Physics Test

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Physics Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

The capacitance of a parallel plate capacitor with air as medium is 3 $μ F$ . With the introduction of a dielectric medium between the plates, the capacitance becomes 15 $μ F$ . The permittivity of the medium is

A
5 C² N^-1m^-2

B
15 C² N^-1m^-2

C
0.44×10^-10 C² N^-1 m^-2

D
8.854×10^-11 C² N^-1 m^-2

Solution

Explanation:
The capacitance of air capacitor
\[
C_{0}=\frac{A \varepsilon_{0}}{d}=3 \mu \mathrm{F}
\]
When a dielectric of permittivity $\varepsilon_{r}$ or dielectric constant $K$ is introduced between the plates of the capacitor, then its capacitance
\[
C=\frac{K A \varepsilon_{0}}{d}=15 \mu \mathrm{F} \ldots(\mathrm{ii})
\]
Dividing Eq. (ii) by Eq. (i)
\[
\begin{aligned}
\frac{C}{C_{0}} &=\frac{\frac{K A_{0}}{d}}{\frac{d}{d}}=\frac{15}{3} \\
& \therefore K=5
\end{aligned}
\]
Permittivity of the medium
\[
\begin{aligned}
\varepsilon=& \varepsilon_{0} K \\
&=8.854 \times 10^{-12} \times 5 \\
&=44.27 \times 10^{-12} \\
=& 0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}
\end{aligned}
\]
Question 2
1 / -0

The half-life period of radium is $1600 y r .$ The fraction of a sample of radium that would remain after $6400 y r$ is

A
1/4

B
1/2

C
1/8

D
1/16

Solution

Relation Between Half Life and Fraction of Nuclei Remaining UndecayedIt is often necessary to determine the fraction of radioactive material (or activity) that remains after a specific elapsed time. If the elapsed time is one half-life, the remaining fraction, f, is always 0.5 . If the elapsed time is not one half-life, the remaining fraction is the fraction 0.5 multiplied by itself the number of times corresponding to the number of half-lives. For example,
1 half-life, f = 0.5
2 half-lives, f = (0.5) x (0.5) = 0.25
3 half-lives, f = (0.5) x (0.5) x (0.5) = 0.125
4 half-lives, f = (0.5) x (0.5) x (0.5) x (0.5) = 0.0625.
This can be expressed as
f = (0.5)^t/T = (0.5)ⁿ where t is the elapsed time, T is the half-life, and n is the number of half-lives.
Question 3
1 / -0

A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at a distance $3 \mathrm{r}$ from the centre is $\mathrm{V}$

then the electric field intensity at distance $3 \mathrm{r}$ from the centre is -

A
$\frac{V}{3 r}$

B
$\frac{V}{4 r}$

C
$\frac{V}{6 r}$

D
$\frac{V}{2 r}$

Solution
Question 4
1 / -0

The resistance of a wire is R ohm. If it is melted and stretched to n times its orginal length, its new resistance will be

A
R/n

B
n²R

C
R/n²

D
nR

Solution
Question 5
1 / -0

An unknown resistance R₁ is connected in series with a resistance of 10 $Ω$ . This combination is connected to one gap of meter bridge while a resistance R₂ is connected in the other gap. The balance point is at 50 cm, Now, when the 10 $Ω$ resistance is removed the balance point shifts to 40 cm. The value of R₁(in ohm) is

A
$20 Ω$

B
$10 Ω$

C
$60 Ω$

D
$40 Ω$

Solution
Question 6
1 / -0

A capacitor has capacitance C and reactance X if capacitance and frequency become double, then reactance will be

A
4X

B
X/2

C
X/4

D
2X

Solution

The reactance of capacitorX=1ωCwhereωis frequency and C is the capacitance of the capacitor.

New reactance x' = x/4
Question 7
1 / -0

In a Young’s double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of interference maxima observed within the central maximum of the single slit diffraction pattern is :

A
3

B
6

C
24

D
12

Solution
Question 8
1 / -0

Electromagnetic waves travel in a medium with a speed of 2 × 10⁸ m s⁻¹.The relative permeability of the medium is 1. Find the relative permittivity.

A
2.25

B
3.25

C
3.70

D
4.20

Solution

Given
Speed of electromagnetic waves, $\mathrm{v}=2 \times 10^{8} \mathrm{m} / \mathrm{s}$
Relative permeability of the medium $=1$ Speed of light, $C=3 \times 10^{8} \mathrm{m} / \mathrm{s}$
Using the formula of speed of electromagnetic wave in a medium,
\[
\begin{array}{l}
\mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}} \\
\mathrm{v}=\frac{1}{\sqrt{\mu_{0} \mu_{r}\left(\varepsilon_{0} \varepsilon_{r}\right)}} \\
\mathrm{v}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \times \frac{1}{\sqrt{\mu_{r} \varepsilon_{r}}}
\end{array}
\]
Therefore,
Relative permittivity,
\[
\begin{aligned}
\varepsilon_{r} &=\frac{c^{2}}{v^{2} \mu_{r}} \\
&=\frac{\left(3 \times 10^{8}\right)^{2}}{\left(2 \times 10^{8}\right)^{2} \times 1} \\
&=2.25
\end{aligned}
\]
Question 9
1 / -0

An $α$ particle of mass m is in equilibrium in vertically upward electric field E. The charge on a particle is

A
mg/E

B
2mg/E

C
3mg/E

D
mg/4E

Solution

mg/E
Question 10
1 / -0

Half life of a radioactive element is 3 months. What will be the fraction of that element leftover after one year?

A
75%

B
50%

C
25%

D
6.25%

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Physics Test - 8

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Physics Test - 8

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Question 1

5 C2 N-1 m-2

15 C2 N-1 m-2

0.44×10-10 C2 N-1 m-2

8.854×10-11 C2 N-1 m-2

Solution

Question 2

1/4

1/2

1/8

1/16

Solution

Question 3

V3r

V4r

V6r

V2r

Solution

Question 4

R/n

n2R

R/n2

nR

Solution

Question 5

20 Ω

10 Ω

60 Ω

40 Ω

Solution

Question 6

4X

X/2

X/4

2X

Solution

Question 7

3

6

24

12

Solution

Question 8

2.25

3.25

3.70

4.20

Solution

Question 9

mg/E

2mg/E

3mg/E

mg/4E

Solution

Question 10

75%

50%

25%

6.25%

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5 C² N^-1m^-2

15 C² N^-1m^-2

0.44×10^-10 C² N^-1 m^-2

8.854×10^-11 C² N^-1 m^-2

$\frac{V}{3 r}$

$\frac{V}{4 r}$

$\frac{V}{6 r}$

$\frac{V}{2 r}$

n²R

R/n²

$20 Ω$

$10 Ω$

$60 Ω$

$40 Ω$