Physics Test - 36

\(1 \mathrm{eV}=1.6 \times 10^{-19}\) \(J\)

\(1 J=\frac{1}{1.6 \times 10^{-19}} \mathrm{eV}\)

\(10^{-20} \mathrm{~J}=\frac{10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\)

\(=0.06 \mathrm{eV}\)

As you know that magnetic field at point on the axis of current carrying ring is:

\({B}_{x}=\frac{\mu_{0} N i R^{2}}{2\left(R^{2}+x^{2}\right)^{\frac{3}{2}}}\)

Where \(x\) is the point on the axis of ring, \(R\) is the radius of ring, \(i\) is the current carrying on ring and \({N}\) is the number of turns at center magnetic field

\({B}_{{c}}=\frac{\mu_{0} {N i}}{2 {R}}\)

Now, According to the question,

Magnetic field at \(x=\) magnetic field at center

\(\frac{\mu_{0} N i R^{2}}{2\left(R^{2}+x^{2}\right)^{\frac{3}{ 2}}}=\frac{\mu_{0} N i}{8 \times 2 R}\)

\(\frac{R^{2}}{2\left(R^{2}+x^{2}\right)^{\frac{3}{ 2}}}=\frac{1}{16 R}\)

\(8 R^{3}=\left(R^{2}+x^{2}\right)^{\frac{3}{2}}\)

This is possible only when 

\(x=\pm \sqrt{3} R\)

Then, \(\sqrt{3} {R}\) distance from the center magnetic field is equal to the magnetic field at the center.

Radius of nuclear given by,

\(R=R O A^{\frac{1 }{ 3}}\) where \(R o=1.25 \mathrm{fm}\)

\(\Rightarrow A=\left(\frac{R}{R o}\right)^{3}=\left(\frac{7.2}{1.25}\right)^{3}=191\) (atomic mass)

Now, no. of protons,

\(Z=\frac{1.28 \times 10^{-17}}{1.6 \times 10^{-19}}=80\)

Thus, no. of neutrons nuclears:

N = A - Z = 191 - 80 = 111

​Positive charge: A body having a deficiency of electrons.

We know that the proton is present in the nucleus so we can't add or remove protons in the element but we can add or remove electrons from the element.

So the body can be positively charged by removing electrons and negatively charged by adding electrons.

A charge cannot be generated by adding or removing protons.

As the coefficient of cubical expansion of liquid equals the coefficient of cubical expansion of the vessel, the level of liquid will not change on heating.

If linear expansion is \(\alpha\), then \(\gamma=3 \alpha\)

Here, \(\alpha=\frac{\gamma}{3}\) so \(\gamma^{\prime}=3 \times \frac{\gamma}{3}\)

Thus, \(\gamma^{\prime}=\gamma\).

By Einstein's photoelectric equation, we have

\(\Rightarrow \mathrm{hf}=\phi+(\mathrm{KE})_{\max }\)

where, \(f\) is the frequency, \(\phi\) is the work-function of metal and \((\mathrm{KE})_{\max }\) is the maximum kinetic energy of emitted photoelectron.

\( \Rightarrow \frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}_0 \)

\( \Rightarrow \mathrm{eV}_0=\frac{\mathrm{hc}}{\lambda}-\phi\)

Here, \(\lambda=280 \mathrm{~nm}\) and \(\phi=2.5 \mathrm{eV}\)

\(\Rightarrow \mathrm{e}\left(\mathrm{V}_0\right)_1=\frac{1240}{280}-2.5 \)

\(\Rightarrow \mathrm{e}\left(\mathrm{V}_0\right)_1=1.93 \mathrm{eV} \)

\(\Rightarrow\left(\mathrm{V}_0\right)_1=1.93 \mathrm{~V}\)

Similarly, stopping potential for the light of wavelength of \(400 \mathrm{~nm}\),

\(\mathrm{e}\left(\mathrm{V}_0\right)_2=\frac{1240}{400}-2.5 \)

\(\Rightarrow\left(\mathrm{V}_0\right)_2=0.6 \mathrm{~V}\)

\(\therefore\) Change in stopping potential,

\(\Delta \mathrm{V}=\left(\mathrm{V}_0\right)_1-\left(\mathrm{V}_0\right)_2 \)

\( =1.93-0.6=1.33 \simeq 1.3 \mathrm{~V}\)

Protons and Neutrons are the components of the nucleus of an atom.

• Protons are the particles that have a positive charge and found within atomic nuclei.
• Neutrons are particles that have no charge and found within atomic nuclei (except for hydrogen-1). Protons are slightly smaller in mass as compared to neutrons.

The heavier the object is harder to move it or the greater the amount of force required to move it hence higher the inertia.Since higher mass has higher inertia. In all the above four options, a cricket ball has maximum mass so it will have maximum inertia.

According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

The inertia of rest: When a body is in rest, it will remain at rest until we apply an external force to move it. This property is called inertia of rest.

The inertia of motion: When a body is in a uniform motion, it will remain in motion until we apply an external force to stop it. This property is called inertia of motion.

The distance \(s_1\) covered by the car during the time it is accelerated is given by \(2 a s_1=v^2\), which gives \(s_1=\frac{v^2}{2 a}\). The distance \(s_2\) covered during the time the car is decelerated is similarly given by \(s_2=\frac{\mathrm{v}^2}{2 \beta}\)

Therefore, the total distance covered is

\(s=s_1+s_2=\frac{v^2}{2}\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots \text { (i) }\)

If \(t_1\) is the time of acceleration and \(t_2\) that of deceleration, then \(v=a t_1=\beta t_2 \Rightarrow \mathrm{t}_1=\frac{\mathrm{v}}{a}\) and \(\mathrm{t}_2=\frac{\mathrm{v}}{\beta}\)

Therefore, the total time taken is

\(t=t_1+t_2=v\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots\)(ii)

From (i) and (ii), the average speed of the car is given by,

\(\frac{\text { Total distance }}{\text { Total time }} \frac{s}{t}=\frac{v}{2}\)

Based on the data provided

\( \mathrm{U}=100-80=20 \mathrm{~cm} \)

\(\mathrm{~V}=180-100=80 \mathrm{~cm}\)

Using \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) or \(f=\frac{u v}{u+v}=\frac{20 \times 80}{20+80}\) or \(f=16 \mathrm{~cm}\)

For error analysis,

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

Differentiating

\(-\frac{\mathrm{Df}}{\mathrm{f}^2}=-\frac{\mathrm{Dv}}{\mathrm{v}^2}-\frac{\Delta \mathrm{u}}{\mathrm{u}^2}\)

To calculate \(\Delta \mathrm{u} \& \Delta \mathrm{v}\)

\(\mathrm{U}=(100 \pm 2)-(80 \pm 0.2)=(20 \pm 0.4) \mathrm{cm}\)

Therefore \(\Delta \mathrm{u}=0.4 \mathrm{~cm}\),

Similarly \(\Delta \mathrm{v}=0.4 \mathrm{~cm}\).

Now \(\frac{\Delta f}{f}=f\left[\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right]\)

\(\frac{\Delta \mathrm{f}}{\mathrm{f}}=16\left[\frac{0.4}{(80)^2}+\frac{0.4}{(20)^2}\right]\)

(Note: every data is in \(\mathrm{cm}\) )

\(\frac{\Delta f}{f}=\frac{16 \times 0.4}{(20)^2}\left[\frac{1}{4^2}+1\right] \)

\(=\frac{16 \times 0.4}{20^2} \times \frac{17}{16}=\frac{17 \times 0.4}{400} \)

\(\% \text { Error : } \frac{\Delta f}{f} \times 100=\frac{17 \times 0.4}{400} \times 1000 \)

\(=1.7\)