Physics Test - 70

The displacement is given as:

\({y}={ut}+\frac{1}{2} {at}^{2}\)

where, \(u\) is the initial velocity and \(a\) is the acceleration.

Comparing with given expression:

\({y}=16 {t}-\frac{2 {t}^{2}}{3}+2\),

We get,

\({u}=16 {~m} / {s},\) and \({a}=-\frac{4}{3} {~m} / {s}^{2}\)

As, \(v=u+at\), where \(v\) is final velocity.

Taking \({v}=0\), the time taken for body to come to rest. \(0=16-\frac{4}{3} {t} \)

\(\Rightarrow {t}=12 {sec}\)

Given:

Potential difference \(( V )=220 V _{1}\)

Power of the bulb \(( P )=100 W\)

Actual voltage \(\left( V ^{\prime}\right)=110 V\)

We know that:

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

The resistance of the bulb can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{100}=484 \Omega\)

The power consumed by the bulb.

\(P=\frac{V^{'2}}{R}=\frac{(110)^{2}}{484}=25 W\)

\(\mathrm F=\) Electrostatic force.
For equilibrium \(\mathrm {\tan \theta_1 =\frac{F}{2 m g} \text { and } \tan \theta_2=\frac{F}{m g}}\)
\(\mathrm \tan \left(\theta_1+\theta_2\right) =\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}\)
\(\mathrm {\tan 45^{\circ}=\frac{\frac{F}{2 m g}+\frac{F}{m g}}{1-\frac{F}{2 m g} \cdot \frac{F}{m g}} }\)
\(\Rightarrow \mathrm { 2 m^2 g^2-F^2=3 m g F \Rightarrow F^2+3 m g F-2 m^2 g^2=0} \)
\(\therefore \mathrm { F=\frac{-3 m g \pm \sqrt{9 m^2 g^2+8 m^2 g^2}}{2}} \)\(\mathrm=\left(\frac{\sqrt{17}-3}{2}\right) \mathrm{mg} \)

Let the density of the liquid be \(\rho\).

Mass of the liquid drops must be equal to the total mass of the droplets.

\(\therefore \rho \times \frac{4 \pi}{3} R ^{3}=64 \times \rho \times \frac{4 \pi}{3} r ^{3}\) 

\( \Rightarrow R =4 r\)

Change in surface area,

\(\Delta A =64 \times 4 \pi r ^{2}-4 \pi R ^{2}\)

\(\therefore \Delta A =64 \times 4 \pi \times r ^{2}-4 \pi(4 r )^{2}\)

\(=192 \pi r ^{2}\)

\(\therefore\) Gain in energy,

\(\quad E = T \Delta A\)

\( =192 \pi r ^{2} T\)

As, we know de-Broglie wavelength,

\(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)

\(\lambda \propto \frac{1}{\mathrm{p}}\)

\(\Rightarrow \frac{\Delta \mathrm{p}}{\mathrm{p}}=\frac{\Delta \lambda}{\lambda}\)

\(\Rightarrow\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|\)

\(\frac{\Delta \mathrm{p}}{\mathrm{p}}=\frac{0.20}{100}\)

\(\frac{\Delta \mathrm{p}}{\mathrm{p}}=\frac{1}{500}\)

\(\mathrm{p}=500 \mathrm\Delta{p}\)

Given: 

\(T _{1}=27+273=300\) K, \(\gamma =\frac{5}{3}, P _{2}=\frac{1}{8} P _{1}\)

Since gas is compressed suddenly, gas is undergone an adiabatic process.

In the adiabatic process,

\(P V^{\gamma}=\) constant

We also know that:

\(T ^{ V } P ^{1- \gamma}=\) Constant

\(T _{1} v P _{1}{ }^{1- \gamma }= T _{2} vP _{2} ^{1- \gamma}\)

Where, \(T_{1}\) is the initial temperature, \(T_{2}\) is the final temperature \(P_{1}\) is the initial pressure, \(P_{2}\) is the final pressure and \(\gamma\) is the adiabatic constant.

\(\frac{P_{1}^{1-\gamma}}{P_{2}^{1-\gamma}}=\frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}\)

\(\left(\frac{P_{1}}{P_{2}}\right)^{1-\gamma}=\left(\frac{T_{2}}{T_{1}}\right)^{\gamma}\)

\(\left(\frac{P_{1} \times 8}{P_{1}}\right)^{1-\left(\frac{5}{3}\right)}=\left(\frac{T_{2}}{300}\right)^{\frac{5}{3}}\)

\(T _{2}=130.58 K\)

\(T _{2}=130.58-273=-142.42^{\circ} C\)

Thus, The final temperature will be \(-142^{\circ} C\).

Young modulus \(Y=\frac{F l}{A \Delta l}\)

\(\text { but } V=A l \)

\(\Rightarrow l=\frac{V}{A}\)

\(\therefore Y=\frac{F V}{A^{2} \Delta l}\)

\(\therefore \text { in } F=\frac{Y A^{2} \Delta l}{V} Y, \Delta l \text { and } \mathrm{V} \text { are equal }\)

\(\therefore F \propto A^{2} \)

\(\therefore \frac{F_{2}}{F_{1}}=\left(\frac{A_{2}}{A_{1}}\right)^{2}\)

\(=\left(\frac{3 A}{A}\right)^{2}=9\)

\(\therefore F_{2}=9 F \quad\left[\because F_{1}=F\right]\)

The condition for maxima in Young's double-slit experiment is \({d} \sin \theta={n} \lambda\) where \({d}\) is the separation between the slits and \(\lambda\) is the wavelength of light used.

The maxima that are farthest from the slits is infinitely up or infinitely down the screen and corresponds to \(\theta=\frac{\pi}{2}\).

So, with the given condition, we have, \(n=\frac{d\sin(90^{\circ})}{\lambda}\)

\(=\frac{d}{ \lambda}\) \(=2\)

Thus, we have two maxima on the screen on either side of the central maxima. Thus, the maximum number of possible maxima observed are \(2+1+2=5\).

Spring force is maximum when the system is released, \(\mathrm {F_{s \max }=k x_0=240 \mathrm{~N}}\)
The limiting friction on \(\mathrm A\) can be \(\mathrm {F_{l A}=\mu_1 M_A g=250 \mathrm{~N}}\)
Hence, block A remains fixed and does not move at all.
The limiting friction on \(\mathrm C\) can be: \(\mathrm {F_{l c}=\mu_2 M_{C }g}\)
Thus maximum acceleration that friction can provide to \(\mathrm C\) is: \(\mathrm {a_{c \max }=\mu_2 \mathrm{~g}=8.5 \mathrm{~m} / \mathrm{s}^2}\)
Just after the release spring force is maximum and it will cause maximum acceleration in \(\mathrm B\). Let us assume that there is no slipping between \(\mathrm B\) and \(\mathrm C\). In that case the maximum acceleration is: \(\mathrm a_{\mathrm c \max }=\frac{240}{28+2}=8.0 \mathrm{~m} / \mathrm{s}^2\)
Friction can easily provide this acceleration to block \(\mathrm {C}\) hence it will not slip over \(\mathrm {B}\).
Speed is maximum when the spring acquires its natural length. \(\mathrm {\frac{1}{2}\left(M_b+M_C\right) v_{\max }^2=\frac{1}{2} k x_0^2}\) \(\mathrm {\Rightarrow v_{\max }=4 \mathrm{~m} / \mathrm{s}}\)

Given, frequency, \(f=500 \mathrm{MHz}=5 \times 10^8 \mathrm{~Hz}\)

\(\mathrm{B}=8.0 \times 10^{-8} \hat{\mathrm{Z}}_{\mathrm{T}}\)

\(\therefore\) Magnitude of peak value of magnetic field is given by

\(\mathrm{B}_0=8 \times 10^{-8} \mathrm{~T}\)

We know that, \(\frac{\mathrm{E}_0}{\mathrm{~B}_0}=\mathrm{C}\)

where, \(\mathrm{E}_0\) is the magnitude of peak value of electric field and \(\mathrm{c}\) is the speed of electromagnetic wave in air (or vacuum).

\(\Rightarrow \mathrm{E}_0=\mathrm{cB}_0 \)

\(=3 \times 10^8 \times 8 \times 10^{-8}=24 \mathrm{~V} / \mathrm{m}\)

Since, the direction of propagation of electromagnetic wave is perpendicular to the direction of \(\mathrm{E}\) and \(\mathrm{B}\) both.

\(\therefore\) Direction of propagation is given by \(\hat{E} \times \hat{B}\).

As, the wave is travelling in \(y\)-direction, and the magnetic field is in z-direction.

\(\Rightarrow \hat{E} \times \hat{z}=\hat{y} \)

\(\hat{E}=-\hat{x}\)

\(\therefore\) The value of electric field will be \(-24 \hat{x} \mathrm{~V} / \mathrm{m}\).