Physics Test-2

Let us consider an elementary ring of radius \(r\) and thickness \(\mathrm{dr}\) in which current \(\mathrm{I}\) is flowing.

Number of turns in this elementary ring \(\mathrm{d} \mathrm{N}=\frac{N}{b-a} d r\)

We know that the magnetic field at the centre of a ring is given by:

\(\mathrm{B}=\frac{\mu_{0} N I}{2 r}\)

Where,

\(\mu_{0}\) is the permeability of free space.

\(N\) is the number of terms.

\(\mathrm{L}\) is currently flowing.

\(r\) is the radius of the ring.

Thus magnetic field at the centre \(O\) due to this ring \(\mathrm{dB}=\frac{\mu_{0} d N I}{2 r}\)

Putting the value of \(\mathrm{dN}\), we get:

\(\mathrm{dB}=\frac{\mu_{0} N I d r}{2(b-a) r}\)

To get the net magnetic field at the centre of the spiral, we will integrate the above equation from 'a' to 'b'. \(\int_{0}^{B} d B=\int_{a}^{b} \frac{\mu_{0} N I d r}{2(b-a) r}=\frac{\mu_{0} N I}{2(b-a)} \int_{a}^{b} \frac{d r}{r}=\frac{\mu_{0} N I}{2(b-a)}[\ln r]_{a}^{b}\)

\(\Rightarrow B=\frac{\mu_{0} N I}{2(b-a)} \ln \frac{b}{a}\)

Therefore, the required magnetic field is 

\(\Rightarrow B=\frac{\mu_{0} N I}{2(b-a)} \ln \frac{b}{a}\)

Given,

\({R}=20~ \mathrm{m}\)

Velocity of the motor cyclist

\(v=72~ \mathrm{km} / \mathrm{hr}\)

\(=72 \times \frac{5}{18}\)

\(=20~ \mathrm{m} / \mathrm{s}\)

Let the motorcyclist bends with respect to vertical plane by angle \(\theta\) To avoid skidding, \(\frac{v^{2}}{g R} \geq \tan \theta\).

\(\tan \theta \leq \frac{20^{2}}{10 \times 20}\)

\(\Rightarrow \theta \leq \tan ^{-1}(2)\)

The loops of electrical current induced within conductors by changing magnetic fields in the conductor are called eddy currents.

• Eddy currents are generated according to Faraday’s law of induction.
• Eddy currents transform useful energy, into heat, which isn’t generally useful. Therefore, option (D) is correct.
• Eddy currents flow in closed loops, in the perpendicular plane of the magnetic field, not in a straight line. Therefore, option (A) is incorrect.
• Eddy currents produce undesirable effects if the large eddy current flows in the transformer or core of a choke coil etc.
• Eddy currents lose the generated electrical energy by generating heat. Therefore, option (B) is incorrect.
• By using a laminated core, the eddy currents produced in the core of a transformer are reduced. Eddy currents cause a loss of energy because they have the tendency to oppose. Therefore, option (C) is incorrect.

We can simplify the network as shown in figure.

Effective resistance through where \(I\) flow is given by

\(R_{1}=\frac{4 \times 4}{414}+4\)

\(=\frac{16}{8}+4\)

\(=6 \Omega\)

This \(6 \Omega\) is in parallel with \(4 \Omega\) and in series with \(1.6 \Omega\). So the effective resistance of the circuit is given by

\(R_{\text {total }}=\frac{6 \times 4}{6+4}+1.6\)

\(=\frac{24}{10}+1.6\)

\(=4 \Omega\)

The current on the battery is given by

\(i=\frac{V}{R}=\frac{4}{4}\)

\(=1 \mathrm{~A}\)

The current \(I\) will be given as:

\(I=\frac{4}{4+6} \times 1=0.4 A\)

Projectile motion: 

When a particle is projected obliquely near the earth's surface, it moves simultaneously in horizontal and vertical directions. The path of such a particle is called projectile and the motion is called projectile motion.

Range of projectile:

The horizontal range of a projectile is the distance along the horizontal plane it would travel, before reaching the same vertical position as it started from.

Formulae in projectile motion:

Range of projectile \(=\frac{u^{2} \sin 2 \theta}{g}\)

Total time of flight \(=\frac{2 u \sin \theta}{g}\)

Maximum Height \(=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

where \(u=\) projected speed, \(\theta=\) angle at which an object is thrown from the ground and \(g=\) acceleration due to gravity \(=9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Given:

The range of a Projectile motion is given by (R):

\(=\frac{u^{2} \sin 2 \theta}{g}\)

For horizontal distance to be maximum:

sin\(2\theta\) = \(1\)

∴ sin \(2\theta\) = sin \(90°\) 

∴ \(\theta = 45°\).

∴ to take the longest possible jump, an athlete should make an angle of \(45°\) with the ground.

Given,

Mass, \(m=1\) tonne \(=1000 \mathrm{~kg}\)

Initial velocity, \(u=30 \mathrm{~m} / \mathrm{s}\)

Final velocity, \(v=0 \mathrm{~m} / \mathrm{s}\)

Here, friction does the work, and therefore work done is negative.

So, we can consider \({F}=-{k} \sqrt{{x}}\)

From the work-energy theorem,

\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x\)

\(\Rightarrow \frac{1}{2} m\left(v^{2}-u^{2}\right)=\int_{0}^{d}-k \sqrt{x} d x\)

\(\Rightarrow \frac{1}{2} \times 1000 \times\left(0^{2}-30^{2}\right)=\int_{0}^{d}-1000 \sqrt{x} d x\)

\(\Rightarrow-450=\left[-\frac{2 x^{\frac{3}{2}}}{3}\right]_{0}^{d}\)

\(\Rightarrow 450=\frac{2 d^{\frac{3}{2}}}{3}\)

\(\mathrm{d}=77 \mathrm{~m}\)

\(\therefore\) The distance covered by car before stopping is \(77 \mathrm{~m}\)

The force exerted by liquid at the other end is the centripetal force due to rotation. The case can be treated as rotation of the center of mass of the liquid at distance \(\frac{\mathrm{L}}{2}\) from the axis of rotation.

So, Force exerted by liquid at the end \(=\mathrm{M} \omega^{2}\left(\frac{\mathrm{L}}{2}\right)=\frac{1}{2} \mathrm{M} \omega^{2} \mathrm{~L}\)

For free-space transmission, it wouldn't be good to use "high" energy radiation such as gamma rays, ultraviolet rays, and x-rays because they could be dangerous for people's health. Also, the air can scatter high-energy radiation (we see a blue sky because of this scattering of blue and violet light for instance).

• Radio waves have a wavelength varying from a millimeter to 10 km.
• The higher the frequency of the electromagnetic wave, the more it penetrates into the atmosphere.
• The frequency of the radio waves is less than \(3 \times 10^{11} \mathrm{~Hz}\).
• The frequency of the microwaves is varies from \(3 \times 10^{11} \mathrm{~Hz}\) to \(10^{13} \mathrm{~Hz}\).
• Since low-frequency radio waves are stopped by the earth's atmosphere, and some radio waves are reflected back by the part of the earth's atmosphere.

So, Microwaves are preferred in order to communicate with the artificial satellites which are orbiting the earth at a particular height from the surface of the earth. Microwaves are preferred because they can penetrate through the Earth's atmosphere very easily and reach the target satellite.

The given relation,

\(p=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)

\(pe^0=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)

By comparing both sides, we get

\(\frac{\alpha z}{k \theta} = 0\)

\(\Rightarrow \alpha=\frac{k \theta}{z}\)

\(\Rightarrow [\alpha]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1} \times \mathrm{K}\right]}{[\mathrm{L}]}\)

\(=\left[\mathrm{MLT}^{-2}\right]\)

And \( p=\frac{\alpha}{\beta}\)

\([\beta]=\left[\frac{\alpha}{p}\right]\)

\(=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}\)

\(=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

Given,

\({E}=5.4 {eV}\) and \(\phi_{0}=3.4 {eV}\)

By Einstein's equation of photoelectric effect:

\({KE}_{\max }={E}-\phi_{0}\)

\(=(5.4-3.4) {eV}\)

\(=(5.4-3.4) {eV}\)

\(=2 {eV}\)

Since \({e}=1.6 \times 10^{-19} {C}\)

\({KE}_{\max }=2 \times 1.6 \times 10^{-19} {~J}\)

\(=3.2 \times 10^{-19} {~J}\)

Suppose at some instant the particle is at radial distance \(r\) from centre of earth \(O .\) Since, the particle is constrained to move along the tunnel, we define its position as distance \(x\) from \(C\). Thus, the equation of motion of the particle is,

\(m a_{x}=F_{x}\)

The gravitational force on mass \(m\) at distance \(r\) is, 

\(F=\frac{G M m r}{R^{3}}(\)towards \(O)\)

Therefore, \(F_{x}=-F \sin \theta\)

\(=-\frac{G M m r}{R^{3}}\left(\frac{x}{r}\right)\)

\(=-\frac{G M m}{R^{3}} \cdot x\)

Since, \(F_{x} \propto-x\), motion is simple harmonic in nature. 

Further,

\( m a_{x} =-\frac{G M m}{R^{3}} \cdot x \)

or \( a_{x}=-\frac{G M}{R^{3}} \cdot x \)

\(\therefore\) Time period of oscillation is,

\(T=2 \pi \sqrt{\left|\frac{x}{a_{x}}\right|}=2 \pi \sqrt{\frac{R^{3}}{G M}}\)

The time taken by particle to go from one end to the other is \(\frac{\mathrm{T}}{2}\).

\(\therefore t=\frac{T}{2}=\pi \sqrt{\frac{R^{3}}{G M}}\)

Given:

\(\overrightarrow{\mathrm{E}}=(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N} / \mathrm{kg}\)

Force acting as unit mass \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{E}}\)

\(=(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N}\)

The motion of unit mass along the line \(4 \mathrm{y}=3 \mathrm{x}+9\)

\(\mathrm{y}=\frac{3 \mathrm{x}+9}{4}\)

At \(x_{2}=1, \quad y_{2}=\frac{12}{4}=3\)

Let the point be \(A\).

At \(\mathrm{x}_{1}=0, \quad \mathrm{y}_{1}=\frac{9}{4}\)

So vector \(\vec{d}\) from \(A\) to \(B\)

\(\overrightarrow{\mathrm{d}} =\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right) \hat{\mathrm{i}}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right) \hat{\mathrm{j}} \)

\(=(1-0) \hat{\mathrm{i}}+\frac{(3-9)}{4} \hat{\mathrm{j}} \)

\(=\hat{\mathrm{i}}+\left(\frac{12-9}{4}\right) \hat{\mathrm{j}} \)

\(=\hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\)

Now work done is moving unit mass a long given line is:

\(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{d}}\)

\(=(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \cdot\left(\hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \)

\(=3 \hat{\mathrm{i}} \times \hat{\mathrm{i}}-4 \hat{\mathrm{j}} \times \frac{3}{4} \hat{\mathrm{j}} \)

\(=3-3=0\)

Given:

M = 6000 kg

\(ω_{0}\) = 0 rad/s,Angular speed (ω) after 2 min = 20 rad/s

time t = 2 min = 120 s

k = 100 cm = 1 m

ω =\(ω_{0}\)+ α t

20 = 0 + α × 120

α =\(\frac{20}{120}\)

= 0.1666 rad/s²

We know that:

I = M × k²

Where, I ismoment of inertia,M is mass, k isRadius of gyration.

I = 6000 ×1²= 6000 kg-m²

Now,T = I × α

Where T is torque, I is moment of inertia,α is angular acceleration.

T =6000 × 0.1666

= 1000 Nm

Let us assume that the charge \(q=\pm 10 \mu {C}=10^{-5} {C}\) is placed at a distance of \(5 {~cm}\) from the square \({ABCD}\) of each side \(10 {~cm}\). The square \({ABCD}\) can be considered as one of the six faces of a cubic Gaussian surface of each side \(10 {~cm}\).

Now, the total electric flux through the faces of the cube as per Gaussian theorem:

\(\phi=\frac{q}{\epsilon_{0}}\)

Therefore, the total electric flux through the square ABCD will be:

\(\phi_{{E}}=\frac{1}{6} \times \phi\)

\(=\frac{1}{6} \times \frac{{q}}{\epsilon_{0}}\)

\(=\frac{1}{6} \times \frac{10^{-5}}{8.854 \times 10^{-12}} \quad\) \((\because \epsilon_0 =8.854 \times 10^{-12})\)

\(=1.88 \times 10^{5} {Nm}^{2} {C}^{-1}\)

Given:

mA = 600 ml, mB = 250 ml,

TA = 60°C, and TB = 26°C

According to the principle of calorimetry,

Heat lost by liquid A = Heat gained by liquid B

⇒ mA SW ΔTA = mB SW ΔTB

⇒ 600 (60 –T) = 250 (T – 26)

⇒ 720 - 12T = 5T - 130

⇒ 850 = 17T

⇒ T = 50° C