Physics Test - 4

Specific heat:  \(\left[ \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\)

Electrical potential: \(\left[M^{1} L^{2} T^{-3} A^{-1}\right]\)

Electrical resistance: \(\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\)

Magnetic flux: \(\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\)

From the above information, it is clear that the specific heat does not have mass in its dimension. 

We know that,

\(\text { Escape velocity } (v _{ e })=\sqrt{\frac{2 G M}{R}} \ldots \ldots . .(1)\)

Where \(G\) is the Universal Gravitational Constant and \(M\) is the mass of the Earth.

Velocity of the object.

\(v=\frac{v_{e} }{ 3}=\sqrt{\frac{2 G M}{9 R}}\)..... (2)

The initial kinetic energy of the object.

\(K E_{1}=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{2 G M}{9 R} \ldots\)... (3)

The initial potential energy of the object.

\(P E_{1}=-\frac{G M m}{R} \ldots \ldots .\) (4)

Here \(m\) is the mass of the object.

Let \(h\) be the maximum height reached by the object. when the object reaches the maximum height, its velocity becomes zero which is why its final kinetic energy \(= KE _{2}=0 \ldots \ldots \ldots\) (5)

At the maximum height, the potential energy of the object \(\left( PE _{2}\right)=-\frac{G M m}{R+h} \ldots \ldots .(6)\)

According to the law of conservation of energy,

\(KE _{1}+ PE _{1}= KE _{2}+ PE _{2}\)

(From \(3,4,5\), and 6)

\(\frac{1}{2} m \frac{2 G M}{9 R}-\frac{G M m}{R}=0-\frac{G M m}{R+h}\)

\(\Rightarrow \frac{1}{9 R}-\frac{1}{R}=-\frac{1}{R+h}\)

\(\Rightarrow -\frac{8}{9 R}=-\frac{1}{R+h} \)

\(\Rightarrow  8 R+8 h=9 R\)

\(\Rightarrow h=\frac{R}{8}\)

Thus, the maximum height reached will be \(\frac{R }{ 8}\).

Dimension of pressure, stress & Young's modulus is \(\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]\).

While dimension of power \(=\frac{\text { Energy }}{\text { Time }}\)

\(=\frac{\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}}{\mathrm{T}}\)

\(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\)

The rms speed of any homogeneous gas sample is given by:

\(V_{rms}=\sqrt{\frac{3 R T}{M}}\) \(\quad \dots\) (1)

Where, \(R =\) universal gas constant, \(T =\) temperature and \(M =\) Molecular mass

Here, \(M\) and \(R\) is constant,

On increasing the value of T by 3 in equation (1) we get,

\(V _{ rms } \propto \sqrt{3 T }\)

If the temperature is increased to \(3\) times, then \(V _{ rms }\) is increased by \(\sqrt{3}\) times.

The direction of EM waves is found from the cross product of the electric field and magnetic field. 

Since both electric and magnetic fields are vectors, the direction of propagation of EM waves is obtained from the right-hand rule.

Let the electric field be denoted by \(\vec{E}\) and magnetic field be denoted by \(\vec{B}\).

According to the rule - If the fingers of the right hand are curled so that they follow a rotation from \(\vec{E}\) to \(\vec{B}\), then the thumb will point in the direction of the vector product i.e., the direction of EM waves.

Given,

Mass of car, \(M=2000 \mathrm{~kg}\)

Mass of bullet, \(m=10 \times 10^{-3} \mathrm{~kg}\)

Velocity of bullet, \(u=500 \mathrm{~m} / \mathrm{sec}\)

The number of bullets fired per second is ten. Then,

\(\frac{N}{t}=10\)

\(F_{\text {avg }}=\frac{\Delta P}{\Delta t}\)

\(=\frac{N m (v_2-v_1)}{t}\)

\(=10 \times 10 \times 10^{-3} \times 5 \times 10^{2}\)

\(=50 \mathrm{~N}\)

Induced \(\mathrm{emf}=(\frac{1}{2}) \omega B R^{2}\)

\(=(\frac{1}{2}) \times 4 \pi \times 0.4 \times 10^{-4} \times(0.5)^{2}\)

\(=6.28 \times 10^{-5} \mathrm{~V}\)

The number of spokes is immaterial because the emf's across the spokes are in parallel.

Given,

Diameter of the sphere \(=2.4\)

\(\therefore\) Radius of sphere, \(r=\frac{2.4}{2}=1.2 m\)

Surface charge density of conducting sphere, \(\sigma=80 \times 10^{-6} C / m ^{2}\)

Therefore,

Charge on sphere will be:

\(q=\sigma A=\sigma 4 \pi r^{2} \)

\(q=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2} \)

\(q=1.45 \times 10^{-3} C\)

Then, the total electric flux leaving the surface of the sphere will be calculated using the gauss formula, i.e.,

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

\(\phi=\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} \quad\left(\because \epsilon_{0}=8.854 \times 10^{-12}\right) \)

\(\phi=1.6 \times 10^{8} Nm ^{2} / C\)

Given, 

\(p \propto T^{3} \quad \ldots\) (\(i\))

In an adiabatic process,

\(T^{\gamma} p^{1-\gamma}=\) constant \(\left[\right.\) as \(\left.\gamma=\frac{C_{p}}{C_{v}}\right]\)

\(T \propto \frac{1}{p^{\frac{(1-\gamma)}{\gamma}}}\)

\(T^{(\frac{\gamma}{\gamma-1})} \propto p \quad \ldots \text { (ii) }\)

Comparing Eqs. (\(i\)) and (\(ii\)),

Since the pressure is same in both the condition, equating the powers of temperature from both sides we get,

\(3 \gamma-3=\gamma\) or \(2 \gamma=3\)

\( \frac{C_{p}}{C_{v}}=\gamma=\frac{3}{2} \)

Suppose the wind strikes the windmill turbines as a cylindrical-shaped structure of area A and length V, which is the velocity of the wind. 

So, the rate of change of volume of this hypothetical cylinder can be written as:

Volume \(=\) Area \(\times\) Velocity 

or \(V^{\prime}=A \times V\)

Here, \(V\) is the velocity while \(V^{\prime}\) is the volume.

We know from Newton’s second law that the force acting on a body is the rate of change of momentum. 

We can write it mathematically as,

\(\vec{F}=\frac{d \vec{P}}{d t}\)

This can be rewritten as,

\(\vec{F}=\frac{d(m \vec{V})}{d t}=m \frac{d v}{d t}+V \frac{d m}{d t}\)

Here, the velocity of the wind is constant, so the term \(\frac{d V}{d t}\) is zero.

So, we can write the force acting on the windmill as, 

\(\vec{F}=\vec{V} \frac{d m}{d t} \quad\quad\) ......(1)

If the air or the wind has a density of \({\rho}\), then the rate of the mass of the wind that hits the turbine can be written as,

\(\frac{d m}{d t}=\rho A V\quad\quad\).......(2)

So, the force acting on the body is, (substituting equation (2) in (1))

\(\vec{F}=V \rho A V\)

\(F=\rho A V^{2}\)

We now have an equation for the force acting on the windmill. So the power of the windmill can be found out by,

Power \(=\) Force \(\times\) Velocity

Power \(=\left(\rho A V^{2}\right) \times(V)\)

Power \(=\rho A V^{3}\)

Suppose the kinetic energy of the windmill is converted into electrical energy without any loss. The electrical power output of the windmill will be proportional to \(V^3\).