Physics Test - 33

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.

At balancing point, we know that emf is proportional to the balancing length. i.e., emf \(\propto\) balancing length

Now, let the emf's be \(3 \varepsilon\) and \(2 \varepsilon\).

\( \Rightarrow 3 \varepsilon=\mathrm{k}(75) \ldots . .(1) \)

\(\text { and } 2 \varepsilon=\mathrm{k}(\mathrm{I}) \ldots . .(2) \)

\(\Rightarrow \mathrm{I}=50 \mathrm{~cm} \)

\(\Rightarrow \text { Difference is }(75-50) \mathrm{cm}=25 \mathrm{~cm} .\)

If a material have less specific heat then it means that less amount of energy is needed to increase its temperature.

The specific heat capacity (S) of a substance is defined as the amount of heat (\(\Delta Q\)) per unit mass of the substance that is required to raise the temperature

\((\Delta T)\) by \(1^{\circ}\) C

i.e., \(S=\frac{\Delta Q}{m \Delta T}\)

And the unit of Specific heat is J/g\(^{\circ}\) C and Cal/g\(^{\circ}\) C

Here,

\(S =\) specific heat of material

\(\Delta Q=\) heat exchange

\(\Delta T =\) change in temperature

\(m =\) unit mass of material

Thus, if specific heat of a material is less it means that the less amount of energy is needed to increase its temperature.

When an object is oscillating in simple harmonic motion in the vertical direction, its maximum speed occurs when the objecthas the maximum net force exerted on it.

Because the total energy is fixed at \(\frac{k A^{2}}{2},\) the maximum kinetic energy occurs at the minimum potential energy, where \(x=0\).

Thus, the maximum speed is at the equilibrium position.

Given that:

Force (F) = 12 N

Displacement (s) = 60 cm = \(\frac{60}{100}\) = 0.6 m

The force (F) and displacement (s) are in the same direction, so θ = 0°

Work done = Fs cos θ = 12 × 0.6 × 1 = 7.2 J

In Rutherford's alpha particle experiment, the scattered alpha particle has large deflection angles. This experiment showed that the positive matter of atom was concentrated in an very small volume and it gave the idea of nucleus of an atom.

In a p-type semiconductor hole are the majority carriers and electrons are the minority carriers. It is obtained by using a trivalent dopant.

• When the trivalent impurity is added to an intrinsic or pure semiconductor (silicon or germanium), it is a p-type semiconductor.
• Trivalent impurity like Al, Ga, In etc., are added.
• The hole density is much greater than the electron density.
• The impurity added in p-type semiconductor provides extra holes known as Acceptor atom.

The property of matter which is responsible for electromagnetic phenomenon is called charge. The space or region around the current carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current. The space or region around the electric charge in which electrostatic force can be experienced by other charge particle is called as electric field by that electric charge.

• When a charge particle is at rest then it only creates electric field around it.
• When a charge particle is moving with a constant velocity then it equivalent to an electric current which only creates a magnetic field around it.
• When a charge particle is moving with an acceleration then it produced an electromagnetic wave around it.

The velocity of body just before touching the lake surface is:

\(v=\sqrt{2 g h}\)

Retardation in the lake:

\(a=\frac{\text { upthrust-weight }}{a}\)

\(\Rightarrow \frac{V \rho g-V \rho g}{V \rho}=\left(\frac{\sigma-\rho}{\rho}\right) g\)

Maximum depth \(d_{\max }=\frac{v_{2}}{2 a}\)

\(\Rightarrow \frac{h \rho}{\sigma-\rho}\)

The length of curved path or orbital radius,

\({R}=\frac{m V}{q B}=\frac{p}{q B}\)

Where,

\(m\) = Mass

\(V\) = Velocity

\(Q\) = Charge

\(B\) = Magnetic field

\(p\) = Momentum

Since \(p\) remains the same for both,

\({R}_{{e}}=\frac{p}{e{B}}\)

\( R_{p}=\frac{p}{e{B}}\)

\(\therefore R_{v}=R_{p}\)

By the above expression this is proved, the length of curved path of electron and proton will be same.