Physics Test

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Physics Test - 30

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is used for making the filaments of electric bulbs?

A
Copper

B
Nichrome

C
Tungsten

D
None of these

Solution

Tungsten has the highest melting point, that's why it is used for making bulb filaments .
Question 2
4 / -1

The expression for exclusive NOR gate is

A
$A . B + A . B$

B
$A . B + A . B$

C
$(A . B) (A . B)$

D
none of these

Solution

Truth table of EX - NOR Gate is

$$\begin{array}{|l|l|l|}
\hline \text { A } & \text { B } & \text { C } \\
\hline 1 & 1 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 0 & 0 & 1 \\
\hline
\end{array}$$

Hence only option 2 satisfies.
Question 3
4 / -1

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t^oC, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is

A
$\frac{4 \pi^{2} \sigma t^{4}}{R^{2}}$

B
$\frac{r^{2} \sigma(t+273)^{4}}{4 \pi R^{2}}$

C
$\frac{16 \pi^{2} r^{2} \sigma^{4}}{R^{2}}$

D
$\frac{r^{2} \sigma(t+273)^{4}}{R^{2}}$

Solution

Power radiated by the sun with its surface temperature $\left(t+273^{\circ} \mathrm{C}\right)$ is given by
$P=\sigma e 4 \pi r^{2}(t+273)^{4}$
Where $r$ is the radius of the sun and emissivity of the sun, $e=1$
Radiant power per unit area received by a surface at a distance $\mathrm{R}$ is
$S=\frac{P}{4 \pi R^{2}}=\sigma e \frac{r^{2}}{R^{2}}(t+273)^{4}$
Question 4
4 / -1

The binary representation of hexadecimal number C3 is

A
1111

B
110011

C
111100

D
11000011

Solution

To convert C3 into hexadecimal, we will write down the respective values of C 12 and 3 into 4-digit binary form.

12 - 1100

3 - 0011

Hence, C3 - 11000011
Question 5
4 / -1

A body thrown vertically upwards reaches a height and comes back. Which of the following is true about the acceleration due to gravity?

A
It varies continuously with maximum height. Maximum at the beginning and is zero at the top.

B
It only changes sign when the body is at the top.

C
It remains constant throughout.

D
It varies but never changes sign.

Solution

Acceleration due to gravity remains constant through out the motion of the object thrown
Question 6
4 / -1

The graph of input characteristics of common emitter amplifier is

A

B

C

D

Solution

The input characteristics of common emitter amplifier is the curve between base current $I_{B}$ and base-emitter voltage $V_{\text {ae }}$ at constant collector-emitter voltage $V_{\mathrm{CE}}$ The input characteristics can be determined by the circuit shown in figure. Keeping $V_{\mathrm{CE}}$ constant (say at $0 \mathrm{V}$ ), note the base current $I_{B}$ for various values of $V_{B E}$

Then, plot the readings obtained on the graph, taking $I_{B}$ along $Y$ -axis and $V_{\text {ex }}$ along $X$ -axis. This gives the input characteristic at $V_{\mathrm{C}}=0 \mathrm{V} .$ Following a similar procedure, a family of input characteristics can be drawn. These curves are similar to as shown in option (2)
Question 7
4 / -1

$\vec{A}$ and $\vec{B}$ are two vectors and $\theta$ is the angle between them. If $|\overrightarrow{\mathrm{A}} \times \vec{B}|=\sqrt{3}(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}),$ the value of $\theta$ is

A
60^o

B
45^o

C
30^o

D
90^o

Solution

$|A||B| \sin \theta=\sqrt{3}|A||B| \cos \theta$
$\sin \theta=\sqrt{3} \cos \theta$
$\tan \theta=\sqrt{3}$
$\Rightarrow \theta=60^{\circ}$
Question 8
4 / -1

Forces of magnitudes 3 and 2 units acting in the direction $5 \hat{i}+3 \hat{j}+4 \hat{k}$ and $3 \hat{i}+4 \hat{j}-5 \hat{k}$ respectively act on a
particle which is displaced from the points (1,-1,-1) to (3,3,1) . The work done by the forces is equal to

A
80√2 units

B
40√2 units

C
$\frac{57}{5} \sqrt{2}$ units

D
8√2 units

Solution

$\overrightarrow{S_{2}}=3 \hat{i}+3 \hat{j}+\hat{k}$
$\overrightarrow{S_{1}}=\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{S_{2}}-\overrightarrow{S_{1}}=2 \hat{i}+4 \hat{j}+2 \hat{k}=\vec{S}$
$\left|\vec{F}_{1}\right|=3\left|\vec{F}_{2}\right|=2$
$A=5 i+3 \hat{j}+4 \hat{k}$
Unit vector along $\mathrm{A}=\left(\frac{5 \hat{i}+3 \hat{j}+4 \hat{k}}{5 \sqrt{2}}\right) \times 3$
$\vec{F}_{1}=\left(\frac{15 \hat{i}+9 \hat{j}+12 \hat{k}}{5 \sqrt{2}}\right)$
Unit vector along $B=\left(\frac{3 i+4 \hat{j}-5 \hat{k}}{5 \sqrt{2}}\right) \times 2$
$\vec{F}_{2}=\left(\frac{6 \hat{i}+8 \hat{j}-10 \hat{k}}{5 \sqrt{2}}\right)$
$\vec{F}_{1}+\vec{F}_{2}=\frac{21 \hat{i}+17 \hat{j}+2 \hat{k}}{5 \sqrt{2}}=\vec{F}$
Work Done $=\vec{F} \cdot \vec{S}=\frac{1}{5 \sqrt{2}}(42+68+4)=\frac{57 \sqrt{2}}{5}$ units
Question 9
4 / -1

The net work done by a net force acting on an object is equal to the change in the kinetic energy of the object. It is known as:

A
Work-Force theorem

B
Work-Kinetic energy theorem

C
Potential energy theorem

D
Force-Kinetic energy theorem

Solution

The Work-Kinetic energy theorem states that the net work done by a net force acting on an object is equal to the change in the kinetic energy of the object.

Kinetic energy is the energy possessed by an object due to its motion or movement. In simple words, it is the energy of motion, observable as the movement of an object, particle, or set of particles.

$W=\Delta K E=\frac{1}{2} m \Delta v^{2}$
Question 10
4 / -1

A river is flowing from West to East with a speed of 5 m/min. A man can swim in still water with a velocity of 10 m/min. In which direction should the man swim, so as to take the shortest possible path to go to South?

A
30^oWest of South

B
60^oWest of South

C
South

D
30^o West of North

Solution

In order to cross the river with shortest possible path, let $\theta$ be the angle made by the

swimmer with South direction towards West.

Then,
$v_{p r} \sin \theta=v_{r}$
$\sin \theta=\frac{v_{r}}{v_{p r}}=\frac{5}{10}$
$\sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}$
Person has to swim at $30^{\circ}$ west of South.

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Physics Test - 30

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Physics Test - 30

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Question 1

Copper

Nichrome

Tungsten

None of these

Solution

Question 2

A.B+A.B

A.B+A.B

A.BA.B

none of these

Solution

Question 3

\(\frac{4 \pi^{2} \sigma t^{4}}{R^{2}}\)

\(\frac{r^{2} \sigma(t+273)^{4}}{4 \pi R^{2}}\)

\(\frac{16 \pi^{2} r^{2} \sigma^{4}}{R^{2}}\)

\(\frac{r^{2} \sigma(t+273)^{4}}{R^{2}}\)

Solution

Question 4

1111

110011

111100

11000011

Solution

Question 5

It varies continuously with maximum height. Maximum at the beginning and is zero at the top.

It only changes sign when the body is at the top.

It remains constant throughout.

It varies but never changes sign.

Solution

Question 6

Solution

Question 7

60o

45o

30o

90o

Solution

Question 8

80√2 units

40√2 units

\(\frac{57}{5} \sqrt{2}\) units

8√2 units

Solution

Question 9

Work-Force theorem

Work-Kinetic energy theorem

Potential energy theorem

Force-Kinetic energy theorem

Solution

Question 10

30o West of South

60o West of South

South

30o West of North

Solution

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$A . B + A . B$

$A . B + A . B$

$(A . B) (A . B)$

60^o

45^o

30^o

90^o

30^oWest of South

60^oWest of South

30^o West of North