Physics Test

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Physics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Maxwell is the unit of

A
Intensity of magnetization

B
Permeability

C
Magnetic flux

D
Magnetic susceptibility

Solution

Maxwell is the unit of magnetic flux.
Question 2
1 / -0

The snow on the mountains does NOT melt all at once when it is heated by the sun because:

A
it becomes very h

B
it reflects most of the heat from the sun

C
it has a low specific heat capacity

D
it has a high latent heat of fusion

Solution

At mountain, there is very large amount of snow i.e. $m$ is very very high.

And for melts of ice, latent heat of fusion is required.

Latent heat of fusion = mass $\times L_{f}$

And since very very large mass is present so very high amount of heat is required which can't be delivered by sun at once.
Question 3
1 / -0

A body of mass m is dropped from a certain height. It has a velocity v when it is at a height h above the ground. Which of the following will remain constant during the free fall?

A
Velocity

B
Mass

C
Density

D
Weight

Solution

Mass will remain constant.

Mass never be changed
Question 4
1 / -0

If the gravitational force varies inversely as the n^th power of distance, then the time period of a planet in a circular orbit of radius 'r' around the sun will be proportional to

A

\begin{equation}R^\frac{n+1}{2}\end{equation}

B

\begin{equation}R^\frac{n-2}{2}\end{equation}

C
Rⁿ

D

\begin{equation}R^\frac{n-1}{2}\end{equation}

Solution

\begin{equation}\mathrm{F}=\frac{G m M}{R^{n}}=m \omega^{2} R \rightarrow \omega=\sqrt{\frac{G M}{R^{n+1}}} \text { hence time period is proportional to } R^\frac{n+1}{2}\end{equation}
Question 5
1 / -0

A closed surface has ‘n’ electric dipole located inside it. The net electric flux emerging out of the surface-

A
ne/ε_0

B
2e/ε_0

C
2ne/ε_0

D
zero

Solution

The net flux will be zero as the electric field lines entering the negative end of the dipole will be exactly cancelled by the electric field lines leaving the positive end of the dipole.
Question 6
1 / -0

One Joule per coulomb is called?

A
Volt

B
Ampere

C
Farad

D
Gauss

Solution

Joule is the unit for work done and coulomb is the unit for the charge. The potential is defined as the ratio of the work done to the charge and the unit of the potential difference is volt.

Therefore, one joule per coulombs is equal to the unit of the potential difference.

We know that,

Energy ${E}={QV}$

$\therefore \frac{{E}}{{Q}}=$ Volt i. e. Joule per Coulomb
Question 7
1 / -0

Let gp be value of acceleration due to gravity at poles and ge be that at equator of earth. Assuming earth to be a sphere of radius 'R' rotating about its axis with angular speed 'w', g_p - g_e is given by

A
$R ω^{2}$

B
${(R ω)}^{3}$

C
$R^{2} ω^{2}$

D
$ω R$

Solution

Acceleration due to gravity at a place of latitude $\lambda$ due to rotation of earth is $g^{\prime}=g-R_{E} \omega^{2} \cos ^{2} \lambda$
At equator, $\lambda=0, \cos 0=1$
$\therefore g^{\prime}=g_{e}=g-R_{E} \omega^{2}$
At poles, $\lambda=90, \cos 90=0$
$\therefore g^{\prime}=g_{p}=g$
$\therefore g_{p}-g_{e}=g-\left(g-R_{E} \omega^{2}\right)=R_{E} \omega^{2}$
Question 8
1 / -0

A charge of magnitude 3e and mass 2m is moving in an electric field, $\vec{E}$ . The acceleration imparted to the charge is

A
2 Ee/3m

B
3 Ee/2m

C
2 m/3Ee

D
3 m/2Ee

Solution

\begin{equation}\text { Acceleration } a=\frac{Q E}{m}=\frac{(3 e) E}{2 m}\end{equation}
Question 9
1 / -0

What is the electric potential at the center of a square in given figure?

A
Zero

B

$k \frac{2 q}{s^{2}}$

C

$k \frac{4 q}{s}$

D

$k \frac{(4 \sqrt{2}) q}{s}$

Solution

Let the distance between each corner and the center of square be $x$

$\therefore s^{2}=x^{2}+x^{2}$

$\Rightarrow x=\frac{s}{\sqrt{2}}$

Potential at the center due to charge at corner $1, V_{1}=\frac{k q}{x}=\frac{\sqrt{2} k q}{s}$

The square is symmetric about its center.

Thus total potential at the center $V=4 V_{1}=k \frac{4 \sqrt{2} q}{s}$
Question 10
1 / -0

A paramagnetic substance of susceptibility 3 x 10^-4is placed in a magnetic field of 4 x 10^-4 Am^-1. What is the intensity (in Am^-1) of magnetisation?

A
1.33 x 10⁸

B
0.75 x 10^-8

C
12 x 10^-8

D
14 x 10^-8

Solution

\begin{aligned}
I &=\chi H=\left(4 \times 10^{-4}\right) \times\left(3 \times 10^{-4}\right) \\
&=12 \times 10^{-8} \mathrm{A} \mathrm{m}^{-1}
\end{aligned}

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Physics Test - 7

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Physics Test - 7

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Question 1

Intensity of magnetization

Permeability

Magnetic flux

Magnetic susceptibility

Solution

Question 2

it becomes very h

it reflects most of the heat from the sun

it has a low specific heat capacity

it has a high latent heat of fusion

Solution

Question 3

Velocity

Mass

Density

Weight

Solution

Question 4

\begin{equation}R^\frac{n+1}{2}\end{equation}

\begin{equation}R^\frac{n-2}{2}\end{equation}

Rn

\begin{equation}R^\frac{n-1}{2}\end{equation}

Solution

Question 5

ne/ε_0

2e/ε_0

2ne/ε_0

zero

Solution

Question 6

Volt

Ampere

Farad

Gauss

Solution

Question 7

Rω2

(Rω)3

R2ω2

ωR

Solution

Question 8

2 Ee/3m

3 Ee/2m

2 m/3Ee

3 m/2Ee

Solution

Question 9

Zero

\(k \frac{2 q}{s^{2}}\)

\(k \frac{4 q}{s}\)

\(k \frac{(4 \sqrt{2}) q}{s}\)

Solution

Question 10

1.33 x 108

0.75 x 10-8

12 x 10-8

14 x 10-8

Solution

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Rⁿ

$R ω^{2}$

${(R ω)}^{3}$

$R^{2} ω^{2}$

$ω R$

1.33 x 10⁸

0.75 x 10^-8

12 x 10^-8

14 x 10^-8