Physics Test - 21

Given,

\(m=10\) kg, \(F_1=4\)N, \(F_2=3\)N, \(θ=90^{\circ}\) and \(t=10\)s

Since the two forces are acting perpendicular to each other, so the resultant force is given as,

\(\mathrm{F}=\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}}\)

\(\Rightarrow \mathrm{F}=\sqrt{4^{2}+3^{2}}\)

\(\Rightarrow \mathrm{F}=5 \mathrm{~N} \ldots(1)\)

By Newton's second law of motion, the force acting on the body is given as,

F=ma

Where m= mass and a= acceleration

\(a=\frac{F}{m}\)

\(\Rightarrow a=\frac{5}{10}\)

\(\Rightarrow a=\frac{1}{2} m / s^{2} \quad \ldots(2)\)

Kinetic energy \(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m}(\text { at })^{2}\)

Undefined control sequence therefore,

\(=\frac{1}{2} \times 10 \times(5)^{2}\)

\(=125 \mathrm{~J}\)

So, the kinetic energy of the body at the end of \(10 \mathrm{~s}\) is \(125 \mathrm{~J}\).

Field emission involves the emission of electrons from a material's surface when subjected to a strong electric field. In this process, electrons are emitted from a material's surface when a strong electric field is applied. It's termed "quick" because it occurs rapidly under the influence of the electric field, without requiring significant heating or other conditions.

Given,

The angle of incident \(=i\)

Prism small angle \(={A}\)

Refractive material of the prism \(=\mu\)

Let the angle of emergence \(={e}\)

Now, it is given that the incident ray emerges normally from the opposite surface of the prism so the angle of emergence becomes zero.

\(\Rightarrow \angle e=0\) .........\((i)\)

Let angle of deviation in the prime be \(\delta\)

So, as we know the relation between the angle of deviation \((\delta),\) small angle of prism \(({A})\) and the refractive index of the prism \(\mu\) which is given as,

\(\Rightarrow \delta=(\mu-1) A\) .........\((ii)\)

Now, as we know in a prism, a small angle of prism plus an angle of deviation in a prism is equal to the sum of angle of incidence and angle of emergence.

Therefore, we have,

\(\Rightarrow \delta+A=i+e\) .......\((iii)\)

Now, substitute the value from equation \((i)\) and \((ii)\) in equation \((iii)\) we have,

\(\Rightarrow(\mu-1) A+A=i+0\)

Now, simplify the above equation we have,

\(\Rightarrow \mu A-A+A=i\)

Now, cancel out the positive, negative same terms we have,

\(\Rightarrow i=\mu A\)

So, this is the required angle of incidence such that the prism has a small angle \({A}\) and emerges normally from the opposite surface.

Let the initial velocity of ball be \(u\).

Time of rise, \(t_{1}=\frac{u}{g+a}\) 

And height reached \(=\frac{u^{2}}{2(g+a)}\) 

Time of fall \(t_{2}\) is given by

\(\frac{1}{2}(g-a) t_{2}^{2}=\frac{u^{2}}{2(g+a)}\)

\(\Rightarrow t_{2}=\frac{u}{\sqrt{(g+a)(g-a)}}\)

\(\Rightarrow t_{2}=\frac{u}{(g+a)} \sqrt{\frac{g+a}{g-a}}\)

\(\therefore t_{2}>t_{1}\) because \(\frac{1}{g+a}<\frac{1}{g-a}\)

Given:

\(\mathrm{y}_{1}=10 \sin (2000 \pi \mathrm{t})\).....(1)

\(\mathrm{y}_{2}=10 \sin (2000 \pi \mathrm{t+\frac{\pi}{2}})\).....(2)

\(\mathrm{y}=a \sin((\omega t+ \phi)\).....(3)

On comparing (1) and (2) with (3).

\(\mathrm{a}_{1}=10, \mathrm{a}_{2}=10\) and \(\phi= \frac{\pi}{2}=90^{\circ}\)

The resultant amplitude A of two waves of amplitudes \(a_{1}\) and \(a_{2}\) at a phase difference of \(\phi\) is:

\(A=\left(a_{1}^{2}+\right.\) \(\left.\mathrm{a}_{2}^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \phi\right)^{\frac{1 }{ 2}}\).....(4)

Substituting all given value in (4).

\(A=\left(10^{2}+\right.\) \(\left.10^{2}+2 ×10 × 10 \cos 90^{\circ}\right)^{\frac{1 }{ 2}}\)

\(=10\sqrt{2}\)

\(=14.1\)

Given,

M.I. of a thin uniform ring about an axis passing through the centre and perpendicular to plane is \(4 \mathrm{~kgm}^{2}=\mathrm{mr}^{2}\)

Using Perpendicular axix theorem,

M.I. of a thin uniform ring about an axis passing through the centre and in the plane of the ring is \(2 \mathrm{~kgm}^{2}=\) \(\frac{1}{2} \mathrm{mr}^{2}\)

Now, Using Parallel axix theorem,

M.I. of ring about an axis tangent to the ring and in a plane of the ring

\(\frac{1}{2} \mathrm{mr}^{2}+\mathrm{md}^{2}=\frac{3}{2} \mathrm{mr}^{2}\)

\(=\frac{3}{2} \mathrm{mr}^{2}=\left(\frac{3}{2}\right) 4 \mathrm{~kgm}^{2}\)

\(=6 \mathrm{~kgm}^{2}\)

At equilibrium, \(m u \cos \alpha=2 m v^{\prime}\)

\(\Rightarrow v^{\prime}=\frac{u \cos \alpha}{2}\)

\(\Rightarrow \frac{1}{2} =\frac{{v}^{\prime}}{u\cos \alpha}.......(1)\)

As we know that,

The coefficient of restitution,

\(e=\frac{{v}^{\prime}}{u\cos \alpha}....(2)\)

From equation (1) and (2), we get, \(e=\frac{1}{2}\)

In the case of a steady electric field, the electric field does not change with time. Therefore, the electric flux density through a closed-loop will be constant.

This results in zero change in electric flux with respect to time i.e.,

\( \frac{d \phi_{E}}{d t}=0\)

The displacement current will be:

\( i_{d}=\epsilon_{0} \frac{d \phi_{E}}{d t}=0\)

Given:

Moment of inertia \((I)=3 \mathrm{~kg}\mathrm{m}^{2}\)

Angular Velocity \((\omega)=2 ~\mathrm{rad} / \mathrm{s}\)

Mass \((\mathrm{m})=12 \mathrm{~kg}\)

To Find velocity (v)

The velocity is given as,

\(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{mv}^{2}\)

\(\frac{1}{2} \times 3(2)^{2}=\frac{1}{2} \times 12 \times \mathrm{v}^{2}\)

\(\mathrm{v}=1 \mathrm{~m} / \mathrm{s}\)

Thus, the speed of the body is \(1 \mathrm{~m} / \mathrm{s}\).

We know that:

Time period of a simple pendulum of length \(l\), is given by:

\(T=2 \pi \sqrt{\frac{l}{g}} \ldots(\mathrm{i})\)

Where, \(g\) is acceleration due to gravity.

When \(g^{\prime}=\frac{g}{4}\)

New, time period is

\(T^{\prime}=2 \pi \sqrt{\frac{l}{\frac{g}{ 4}}} \ldots(\mathrm{ii})\)

Dividing Eq. (ii) by Eq. (i), we get

\(\frac{T^{\prime}}{T}=\sqrt{\frac{g}{\frac{g}{ 4}}}=2\)

\(\Rightarrow T^{\prime}=2 T\)

Therefore, new time period becomes twice of the original value.