Physics Test - 34

Let the car stops in S distnace 

When force \(=F\), then retardation on car \(a=\frac{-F}{m}\) 

The final velocity is \(O\) 

Using \(V ^{2}- U ^{2}=2as\) also \(U = v , V =0\)

We get \(v^{2}=\frac{2 s F}{m}\)....(i)

Now let \(F _{1}\) is applied to stop car in \(S\) distance with initial velocity \(U =3 v\),we get

\((3 v)^{2}=\frac{2 F_{1} s}{m}\)....(ii)

from (i) and (ii)

\(F _{1}=9 F\)

Sound is a mechanical wave. It requires a medium for transmission. There is no transmission of sound in a vacuum. The transmission of sound in the air is in the form of a longitudinal wave. The transmission of sound in the air is in the form of compression and deceleration caused by the vibration of air particles. So,the phenomenon of sound propagation in the air is also an adiabatic process.

Given, electric field,

\(\mathrm{E}=20 \cos \left(2 \times 10^{10} \mathrm{t}-200 \mathrm{x}\right) \mathrm{V} / \mathrm{m}\)

Comparing with the standard equation, \(\mathrm{E}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kx}) \mathrm{V} / \mathrm{m}\), we get

Wave constant, \(\mathrm{k}=200\)

Angular frequency, \(\omega=2 \times 10^{10} \mathrm{rad} / \mathrm{s}\)

Speed of the wave, \(\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{2 \times 10^{10}}{200}=10^8 \mathrm{~m} / \mathrm{s}\)

Refractive index, \(\mu=\frac{\mathrm{C}}{\mathrm{V}}=\frac{3 \times 10^8}{10^8}=3\)

As we know the relation between the refractive index and dielectricconstant,

\(\mu=\sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}}\)

Substituting the value in the above equations, we get

\(3  =\sqrt{\varepsilon_{\mathrm{r}}(1)} \)

\(\varepsilon_{\mathrm{r}}  =9\)

Thus, the dielectric constant of the medium is 9 .

The temperature determines the direction of net change of intermolecular kinetic energy.

On heating, the temperature rises, particles gain energy. As the kinetic energy increases, intermolecular space increases as they get separated from each other and the force of attraction decreases as the particles go far away from each other.

The two forces acting on the rocket immediately after leaving the launching pad are gravitational force and frictional force. The gravitational force acts downwards towards the earth and frictional force acts due to surrounding air.

Given:

Atomic weight of element = M gram/mol

So, no of mole \(=\frac{M}{M_{\text {atomic }}}\)

Now no of atoms \(=\) mole \(\times\) Avogadro No.

\(N=\frac{m}{M} \times N_{A}\)

We know,

Radioactivity \(=\lambda \mathrm{N}\)

\(R=\frac{\lambda m}{M} \mathrm{N_{A}}\)

\(N=\) no of atom

\(R=\left(\frac{\mathrm{N}_{\mathrm{a}} m}{M}\right) \lambda\)

The force/load per unit cross-sectional area is called stress.

Stress \((S)=\frac{\text { Force }(F)}{\text { Area }(A)}\)

Given that:

Force /load (F) on both the wires is the same.

Let the cross-sectional area of wire B is equal to A.

Cross-sectional area of wire A (A₁) = 2 × Cross-sectional area of wire B (A₂) = 2 A

Stress on wire \(A(S)=\frac{\operatorname{Force}(F)}{\text { Area of wires } A\left(A_{1}\right)}=\frac{F}{2 A}\) .....(i)

Stress on wire \(B\left(S^{\prime}\right)=\frac{\text { Force }(F)}{\text { Area of wires } B\left(A_{2}\right)}=\frac{F}{A}\) .....(ii)

On dividing equations (i) and (ii), we get

\(\Rightarrow \frac{\text { Stress on wire } A(S)}{\text { Stress on wire } B\left(S^{\prime}\right)}=\frac{1}{2}\)

\(\Rightarrow\) Stress on wire \(B\left(S^{\prime}\right)=2 \times\) Stress on wire \(A(S)\)

\({ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{H} e^4 \text { Total binding energy of two deuterium nuclei } \)

\(=1.1 \times 4=4.4 \mathrm{MeV}\)

Binding energy of a \(\left({ }_2 \mathrm{He}^4\right)\) nuclei \(=4 \times 7=28 \mathrm{MeV}\)

Energy released in this process \(=28-4.4=23.6 \mathrm{MeV}\)

The type of electron emission in which a very strong electric field applied to a metal pulls electrons out of the metal surface is called field emission.

• Electron emission: It is the phenomenon when electrons escape from a metal surface. There are three types of electron emission.
• Thermionic Emission: It is the process in which the electrons are ejected from a metal surface when the metal is heated to a sufficient temperature.
• Field Emission: It is the process in which electrons are pulled out of the metal surface due to a very strong electric field.
• Photoelectric effect: The photoelectric effect is a phenomenon where electrons are ejected from a metal surface when the light of sufficient frequency is incident on it.
• Electrons emitted due to the electric field applied to a metal surface is a result of field emission. Therefore, the correct answer is field emission.