Physics Test - 9

Given:

\(\mathrm{y}_{1}=10 \sin (2000 \pi \mathrm{t})\).....(1)

\(\mathrm{y}_{2}=10 \sin (2000 \pi \mathrm{t+\frac{\pi}{2}})\).....(2)

\(\mathrm{y}=a \sin((\omega t+ \phi)\).....(3)

On comparing (1) and (2) with (3).

\(\mathrm{a}_{1}=10, \mathrm{a}_{2}=10\) and \(\phi= \frac{\pi}{2}=90^{\circ}\)

The resultant amplitude A of two waves of amplitudes \(a_{1}\) and \(a_{2}\) at a phase difference of \(\phi\) is:

\(A=\left(a_{1}^{2}+\right.\) \(\left.\mathrm{a}_{2}^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \phi\right)^{\frac{1 }{ 2}}\).....(4)

Substituting all given value in (4).

\(A=\left(10^{2}+\right.\) \(\left.10^{2}+2 ×10 × 10 \cos 90^{\circ}\right)^{\frac{1 }{ 2}}\)

\(=10\sqrt{2}\)

\(=14.1\)

Given:

\(y_{1}=a \cdot \cos (\omega t)\) and \(y_{2}=2 a \cdot \cos (\omega t)\)

If the two waves undergo constructive interference, then the resultant displacement is given as,

\(y=y_{1}+y_{2}\)

\(\Rightarrow y=a \cos (\omega t)+2 a \cos (\omega t)\)

\(\Rightarrow y=3 a \cdot \cos (\omega t) \quad \ldots\) (1)

By equation (1), the amplitude 'A' of the resultant wave is given as,

A \(=3{a}\)

Let the charge distribution be as shown in figure:

\(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

or, \( \left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

\( \left(V_{A}-V_{B}\right)+E=\frac{{q}\left({C}_{2}+C_{1} \right)}{{C}_{1} {C}_{2}}\)

\(\therefore  {q}=\frac{\left[\left({V}_{{A}}-{V}_{{B}}\right)+{E}\right] {C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\)

Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 2.0}{1.0+2.0}\)

\(=10\) Volt

Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{1}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 1.0}{1.0 \times 2.0}\)

\(=5\) Volt

A balloon is rising with a velocity of \(10\) m/s. So when a packet is released its initial velocity will be the same as the ballon i.e., \(15\) m/s. So,

\(u =10\) ms\(^{-1}, s =-40\) m, \(a =- g =-10\) ms\(^{-2}\)

We know that:

\(s = ut +\left(\frac{1}{2}\right) at ^{2}\)

Where, \(u\) is initial velocity, \(a\) is constant acceleration, \(t\) is time and \(s\) is displacement.

\(-40=10 \times t +\left(\frac{1}{2}\right)(-10) t ^{2}\)

Solving equation \(t=-2\) (not possible) thus, we will take 4.

Therefore, \(t=4\) s

Initial flux through the coil,

\(\Phi_{B \text { (initial) }}=B A \cos \theta\)

\(=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 0^{\circ}\)

\(=3 \pi \times 10^{-7} ~Wb\)

Final flux after the rotation,

\(\Phi_{B(\text { final) }}=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 180^{\circ}\)

\(=-3 \pi \times 10^{-7} ~Wb\)

Therefore, estimated value of the induced emf is,

\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)

\(=500 \times\frac{\left(6 \pi \times 10^{-7}\right)}{0.25}\)

\(=3.8 \times 10^{-3} ~V\)

\(I=\frac{\varepsilon}{R}\)

\(=\frac{3.8 \times 10^{-3}}{2}\)

\(=1.9 \times 10^{-3} ~A\)

Given the Radius of the path = r

From the diagram, if we consider that the particle has started from point A and moves on the circumference of a circle.

According to the question it reaches its diametric end so if the particle starts from point A then it will reach to point B.

For this motion the minimum distance between point A and point B is 2r.

Displacement \(=2 r\)

So, \(v=\frac{2 r}{t}\)

Where, v is average velocity and t is time.

Given:

Potential difference \(( V )=220 V\)

Power of the bulb \(( P )=40 W\)

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 can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{40}=1210 \Omega\)

Given:

\(N=200, a=0.16\) m^\(2\)

Change in magnetic flux density \((dB)=0.5-0.1=0.4\) Wb/m^\(2\) 

\(t=0.02\) sec

Induced emf \(e=\frac{-N d \phi}{d t}\)

\(d \phi=(dB) \cdot a\)

\(=0.4 \times 0.16\)

\(=0.064\) Wb

\(e=\frac{-200 \times(0.064)}{0.02}\)

\(=-640\) V

A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to it.

The shunt resistance should have very low resistance. So, the ammeter (the parallel combination of galvanometer and shunt resistance) will have low resistance.

To convert a galvanometer into an ammeter of current rating \(‘I’\), a small resistance \(‘S’\) (shunt resistance) is connected in parallel across the galvanometer.

\(V_g=\left(I-I_g\right) S=I_g R_g\)

Where \(V_g\) is the voltage across the galvanometer, I is current in the circuit, \(I_g\) is the current is galvanometer, \(R _{ g }\) is the resistance of galvanometer and \(S\) is the resistance of the shunt.

A moving coil galvanometer can be converted into a ammeter by connecting a low resistance in in parallel to the moving coil galvanometer.

Given:

\(1^{\text {st }}\) orbital radius \(\left(R_{1}\right)=R\), and \(2^{\text {nd }}\) orbital radius \(\left(R_{2}\right)=4 R\)

According to the law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semimajor axis of the orbit i.e., \(T^{2} \propto R^{3}\)

\( \mathrm{T}^{2} \propto \mathrm{R}^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R_{1}}{R_{2}}\right)^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R}{4 R}\right)^{3}\)

\(=\frac{1}{64} \)

\(\Rightarrow \frac{T_{1}}{T_{2}}=\left(\frac{1}{64}\right)^{\frac{1}{2}}\)

\(=\frac{1}{8}\)