ F. The ratio between maximum and minimum capacities is $ rac{50}{0.5} = 100:1$.
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.