Physics Test - 32

Result

Physics Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
1 / -0

The maximum energy in thermal radiation from a source occurs at the wavelength 4000 Å. The effective temperature of the source is (Wien's constant, b = 2.93 × 10⁻³)

A
7325 K

B
800 K

C
10⁴ K

D
10⁶ K

Solution

According to Wien's displacement law

\(\lambda_{m}=\frac{b}{T} \Rightarrow T=\frac{b}{\lambda_{m}}=\frac{2.93 \times 10^{-3}}{4000 \times 10^{-10}}=7325 \mathrm{K}\)

KEY CONCEPTS

Wien's Displacement Law According to Wien's law the product of wavelengths corresponding to maximum intensity of radiation and temperature of body (in Kelvin) is constant, ie., \(\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}=\) constant where \(\mathrm{b}\) is Wien's constant and has value \(2.89 \times 10^{-3} \mathrm{m}-\mathrm{K}\)

Wien's displacement Law According to Wien's law the product of wavelength corresponding to maximum intensity of radiation and temperature of body (in Kelvin) is constant, i.e.
\(\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}=\) constant
Where b is Wien's constant and has value \(2.89 \times 10^{-3} \mathrm{mk}\)
As the temperature of the body increases, the wavelength at which the spectral intensity \(\left(\mathrm{E}_{\lambda}\right)\) is maximum shifts towards left. Therefore it is also called Wien's displacement law.

\(\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}\)
\(\lambda_{\mathrm{m}_{2}}<\lambda_{\mathrm{m}_{2}}<\lambda_{\mathrm{m}_{2}}\)
This law is of great importance in 'Astrophysics' as through the analysis of radiations coming from a distant star, by finding \(\lambda_{\mathrm{m}}\) the temperature of the star \(\mathrm{T}\left(=\frac{\mathrm{b}}{\lambda_{\mathrm{m}}}\right)\) is determined
Examples on Wien's law The sun has a temperature of \(5800 \mathrm{K}\). What is the sun's wavelength of maximum emission?
Wavelength of maximum emission \(=2900 /\) Object's temperature in Kelvin
Sun's wavelength of maximum emission \(=2900 / 5800=0.5\) microns
Question 2
1 / -0

In the given figure, if velocity of block \(\mathrm{C}\) at a particular instant is \(-20 \mathrm{m} / \mathrm{s} \hat{\mathrm{j}}\), then velocity of rod (A) at that instant

will be : (string is attached to wedge)

A

B

C

D

Solution
Question 3
1 / -0

If a simple harmonic motion is represented by \(\frac{d^{2} x}{d t^{2}}+\alpha x=0,\) its time period is

A
2π α

B
2π √α

C
2π / α

D
2π / √α

Solution

Standard differential equation of SHM is \(\frac{d^{2} x}{d t^{2}}+\omega^{2} x=0\)
Given equation is \(\frac{d^{2} x}{d t^{2}}+\alpha x=0\)
\(\therefore \quad \omega^{2}=\alpha\)
or \(\omega=\sqrt{\alpha}\)
\(\therefore \quad T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\alpha}}\)
Question 4
1 / -0

An isosceles prism of angle 120^° has a refractive index 1.44. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite faces

A
are parallel to each other

B
are diverging

C
make an angle of 2[sin⁻¹(0.72)−30^o] with each other

D
make an angle of 2sin⁻¹(0.72) with each other

Solution

Applying Snell's law at \(P, \mu=\frac{\sin r}{\sin 30^{\circ}}\)
\(\Rightarrow \quad \sin r=\frac{1.44}{2}=0.72\)
\(\therefore \quad \delta \quad=r-30^{\circ}=\sin ^{-1}(0.72)-30^{\circ}\)
\(\therefore\) The rays make an angle of
\(2 \delta=2\left[\sin ^{-1}(0.72)-30^{\circ}\right]\) with each other
Question 5
1 / -0

A waveform shown when applied to the following circuit will produce which of the following output waveform. Assuming ideal diode configuration and R₁=R₂

A

B

C

D

Solution

The P-N junction will conduct only when it is forward biased i.e. when \(-5 \mathrm{V}\) is fed to it, so it will conduct only for 3 rd quarter part of signal shown and when it conducts potential drop 5 volt will be across both the resistors, so output voltage across \(\mathrm{R}_{2}\) is \(2.5 \mathrm{V}\)
\(\therefore V_{0}=-2.5 \mathrm{V}\)
Question 6
1 / -0

A wavefront is represented by the plane y=3−x. The propagation of wave takes place at

A
45^°with x - direction

B
135^° with x - direction

C
60^° with x - direction

D
No sufficient data

Solution

The propagation of ray takes place in
perpendicular direction of wavefront
Here, \(m_{1} m_{2}=-1\)
\[
\begin{array}{c}
m_{1}=\text {slope of wavefront}=-1 \\
m_{2}=\text {Slope of } \operatorname{ray} \\
(-1) m_{2}=-1 \Rightarrow m_{2}=1 \\
\tan \theta=1 \Rightarrow \theta=45^{\circ}
\end{array}
\]
Question 7
1 / -0

The ratio of the velocity of sound in hydrogen (γ=7 / 5) to that helium (γ=5 / 3) at the same temperature is

A

B

C

D

Solution

Velocity of sound in a gas
\(v=\sqrt{\frac{\gamma p}{d}}\)
\(\therefore \frac{v H_{2}}{v H_{e}}=\sqrt{\frac{\gamma H_{2} \times d_{H e}}{d_{H_{2}} \times \gamma_{H e}}}\)
\(\frac{v_{H_{2}}}{v_{H e}}=\sqrt{\frac{7 \times 3 \times 2}{5 \times 5}} \quad\left[A s \frac{d_{H e}}{d_{H_{2}}}=2\right]\)
\(\therefore \frac{v_{H_{2}}}{v_{H e}}=\frac{\sqrt{42}}{5}\)
Question 8
1 / -0

A flywheel rotates about a fixed axis and slows down from 300 rpm to 100 rpm in 2 min. Then its angular retardation in rad/min² is

A
100 / π

B
100

C
100 π

D
200 π

Solution

Because of uniform angular acceleration. we can use equation of motion in rotation. \(\alpha=\frac{\omega_{2}-\omega_{1}}{t}=\frac{2 \pi(100-300)}{2}=-200 \pi \mathrm{rad} / \mathrm{m}^{2}\)
KEY CONCEPTS
Angular acceleration In the study of linear motion, we found that the simplest form of accelerated motion to analyze is motion under constant linear acceleration. Likewise, for rotational motion about a fixed axis, the simplest accelerated motion to analyze is motion under constant angular acceleration.
Kinematical expressions for rotational motion under constant angular acceleration are of the same form as those for linear motion under constant linear acceleration. We shall not derive them here, but simply, write them from the corresponding linear equations, substituting equivalent angular quantities for the linear ones. Table below displays both sets of equations. For simplicity, we let \(\mathrm{x}_{0}=0\) and \(\theta_{0}=0\) in these equations. The expressions are valid for both rigid-body rotational and particle motion about a fixed axis.
Equation of motion for constant linear and for constant anqular acceleration
Question 9
1 / -0

Two springs of force constants K₁ and K₂ are connected to a mass mm as shown. The frequency of oscillation of the mass is f. If both K₁ and K₂ are made four times their original values, the frequency of oscillation becomes?

A
f/2

B
f/4

C
4f

D
2f

Solution

In the given figure two springs are connected in parallel. Therefore the effective spring constant is given by
\[
k_{e f f}=k_{1}+k_{2}
\]
Frequency of oscillation,
Question 10
1 / -0

In a wave motion y= a sin(kx−ωt), y can represent

A
Electric field

B
Pressure

C
Displacement

D
All of these

Solution

In case of sound wave, y can represent pressure and displacement, while in case of an electromagnetic wave it represents electric and magnetic fields.

(In general y is any general physical quantity which is made to oscillate at one place and these oscillations are propagated to other places also).