Physics Test - 14

Given:

The angle of the prism \(A=60^{\circ}\)

The refractive index of the prism \((\mu)=1.5\)

\(\sin^{-1} 0.75=48^{\circ}\)

The Refractive index \((\mu)\) of a prism for minimum deviation \(\left(\delta_{m}\right)\):

\(\mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}}\)

Where, \(A=\) angle of the prism

The relation between the angle of the prism \(A\) and incident angle \((i)\):

\(i=\frac{\left(A+\delta_{m}\right)}{2}\)

From the formula of refractive index,

\(1.5=\frac{\sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}\)

\(1.5×\sin 30^{\circ}= \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)\)

\(1.5×\frac{1}{2}= \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)\)

\(0.75= \sin \left(\frac{60^{\circ}+\delta_{m}}{2}\right)\)

\(\sin^{-1}0.75= \frac{60^{\circ}+\delta_{m}}{2}\)

\(48^{\circ}= \frac{60^{\circ}+\delta_{m}}{2}\)

\(\delta_{m}=96^{\circ}-60^{\circ}\)

\(\delta_{m}=36^{\circ}\)

Angle of incidence \((i)\)

\(i=\frac{(60+36)}{2}\)

\(i=48^{\circ}\)

So, the angle of deviation and angle of incidence is \(36^{\circ}\) and \(48^{\circ}\) respectively.

Given,

No. of vibrations \(= 32\)

Frequency of tuning fork, \(ν = 256\) Hz

Speed of sound in air, \(ν = 320\) m/s

Distance d travelled by a wave in a single vibration is given as

\(d=\frac{v}{v}=\frac{320}{256}=1.25\) m

total distance travelled by wave in \(32\) vibrations \(= 1.25 × 32 = 40\) m

According to the first law of Thermodynamics:

\(\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}\)

Where \(\Delta \mathrm{Q}=\) Heat supplied to the system, \(\Delta \mathrm{W}=\) work done by the system, and \(\Delta \mathrm{U}=\) change in internal energy of system

Isothermal: The thermodynamic process in a system, during which the temperature remains constant is called the isothermal process.

\(\Delta \mathrm{T}=0\)

Isochoric: The thermodynamic process in a system, during which the Volume remains constant is called the isochoric process.

\(\Delta \mathrm{V}=0\)

Adiabatic: The thermodynamic process in a system, during which no heat or mass transfer occurs between thermodynamic systems is called an adiabatic process. Most rapid processes are adiabatic.

\(\Delta \mathrm{Q}=0\)

Expansion means volume is increasing while in the case of compression volume is decreasing.

In the case of adiabatic compression:

\(\Delta \mathrm{Q}=0\) and Volume decreases, so work done \((\Delta \mathrm{W})\) is negative.

Since \(\Delta \mathrm{Q}=\Delta \mathrm{W}+\Delta \mathrm{U}=0\)

\(\mathrm{\Rightarrow \Delta U > 0}\)

or \(\mathrm{mC\Delta T > 0}\)

or \(\mathrm{\Delta T > 0}\)

or \(\mathrm{T_2 - T_1 > 0}\)

or \(\mathrm{T_2 > T_1 }\)

Hence,the correct option is (C).

When the lift is at rest: T = mg

When lift is accelerating upward: \(R_{1}=T_{U}=m g+m a\)

When lift is accelerating downward: \(R_{2}=T_{D}=m g-m a\)

\(T_{U}=2 T_{D}\)

\(m g+m a=2 m g-2 m a\)

\(a=\frac{g}{3}\)

From Wien's law,

\(\lambda_{\mathrm{m}} \mathrm{T}=\) constant

Where \(\mathrm{T}\) is the temperature of black body

\(\lambda_{\mathrm{m}}\) is the wavelength corresponding to maximum energy of emission.

\(\lambda_{\mathrm{m}} =\frac{b}{T}\)

\(=\frac{2.88 \times 10^{6} \mathrm{nmK}}{2880 \mathrm{K}}\)

\(=1000nm\)

So, U₂ is maximum since all wavelength ranging from 0 to infinity. Thus U₁ and U₂ are nor zero.

Energy distribution of black body radiation is given below:

• \(U_{1}\) and \(U_{2}\) are not zero because a black body emits radiations of nearly all wavelengths.
• Since \(U_{1}\) corresponding to lower wavelength, \(\mathrm{U}_{3}\) corresponds to higher wavelength and \(\mathrm{U}_{2}\) corresponds to medium wave length hence \(\mathrm{U}_{2}>\) \(\mathrm{U}_{1}\).

Given that:

Dipole Moment, \(P=3 \times 10^{-3} cm\)

We know that:

In rotating the dipole from the position of stable equilibrium by an angle \(\theta\), the amount of work done is given by,

\(W=P E(1-\cos \theta)\)

For unstable equilibrium, \(\theta=180^{\circ}\)

\(\therefore W=P E\left(1-\cos 180^{\circ}\right) \quad\left[\because \cos 180^{\circ}=-1\right]\)

\(=2 P E\)

\(=2 \times 3 \times 10^{-3} \times 10^{4} J\)

\(=60 J\)

The value of \(D\) can be found by,

\(2 D = n \times\lambda+\Delta \times \lambda\)

On substitution, we get

\(2 D =40 \times\left[2+\left(\frac{180-0}{360}\right)\right]\)

\(D =50~ m\)

LED works on the principal of spontaneous emission.

In the avalanche photo diode due to the avalanche effect there is large current gain.

Tunnel diode has very large doping.

LASER diode is used for coherent radiation.

Given a mass of copper slab \(=2 g\) contains \(2 \times 10^{22}\) atoms.

We have to find the fraction of electrons that must be removed from sphere to give it \(+2 \mu C\) charge.

Charge on nucleus of each atom \(=29 e\)

\(\therefore\) Net charge on \(2 g\) sphere

\(=(29 e ) \times\left(2 \times 10^{22}\right)=5.8 \times 10^{23} ec\)

\(\therefore\) No of electrons on sphere \(=5.8 \times 10^{23}\)

\(\therefore\) Number of electrons removed to give \(2 \mu c\) charge \(=\frac{q}{e}\)

\(=\frac{2 \times 10^{-6}}{1.6 \times 10^{-19}}\)

\(=1.25 \times 10^{13}\)

Fraction of electrons removed \(=\frac{1.25 \times 10^{13}}{\text { Total number of electrons in sphere }}\)

\(=\frac{1.25 \times 10^{13}}{29 \times 2 \times 10^{22}}\)

\(=2.16 \times 10^{-11}\)

Given that, mass = 22 kg, velocity = 5 m/s

We know that,

K.E =$\frac{1}{2}\times$m$\times$v²

∴ K.E =$\frac{1}{2}\times$22$\times$5²

K.E = 275 J

Therefore, the kinetic energy of the object is 275 J.