Physics Test

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Physics Test - 49

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Question 1
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Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are) correct ?

A
Both cylinders P and Q reach the ground at the same time.

B
Cylinders P has larger linear acceleration than cylinder Q.

C
Both cylinders reach the ground with same translational kinetic energy.

D
Cylinder Q reaches the ground with larger angular speed.

Solution

a=MgsinθM+1R₂

ap=MgsinθM+MR₂R₂≈g2sinθ

aQ=gsinθasIQ∼0

mgh=12Iω₂+12mv₂

v=Rω

Undefined control sequence \therefore

Undefined control sequence \therefore

Undefined control sequence \therefore

Concepts :
Main Concept :
Calculation of moment of inertiaSince the moment of inertia of an ordinary object involves a continuous distribution of mass at a continually varying distance from any rotation axis, the calculation of moments of inertia generally involves calculus, the discipline of mathematics which can handle such continuous variables. Since the moment of inertia of a point mass is defined by
$I = m r^{2}$
Then the moment of inertia contribution by an infinitesimal mass element dm has the same form. This kind of mass element is called a differential element of mass and its moment of inertia is given by

$d I = r^{2} d m$
The “d” preceding any quantity denotes a vanishingly small or “differential” amount of it
Note that the differential element of moment of inertia dI must always be defined with respect to a specific rotation axis. The sum over all these mass elements is called an integral over the mass.
$\mathrm{I}=\int \mathrm{dI}=\int_{0}^{\mathrm{M}} \mathrm{r}^{2} \mathrm{dm}$
Hence, the correct option is (D)
Question 2
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Two slits, 4 mm apart, are illuminated by light of wavelength 6000 A. What will be the fringe width on a screen placed 2 m from the slits?

A
0.12 mm

B
0.3 mm

C
3.0 mm

D
4.0 mm

Solution

$\begin{aligned} \beta &=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{6000 \times 10^{-10} \times 2}{4 \times 10^{-3}} \\ &=0.3 \times 10^{-3} \mathrm{m}=0.3 \mathrm{mm} \end{aligned}$

Concepts :
Main Concept :
Fringe Width in YDSE

Fringe width (β): The separation between any two consecutive bright or dark fringes is called fringe width. In YDSE all fringes are of equal width. Fringe widthβ =λD/d

and angular width $\theta=\frac{\lambda}{d}=\frac{\beta}{D}$
Hence, the correct option is (B)
Question 3
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A coil of inductive reactance $31 Ω$ has a resistance of $8 Ω$ .It is placed in series with a condenser of capacitive reactance $25 Ω$ .The combination is connected to an ac source of 110 V. The power factor of the circuit is :

A
0.33

B
0.56

C
0.64

D
0.80

Solution

$\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{X}_{\mathrm{C}}=25 \Omega, \mathrm{R}=8 \Omega$Impedance of series LCR is$\mathrm{Z}=\sqrt{\left(\mathrm{R}^{2}\right)+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$$=\sqrt{(8)^{2}+(31-25)^{2}}=\sqrt{64+36}=10 \Omega$Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}=0.8$
Hence, the correct option is (D)
Question 4
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In an N-P-N transistor as common emitter amplifier 10¹⁰ electrons enter the emitter in 10^-6 s. If 2% of the electrons are lost in the base. Then Calculate the current transfer ratio and current amplification factor.

A
0.76 ; 38

B
0.98 ; 49

C
1.08 ; 57

D
1.26 ; 68

Solution

Emitter current $I_{e}=\frac{N e}{y}=\frac{10^{10} \times 1.6 \times 10^{-19}}{10^{-6}}=1.6 \mathrm{mABase}$ current $I_{b}=\frac{2}{100} \times 1.6=0.032 \mathrm{mA}$

but $I_{e}=I_{c}+I_{b} \therefore I_{c}=I_{c}=I_{b}=1.6=1.6-0.032=1.568 \mathrm{mA}$

$\therefore \alpha=\frac{I_{c}}{I_{c}}=\frac{1.568}{1.6}=0.98$ and $\beta=\frac{I_{c}}{I_{b}}=\frac{1.568}{0.032}=49$)
Concepts:
Main Concept :Emitter, base and collector terminals of a transistor

A junction transistor consists of a thin piece of one type of semiconductor material between two thicker layers of the opposite type. For example, if the middle layer is p-type, the outside layers must be n-type. Such a transistor is an NPN transistor. One of the outside layers is called the emitter, and the other is known as the collector. The middle layer is the base. The places where the emitter joins the base and the base joins the collector are called junctions.
Manufacturers also make PNP junction transistors. In these devices, the emitter and collector are both a p-type semiconductor material and the base is n-type.
Hence, the correct option is (B)
Question 5
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Work done by the system in closed path ABCA, as shown in figure is :

A
Zero

B
(V₁ - V₂) (P₁ - P₂)

C
$\frac{\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)}{2}$

D
$\frac{\left(\mathrm{P}_{2} \mathrm{P}_{1}\right)\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)}{2}$

Solution

We know that the current = charge/time The emitter current $(\mathrm{x})$ is given by

$\mathrm{I}_{\mathrm{E}}=\frac{\mathrm{Ne}}{\mathrm{t}}=\frac{10^{10} \times\left(1.6 \times 10^{-19}\right)}{10^{-6}}=1.6 \mathrm{mA}$

The base current ( $\mathrm{L})$ is given by

$\mathrm{I}_{\mathrm{B}}=\frac{2}{100} \times 1.6=0.032 \mathrm{mA}$

In a transistor,

$\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}$

$\therefore \quad \mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{B}}=1.6-0.032=1.568 \mathrm{mA}$

Current transfer ratio $=\frac{\mathrm{I} \mathrm{c}}{\mathrm{I}_{\mathrm{B}}}=\frac{1.56 \mathrm{B}}{1.6}=0.98$

Current amplification factor $=\frac{\mathrm{Ic}}{\mathrm{I}_{\mathrm{B}}}=\frac{1.56 \mathrm{B}}{0.032}=49$

Hence, the correct option is (C)
Question 6
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Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.

A
330.94 m s^-1

B
187.6 m s^-1

C
324.5 m s^-1

D
246.2 m s^-1

Solution

At N.T.P., $P=1.013 \times 10^{5} \mathrm{N} \mathrm{m}^{-2}$
volume of air, $V=22400 \mathrm{cm}^{3}=2.24 \times 10^{-2} \mathrm{m}^{3}$
mass of 1 mole of air, $M=29.0 \times 10^{-3} \mathrm{kg}$
Therefore, density of air at N.T.P.
$$
\rho=\frac{M}{V}=\frac{29.0 \times 10^{-3}}{2.24 \times 10^{-2}}=1.295 \mathrm{kg} \mathrm{m}^{-3}
$$
$\mathrm{Now}, v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{1.4 \times 1.013 \times 10^{5}}{1.295}}$
$$
=330.94 \mathrm{m} \mathrm{s}^{-1}
$$
Concepts:
Main Concept :
Kinetic Theory of Gases Different equationsKinetic Theory of Gases Different equations used in kinetic theory of gases are listed below.
(i) $\mathrm{pV}=\mathrm{nRT}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$(\mathrm{m}=$ mass of gas in gms)
(ii) Density $\rho=\frac{\mathrm{m}}{\mathrm{V}}$
(general)
$=\frac{p M}{R T}$
(for ideal gas)Kinetic Theory of Gases Different equations used in kinetic
theory of gases are listed below.
(i) $\mathrm{pV}=\mathrm{nRT}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad(\mathrm{m}=$ mass of gas in gms $)$
(ii) Density $\rho=\frac{\mathrm{m}}{\mathrm{V}}$
(general)
$=\frac{\mathrm{pM}}{\mathrm{RT}}$
(for ideal gas)
Hence, the correct option is (A)
Question 7
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Prism angle & refractive index for a prism are 60^o & 1.414. Angle of minimum deviation will be :

A
15^o

B
30^o

C
45^o

D
60^o

Solution

$\mu=\frac{\sin \frac{\left(\Lambda+\delta_{\mathrm{m}}\right)}{2}}{\sin 30^{\circ}}$
$\Rightarrow 1.414=\frac{\sin \frac{\left(60+\delta_{\mathrm{m}}\right)}{2}}{\sin 30^{\circ}}$
$\Rightarrow \quad \sin \left(\frac{60^{\circ}+\delta_{\mathrm{m}}}{2}\right)=0.707=\sin 45^{\circ}$
$\Rightarrow \quad \frac{60+\delta_{\mathrm{m}}}{2}=45^{\circ} \Rightarrow \delta_{\mathrm{m}}=30^{\circ}$
Hence, the correct option is (B)
Question 8
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An electron gun G emits electron of energy 2 kev travelling in the (+)ve x-direction. The electron are required to hit the spot S where GS = 0.1m & the line GS makes an angle of $60^{0}$ with the x-axis, as shown in the fig. A uniform magnetic field $\underset{B}{\to}$ parallel to GS exists in the region outsides to electron gun. Find the minimum value of B needed to make the electron hit S.

A

B

C

D

Solution

$\mathrm{GS}=\mathrm{V} \cos \theta \times \mathrm{T}=\mathrm{V} \cos \theta \times \frac{2 \pi \mathrm{m}}{9 \mathrm{B}}=\frac{2 \pi}{98} \mathrm{mV} \cos 60^{\circ}=\frac{\pi \mathrm{mV}}{\mathrm{qB}}=\frac{\pi}{9 \mathrm{B}} \sqrt{2 \mathrm{mE}}$
$B=\frac{\pi}{q(G S)} \sqrt{2 m E}=\frac{3.14}{1.6 \times 10^{-19} \times 0.1} \times \sqrt{2 \times 9 \cdot 1 \times 10^{-31} \times 2 \times 10^{3} \times 1 \cdot 6 \times 10^{-19}}$
$\mathrm{B}=31.4 \sqrt{\frac{4 \times 9 \times 10^{-28}}{1.6 \times 10^{-19}}}=31.4 \times 1.5 \times 10^{-4}=4.71 \times 10^{-3} \mathrm{T}$
Hence, the correct option is (A)
Question 9
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A body executes S.H.M. of period 20 seconds. Its velocity is 5 cm s^-1, 2 seconds after it has passed the mean position. Find amplitude of the bob :

A
21.45 cm

B
16.56 cm

C
19.67 cm

D
15.34 cm

Solution

Here, $T=20 \mathrm{s}$
Also, when $\mathrm{t}=2 \mathrm{s}, \mathrm{v}=5 \mathrm{cm} \mathrm{s}^{-1}$
Now, $\mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t}=\mathrm{r} \times \frac{2 \pi}{\mathrm{T}} \cos \frac{2 \pi}{\mathrm{T}} \mathrm{t}$
$\therefore \quad 5=\mathrm{r} \times \frac{2 \pi}{20} \cos \frac{2 \pi}{20} \times 2$ or $\frac{\mathrm{r} \pi}{10} \cos \frac{\pi}{5}=5$
or $\frac{\pi \mathrm{r}}{10} \cos 36^{\circ}=5$ or $\frac{\pi \mathrm{r}}{10} \times 0.8090=5$
or $\mathrm{r}=19.67 \mathrm{cm}$
Concepts:
Main Concept:
General equation of SHM The necessary and sufficient condition for SHM is $F=-k x$
we can write above equation in the following way:
$\mathrm{ma}=-\mathrm{kx}$
$m \frac{d^{2} x}{d t^{2}}=-k x$
$\frac{d^{2} x}{d t^{2}}+\frac{k}{m} x=0$
Equation (1) is Double Differential Equation of SHM.
Now $\frac{d^{2} x}{d t^{2}}+\omega^{2} x=0$
It's solution is $\mathbf{x}=\mathbf{A} \sin (\omega t+\phi)$
where $\omega=$ angular frequency $=\sqrt{\frac{k}{m}}$
$\mathrm{x}=$ displacement from mean position
$\mathrm{k}=\mathrm{SHM}$ constant.
The equality $(\omega t+\phi)$ is called the phase angle or simply the phase of the SHM and $\phi$ is the initial phase i.e., the phase at $t=0$ and depends on initial position and direction of velocity at $t=0$.
Hence, the correct option is (C)
Question 10
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A uniform rod of mass m and length ℓ is rotating with constant angular velocity ω about an axis which passes through its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its Young's modulus is Y. Neglect gravity. The strain at the mid point of the rod is :

A
$\mathrm{m} \omega^{2} \mathrm{Y} / 8 \mathrm{AY}$

B
$\frac{3 m \omega^{2} \ell}{8 A Y}$

C
$\frac{3 m \omega^{2} \ell}{4 A Y}$

D
$\frac{m \omega^{2} \ell}{4 A Y}$

Solution

Young's Modulus of Elasticity $\mathbf{Y}=\frac{\overline{\mathrm{F}} / \mathbf{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$
$\mathbf{F}=\mathbf{m g}$ and $\mathbf{A}=\pi \mathbf{r}^{2}$
It is defined as the ratio of the normal stress to the longitudinal strain.
Concepts:
Main Concept:
Young's Modulus of Elasticity
Young's Modulus of Elasticity $\mathrm{Y}=\frac{\text { F/A }}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$
$\mathbf{F}=\mathbf{m g}$ and $\mathbf{A}=\pi \mathbf{r}^{2}$
It is defined as the ratio of the normal stress to the longitudinal strain.
Hence, the correct option is (B)