Physics Test - 22

The force that binds the atmosphere around the Earth is the force of gravity.

Pressure is constant \(P _{2}= P _{1}\)

Volume is doubled \(V _{2}=3 V _{1}\)

\(T _{1}=27^{\circ}\) C \(=273+27=300 K\), \(T _{2}=?\)

Ideal gas equation,

\(PV = nRT\), \(n =\frac{P V}{R T}\)

During the whole process, no. of moles will not change.

So, \(n_{1}=n_{2}\)

\(\frac{P_{1} V_{1}}{R T_{1}}=\frac{P_{2} V_{2}}{R T_{2}}\)

\(\frac{P_{1} V_{1}}{300 R}=\frac{\left(P_{1}\right)\left(3 V_{1}\right)}{R T_{2}}\)

\(\frac{1}{300}=\frac{3}{T_{2}}\)

\(T_{2}=900 K\)

Given:

A point particle of mass \(0.1 \mathrm{~kg}\) is executing SHM of amplitude \(0.1 \mathrm{~m}\).

\(\mathrm{KE}\) = \(8 \times 10^{-3} J\)

Initial phase of oscillation = \(45^{\circ}=\frac{\pi}{4}\)

We know that:

\(\mathrm{KE}\) at mean position \(=\frac{1}{2} m \omega^{2} a^{2}\)

According to the question:

\(\frac{1}{2} m \omega^{2} a^{2}=8 \times 10^{-3}\).....(1)

Put all the given values in (1).

\(\omega=\left(\frac{2 \times 8 \times 10^{-3}}{m a^{2}}\right)^{\frac{1}{2}}\)

\(=\left[\frac{2 \times 8 \times 10^{-3}}{0.1 \times(0.1)^{2}}\right]^{\frac{1 }{2}}\)

\(=4\)

Equation of SHM is,\(y=a \sin (\omega t+\theta)\)

\(=0.1 \sin \left(4 t+\frac{\pi}{4}\right)\)

The resistance of the given resistors is, \(\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega \),

1. Equivalent resistance, \(\mathrm{R}=\frac{11}{3} \Omega\)

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

\(R=\frac{2 \times 1}{2+1}+3=\frac{2}{3}+3=\frac{11}{3} \Omega\)

2. Equivalent resistance, \(\mathrm{R}=\frac{11}{5} \Omega\)

Consider the following combtnatlon of the resistors.

Equivalent resistance of the circuit is given by,

\(\mathrm{R}=\frac{2 \times 3}{2+3}+1=\frac{6}{5}+1=\frac{11}{5} \Omega\)

3. Equivalent resistance, \(\mathrm{R}=6 \Omega\)

Consider the series combination of the resistors, as shown in the given circuit. Equivalent resistance of the circuit is given by the sum.

\(\mathrm{R}=1+2+3=6 \Omega\)

4. Equivalent resistance, \(\mathrm{R}=\frac{6}{11} \Omega\)

Consider the series combination of the resistors, as shown in the given circuit. Equivalent resistance of the circuit is given by,

\(\mathrm{R}=\frac{1 \times 2 \times 3}{1 \times 2+2 \times 3+3 \times 1}=\frac{6}{11} \Omega\)

Lenz's law: It states that the direction of the induced current in the coil will be always in such a way that it opposes the change which produces the current.

It is given, the energy required to dissociate a carbon monoxide molecule into catbon and oxygon atoms is \(E = 11 eV\)

We know that, \(E=h f=6.62 \times 10^{-34} J - s\)

\(f=\) frequency

\(\Rightarrow 11 eV =h f\)

\(f =\frac{11 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=2.65 \times 10^{15} Hz\)

Hence, the frequency required for the dissociation is 2.656×1015 Hz and it lies in the range of UV region. 

Rydberg's formula is given as:

\(\frac{h c} {\lambda}=21.76 \times 10^{-19}\left[\frac{1 }{\left(n_{1}\right)^{2}}-\frac{1 }{\left(n_{2}\right)^{2}}\right]\)

Where,

\(\mathrm{h}=\) Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)

\(\mathrm{c}=\) Speed of light \(=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

\(\left(\mathrm{n}_{1}\right.\) and \(\mathrm{n}_{2}\) are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values,

\(\mathrm{n}_{1}=3\)

\(\mathrm{n}_{2}=\infty\)

By putting the values, we get

\(\frac{hc }{ \lambda}=21.76 \times 10^{-19}\left[\frac{1}{ (3)^{2}}-\frac{1 }{(\infty)^{2}}\right]\)

\(\therefore \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 9} { 21.76 \times 10^{-19}}\)

\(=8.189 \times 10^{-7} \mathrm{~m}\)

\(=818.9 \mathrm{~nm}\)

As we know if we pull the chain, each part of the chain will not cover same height, therefore will not contribute the same potential energy, so let's take mass of the chain \(L^{\prime}=\frac{L}{2 n}\)

Now we can write

\(W=\frac{m g L}{2 n^2}\)

Given, 

length = \(2 \mathrm{~m}\)

mass = \(4 \mathrm{~kg}\)

After putting all the values and solving

\(=\frac{4 \times 10 \times 2}{2 \times\left(\frac{10}{3}\right)^2}=3.6 \mathrm{~J}\)

Given,

Mass \((m)=2\) kg

Coefficient of limiting friction \((\mu)= 0.7\)

Inclined plane making angle \((\theta)= 30^\circ\)

As we know,

The force applied on the body that is on the inclined plane is given as,

\(F = mg \sin \theta\)

\(\therefore F =2 \times 9.8 \times \sin 30^{\circ}\)

\(=2 \times 9.8 \times \frac{1}{2}\)

\(=9.8\) N

The limiting friction force between the block and the inclined plane is given as,

\(f =\mu mg \cos \theta\)

\(\therefore f =0.7 \times 2 \times 9.8 \cos 30^{\circ}\)

\(= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}\)

\(=11.88\) N

Since the limiting friction force is greater than the force that tends to slide the body.

Thus, the body will be at rest and the force of friction on the block is \(9.8\) N.

Displacement is the distance between the initial and final position after the motion or journey of an object.

Displacement is a vector quantity and can be obtained from the difference of position coordinates of the initial and final positions.

The lady starts the journey from her home (initial point) and reaches back to her home (final point) upon seeing the shop closed. Here, the initial and final point at the end of the journey is the same. So, the displacement is zero.