Physics Test

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Physics Test - 20

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Weekly Quiz Competition

Question 1
4 / -1

The resistance of a copper wire is 19.5 ohm. Its length is 8 km and diameter is 3 mm. What is the specific resistance of copper?

A
1.35 x 10-8ohm metre

B
1.72 x 10-8ohm metre

C
4.05 x 10-6ohm metre

D
8.10 x 10-5ohm metre

Solution

Resistance $(\mathrm{R})=19.5 \Omega$
Length $(\ell)=8 \mathrm{km}=8000 \mathrm{m}$
Diameter $=3 \mathrm{mm}=3 \times 10^{-3} \mathrm{m}$
Radius $(r)=3 / 2 \times 10^{-3} \mathrm{m}$
Resistance $(\mathrm{R})=\mathrm{P} \frac{\ell}{\mathrm{A}}=\mathrm{p} \frac{\ell}{\pi \mathrm{r}^{2}}$
Here, $\rho$ is the specific resistance $19.5 \Omega=\rho \frac{(8000)}{\frac{22}{7} \times\left(\frac{3}{2} \times 10^{-3}\right)^{2}}$
$\mathrm{p}=0.01723 \times 10^{-6}$ or $1.72 \times 10^{-8}$ ohm metre.
Question 2
4 / -1

Directions:For the given statements of Assertion and Reason, mark the correct answer as:

(a) if both assertion and reason are true and reason is the correct explanation of assertion

(b) if both assertion and reason are true, but reason is not the correct explanation of assertion

(c) if assertion is true, but reason is false

(d) if both assertion and reason are false

Assertion:Reversible systems are difficult to find in real world.

Reason:Most processes are dissipative in nature.

A
A

B
B

C
C

D
D

Solution

No loss of energy occurs in a perfectly reversible system. There is no such system in the world which calls for zero energy loss due to dissipative nature of most of the natural processes.
Question 3
4 / -1

With the two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

A
$(2+\sqrt{2}) \mathrm{N}$ and $(2-\sqrt{2}) \mathrm{N}$

B
$(2+\sqrt{3}) \mathrm{N}$ and $(2-\sqrt{3}) \mathrm{N}$

C
$\left(2+\frac{1}{2} \sqrt{2}\right) \mathrm{N}$ and $\left(2-\frac{1}{2} \sqrt{2}\right) \mathrm{N}$

D
$\left(2+\frac{1}{2} \sqrt{3}\right) \mathrm{N}$ and $\left(2-\frac{1}{2} \sqrt{3}\right) \mathrm{N}$

Solution

Let $\mathrm{P}$ and $\mathrm{Q}$ are forces. We know that
\[
\mathrm{R}=\sqrt{\mathrm{p}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta}
\]
When $\theta=0^{\circ}, R=4 N$
\[
\begin{array}{l}
\mathrm{R}=4 \mathrm{N}=\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ}} \\
\mathrm{P}+\mathrm{Q}=4
\end{array}
\]
when $\theta=90^{\circ}, \mathrm{R}=3 \mathrm{N}$
\[
P^{2}+Q^{2}=9
\]
From equation (1) $(\mathrm{P}+\mathrm{Q})^{2}=16$
\[
\begin{array}{l}
\mathrm{P}^{2}+\mathrm{Q}^{2}=2 \mathrm{PQ}=16 \\
9+2 \mathrm{PQ}=16 \\
9+2 \mathrm{PQ}+16 \\
2 \mathrm{PQ}=7
\end{array}
\]
Now,
\[
\begin{array}{l}
(P-Q)^{2}=P^{2}+Q^{2}-2 P Q \\
(P-Q)^{2}=9-7 \\
P-Q=\sqrt{2}
\end{array}
\]
On solving equation (i) and equation (iii)
\[
P=\left(2+\frac{1}{2} \sqrt{2}\right) \mathrm{N}
\]
and
\[
Q=\left(2-\frac{1}{2} \sqrt{2}\right) \mathrm{N}
\]
Question 4
4 / -1

The values of the horizontal component of Earth's magnetic field and angle of dip are 1.8 × 10−5Wb/m2and 30o, respectively, at some place. The total intensity of Earth's magnetic field at that place will b

A
3.18 × 10−3Wb/m2

B
3.67 × 10−5Wb/m2

C
5 × 10−3Wb/m2

D
2.08 × 10−5Wb/m2

Solution

Given : $H=1.8 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}$
Angle of dip $\theta=30^{\circ}$ Using the formula
Or
\[
\begin{array}{l}
H=V \cos \theta \\
V=\frac{H}{\cos \theta} \\
=\frac{1.8 \times 10^{-5}}{\cos 30^{\circ}} \\
=\frac{1.8 \times 10^{-5}}{\frac{\sqrt{3}}{2}} \\
V=\frac{1.8 \times 10^{-5} \times 2}{\sqrt{3}}
\end{array}
\]
\[
=2.08 \times 10^{-5} \mathrm{Wb} / \mathrm{m}^{2}
\]
Question 5
4 / -1

The equation of displacement of two waves are given as: $y 1=10 \sin \left(3 \pi t+\frac{\pi}{3}\right), y 2=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) .$ What is the ratio of their amplitudes?

A
1 : 2

B
2 : 1

C
1 : 1

D
None of these

Solution

As $y_{2}=5 \sin 3 \pi t+5 \sqrt{3} \cos 3 \pi t$
This is super position of two waves having phase difference of $\frac{\pi}{2}$ in between them, resultant amplitude of these is given by
$\mathrm{R}=\sqrt{5^{2}+(5 \sqrt{3})^{2}}=\sqrt{25+75}$
$\mathrm{R}=\sqrt{100}=10$ Units and
$y_{1}=10 \sin \left(3 \pi t+\frac{\pi}{3}\right)$
So the ratio of the amplitudes of $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$ is 1: 1
Question 6
4 / -1

Four identical spheres, each of mass 1 kg and radius 10 cm, are placed on a horizontal surface touching one another such that their centres are located at the corners of a square of side 20 cm. What is the distance of the centre of mass from the centre of sphere?

A
5 cm

B
10 cm

C
15 cm

D
$10 \sqrt{2}$ cm

Solution

Applying Pythagorean theorem, As length of side is given $20 \mathrm{cm},$ so it will become height and base. We can find the length of the diagonal. (Diagonal) $^{2}=(\text { Side })^{2}+(\text { Side })^{2}$
Diagonal $=\sqrt{2 \times(20)^{2}}=20 \sqrt{2}$
Distance of CM from any corner is half of the diagonal $=\frac{20 \sqrt{2}}{2}=10 \sqrt{2} \mathrm{cm}$
Question 7
4 / -1

The equation of a wave travelling on a string is $y=4 \sin \left\{\frac{\pi}{2}\left(8 t-\frac{x}{8}\right)\right\}$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in seconds. The velocity of the wave is

A
64 cm/s in - x direction

B
32 cm/s in - x direction

C
32 cm/s in + x direction

D
64 cm/s in + x direction

Solution

Comparing the given equation with standard wave equation travelling in the positive $x$ - axis, we have
$y=A \sin (\omega t-k x)$
Wave velocity
$v=\frac{\omega}{k}$
$v=\frac{4 \pi}{\left(\frac{\pi}{16}\right)}=64 \mathrm{cm} / \mathrm{s}$
Question 8
4 / -1

When a sound wave of frequency 300 Hz passes through a medium, the maximum displacement of a particle of the medium is 0.1 cm. The maximum velocity of the particle is equal to

A
10π cm/s

B
30π cm/s

C
30π cm/s

D
60π cm/s

Solution

frequency of sound wave = 300Hz
amplitude of a particle of the medium = 0.1 cm

we know, formula of maximum velocity = angular frequency × amplitude

angular frequency = 2π × frequency
= 2π × 300 rad/s
= 600π rad/s

now, maximum velocity of wave = 600π rad/s × 0.1cm
= 600π × 0.1/100 m/s
= 0.6π m/s
Question 9
4 / -1

In a LR circuit of $3 \mathrm{mH}$ inductane and $4 \Omega$ resistance, $\operatorname{emf} \mathrm{E}=4 \cos (1000 \mathrm{t})$ volt is applied. The amplitude of current is:

A
0.8 A

B
(4/7) A

C
1.0 A

D
(4/√7)A

Solution

Given:

$ \mathrm{L}=3 \mathrm{mH}=3 \times 10^{-3} \mathrm{H} $

$ \mathrm{R}=4 \Omega$

Emf, $ \mathrm{E}=4 \cos 1000 \mathrm{t}$

$\therefore \mathrm{E}_{\mathrm{o}}=4$ volt and $\omega=1000$

Inductive reactance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=1000 \times 3 \times 10^{-3}=3 \Omega$

Impedance, $ \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{4^{2}+3^{2}}=5 \Omega$

Current amplitude, $ \mathrm{I}=\frac{\mathrm{E}_{\mathrm{o}}}{7}=\frac{4}{5}=0.8 \mathrm{~A}$
Question 10
4 / -1

If a body travels half the distance with velocity v1and the next half with velocity v2, then its average velocity will be given by

A
$v=\sqrt{v_{1}+v_{2}}$

B
$v=\frac{v_{1}+v_{2}}{2}$

C
$v=\frac{v_{1}}{v_{2}}$

D
$\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}$

Solution

Average velocity is defined as the ratio of the total distance travelled to the total time taken.
$v=\frac{2 d}{t_{1}+t_{2}}$
where $t_{1}=\frac{d}{v_{1}}, t_{2}=\frac{d}{v_{2}}$
Substituting the values of $\mathrm{t}_{1}$ and $\mathrm{t}_{2}$, we get
$v=\frac{2 d}{\frac{d}{v_{1}}+\frac{d}{v_{2}}} \Rightarrow v=\frac{2}{\frac{1}{v_{1}}+\frac{1}{v_{2}}}$
So, the average velocity of the two given velocities is the harmonic mean of the two. So,
option 4 is the answer.

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Physics Test - 20

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Physics Test - 20

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Question 1

1.35 x 10-8ohm metre

1.72 x 10-8ohm metre

4.05 x 10-6ohm metre

8.10 x 10-5ohm metre

Solution

Question 2

A

B

C

D

Solution

Question 3

\((2+\sqrt{2}) \mathrm{N}\) and \((2-\sqrt{2}) \mathrm{N}\)

\((2+\sqrt{3}) \mathrm{N}\) and \((2-\sqrt{3}) \mathrm{N}\)

\(\left(2+\frac{1}{2} \sqrt{2}\right) \mathrm{N}\) and \(\left(2-\frac{1}{2} \sqrt{2}\right) \mathrm{N}\)

\(\left(2+\frac{1}{2} \sqrt{3}\right) \mathrm{N}\) and \(\left(2-\frac{1}{2} \sqrt{3}\right) \mathrm{N}\)

Solution

Question 4

3.18 × 10−3Wb/m2

3.67 × 10−5Wb/m2

5 × 10−3Wb/m2

2.08 × 10−5Wb/m2

Solution

Question 5

1 : 2

2 : 1

1 : 1

None of these

Solution

Question 6

5 cm

10 cm

15 cm

102cm

Solution

Question 7

64 cm/s in - x direction

32 cm/s in - x direction

32 cm/s in + x direction

64 cm/s in + x direction

Solution

Question 8

10π cm/s

30π cm/s

30π cm/s

60π cm/s

Solution

Question 9

0.8 A

(4/7) A

1.0 A

(4/√7)A

Solution

Question 10

\(v=\sqrt{v_{1}+v_{2}}\)

\(v=\frac{v_{1}+v_{2}}{2}\)

\(v=\frac{v_{1}}{v_{2}}\)

\(\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}\)

Solution

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$10 \sqrt{2}$ cm