Physics Test - 10

For a particle executing SHM, we can say that

\(x=A \sin (\omega t+\phi)\)

Given two particles are executing SHM, we can write that \(x_{1}=A \sin \left(\delta_{1}\right)\) and \(x_{2}=A \sin \left(\delta_{2}\right)\)

Where \(\delta_{1}\) and \(\delta_{2}\) are the phase of two particles.

From the data given in the question, two particles pass one another going in opposite directions when their displacements are half of their amplitudes.

\(\therefore {x}_{1}={x}_{2}=\frac{\mathrm{A}}{2}\), we get

For particle 1:

Assuming particle is moving away from mean position.

\(\Rightarrow \frac{\mathrm{A}}{2}=\mathrm{A} \sin \left({\delta}_{1}\right)\) \(\Rightarrow \sin \left({\delta}_{1}\right)=\frac{1}{2}\) \(\Rightarrow \delta_{1}=\frac{\pi}{6}\) Similarly,

For particle 2 :

Assuming particle is moving towards mean position.

\(\Rightarrow \frac{A}{2}=A \sin \left(\delta_{2}\right)\)

\(\Rightarrow \sin \left(\delta_{2}\right)=\frac{1}{2}\)

\(\Rightarrow \delta_{2}=\frac{5 \pi}{6}\)

Phase difference between the two particles is \(\delta=\left|\delta_{1}-\delta_{2}\right|=\frac{5 \pi}{6}-\frac{\pi}{6}=\frac{2 \pi}{3}\) or \(120^{\circ}\)

The correct sentence is that work has only magnitude but no direction.

• Work is donewhen the body moves with the application of external force or the moving body stop after theexternal forceapplied.
• Work done is thedot productof force and displacement.
• W = F ×d
  1. Where F = force applied
  2. d = displacement
• The dot product of vector quantities is always scalarwhich means work is having only magnitude but no direction.

When a horizontal disc is rotated about its vertical axis and sand is dropped on it, then its angular velocity will decrease.

According to the Law of conservation of angular momentum, the total angular momentum of the body about that axis remains constant i.e., if the moment of inertia of the body increases then angular velocity and vice-versa.

When the sand is dropped at the center, the total mass of the system increase hence the moment of inertia (I) increases as \({I}={mr}^{2}\).

So, the angular velocity of the system \((\omega)\) decreases, but as soon as the starts falling, the moment of inertia of the system starts decreasing, and hence angular velocity \((\omega)\) starts increasing again.

Mutual induction between the two coils of area A, number of turns N₁ and N₂ with the length of secondary or primary l is given by,

$M=\frac{{\mu }_o{N}_1{N}_{2}A}{l}$

From the above formula, the mutual inductance of two coils can be increased by increasing the number of turns in the coils.

Let \(\mathrm{M}\) and \(\mathrm{R}\) be the mass and radius of the disc respectively.

Moment of inertia about \(\mathrm{AB}\) is \(\mathrm{I}\).

Moment of inertia about an axis passing through \(\mathrm{O}\) and perpendicular to the plane,

\(\mathrm{I}_{\mathrm{O}}=\frac{1}{2} \mathrm{M} \mathrm{R}^{2}\)

Moment of inertia about an axis passing through \(\mathrm{C}\) and perpendicular to the plane,

\(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{O}}+\mathrm{M} \mathrm{R}^{2}\)

\(\mathrm{I}_{\mathrm{C}}=\mathrm{3 I}_{\mathrm{O}}\)

Using perpendicular axis theorem,

\(\Rightarrow \mathrm{I}_{O}=2 \mathrm{I}\)

Thus \(\mathrm{I}_{\mathrm{C}}=\mathrm{6 I}\)

As we know that from mirror formula,

\(\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \).....(1)

\(u= \) distance of object

\(v=\) distance of object

\(f=\) focal length

Differentiating (1), we get

\(-v^{-2} \mathrm{dv}-\mathrm{u}^{-2} \mathrm{du}=0\)

\(|\mathrm{d} v|=\left|\frac{v^{2}}{\mathrm{u}^{2}}\right| \mathrm{du}\).....(2)

Here \(|\mathrm{dv}|=\) size of image,

\(|\mathrm{du}|=\) size of object \((\mathrm{b})\)

From the equation (1), we write

\(\frac{\mathrm{u}}{\mathrm{v}}+1=\frac{\mathrm{u}}{\mathrm{f}}\)

Squaring both sides, we get

\(\frac{v^{2}}{\mathrm{v}^{2}}=\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}\).....(3)

Substituting equation (3) in equation (2) we get,

Size of the image \(\mathrm{dv}=\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}~\mid du \mid\)

Let \(|\mathrm{du}|=\) size of object \((\mathrm{b})\)

Size of the image \(\mathrm{dv}=\mathrm{b}\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)^{2}\)

In practice, the work output of a machine is always less than the work input due to the effect of friction.

The work output of a machine is never equal to the work input because some of the work done by the machine is used to overcome the friction created by the use of the machine.

This is the reason why the efficiency of a machine can never be 100%.

The efficiency is the work output, divided by the work input, and expressed as a percentage.

The electric field of a plane electromagnetic wave is given by

\(\vec{E}=E_{0}(\hat{x}+\hat{y}) \sin (k z-\omega t)\)

Direction of propagation \(=+\hat{k}\)

\(\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

\(\hat{k}=\hat{E} \times \hat{B}\)

\(\hat{k}=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times \hat{B}\)

\(\Rightarrow \hat{B}=\frac{-\hat{i}+\hat{j}}{\sqrt{2}}\)

\(\hat{B}=\frac{E_{0}}{c}(-\hat{x}+\hat{y}) \sin (k z-\omega t)\)

Given,

The percentage error in the measurement of mass, \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=2 \%\)

The percentage error in the measurement of speed, \(\frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=3 \%\)

The kinetic energy of the particle is given by

\(\mathrm{KE}=\frac{\mathrm{mv}^{2}}{2}\)

The percentage error in kinetic energy will be,

\(=\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100+2 \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100\)

\(=2+6\)

\(=8 \%\)

Given,

\({W}\) (load in the wire \()={Mg}\), and \({l}=\) increment in the length

Let \(L=\) initial length of the wire

The developed stress in the wire is given as,

\(\sigma=\frac{W}{A}\)

Where \(A=\) cross-sectional area of the wire

\(\sigma=\frac{M g}{A}\)

The strain in the wire is given as,

\(\epsilon=\frac{\text { Change in length }}{\text { Initial length }}\)

\(\epsilon=\frac{l}{L}\)

The volume of the wire is given as,

\(V=A L\)

We know that when a body is stretched by a load, then the work done by the load on the body gets stored in the form of strain energy. So work done on the wire is given as,

Strain Energy: The energy stored in a body when it is strained within its elastic limit is known as strain energy. The strain energy is given as,

Work Done \(=\) Strain energy

Work Done =\(E=\frac{1}{2} \times \sigma \times \epsilon \times V\)

Work Done \(=\frac{1}{2} \times \frac{M g}{A} \times \frac{l}{L} \times A L\)

Work Done \(=\frac{M g l}{2}\)