Physics Test

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Physics Test - 11

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Question 1
4 / -1

The maximum value of the function is?
f(x, y) = sin(x).cos(2y).cos(x + 2y) + sin(2y).cos(x + 2y).cos(x) in the region x=0; y=0; x+2y = 3

A
90

B
cos(1)

C
sin(1).cos(1)

D
sin(3).cos(3)

Solution

Rewrite the function as
f(x, y) = cos(x + 2y) * (sin(x).cos(2y) + cos(x).sin(2y))
f(x, y) = cos(x + 2y).sin(x + 2y)
Put x+2y = 3
= sin(3).cos(3).
Question 2
4 / -1

Of the diodes shown in the following diagrams, which one is reverse biased?

A

B

C

D

Solution

A diode is said to be reverse biased if p-type semiconductor of p-n junction is at low potential with respect to n-type semiconductor of p-n junction. It is so for circuit (3).
Question 3
4 / -1

The water filled in a tank has a hole 'h' metres from the top. Find the distance 'x' where the stream of water will touch the ground from the bottom.

A
$\sqrt{2 g \mathrm{H}}$

B
$\sqrt{2 g(H-h)}$

C
$2 \sqrt{h H}$

D
$2 \sqrt{h(H-h)}$

Solution

The pressure on the water at the level of the hole $=[H-(H-h)] \rho g=h \rho g$. The time taken by the water to fall through a height $H-h$ is $t=\sqrt{\frac{2(H-h)}{g}}$
The velocity of efflux, $v_{e l f}=\sqrt{2 g h}$
$\therefore$ Horizontal distance covered, $x=v_{e f f} \times t$ $=\sqrt{2 g h} \times \sqrt{\frac{2(H-h)}{g}}=2 \sqrt{h(H-h)}$
Question 4
4 / -1

A force of 50 dynes is acted on a body of mass $5 \mathrm{~g}$ which is at rest for an interval of 3 seconds, then impulse is (in $\mathrm{Ns}$ ):

A
$1.5 \times 10^{-3} $

B
$1.5 \times 10^{-2} $

C
$2.5 \times 10^{-3}$

D
$1.5 \times 10^{3}$

Solution

Force acting on the body $\mathrm{F}=50$ dynes
$=50 \times 10^{-5} \mathrm{~N}$
Time of action of force $t=3 \mathrm{~s}$
Thus impulse $\mathrm{I}=\mathrm{F} \mathrm{t}$
$\therefore \mathrm{I}=50 \times 10^{-5} \times 3$
$=1.5 \times 10^{-3} \mathrm{~N} \mathrm{~s}$
Question 5
4 / -1

When an electron-positron pair annihilates, how much energy is released?

A
0.8 x 10-13

B
1.6 x 10-13J

C
3.2 x 10^-13 J

D
4.6 x 10-13J

Solution

$+1 \beta^{0}+-1 \beta^{0}=h v+h v$
(Y-photon) (Y-photon)
$\begin{aligned} E &=2 \times 9 \cdot 1 \times 10^{-31} \times\left(3 \times 10^{8}\right)^{2} \\ &=16 \cdot 4 \times 10^{-14} \\ &=1.64 \times 10^{-13} \\ &=16 \times 10^{-13} \mathrm{J} \end{aligned}$
Question 6
4 / -1

For a transistor, the current gain $\alpha$ is $0.96 .$ It is used as an amplifier in a common base circuit with a load resistance of $4 \mathrm{k}^{\Omega}$. If the dynamic resistance of the emitter-base junction is $48^{\Omega}$, then the voltage gain is

A
40

B
80

C
120

D
160

Solution

Voltage gain $=\alpha \times \frac{R_{L}}{R_{D}}=0.96 \times \frac{4000}{48}=80$
Question 7
4 / -1

An electron moves at right angle to a magnetic field of 1.5 x 10^-2T with a speed of 6 x 10⁷m/s. If the specific charge of the electron is 1.7 x 10¹¹C/kg, the radius of the circular path will be :

A
2.9 cm

B
3.9 cm

C
2.35 cm

D
2 cm

Solution

Force due to magnetic field provide the necessary centripetal force.

When electron enters a magnetic field $B$ the force acting is

\[

F=e v B \sin \theta

\]

Given, $\theta=90^{\circ}$
$\therefore \quad F=e v B$
where $\vec{v}$ is velocity of electron and $e$ the charge on it.
since, the electron describes a circular path the centripetal force acting on it is given by
\[
F=\frac{m v^{2}}{r}
\]
Equating Eqs. (i) and (ii), we get
\[
\begin{aligned}
& e v B=\frac{m v^{2}}{r} \\
\Rightarrow \quad r &=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{M}\right) B}
\end{aligned}
\]
Given, $v=6 \times 10^{7} \mathrm{m} / \mathrm{s}, \frac{e}{m}=1.7 \times 10^{11} \mathrm{C} / \mathrm{kg}$
\[
B=1.5 \times 10^{-2} \mathrm{T}
\]
\[
\begin{aligned}
\therefore \quad r &=\frac{6 \times 10^{7}}{1.7 \times 10^{11} \times 1.5 \times 10^{-2}} \\
r &=2.35 \times 10^{-2} \mathrm{m} \\
r &=2.35 \mathrm{cm}
\end{aligned}
\]
When magnetic field is acting into the plane of paper, the force is downward and the negatively charged particle describes a clock wise circle.
Question 8
4 / -1

A stone is dropped in a lake by a person from a tower 500 m high. The sound of splash will be heard by the man after approximately, (speed of the sound in air is 340 m/s , g =10m/s2)

A
11.6 sec

B
9.4 sec

C
10 sec

D
1.6 sec

Solution

Let the time taken by the stone to reach the water level is t
Usings $=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} ; \mathrm{u}=0$ and $\mathrm{a}=10 \mathrm{m} / \mathrm{s}^{2},$ put $\mathrm{s}=500 \mathrm{m}$
$500=\frac{1}{2} \times 10 \times \mathrm{t}_{1}^{2} \Rightarrow \mathrm{t}_{1}=10 \mathrm{sec}$
Speed of the sound in air $=340 \mathrm{m} / \mathrm{s},$ let $\mathrm{t} 2$ is the time taken by the sound to travel $500 \mathrm{m}$ in upward direction.
$500=330 \times t_{2} \Rightarrow t_{2}=\frac{500}{340}=1.6 \mathrm{sec}$
Total time taken for the splash is $t=t_{1}+t_{2}=10+1.6=11.6 \mathrm{sec}$
Question 9
4 / -1

An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be

A
straight line along the x-direction

B
a circle in the xz-plane

C
a circle in the yz-plane

D
a circle in the xy-plane

Solution

For the charge moving in the magnetic field, the force is given by
$\vec{F}=q(\vec{v} \times \vec{B})$
Hence Force exerted by the magnetic field is always perpendicular to the velocity vector and the magnetic field and it will provide the necessary centripetal force and the charge particle will move in the xz plane.
Question 10
4 / -1

If the forces of magnitude 12 Kg wt, 5 Kg wt and 13 Kg wt acting at a point are in equilibrium, then the angle between the two forces is

A
$30 °$

B
$90 °$

C
$60 °$

D
$45 °$

Solution

Since (13)²= (12)²+ (5)²
So two of them are inclined at 90 to each other.