Physics Test - 19

Visible spectrum lies betweeninfrared spectrum and ultra violet rays.

• The increasing order of frequency of EM waves are given as Radio, Microwave, Infrared, Visible, Ultraviolet, X Rays, and Gamma Rays.
• From the above, it is clear that the visible spectrum lies between infrared spectrum and ultraviolet rays.

From the given graph:

At time (t) = 0 the displacement (x) of the body is equal to A.

X = A Sin (ω t + θ)

X = A Sin (ω × 0 + θ) = A

A Sin θ = A

Sin θ = 1

So Initial phase angle (θ) = 90°

By Pascal's law,

\(\frac{F}{A}=\frac{f}{a}\)

or \(f=\frac{F a}{A}\)

\(\Rightarrow f=\frac{100 g \times(\pi r)^{2}}{(\pi \times 4 r)^{2}}\)

\(\Rightarrow f=6.25 g\)

\(\Rightarrow f=62.5 N\)

Potential gradient (k):Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e.,

\(k=\frac{V}{l}\)

Where, \(V=\) Potential difference and \(l =\) length of the wire

Potential gradient directly depends upon:

• The resistance per unit length (R/L) of potentiometer wire.
• The radius of potentiometer wire ( i.e., Area of crosssection)
• The specific resistance of the material of potentiometer wire (i.e., ρ)
• The current flowing through potentiometer wire (I).

\(\begin{aligned} & \mathrm{B}_{\mathrm{H}}=3.5 \times 10^{-5} \mathrm{~T} \\ & \mathrm{~F}=i \ell \mathrm{B} \sin \theta, \quad \mathrm{i}=\sqrt{2} \mathrm{~A} \\ & \frac{\mathrm{F}}{\ell}=i \mathrm{~B} \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \\ & =35 \times 10^{-6} \mathrm{~N} / \mathrm{m}\end{aligned}\)

Given,

Error in measurement of radius of a sphere = 1%

As we know,

Volume\(=\frac{4 \pi \mathrm{R}^{3}}{3}\)

Then the percentage error of volume,

\(\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3 \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100\)

\(=3 \times \frac{1}{100} \times 100\)

\(=3 \)%

Paschen series: When an electron in a Hydrogen atom transit from higher energy orbit to 3rd orbit. (outer orbit \(n_{2}\) = n > 3 to the orbit \(n_{1}\) = 3) known as Paschen Series.

So the empirical formula for the observed wavelengths (λ) for hydrogen (Z = 1) is,

\(R_{\infty}=1.1\)

\(\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\).....(i)

For \(\lambda_{\min },\) \(n_{1}=3\) and \( \mathrm{n}_{2}=\infty\)

Put above values in (i),

\(\frac{1}{\lambda_{\min }}=R_{\infty}\left(\frac{1}{3^{2}}-0\right)\)

\(\Rightarrow \frac{1}{\lambda_{\min }}=1.1 \times 10^{7}\left(\frac{1}{3^{2}}\right)\)

\(\Rightarrow \lambda_{\min }=0.818 \mu \mathrm{m}\)

For \(\lambda_{\max },\)  \(n_{1}=3\) and \( n_{2}=4\)

Put above values in (i),

\(\frac{1}{\lambda_{\max }}=R_{\infty}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)\)

\(\Rightarrow \frac{1}{\lambda_{\max }}=1.1 \times 10^{7}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)\)

\(\Rightarrow \lambda_{\max }=1.89 \mu \mathrm{m}\)

\(\lambda_{\min }=0.818 \mu \mathrm{m}\) and \(\lambda_{\max }=1.89 \mu \mathrm{m}\)

Concept:

• The coil which stores magnetic energy in a magnetic field is called an inductor.
• The property of an inductor that causes the emf to generate by a change in electric current is called as the inductance of the inductor.
• The SI unit of inductance is Henry (H).

The magnetic potential energy stored in an inductor is given by:

Magnetic potential energy \((U)=\frac{1}{2} L I^{2}\)

Where L is the inductance of the inductor and I is the current flowing.

Given that:

Magnetic potential energy (U) = 25 mJ = 25 × 10^-3 J

Current (I) = 60 mA = 60 × 10^-3 A

Use the formula:

Magnetic potential energy \((U)=\frac{1}{2} L I^{2}\)

\(25 \times {10^{ - 3}} = \frac{1}{2} \times L \times {\left( {60 \times {{10}^{ - 3}}} \right)^2}\)

Inductance \((L)=\frac{2 \times 25 \times 10^{-3}}{3600 \times 10^{-6}}=13.89 \mathrm{H}\)

If wavelength is \(\lambda\) then the speed of the electron is \(v\). 

Let us assume that when wavelength is \(\frac{3 \lambda}{4}\) then the speed of electron is \(v_{1}\).

According to the equation of photoelectric emission:

\(\frac{\mathrm{hc}}{\lambda}=\frac{1}{2} \mathrm{mv}^{2}+\phi \quad \ldots . .\)(1)

 When the wavelength \(\frac{3 \lambda}{4}\). Then the speed of electron is \(v_{1}\).

\(\frac{4 \mathrm{hc}}{3 \lambda}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\phi \quad \ldots . .\)(2)

On dividing (2) by (1)

\(\frac{4}{3}=\frac{\frac{1}{2} m v_{1}^{2}+\phi}{\frac{1}{2} m v^{2}+\phi}\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{4\phi}{3}-\phi\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{\phi}{3}\)

or \(v_{1}=\) greater than \(v\left(\frac{4}{3}\right)^{\frac{1 }{ 2}}\)

As we know,

(1) The formula of the volume of the cylinder is given by

\(V=\pi r^{2} l\)

Where \(V\) is the volume of the cylinder, \(r\) is the radius of the cylinder and \(l\) is the length of the cylinder.

(2) The formula of the Poisson’s ratio is given by

\(\sigma=\frac{\Delta r}{\Delta l}\)

Where \(\sigma\) is the Poisson's ratio, \(\Delta r\) is the change in the radius after stretching and \(\Delta l\) is the change in the length after stretching.

It is given that there is no change in the volume of the wire when stretching.

Since the wire is in the form of the cylinder, the formula of the cylinder is used.

\(V=\pi r^{2} l\)

Differentiating the above formula with respect to the radius and the length of the wire, we get

\(\Rightarrow d V=2 \pi r l d r+\pi r^{2} d l\)

Since there is no change in the volume, \(dV=0\)

\(\Rightarrow 2 \pi r l d r+\pi r^{2} d l=0\)

By solving the above equation, we get

\(\Rightarrow \frac{d r}{r}=-\frac{1}{2} \frac{d l}{l}\)

\(\Rightarrow \frac{\frac{d r}{d l}}{\frac{r}{l}}=-\frac{1}{2}\)

From the formula of the Poisson’s ratio, we get

\(\Rightarrow \sigma=-\frac{1}{2}\)

The negative value of the Poisson’s ratio indicates that when the length of the wire is increased by stretching, the radius of the wire decreases. From the above explanation, we get

\(\Rightarrow \sigma=0.5\)