Physics Test - 9

Given:

\(t_{1}=2 \mathrm{~s} \)

\(\overrightarrow{a_{1}}=(6 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}^{2} \)

\(t_{2}=5 \mathrm{~s} \)

\(\overrightarrow{a_{2}}=(4 \hat{i}+(-6) \hat{j}) \mathrm{m} / \mathrm{s}^{2}

\)

The two values of the acceleration vector are perpendicular to each other.

We can verify that by evaluating their dot product:

\(\overrightarrow{a_{1}} \cdot \overrightarrow{a_{2}}=(6 \hat{i}+4 \hat{j}) \cdot(4 \hat{i}-6 \hat{j})=0\)

Since, the radius vector in circular motion is anti-parallel to the centripetal acceleration,

Radius vectors at each instant \(\left(t_{1}=2 \mathrm{~s}\right.\) and \(\left.t_{2}=5 \mathrm{~s}\right)\) are also perpendicular to each other.

Therefore, \(\vec{r}_{1}\) points to the east while \(\vec{r}_{2}\) points to the south.

Since, the movement is counterclockwise,

therefore, between the two instants is a three-quarter cycle.

Therefore, the period is given by the form:

\(t_{2}-t_{1}=\frac{3 T}{4}=3 \mathrm{~s} \)

\(\Rightarrow T=4 \mathrm{~s}\)

The magnitude of the acceleration is:

\(a =\sqrt{a_{x}^{2}+a_{y}^{2}} \)

\(=\sqrt{\left(6 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}+\left(4 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}} \)

\(=7.21 \mathrm{~m} / \mathrm{s}^{2}\)

Now, we want to know the value of the radius, it can be obtained from:

\(a_{c}=\frac{V^{2}}{r}\)

\(=\frac{(2 \pi r / T)^{2}}{r} \)

\(=\frac{4 \pi^{2} r^{2}}{T^{2} r} \)

\(\Rightarrow r=\frac{a_{c} T^{2}}{4 \pi^{2}} \)

\(=\frac{(7.21)(4)^{2}}{4 \pi^{2}} \)

\(=2.92 \mathrm{~m} .\)

Upon catching a ball, a cricket fielder swings his hands backwards. The concept behind this is explained byNewton's second law of motion.

Newton's second law of motion: It states that the rate of change of momentum of a body over time is directly proportional to the force applied, and occurs in the same direction as the applied force.The rate of change of momentum is directly proportional to the force applied by the moving body.

\(\frac{d p}{d t} \propto \mathrm{F}\)

\(\frac{d p}{d t}=\frac{d(m v)}{d t}\)

\(=m \frac{d v}{d t}\)

\(=m a=F\)

\(\mathrm{I}=10 \mathrm{~A}\)

\(\mathrm{f}=50 \mathrm{~Hz}\)

\(\mathrm{I}_{\mathrm{rms}}=\left(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\right)\)

where \(I_{m}\) is peak value \(I_{m}=\sqrt{2} \cdot I_{\text {rms }}\)

\(\therefore I_{m}=10 \sqrt{2} A=10 \times 1.41=14.1 \mathrm{~A}\)

Time taken to reach form \(\mathrm{0}\) to \(\mathrm{I}_{\mathrm{m}}\) is \([(\frac{\mathrm{T}} { 4})-0]=(\frac{\mathrm{T}}{4})\)

\(\mathrm{t=(\frac{T}{ 4})=(\frac{1} { 4 f})}\)

\(\therefore \mathrm{t}=[\frac{1}{ \{4 \times 50}\}]=[\frac{1} { 200}] \mathrm{sec}\)

\(\therefore \mathrm{t}=0.5 \times 10^{-2} \mathrm{sec}\)

\(\mathrm{t}=5 \times 10^{-3} \mathrm{sec}\)

Newton's first law defines inertia and is rightly called the law of inertia. An inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called inertia.

If we have two bodies of equal mass, one in motion and another is at rest, possess the same inertia because it is a factor of mass only and does not depend upon the velocity. The mass of an object is a measure of its inertia.SI unit of mass is the kilogram (kg).

The interaction which changes or tries to change the position of anybody is called force.Mathematically force can be written as:

F = ma

Where m = mass and a = acceleration of the object.

Potential energy: The energy due to the position of the particle is called the potential energy of the particle.

The relation between force and potential energy is given by:

Force (F)$=-\frac{dU}{dx}$

Where U is potential energy and x is the position.

It is given that force = F(x) = kX²- By

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}={\mathrm{kX}}^2-\mathrm{By}$

dU = - (kX²- By) dx

Integrating both sides of the above equation,

$\int dU=-\int ({\mathrm{kX}}^2-\mathrm{By})\mathrm{dx}$

Potential energy (U)$=-\frac{k{x}^3}{3}+Byx=-\frac{x}{3}(k{x}^2-3By)$

The cutoff voltage in photoelectric emission is the minimum negative voltage that must be applied to the cathode in order to stop the photoelectric emission.

The relationship between the cutoff voltage and frequency is given by:

\(\mathrm{eV}_{0}=\mathrm{h}\left(\mathrm{V}-\mathrm{V}_{0}\right)\)

Where,

\(\mathrm{V}_{0}=\) Cutoff voltage

\(v=\) frequency

\(v_{0}=\) Threshold frequency

The above equation can be rewritten as:

\(\mathrm{hv}=\mathrm{eV}_{0}+\mathrm{hv} \mathrm{V}_{0}\)

If the frequency is doubled \(v=2 v\) in the above equation, and assume new cutoff voltage as \(v_{1}\), then the above equation becomes:

\(\mathrm{eV} \mathrm{V}_{1}=\mathrm{h}\left(2 \mathrm{v}-\mathrm{V}_{0}\right)\)

\(\Rightarrow \mathrm{eV}_{1}=2 \mathrm{hv}-\mathrm{hv}_{0}\)

\(\Rightarrow \mathrm{eV} \mathrm{V}_{1}=2\left(\mathrm{eV} \mathrm{V}_{0}+\mathrm{hV}_{0}\right)-\mathrm{h \textrm {V } _ { 0 }}\)

\(\Rightarrow \mathrm{eV}=2 \mathrm{eV}_{0}+2 \mathrm{hv}_{0}-\mathrm{hv}_{0}\)

\(\mathrm{eV}_{1}=2 \mathrm{eV}_{0}+\mathrm{hV}_{0}\)

From the above equation, it is clear that the stopping potential is doubled.

According to kinetic theory of gases, at absolute zero, Kinetic energy of molecules reduces to zero.

The following are the basic assumptions of the Kinetic Molecular Theory:

• The volume occupied by the individual particles of gas is negligible compared to the volume of the gas itself.
• The particles of an ideal gas exert no attractive forces on each other or on their surroundings.
• Gas particles are in a constant state of random motion and move in straight lines until they collide with another body.
• The collisions exhibited by gas particles are completely elastic; when two molecules collide, total kinetic energy is conserved.
• The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.

\(\mathrm{k}=2 \mathrm{~N} / \mathrm{m}\)

\(\mathrm{x}=4 \mathrm{~m}\)

\(\mathrm{M}=0.25\)

Assume block hits with velocity \({u}~ \mathrm{m} / \mathrm{s}\)

The spring will compress till the point the block stops.

Also friction is acting all this while.

Applying work energy theorem:

(Only friction and spring forces are acting)

\(W_{f r}+W_{s p}=\Delta K E\)

\(-\mu M g -\frac{1}{2} k x^{2}=0-\frac{1}{2} M u^{2}\)

\(\mu M g +\frac{1}{2} k x^{2}=\frac{1}{2} M u^{2}\)

Substituting the given values in the above equation

\(2× 0.25 ×10 +\frac{1}{2}× 2× 4^{2}=\frac{1}{2}× 0.25 u^{2}\)

\(u=\sqrt{52} \mathrm{~m} / \mathrm{s}\)

Faraday's Laws of Electromagnetic Induction:

• First law: Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is change or cutting of flux.
• Second law: The induced emf is given by the rate of change of magnetic flux linked with the circuit i.e.,

\(e=-\frac{d \phi}{d t}\)

Where \(d \phi=\) change in magnetic flux, \(dt =\) change in time and \(e =\) induced emf.

The negative sign indicates that induced emf (e) opposes the change of flux.

From the above equation, it is clear that induced e.m.f. in a coil is depends on the change in the flux and time and independent of the resistance of the circuit.

The above logic gate is NAND gate.

Logic gate: The digital circuit that can be analyzed with the help of Boolean algebra is called a logic gate or logic circuit. A logic gate has two or more inputs but only one output.

NAND gate: It a combination of an AND and a NOT gate.

It is obtained by connecting the output fo an AND gate to the input of a NOT gate. It is described by the Boolean expression: \(Y=\overline{A \cdot B}\)The above logic gate is the NAND gate.