Physics Test

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Physics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B as shown in the figure. Then the capacitance of the system is :

A
_ϵ0A/d

B
2_ϵ0A/d

C
3_ϵ0A/d

D
4_ϵ0A/d

Solution

There are two capacitors with capacitance \(C_{1}=C_{2}=\frac{A \epsilon_{0}}{d}\)
As they are in parallel so the net capacitance \(C=C_{1}+C_{2}=\frac{A \epsilon_{0}}{d}+\frac{A \epsilon_{0}}{d}=\frac{2 A \epsilon_{0}}{d}\)
Hence, the correct option is (B)
Question 2
1 / -0

n equal capacitors are first connected in series and then in parallel. The ratio of capacitances in series and parallel arrangements will be :

A
n

B
n²

C
1/n

D
1/n²

Solution

When n capacitors are connected in series, the equivalent capacitance is \(C_{\text {series}}=\frac{C}{n}\) and when n capacitors are connected in parallel, the equivalent capacitance is \(C_{\text {parallel }}=n C\) \(\frac{C_{\text {series}}}{C_{\text {parallel}}}=\frac{\frac{C}{n}}{n C}=\frac{1}{n^{2}}\)
Hence, the correct option is (D)
Question 3
1 / -0

The minimum number of capacitors each of 3 μF required to make a circuit with an equivalent capacitance 2.25 μF is :

A
3

B
4

C
5

D
6

Solution

There are four capacitors in which 3 capacitors are parallel and in series with 1 capacitor. Net Capacitance for 3 parallel capacitors \(=9 \mu F\) Equivalent Capacitance for 3 parallel capacitors and in series with 1 capacitor \(=\frac{9 \times 3}{9+3}=2.25 \mu F\)
Hence, the correct option is (B)
Question 4
1 / -0

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is 3.75μF. The capacity of each capacitor is

A
5μF

B
6μF

C
7μF

D
8μF

Solution

Let the capacitance be \(C\) when connected parallel, \(C_{1}=C+C+C=3 C\) when other was connected in series So \(\frac{1}{C_{f}}=\frac{1}{3.75}=\frac{1}{3 C}+\frac{1}{C}\)
\(C=5 \mu F\)
Hence, the correct option is (A)
Question 5
1 / -0

A spherical drop of capacitance 1 μF is broken into 8 drops of equal radius. Then, the capacitance of each small drop is:

A
1/2μF

B
1/4μF

C
1/8μF

D
8μF

Solution

Potential of a charged sphere is given by the formula
\(V=\frac{k Q}{R}\)
Hence, by \(Q=C V\),
\(C=\frac{Q}{V}=\frac{R}{k}\)
As per given information, \(\frac{R}{k}=1 \mu F\)
After splitting,
\(8\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}\)
\(r=\frac{R}{2}\)
Thus, capacitance of each small drop is \(V=\frac{r}{k}=\frac{1}{2} \frac{R}{k}=\frac{1}{2} \mu F\)
Hence, the correct option is (A)
Question 6
1 / -0

Three capacitors of capacitance 1.0,2.0 and 5.0μF are connected in series to a 10V source. The potential difference across the 2.0μF capacitor is

A
100/17 V

B
20/17 V

C
50/17 V

D
10 V

Solution

\(C_{e q}=\frac{10}{17} \mu F, Q=\frac{100}{17} \mu C\)
\(\therefore\) Potential difference across \(2 \mu F\) capacitor \(=\frac{\frac{100}{17} \mu C}{2 \mu F}=\frac{50}{17} V\)
Hence, the correct option is (C)
Question 7
1 / -0

A person has only two capacitors. By using them singly, in parallel or in series, he is able to obtain the capacitance 2μF,3μF,6μF and 9μF. What are the capacitance of the capacitors?

A
2μF and 3μF

B
3μF and 6μF

C
6μF and 9μF

D
2μF and 9μF

Solution

For \((\mathrm{A}), C_{e q}^{s}=\frac{2 \times 3}{2+3}=\frac{6}{5} \mu F\)
\(C_{e q}^{p}=2+3=5 \mu F\)
For \((\mathrm{B}), C_{e q}^{s}=\frac{3 \times 6}{3+6}=2 \mu F\)
\(C_{e q}^{p}=3+6=9 \mu F\)
For (C), \(C_{e q}^{s}=\frac{6 \times 9}{6+9}=\frac{18}{5} \mu F\)
\(C_{e q}^{p}=6+9=15 \mu F\)
For (D), \(C_{e q}^{s}=\frac{2 \times 9}{2+9}=\frac{18}{11} \mu F\)
\(C_{e q}^{p}=2+9=11 \mu F\)
Thus only (B) will give the capacitances which are expectable.
Hence, the correct option is (B)
Question 8
1 / -0

Two condensers each of capacitance 2μF are connected in parallel and this combination is connected in series with a 12μF capacitor. The resultant capacity of the system will be:

A
16μF

B
13μF

C
6μF

D
3μF

Solution

When two capacitors of capacitance \(2 \mu F\) are in parallel, the equivalent capacitance is \(C_{e q}=2+2=4 \mu F .\) Now it is connected to \(12 \mu F\) in series. So the capacitance of system is \(C_{s y s}=\frac{4 \times 12}{4+12}=3 \mu F\)
Hence, the correct option is (D)
Question 9
1 / -0

A charge Q is divided into two parts. The two charges kept at a distance apart have a maximum columbian repulsion. Then the ratio of Q and one of the parts is given by :

A
1:4

B
1:2

C
2:1

D
4:1

Solution

Total charge is \(\mathrm{Q}\) Now let it is divided into q and (Q-q) Force of repulsion is \(F=\frac{q(Q-q)}{r^{2}}\)
For \(\mathrm{F}\) to be maximum \(\frac{d F}{d q}=0\) \(\Rightarrow \frac{Q-q}{r^{2}}-\frac{q}{r^{2}}=0\)
\(\frac{Q}{q}=2\)
Hence, the correct option is (C)
Question 10
1 / -0

Two equally charged identical metal spheres A and B repel each other with a force F. Another identical uncharged sphere C is touched to A and then placed midway between A and B. The net force on C is in the direction:

A
F towards A

B
F towards B

C
2F towards A

D
2F towards B

Solution

The spheres are equally charged. Let the charge be \(q\). Now, using coulombs law, \(k q^{2} / r^{2}=F(r\) is the distance between the two spheres ) When the uncharged sphere touches \(A, q / 2\) charge gets transferred to C Now, force on \(C\) \(=\frac{k(q / 2)(q / 2)}{(r / 2)^{2}}-\frac{k(q / 2)(q)}{(r / 2)^{2}}=k q^{2} / r^{2}-2 k q^{2} / r^{2}=-F\)
Hence, the correct option is (A)

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Physics Test - 32

Result

Physics Test - 32

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Question 1

ϵ0A/d

2ϵ0A/d

3ϵ0A/d

4ϵ0A/d

Solution

Question 2

n

n2

1/n

1/n2

Solution

Question 3

3

4

5

6

Solution

Question 4

5μF

6μF

7μF

8μF

Solution

Question 5

1/2μF

1/4μF

1/8μF

8μF

Solution

Question 6

100/17 V

20/17 V

50/17 V

10 V

Solution

Question 7

2μF and 3μF

3μF and 6μF

6μF and 9μF

2μF and 9μF

Solution

Question 8

16μF

13μF

6μF

3μF

Solution

Question 9

1:4

1:2

2:1

4:1

Solution

Question 10

F towards A

F towards B

2F towards A

2F towards B

Solution

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_ϵ0A/d

2_ϵ0A/d

3_ϵ0A/d

4_ϵ0A/d

n²

1/n²