Physics Test - 22

Brewster Law \(\tan i _{ p }=\mu\) but \(\mu \propto \frac{1}{\lambda}\)

\(\therefore \tan i _{ p } \propto \frac{1}{\lambda}\)

where \(i _{ p }, \mu\) and \(\lambda\) are angle of polarisation (incidence), refractive index of material and wavelength of light respectively

The de-Broglie wavelength of a particle of mass \(\mathrm{m}\) and moving with velocity \(\mathrm{v}\) is given by,

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})\)

de-Broglie wavelength of a proton of mass \(\mathrm{m}_{1}\) and kinetic energy \(\mathrm{k}\) is given by,

\(\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{k}}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mk}})\)

\(\Rightarrow \lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{qV}}} \ldots . .\) (i) \(\quad[\because \mathrm{k}=\mathrm{qV}]\)

For an alpha particle mass \(\mathrm{m}_{2}\) carrying charge \(\mathrm{q}_{0}\) is accelerated through potential \(\mathrm{V}\), then,

\(\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{q}_{0} \mathrm{~V}}}\)

\(\because\) For \(\alpha-\) particle \(\left({ }_{2}^{4} \mathrm{He}\right): \)

\(\mathrm{q}_{0}=2 \mathrm{q}\) and \(\mathrm{m}_{2}=4 \mathrm{~m}_{1}\)

\(\therefore \lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{1} \times 2 \mathrm{q} \times \mathrm{V}}}.....\) (ii)

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{q} \mathrm{V}}} \times \frac{\sqrt{2 \times \mathrm{m}_{1} \times 4 \times 2 \mathrm{q} \mathrm{V}}}{\mathrm{h}}\)

\(=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

\(\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=2 \sqrt{2}\)

Rotational motion: When a block is moving about a fixed axis on a circular path then this type of motion is called rotational motion.

Torque (τ):

• It is the twisting force that tends to cause rotation.
• The point where the object rotates is known as the axis of rotation.
• Mathematically it is written as,

τ = rFsin θ

• Torque (τ)​ is a physical quantity, similar as force that causes the rotational motion.

⇒ Torque (τ) = I × α

Where I = moment of inertia, α = angular acceleration

As it is given that τ = 0, therefore

⇒ I × α = 0

As we know I ≠ 0, therefore

⇒ α = 0

We know that \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\) and \(\mathrm{E}=\frac{\mathrm{I}}{\sigma \mathrm{A}}\);

\(\therefore \mathrm{E}=\frac{\mathrm{J}}{\sigma}=\rho \mathrm{J}=\mathrm{kJ}\)

In given experiment, a composite wire is stretched by a force$F$.

Net elongation in the wire = elongation in brass wire + elongation in steel wire .. . (i)

Now, Young’s modulus of a wire of cross-section$\left(A\right)$when some force$(F$is applied,

\(Y=\frac{F I}{A \Delta I}\)

We have,

\(\Delta {I}=\) elongation \(=\frac{{FI}}{{AY}}\)

So, from relation (i), we have

\(\Delta {I}_{\text {net }}=\Delta {I}_{\text {brass }}+\Delta {I}_{\text {steel }}\)

\(\Rightarrow \Delta {I}_{\text {net }}=\left(\frac{{FI}}{{AY}}\right)_{\text {brass }}+\left(\frac{{FI}}{{AY}}\right)_{\text {steel }}\)

As wires are connected in series and they are of same area of cross-section, length and subjected to same force, so

\(\Delta {I}_{\text {net }}=\frac{{F}}{{A}}\left(\frac{{I}}{{Y}_{\text {brass }}}+\frac{{I}}{{Y}_{\text {steel }}}\right)\)

Here,

\(\Delta {I}_{\text {net }}=0.2 {~mm}\)

\(=0.2 \times 10^{-3} {~m}\)

\(\text { and } {I}=1 {~m}\)

\({Y}_{\text {brass }}=60 \times 10^9 {Nm}^{-2}\)

\({Y}_{\text {steel }}=120 \times 10^9 {Nm}^{-2}\)

On putting the values, we have

\(0.2 \times 10^{-3}=\frac{F}{A}\left(\frac{1}{60 \times 10^9}+\frac{1}{120 \times 10^9}\right)\)

\(\Rightarrow \text { Stress }=\frac{F}{A}=8 \times 10^6 {Nm}^{-2}\)

The power of the lens should be increased to increase the angular magnification of a simple microscope.

when the final image is formed at the near point,

\(m_{e}=\left(1+\frac{D}{f_{e}}\right)\)

When the final image is formed at infinity,

\(m_{e}=\left(\frac{D}{f_{e}}\right)\)

Here Increament and decreament of angular magnification depends upon value of \(\left(\frac{1}{f_{e}}\right)\) which is the power of the lens.

Work done during isothermal process,

\(\begin{aligned} & \mathrm{W}_{\mathrm{AB}}=n R T \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & \mathrm{W}_{\mathrm{AB}}=\mathrm{nRT} \ln \left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)=\mathrm{nRT} \ln 2 \\ & \mathrm{~W}_{\mathrm{BC}}=\mathrm{nR}\left(\mathrm{T}_{\mathrm{C}}-\mathrm{T}_{\mathrm{B}}\right)=\mathrm{nR}\left(\frac{\mathrm{T}}{2}-\mathrm{T}\right)=-\mathrm{nR} \frac{\mathrm{T}}{2} \\ & \mathrm{~W}_{\mathrm{CA}}=\mathrm{P} \Delta \mathrm{V}=0(\because \Delta \mathrm{V}=0) \\ & \Rightarrow \mathrm{W}_{\text {net }}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}} \\ & \Rightarrow \mathrm{W}_{\text {net }}=\mathrm{nRT} \ln 2-\frac{\mathrm{nRT}}{2} \\ & \Rightarrow \mathrm{W}_{\text {net }}=\mathrm{nRT}\left(\ln 2-\frac{1}{2}\right)\end{aligned}\)

Using lens maker's formula

\(\frac{1}{\mathrm{f}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]\)

Here, \(\mu_{\mathrm{g}}\) and \(\mu_{\mathrm{a}}\) are the refractive index of glass and air respectively

\(\Rightarrow \frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)

When immersed in liquid

\(\frac{1}{\mathrm{f}_1}=\left(\frac{\mu_{\mathrm{g}}}{\mu_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

[Here, \(\mu_1=\) refractive index of liquid]

\(\Rightarrow \frac{1}{f_1}=\left(\frac{1.5}{1.42}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Dividing (i) by(ii)

\(\Rightarrow \frac{\mathrm{f}_1}{\mathrm{f}}=\frac{(1.5-1) 1.42}{0.08}=\frac{1.42}{0.16}=\frac{142}{16} \approx 9\)

It causes a twisting of such elements due to an external applied torque. Specifically, torsional vibrations (or torsional oscillations) can be defined as the periodical movement of a shaft. Therefore, such shaft twists ups around itself, alternating its turning direction.

We know that, the torsional oscillation frequency is given by the formula:

\(\omega=\frac{\mathrm{k}}{\sqrt{\mathrm{I}}}\)

Where,

\(\mathrm{k}=\) Torsion constant

I = Moment of Inertia

Rod without mass:

The moment of inertia of rod is:

\(\mathrm{I}=\frac{\mathrm{ML}^{2}}{12}\)

Where,

\(M=\) Mass of \(\operatorname{rod}=\mathrm{M}\) (given)

\(\mathrm{L}=\) Length of \(\operatorname{rod}=2 \mathrm{~L}\) (given)

Now,

\(\Rightarrow \mathrm{I}=\frac{\mathrm{M}(2 \mathrm{~L})^{2}}{12}\)

\(\Rightarrow \mathrm{I}=\frac{\mathrm{M}\left(4 \mathrm{~L}^{2}\right)}{12}\)

\(\therefore \mathrm{I}=\frac{\mathrm{ML}^{2}}{3}\)

Now, the torsional oscillation frequency of rod is:

\(\therefore \omega_{1}=\frac{\mathrm{k}}{\sqrt{\frac{\mathrm{ML}^{2}}{3}}}\)

Rod with mass:

The moment of inertia of rod and two masses is:

\(\mathrm{I}=\frac{\mathrm{ML}^{2}}{12}+2 \mathrm{~m}(\mathrm{l})^{2}\)

Where,

\(\mathrm{m}=\) Attached mass \(=\mathrm{m}\) (given)

\(\mathrm{I}=\) Length from which the mass is attached \(=\mathrm{L} / 2\) (given)

\(M=\) Mass of \(\operatorname{rod}=M\) (given)

\(\mathrm{L}=\) Length of \(\operatorname{rod}=2 \mathrm{~L}\) (given)

Now,

\(\Rightarrow \mathrm{I}=\frac{\mathrm{M}(2 \mathrm{~L})^{2}}{12}+2 \mathrm{~m}\left(\frac{\mathrm{L}}{2}\right)^{2}\)

\(\Rightarrow \mathrm{I}=\frac{\mathrm{ML}^{2}}{3}+\frac{2 \mathrm{~mL}^{2}}{4}\)

\(\therefore \mathrm{I}=\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\)

Now, the torsional oscillation frequency of rod with mass is:

\(\therefore \omega_{2}=\frac{\mathrm{k}}{\sqrt{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}}\)

From question, after the mass is attached its frequency which is reduced by \(20 \%(100 \%-20 \%=80 \%)\)

\(\omega_{2}=80 \%\) of \(\omega_{1}\)

\(\omega_{2}=\frac{80}{100} \omega_{1}\)

\(\Rightarrow \omega_{2}=0.8 \omega_{1}\)

Now, substituting the torsional oscillation frequencies,

\(\Rightarrow \frac{\mathrm{k}}{\sqrt{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}}=(0.8) \frac{\mathrm{k}}{\sqrt{\frac{\mathrm{ML}^{2}}{3}}}\)

\(\Rightarrow \frac{1}{\sqrt{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}}=\frac{0.8}{\sqrt{\frac{\mathrm{ML}^{2}}{3}}}\)

Squaring on both sides,

\(\Rightarrow \frac{1}{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}=\frac{0.64}{\frac{\mathrm{ML}^{2}}{3}}\)

\(\Rightarrow \frac{\mathrm{ML}^{2}}{3}=0.64\left(\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\right)\)

\(\Rightarrow \frac{\mathrm{ML}^{2}}{3}=\left((0.64) \frac{\mathrm{ML}^{2}}{3}\right)+\left((0.64) \frac{\mathrm{mL}^{2}}{2}\right)\)

\(\Rightarrow \frac{\mathrm{ML}^{2}}{3}-(0.64) \frac{\mathrm{ML}^{2}}{3}=(0.64) \frac{\mathrm{mL}^{2}}{2}\)

\(\Rightarrow(0.36) \frac{\mathrm{ML}^{2}}{3}=(0.64) \frac{\mathrm{mL}^{2}}{2}\)

\(\Rightarrow \frac{\mathrm{m}}{\mathrm{M}}=\frac{2}{3} \times \frac{0.36}{0.64}\)

\(\therefore \frac{\mathrm{m}}{\mathrm{M}}=0.375\)

When the lift is stationary spring force balances weight:

\({kx}=\mathrm{mg}=49 \mathrm{~N}...(1)\)

Where,

\(k=\) force constant of spring

\(x=\) elongation

True weight \(=49 \mathrm{~N}\)

From equation (1), we get

\(k=\frac{49}{x}\)

\(\mathrm{m}=\frac{49}{9.8}=5 \mathrm{~kg}\)

when lift moves with acceleration \(5 \mathrm{~m} / \mathrm{s}^{2}\) downward we have:

\({kx}_{2}=\mathrm{mg}-5 \times \mathrm{m}\)

Where,

Pseudo force in lift frame \(=5m\)upward

\({x}_{2}=\) new elongation

\(\Rightarrow \mathrm{kx}_{2}=49-5 \times 5=24 \mathrm{~N}\)

So new reading in spring balance \(=24\) N