Physics Test-34

It is given that,

Water flows at a rate of \(3.0\) litre \(/ min =3 \times 10^{-3} m ^{3} / min\)

Density of water, \(\rho=10^{3} kg / m ^{3}\).

Clearly, mass of water flowing per minute \(=3 \times 10^{-3} \times 10^{3} kg / min =3 kg / min\)

The geyser heats the water, raising the temperature from \(27^{\circ} C\) to \(77^{\circ} C\).

Initial temperature, \(T_{1}=27^{\circ} C\)

Final temperature, \(T_{2}=77^{\circ} C\)

Thus, rise in temperature,

\(\Delta T=T_{2}-T_{1}\)

\(\Delta T=77^{\circ} C -27^{\circ} C \)

\(\Delta T=50^{\circ} C\)

Now, heat of combustion \(=4 \times 10^{4} J / g =4 \times 10^{7} J / kg\)

Specific heat of water \(=4.2 J / g^{\circ} C\)

It is known that total heat used, \(\Delta Q=m c \Delta T\)

\(\Delta Q=3 \times 4.2 \times 10^{3} \times 50 \)

\(\Delta Q=6.3 \times 10^{5} J / min\)

Now, consider \(m kg\) of fuel to be used per minute.

Thus, the heat produced \(=m \times 4 \times 10^{7} J / min\)

However, the heat energy taken by water \(=\) heat produced by fuel

Thus, equating both the sides,

\(\Rightarrow 6.3 \times 10^{5}=m \times 4 \times 10^{7} \)

\(\Rightarrow m=\frac{6.3 \times 10^{5}}{4 \times 10^{4}} \)

\(\Rightarrow m=15.75 g / min\)

Clearly, the rate of consumption of the fuel when its heat of combustion is \(4.0 \times 10^{4} J / g\) supposing the geyser operates on a gas burner is \(15.75 g / min\).

Given that,

Mass of train, \(\mathrm{m}=2.05 \times 10^{6} \mathrm{Kg}\)

Time, \(\mathrm{t}=5\) minutes \(=300 \mathrm{~s}\)

\(\mathrm{v}=25 \mathrm{~m} / \mathrm{s}\)

\(\mathrm{u}=5 \mathrm{~m} / \mathrm{s}\)

Acceleration; \(a=\frac{v-u}{t}\)

\(a=\frac{25-5}{300}\)

\(a=\frac{2}{30}=\frac{1}{15} \mathrm{~m} / \mathrm{s}^{2}\)

Using equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\)

\((25)^{2}-(5)^{2}=2 \times \frac{1}{15} \times s\)

\(\mathrm{s}=4500 \mathrm{~m}\)

Power,\(\mathrm{P}=\frac{\text { work done }}{\text { time }}\)\(=\frac{\mathrm{F} \times \mathrm{d}}{\mathrm{t}}\)\(=\frac{m \times a \times s}{t}\)

\(=\frac{2.05 \times 10^{6} \times 1 \times 4500}{15 \times 300}\)

\(\mathrm{P}=2.05 \times 10^{6} \mathrm{~W}\)

\(\mathrm{P}=2.05 \mathrm{MW}\)

The expression \(\frac{I \omega^2} { 2}\) represents rotational kinetic energy where I = moment of inertia and ω = angular velocity.

When electrodes are immersed in water and electricity passed, the bubbles formed on the negative terminal is actually hydrogen gas.

When an electric current is passed through acidulated water, it will produce a reaction called electrolysis. If an electric current is passed through acidified water, then bubbles of oxygen gas(O₂) and hydrogen(H₂) gas are produced at the two electrodes immersed in it.

The Electromagnetic waves aretransverse waves.

Transverse waves:

• Those waves whose direction of propagation and direction of disturbance is always perpendicular, are known as transverse waves.
• These waves produced in a medium that can sustain shearing strain.
• Example: Electromagnetic Waves, Ripples on the surface of water, Vibrations in a guitar string.

On increasing the temperature of a semiconductor diode, its resistance decreases.

• Compared to insulators, the energy band gap is very less in semi-conductors. Thus, the electrons require very little energy to jump to the conduction band.
• When the temperature is increased, most of the electrons acquire a sufficient amount of energy faster and jumps to the conduction band.

We know that the voltage sensitivity is given as,

\(V_{s}=\left(\frac{I_{s}}{R}\right) \quad \ldots\)(1)

Where, \(V _{ s }=\) voltage sensitivity, \(I _{ s }=\) current sensitivity, and \(R =\) resistance

When the resistance is increased by factor 2,

\(R^{\prime}=2 R\quad \ldots\)(2)

\(I_{s}^{\prime}=I_{s}+20 \% \text { of } I_{s}\)

\(\Rightarrow I_{s}^{\prime}=I_{s}+\frac{20}{100} I_{s}\)

\(\Rightarrow I_{s}^{\prime}=1.2 I_{s}\quad \ldots\)(3)

From equation (1) we get,

\( V_{s}^{\prime}=\frac{I_{s}^{\prime}}{R^{\prime}}\)

\(\Rightarrow V_{s}^{\prime}=\frac{1.2 I_{s}}{2 R}\quad \ldots\)(4)

By equation (1) and equation (4)

\(V_{s}^{\prime}=0.6 V_{s}\quad \ldots\)(5)

The change in voltage sensitivity is given as,

\(\% \Delta V=\frac{V_{s}^{\prime}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=\frac{0.6 V_{s}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=-40 \%\)

In fluid dynamics, Bernoulli’s theorem gives the relation among the elevation, velocity, and pressure in a moving fluid, for which, the viscosity and compressibility are negligible and the flow is laminar or steady.

\(\frac{ P _{1}}{\rho}+ gh _{1}+\frac{1}{2} v _{1}^{2}=\frac{ P _{2}}{\rho}+ gh _{2}+\frac{1}{2} v _{2}^{2}\)

\(\frac{ P }{\rho}+ gh +\frac{1}{2} v ^{2}=\) constant

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., \(T^{2} \propto R^{3}\)

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

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

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

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

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

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

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).

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

\(=10\sqrt{2}\)

\(=14.1\)