Physics Test - 72

The species and their number of electrons, protons, and neutrons are:

Given that,

\({V}_{{D}} = 0.3 {V}\)

\({I}=6 {mA} = 0.006 {A}\)

\({R} = 200 \Omega\)

Here we can write,

\({E} - {V}_{{D}} = {I} {R}\)

Where, \({E}\) is the emf of battery

\({E} = {I} {R} + {V}_{{D}}\)

\({E}=(0.006 \times 200)+0.3\)

\({E}=1.2+0.3\)

\(=1.5 {V}\)

An adiabatic expansion has less work done and no heat flow, thereby a lower internal energy comparing to an isothermal expansion which has both heat flow and work done.For an adiabatically expanding ideal monatomic gas which does work on its environment (W is positive), internal energy of the gas should decrease.

According to lenz's law, the rate of charge of flum produced by bar magnet will be approused by the conducting loops.

A girl swings on the cradle in a sitting position. If she stands then the time period of girl and cradle will decrease.

The time period, \(T=2 \pi \sqrt{\frac{l}{g}}\) where \({l}\) is the length of simple pendulum. On standing the effective length of the pendulum measured from the point of suspension decreases. So, the time period of the cradle decreases.

Given:

Two photons of same frequency are produced due to the annhiliation of a proton and antiproton.

We know that:

\(c=3 \times 10^{8}\)

Mass of a proton = \(1.67 \times 10^{-27}\)

\(E=m c^{2}\)

Put the values in above formula.

\(E=\left(2 \times 1.67 \times 10^{-27}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)

\(=3.006 \times 10^{-10} \mathrm{~J}\)

Also We know that:

\(2 h \nu=E\) or \(2 h \frac{c}{\lambda}=E\)

\(\therefore \lambda=\frac{2 h c}{E}\)

Put the values in above formula.

\(=\frac{2 \times 6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.006 \times 10^{-10}} \mathrm{~m}\)

\(=1.323 \times 10^{-15} \mathrm{~m}\)

Magnetic field produced by a current \(i\) in a large square loop at its centre \(B \propto \frac{i}{L}\) say \(B=K \frac{i}{L}\)

\(\therefore\) Magnetic flux linked with smaller loop,

\(\phi=B . S \phi=\left(K \frac{i}{L}\right)\left(l^{2}\right)\)

Therefore, the mutual inductance

\(M=\frac{\phi}{i}\)\(=K \frac{l^{2}}{L}\)

or \(M \propto \frac{l^{2}}{L}\)

Given:

\(C=3 \times 10^{8} \mathrm{~ms}^{-1}\)

\(\Delta \lambda=0.02 \%\)

We know that wavelength is expressed as \(\lambda\). So now the change in the wavelength is given as follows:

\(\frac{\Delta \lambda}{\lambda}=\frac{v}{C}\)

Here \(v\) is the velocity and \(C\) represents the speed of light. Hence we can express \(\mathrm{v}\) as:

\(v=\frac{\Delta \lambda}{\lambda} C\)

Now we have to put the values in the above expression, to get:

\(v=\frac{0.02}{100} \times 3 \times 10^{8} \mathrm{~ms}^{-1} \)

\(v=60 \mathrm{kms}^{-1}\)

 When both move in the same direction with the same speed,  no current induced in the coil with its plane perpendicular to the page as shown in the question figure.

Current is induced in the loop due to electromagnetic induction when the magnetic flux through it changes. When they move with the same speed or velocity, their relative displacement is zero. Therefore, magnetic flux does not change and no current is induced. The direction does not change as it reaches to the same point. Reference can be taken from the figure.

In this process on the system no external force acts on the system, hence net torque acting on the system is zero. Now we know the relationship between torque and angular momentum is given by,

\(\frac{d L}{d t}=\tau\)

Since \(\tau=0\), this means that angular momentum is constant.

So, by conservation of angular momentum, total momentum before collision must be equal to the total momentum after collision.

Momentum before collision= momentum after collision

\(L_{1}=L_{2}\)

Therefore, taking the same mass of sphere, if radius is increased, then moment of inertia, rotational kinetic energy and angular velocity will change but according to law of conservation of momentum, angular momentum will not change.