Physics Test - 4

Given,

Atmospheric Pressure = 10⁶ dyne/cm²

As we know, the SI unit for atmospheric pressure is Newton/Meter². 

And

1 dyne = 1 g × 1 cm/sec²

∵ 1 m = 100 cm and 1 kg = 1000 g

⇒ 1 dyne = 10^-3 kg × 10^-2 m/s2

⇒ 1 dyne = 10^-5 N

⇒ 10⁶ dyne/cm² = \(\frac{10^{6} \times 10^{-5} \mathrm{~N}}{\left(10^{-2}\right)^{2} \mathrm{~m}^{2}}=10^{5}\) Newton/meter²

Given, 

Radius \((r)=10 m\)

Velocity \((v)=10 m / s \)

\( g =10 m / s ^{2}\)

Free body diagram of the given condition,

So, \(F _{\text {net }}=0\)

\(T \sin \theta=\frac{m v^{2} }{ r} \)...(1) 

\(T \cos \theta=m g \)...(2) 

Dividing the equation (1) and (2), we get

\(\frac{T \sin \theta}{T \cos \theta}=\frac{v^{2}}{r g}\)

Substituting all the values in the above equation, we get

\(\frac{T \sin \theta}{T \cos \theta}=\frac{(10)^2}{10 \times 10} \)

\(\Rightarrow \frac{T \sin \theta}{T \cos \theta}=\frac{100}{100} \)

\(\Rightarrow \tan \theta=1\)

\(\Rightarrow \tan \theta=\tan 45^\circ\)

\(\therefore \theta= 45^\circ\)

The angle subtended by the rod from the path is \(45^{\circ}\).

Kinetic energy of projectile before collision \({K}_{i}=\frac{1}{2} {~m}_{1} {u}_{1}^{2}\)

Kinetic energy of projectile after collision \({K}_{{f}}=\frac{1}{2} {~m}_{1} {v}_{1}^{2}\)

Kinetic energy transferred from projectile to target \(D K=\) decrease in kinetic energy in projectile

\(\Delta {K}=\frac{1}{2} {~m}_{1} {u}_{1}^{2}-\frac{1}{2} {~m}_{1} {v}_{1}^{2}=\frac{1}{2} {~m}_{1}\left({u}_{1}^{2}-{v}_{1}^{2}\right)\) 

Fractional decrease in kinetic energy

\(\frac{\Delta {K}}{{K}}=\frac{\frac{1}{2} {~m}_{1}\left({u}_{1}^{2}-{v}_{1}^{2}\right)}{\frac{1}{2} {~m}_{1} \mathrm{u}_{1}^{2}}=1-\left(\frac{{v}_{1}}{{u}_{1}}\right)^{2}\)....(i)

We can substitute the value of \(v_{1}\) from the equation \(v_{1}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1}+\frac{2 m_{2} u_{2}}{m_{1}+m_{2}}\)

If the target is at rest i.e. \(u_{2}=0\) then \(v_{1}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1}\)

From Equation (i) 

\(\frac{\Delta K}{K}=1-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2}\)

Or

\(\frac{\Delta K}{K}=\frac{4 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)^{2}}\) 

Percentage loss of energy \(=\frac{4 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)^{2}} \times 100\) \(=\frac{4 m \times 2 m}{(2 m+m)^{2}} \times 100=\frac{800}{9}\)

\(=88.9 \%\)

Given,

\({E}_{{rms}}=6 {~V} / {m}\)

\({E}_{{rms}}={E}_{{rms}} \times \sqrt{2}=6 \sqrt{2} {~V} / {m}\)

Now, \({E}={CB}\)

\({B}=\frac{{E}}{{C}}=\frac{6 \sqrt{2}}{3 \times 10^{8}}\)

\(=2.25 \times 10^{-8} {~T}\)

\(=2.828 \times 10^{-8} {~T}\)

Surface tension is the property of a fluid resisting an external force on its surface. Surface tension is caused by cohesive forces between the molecules of the fluid on the surface.

Mathematically,

Surface tension (S) \(= \frac{F}{l}\)

Where, \(F\) = force, and \(l\) = length of the film of the fluid.

Surface tension is dimensionally equal to the ratio of force to length.

Therefore, the SI unit of surface tension is newton/metre.

In telecommunication and information theory, broadcasting is a method of transferring a message to all recipients simultaneously. Broadcasting can be performed as a high-level operation in a program, for example, broadcasting in Message Passing Interface, or it may be a low-level networking operation, for example broadcasting on Ethernet.

In the broadcast mode, there are a large number of receivers corresponding to a single transmitter. Radio and television are examples of broadcast modes of communication.

The equation of the two superimposing waves can be give by:

\(\mathrm{y}_{1}=\mathrm{a}_{1} \sin (\omega \mathrm{t})\)

\(\mathrm{y}_{2}=\mathrm{a}_{2} \sin (\omega \mathrm{t}+ \frac{\pi}{ 2})=\mathrm{a}_{2} \cos (\omega \mathrm{t})\)

So, the equation of the resultant wave is:

\(\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}\)

\(=\mathrm{a}_{1} \sin (\omega \mathrm{t})+\mathrm{a}_{2} \cos (\omega \mathrm{t})\).....(1)

Assuming \(\omega t=\phi, \mathrm{a}_{1}=\mathrm{a} \cos (\phi)\) and \(\mathrm{a}_{2}=\mathrm{a} \sin (\phi)\),

Put all values in (1).

\(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}=\mathrm{a}^{2}\left(\sin ^{2}(\phi)+\cos ^{2}(\phi)\right)=\mathrm{a}^{2}\)

\(\mathrm{a}=\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}}\)

\(\therefore\) The resultant wave will be \(\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}}\).

When force F is applied at the centre of roller of mass m as shown below

Its acceleration is given by

\(\frac{F-f}{m}=\alpha\) ......... (1)

Where, \(f=\) force of friction and \(m=\) mass of roller.

Torque on roller is provided by friction \(f\) and it is torque is given by,

\(\mathrm{T}={f R}=\mathrm{la}\)

Where, \(\mathrm{I}=\) moment of inertia of solid cylindrical roller \(=\frac{m R^{2}}{2}\)

\(\alpha=\) angular acceleration of the cylinder \(=\frac{a}{R}\)

So,

\(f=\frac{m R^{2}}{2} \cdot \frac{a}{R}\)

\(f=\frac{m a R}{2}\)

\(\therefore f=\frac{m \alpha}{2} \quad \ldots\)

\(\left[\alpha=\frac{a}{R}\right]\) ......... (2)

Substituting equations (2) in (1), we get

\(\left(F-\frac{m a}{2}\right)=m \alpha\)

\(F=m \alpha+\frac{m a}{2}\)

\(F=\frac{m a}{R}+\frac{m a}{2}=m a+\frac{m a}{2}\)

\(F=\frac{2 m a+m a}{2}\)

\(F=\frac{3 m a}{2}\)

\(F=\frac{3}{2} m a\)

Given,

Radius of the roller \(=\mathrm{R}\)

Mass of the roller \(=\mathrm{m}\)

To find the value of a

\(F=\frac{3}{2} m a\)

\(a=\frac{2 F}{3 m}\)

Hence, the value of \(\alpha\) is,

\(a=\frac{2 F}{3 m}\)

\(\left[\alpha=\frac{a}{R} \Rightarrow a=R \alpha\right]\)

\(R \alpha=\frac{2 F}{3 m}\)

\(\therefore \alpha=\frac{2 F}{3 m R}\)

Therefore, the angular acceleration of the cylinder is \(\frac{2 F}{3 m R}\).

In forward biasing, the applied voltage V of the battery mostly drops across the depletion region and the voltage drops across the p-side and n-side of the p-n junction is negligibly small.

It is due to the fact that the resistance of the depletion region is very high as it has no free charge carriers.

In forward biasing the forward voltage opposes the potential barrier Vb. As a result of it, the potential barrier height is reduced and the width of the depletion layer decreases.

As forward voltage is increased, at a particular value the depletion region becomes very much narrow such that large number of majority charge carrier can cross the junction.

​The number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant. It is equal to 6.022 ×10²³.

The units may be electrons, ions, atoms, or molecules, depending on the character of the reaction and the nature of the substance.