Physics Test

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Physics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

A parallel plate capacitor has a capacity 80×10⁻⁶ F when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 V by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :

A
45.6×10⁻³ C

B
25.3×10⁻³ C

C
120×10⁻³ C

D
12×10⁻³ C

Solution

Charge passing through the wire, \(\Delta q=\Delta C V\)
\(=\left(C^{\prime}-C\right) V\)
\(=(k-1) C V\)
\(=(20-1)\left(80 \times 10^{-6}\right)(30)\)
\(=45.6 \times 10^{-3}\)
So, charge passing through the wire \(=45.6 \times 10^{-3} \mathrm{C}\)
Hence, the correct option is (A)
Question 2
1 / -0

Consider the circuit given here. The potential difference VBC between the points B and C is :

A
1 V

B
0.5 V

C
-1 V

D
0

Solution

\(i=\frac{3}{6 \times 10^{3}}=0.5 \times 10^{-3}\)
\(V_{A D}=i R=0.5 \times 10^{-3} \times 3 \times 10^{3}=1.5 \mathrm{V}\)
\(Q=\frac{2 \times 1.5}{3}=2 \times 0.5=1 \mu C\)
\(\mathrm{KVL}\) from \(B\) to \(C\) \(V_{B}-0.5 \times 10^{-3} \times 10^{3}+\frac{1}{2}=V_{C}\)
\(V_{B}-V_{C}=1-\frac{1}{2}=0.5 V\)
Hence, the correct option is (B)
Question 3
1 / -0

Four metallic plates are arranged as shown in the figure. If d is the distance between each plate then capacitance of the given system between points A and B is :

(given d<<A}

A
_ϵ0A/d

B
2_ϵ0A/d

C
3_ϵ0A/d

D
4_ϵ0A/d

Solution

There will have two capacitors in between A and B and they are in parallel combination. The capacitance of parallel plate capacitor is \(C=\frac{A \epsilon_{0}}{d}\)
The capacitance in between \(\mathrm{A}\) and \(\mathrm{B}\) is \(C_{A B}=C+C=2 C=\frac{2 A \epsilon_{0}}{d}\)
Hence, the correct option is (B)
Question 4
1 / -0

Charge Q taken from the battery of 12V in the circuit is :

A
72μC

B
36μC

C
156μC

D
20μC

Solution

The equivalent capacitance of the circuit is, \(C=4+\frac{3 \times 6}{3+6}=4+2=6 \mu F\)
Thus, Charge taken from the battery is, \(Q=C V=6 \times 12=72 \mu C\)
Hence, the correct option is (A)
Question 5
1 / -0

The potential difference between the plate of a capacitor is increased by 20%. The energy stored on the capacitor increases by:

A
20%

B
22%

C
40%

D
44%

Solution

Energy stored in a capacitor with potential difference \(V\) applied across it is \(\frac{1}{2} C V^{2}=E_{0}\) When potential difference increases by \(20 \%\), \(V^{\prime}=1.2 V\)
\(\Longrightarrow E^{\prime}=\frac{1}{2} C V^{\prime 2}=\frac{1}{2} C\left(1.2^{2} V^{2}\right)\)
\(\Longrightarrow E^{\prime}=1.2^{2} E_{0}=1.44 E_{0}\)
Thus the energy stored increases by \(44 \%\)
Hence, the correct option is (D)
Question 6
1 / -0

Area of each of the conducting plates 1, 2, 3, 4, 5 and 6 is A. The system is kept in air. Find the the capacitance between A and B (1 & 4, 2 & 5 and 3 & 6 are pairs of parallel plates).

A
3_ε0A/d

B
23_ε0A/d

C
32_ε0A/d

D
_ε0A3/d

Solution

The given figure can be simplified as shown in the figure . Hence, \(C_{A B}=\frac{1}{\frac{1}{C}+\frac{1}{C+C}}\)
\(\frac{C \times 2 C}{C+2 C}=\frac{2}{3} C\)
\(=\frac{2}{3} \frac{\varepsilon_{0} A}{d}\)
Hence, the correct option is (B)
Question 7
1 / -0

The equivalent capacitance between x and y is:

A
5/6 μF

B
7/6μF

C
8/3μF

D
4μF

Solution

We have, Series capacitance, \(\frac{1}{C_{S}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots+\frac{1}{C_{n}}\)
Parallel capacitance, \(C_{P}=C_{1}+C_{2}+\ldots+C_{n}\)
Hence, the correct option is (D)
Question 8
1 / -0

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is-

A
E²Ad/_ε0

B
12_ε0E²Ad

C
_ε0EAd

D
1/2_ε0E²

Solution

For a capacitor, Total energy \(=\frac{1}{2} C V^{2}=\) energy density \(\times\) volume where \(\mathrm{C}=\frac{\epsilon_{0} A}{d}, V=E d\)
\(=\left(\frac{1}{2} \in_{0} E^{2}\right)(A d)\)
Hence, the correct option is (C)
Question 9
1 / -0

A parallel plate air capacitor has capacity ′C′ farad, potential ′V′ volt and energy ′E′ joule. When the gap between the plates is completely filled with dielectric.

A
Both V and E increase

B
Both V and E decrease

C
V decreases, E increases

D
V increases, E decreases

Solution

A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely proportional to dielectric V′=V/K where K is dielectric constant.
Hence, the correct option is (B)
Question 10
1 / -0

A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be:

A
(K+3)C/4

B
(K+2)C/4

C
(K+1)C/4

D
KC/4

Solution

The two condensers with \(K\) and with air are in parallel. With air, \(C_{1}=\frac{\epsilon_{0}}{d}\left(\frac{3 A}{4}\right)=\frac{3 \epsilon_{0} A}{4 d}\)
With medium, \(C_{2}=\frac{\epsilon_{0} K}{d}\left(\frac{A}{4}\right)=\frac{\epsilon_{0} A K}{4 d}\)
\(\therefore C^{\prime}=C_{1}+C_{2}\)
or \(C^{\prime}=\frac{3 \epsilon_{0} A}{4 d}+\frac{\epsilon_{0} A K}{4 d}=\frac{\epsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]\)
or \(C^{\prime}=\frac{C}{4}(K+3)\left[\because C=\frac{A \epsilon_{0}}{d}\right]\)
Hence, the correct option is (A)