Physics Test - 21

When wire CD is made to slide on wires PQ and ST, the flux linked with the circuit changes with time and hence an emf is induced in the circuit, which is given by:

\(| e |=\frac{ d \phi}{ dt }=\frac{ d }{ dt }( BA )= B \frac{ dA }{ dt }\)

The induced current is: \(I=\frac{e}{R}=\frac{B w v}{R}\)

This current is caused by the motion of wire CD. From Lenz's law, the current I opposes the motion of wire \(C D\). Therefore, work has to be done to slide the wire CD. Now, the magnetic force on wire CD (of length \(w\) ) is

\(F = BIw = B \left(\frac{ Bw }{ R }\right) w =\frac{ B ^2 w ^2 v }{ R }\)

Work done is sliding wire \(C D\) through a small distance \(d x\) in time \(d t\) is

\(dW = Fdx\)

Therefore, the work done per second is

\(P =\frac{ dW }{ dt }= F \frac{ dx }{ dt }= Fv\)

Using equation (1), we get

\(P =\frac{ B ^2 w ^2 v ^2}{ R }\)

\(P = \frac{(B w v)^2}{R}\)

\(\text { Given, } \mathrm{M}=1000 \mathrm{~kg}, \mathrm{a}=20 \mathrm{~m} / \mathrm{s}^2 \)

\(\mathrm{v}_{\text {relative }}=500 \mathrm{~m} / \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)

The given situation is shown below

\(\mathrm{F}_{\text {thrust }}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{Mv}_{\text {relative }}\right) \)

\(\Rightarrow \mathrm{F}_{\text {thrust }}=v_{\text {relative }}\left(\frac{\mathrm{dM}}{\mathrm{dt}}\right)\)

By Newton's second law of motion,

\(\Rightarrow \mathrm{F}_{\text {thrust }}-\mathrm{Mg}=\mathrm{Ma}\)

\(\Rightarrow \mathrm{v}_{\text {relative }}\left(\frac{\mathrm{dM}}{\mathrm{dt}}\right)-\mathrm{Mg}=\mathrm{Ma} \)

\(\Rightarrow 500\left(\frac{\mathrm{dM}}{\mathrm{dt}}\right)-1000 \times 10=1000 \times 20 \)

\(\Rightarrow 500\left(\frac{\mathrm{dM}}{\mathrm{dt}}\right)=1000(20+10)\)

\(\Rightarrow 500\left(\frac{\mathrm{dM}}{\mathrm{dt}}\right)=1000 \times 30 \)

\(\Rightarrow \frac{\mathrm{dM}}{\mathrm{dt}}=\frac{1000 \times 30}{500}=60\)

\(\Rightarrow \frac{\mathrm{dM}}{\mathrm{dt}}=60 \mathrm{~kg} / \mathrm{s}\)

An ion is an atom or a group of atoms that does not equal the number of electrons to the number of protons.Electrons have a negative charge, whereas there is a positive charge for protons.This results in a negative charge when an atom gains electrons. An anion is called this form of ion.This results in a positive charge when an atom loses electrons. A cation is considered a positively-charged ion.Usually, positive ions are metals or behave like metals.Just like atoms may lose electrons to become cations, others can absorb electrons and become anions that are negatively charged.

Two identical metal wires of thermal conductivities \(\mathrm{K}_1\) and \(\mathrm{K}_2\) respectively are connected in series are represented as follows

The above figure can also be represented as

Therefore, the effective thermal conductivity of the combination will be given by

\(\begin{aligned} & R_{e f f}=\frac{I}{\mathrm{~K}_1 A}+\frac{I}{\mathrm{~K}_2 \mathrm{~A}}=\frac{21}{\mathrm{~K}_{e q} \mathrm{~A}} \\ & \Rightarrow \frac{2 \mathrm{I}}{\mathrm{K}_{e q} \mathrm{~A}}=\frac{1}{\mathrm{~A}}\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}\right) \Rightarrow \frac{2 \mathrm{I}}{\mathrm{K}_{e q} \mathrm{~A}}=\frac{I}{\mathrm{~A}}\left(\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right) \\ & \Rightarrow \frac{2}{\mathrm{~K}_{e q}}=\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2} \Rightarrow \mathrm{K}_{\mathrm{eq}}=\frac{2 \mathrm{~K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\end{aligned}\)

Moment of inertia:

• Moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• Moment of inertia of a particle is

I = mr²

Where r = the perpendicular distance of the particle from the rotational axis.

Moment of inertia of a body made up of a number of particles (discrete distribution)

I = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄² + -------

Given \(-I_{s}=I_{h}=I\) (let) and Mass of solid sphere \(=\) mass of hollow sphere

- The moment of inertia of a solid sphere of mass 'M' and radius \({ }^{\prime} R_{1}\) ' about its diameter is

\(\Rightarrow I_{s}=\frac{2}{5} M R_{1}^{2}\) ........... (1)

- The moment of inertia of a hollow sphere of mass 'M' and radius \({ }^{\prime} R_{2}{ }^{\prime}\) about its diameter is

\(\Rightarrow I_{h}=\frac{2}{3} M R_{2}^{2}\) .......... (2)

On equating equation 1 and 2, we get

\(\Rightarrow I_{S}=I_{h}\)

\(\Rightarrow \frac{2}{5} M R_{1}^{2}=\frac{2}{3} M R_{2}^{2}\)

\(\Rightarrow \frac{R_{2}}{R_{1}}=\sqrt{\frac{3}{5}}\)

Given,

\(KE _{2}=9 KE _{1}\)

We know that the kinetic energy of a body is given as,

\( K E=\frac{1}{2} m v^{2}\)...(1)

The linear momentum of a body is given as,

\( P = mv\)...(2)

By equation (1) and equation (2) the relation between the kinetic energy and the momentum is given as,

\( K E=\frac{P^{2}}{2 m}\)...(3)

By equation (3) for the initial position,

\( K E_{1}=\frac{P_{1}^{2}}{2 m}\)...(4)

By equation (3) for the final position,

\( K E_{2}=\frac{P_{2}^{2}}{2 m} \)

\(\Rightarrow 9 \times K E_{1}=\frac{P_{2}^{2}}{2 m}\)...(5)

By equation (4) and equation (5),

\(\frac{P_{2}^{2}}{2 m} \times \frac{2 m}{P_{1}^{2}}=\frac{9 \times K E_{1}}{K E_{1}}\)

\(\Rightarrow \frac{P_{2}^{2}}{P_{1}^{2}}=9 \)

\(\Rightarrow \frac{P_{2}}{P_{1}}=3 \)

\(\Rightarrow P_{2}=3 P_{1}\)

So, if the kinetic energy of a body is increased nine times then the momentum of the body will be increased by three times.

Archimedes principle:

• It states that a body when wholly or partially immersed in liquid experiences an upward thrust which is equal to the volume of the liquid.
• Archimedes Principle is also known as thephysical law of buoyancy.
• Purity of gold in the king's crown,Archimedes determine after discovering the Archimedes principle.
• Hydrometer, Ships, and Submarines work on the Archimedes Principle.
• A Ship/boat floats on the basis of theArchimedes Principle.

Human eyecannot detect polarization of light.

Polarization changes when plane of vibration of polarized light changes.Human eye is insensitive to change in polarization and hence, cannot detect polarization of light.

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be F₁

The net force, \(\mathrm{F}_{\text {net }}=\frac{\mathrm{Mv}^2}{\mathrm{R}}\)

Along the centre of circle,

\(\sqrt{2} \mathrm{~F}+\mathrm{F}_1=\frac{\mathrm{Mv}^2}{\mathrm{R}} \)

\(\sqrt{2}\left(\frac{\mathrm{GMM}}{(\sqrt{2} \mathrm{R})^2}\right)+\left(\frac{\mathrm{GMM}}{(2 \mathrm{R})^2}\right)=\frac{\mathrm{Mv}^2}{\mathrm{R}} \)

\(\Rightarrow \frac{\mathrm{GM}}{\mathrm{R}^2}\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)=\frac{\mathrm{v}^2}{\mathrm{R}} \)

\(\Rightarrow \mathrm{v}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}(2 \sqrt{2}+1)}{\mathrm{R}}}\)

Hence, the speed of each particle is \(\frac{1}{2} \sqrt{\frac{\mathrm{GM}(2 \sqrt{2}+1)}{R}}\).