Physics Test-5

Given:

Initial velocity \(({u})=0 {~m} / {sec}\)

The distance covered in \(20~ {sec}\) is,

\(s=\frac{1}{2} a(20)^{2}\)

\(\Rightarrow s=200~ a\quad\dots(1)\)

The distance covered in \(10~ sec\) is,

\(s_{1}=\frac{1}{2} a(10)^{2}\)

\(\Rightarrow s_{1}=50 a \quad\dots(2)\)

The distance covered in the next \(10~ sec\) is,

\(s_{2}=s-s_{1}\)

\(=200 a-50 a\)

\(=150 a\)

\(=3 \times 50 a\)

\(=3 s_{1} \quad\left[\because s_{1}=50 a\right]\)

Intensity is proportional to the area of the slit.

As slit widths are in the ratio of \(1: 9\).

The areas are also in the ratio \(1: 9\).

Thus Intensities are in the ratio \(1: 9\).

amplitudes are square root of Intensities.

Thus amplitudes are in ratio \(1: 3\).

Let amplitudes be \(x\) and \(3 x\).

At maxima the amplitudes get added up \(x +3 x\) \(=4 x\)

At minima they become \(x -3 x =-2 x\)

Intensity of maxima to minima is \(\frac{16 x^2}{4 x^2}=\frac{4}{1}=4:1\)

The work done in an isothermal expansion of a gas depends upon both temperature and expansion ratio.

The work done in an isothermal process is given by:

\(W = nRT \ln \left(\frac{V_{1}}{V_{0}}\right) \Rightarrow W \propto T , \ln \left(\frac{V_{1}}{V_{0}}\right)\)

\(V_{1}\) and \(V_{0}\) are the final and initial volumes respectively, thus it represents the change in volume during expansion. Therefore, work done in an isothermal change of a gas depends on both the temperature and volume expansion ratio.

We know that, joule, newton metre and kilowatt hour are the units of energy and the kilowatt is the unit of power.

Work done \(=\) Force \(\times\) Distance \(=\) Newton - metre, Therefore it is the unit of Work or Energy.

Energy \(=\) Power \(\times\) time \(=\) kilowatt - hour, in fact it is the commercial unit of Energy.

In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference patternthe intensities of both the maxima and the minima increase.

• As one slit is made twice as wide as the other the intensity of one slit becomes four times that of the other.
• As the intensities of the input increasing ,the intensities of both maxima and minima increases.
• It is because of sum of the input intensities is equal to the average of the fringe intensities.

The magnetic field \(B\) inside the toroid is given as,

\(B=\frac{\mu_{o} N I}{2 \pi r}\quad \ldots\) (1)

Where, \(N =\) number of turns, \(I=\) current, and \(r =\) average radius of the toroid

When the current in the toroid is doubled, then:

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

So, the magnetic field in the toroid when the current is doubled is given as,

\(B^{\prime}=\frac{\mu_{o} N I^{\prime}}{2 \pi r}\)

\(\Rightarrow B^{\prime}=\frac{2 \mu_{o} N I}{2 \pi r}\quad \ldots\) (3)

By equation (1) and equation (3),

\(B^{\prime}=2B\)

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.

Film coefficient is defined as Inside diameter of tube Thermal conductivity Equivalent thickness of film Specific heat × Viscosity.

The heat transferred by convection per unit area per degree temperature difference between the surface and the fluid called Film coefficient or unit convection conductance surface coefficient.

The position of block executing SHM is \({x}={A} \sin \left(\frac{2 \pi {t}}{{T}}+\phi\right)\), where \(\phi\) is the initial phase
given at \({t}=0, {x}=6 {~cm}\) and \({A}=6 {~cm}\)
therefore by above equation \(6=6 \sin (0+\phi)\)
or \(1=\sin \phi\) or \(\sin \frac{\pi}{2}=\sin \phi\) or \(\phi=\frac{\pi}{2}\)
now time period of system \({T}=2 \pi \sqrt{\frac{{m}}{{k}}}\) \(=2 \pi \sqrt{\frac{4}{400}}\) \(=\pi / 5\)
so position of block \(x=6 \sin \left(10 t+\frac{\pi}{2}\right)\)
as the initial phase is in positive \(x\) -direction therefore \(+\) sign is taken there.