NEET-UG 2023

Error arise due to unpredictable fluctuation in temperature and voltage supply are $\to$ random errors.

$$\rho =\frac{M}{V}$$

$$\rho =\frac{M}{\pi r^2\ell }$$

$$\frac{\Delta \rho }{\rho }=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{2\Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \ell }{\ell }$$

$$\frac{\Delta \rho }{\rho }\%=\left[\frac{0.002}{0.4}+\frac{2(0.001)}{(0.3)}+\frac{0.02}{5}\right]\times 100\%$$

$$=\frac{1}{2}\%+\frac{2}{3}\%+\frac{2}{5}\%$$

$$=1.6\%$$

$$V_{\text{avg }}=\frac{2v_1{v}_2}{v_1+v_2}=\frac{2(v)(2v)}{v+2v}=\frac{4v^2}{3v}=\frac{4v}{3}$$

$${\mathrm{H}}_{\max }=\frac{{\mathrm{u}}^2{\sin }^2\theta }{2g}$$

$$=\frac{(280{)}^2{\left(\sin 30^{\circ }\right)}^2}{2(9.8)}$$

$$=1000\mathrm{m}$$

$$S=ut+\frac{1}{2}a{t}^2$$

$$-\mathrm{H}=4\times 4-\frac{1}{2}\times 10\times 4^2$$

$$-\mathrm{H}=16-80$$

$$-\mathrm{H}=-64$$

$$\mathrm{H}=64\mathrm{m}$$

${\vec{V}}_{\mathrm{i}}=(\mathrm{V})$ Southward

${\overrightarrow{\mathrm{V}}}_{\mathrm{F}}=(\mathrm{V})$ Eastward

$$\overrightarrow{\Delta \mathrm{V}}={\overrightarrow{\mathrm{V}}}_{\mathrm{F}}-{\overrightarrow{\mathrm{V}}}_{\mathrm{i}}$$

= Along North - East

To find the maximum acceleration ($a_{\max }$) of the car that allows a body to stay stationary relative to the car, we use the concept of static friction. Static friction ($F_s$) is what keeps the body from sliding on the car's floor. It acts in the opposite direction of the potential movement of the body.

The maximum static friction force is given by $F_{s_{\max }}={\mu }_s\times N$ where ${\mu }_s$ is the coefficient of static friction and $N$ is the normal force. In this scenario, the normal force is equal to the gravitational force on the body ($mg$), where $m$ is the mass of the body and $g$ is the acceleration due to gravity.

Since the maximum force of static friction equals the product of the mass and the maximum acceleration ($m{a}_{\max }$), we have:

$${\mu }_{s}mg=m{a}_{\max }$$

By canceling out the mass $m$ on both sides, we get:

$$a_{\max }={\mu }_{s}g$$

Substituting the given values (${\mu }_s=0.15$ and $g=10{m/s}^2$):

$$a_{\max }=0.15\times 10=1.5{m/s}^2$$

Therefore, the maximum acceleration of the car to ensure the body remains stationary with respect to the car's floor is $1.5{m/s}^2$.

$$By{v}^2=u^2+2as$$

$${\left(\frac{u}{3}\right)}^2=u^2-2ax$$

$$2ax=u^2-\frac{u^2}{9}$$

$2\mathrm{ax}=\frac{8{\mathrm{u}}^2}{9}$ ..... (1)

Similarly from starting

$$v^2=u^2+2ax$$

$$0=u^2-2a{x}_2$$

$2{\mathrm{ax}}_2={\mathrm{u}}^2$ ..... (2)

$$\mathrm{By}(1)/(2)$$

$$\frac{x}{x_2}=\frac{8}{9}$$

$$\frac{24}{x_2}=\frac{8}{9}$$

$${\mathrm{x}}_2=27\mathrm{cm}$$

The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula:

$$U=\frac{1}{2}k{x}^2$$

where:

• $k$ is the spring constant,
• $x$ is the displacement from the spring's equilibrium position (i.e., how much the spring is stretched or compressed).

If the spring is initially stretched by $2$ cm (or $0.02$ meters, since we generally use SI units for these calculations), the potential energy stored can be represented as:

$$U=\frac{1}{2}k(0.02{)}^2$$

When the spring is stretched by $8$ cm (or $0.08$ meters), the new potential energy $U^{\prime }$ becomes:

$$U^{\prime }=\frac{1}{2}k(0.08{)}^2$$

To find the relation between $U^{\prime }$ and $U$, we can divide the expression for $U^{\prime }$ by that of $U$:

$$\frac{U^{\prime }}{U}=\frac{\frac{1}{2}k(0.08{)}^2}{\frac{1}{2}k(0.02{)}^2}$$

Simplifying this quotient gives:

$$\frac{U^{\prime }}{U}={\left(\frac{0.08}{0.02}\right)}^2={\left(4\right)}^2=16$$

Thus, if the spring's displacement is increased from $2$ cm to $8$ cm, the potential energy stored in the spring increases by a factor of $16$, meaning:

$$U^{\prime }=16U$$

To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere (K₁) to the radius of gyration for a thin hollow sphere (K₂) of the same mass M and radius R.

The moment of inertia (I) of a solid sphere about its own axis is given by :

$$I_{solid}=\frac{2}{5}M{R}^2$$

The radius of gyration (K) is related to the moment of inertia (I) and mass (M) by the formula :

$$I=M{K}^2$$

So for the solid sphere, we can find K₁ using :

$K_1^2=\frac{I_{solid}}{M}=\frac{2}{5}{R}^2$

$K_1=R\sqrt{\frac{2}{5}}$

Now, for a thin hollow sphere, the moment of inertia about its axis is given by :

$$I_{hollow}=\frac{2}{3}M{R}^2$$

We can find K₂ using :

$K_2^2=\frac{I_{hollow}}{M}=\frac{2}{3}{R}^2$

$K_2=R\sqrt{\frac{2}{3}}$

Now, we need to find the ratio K₁ : K₂ :

$$\frac{K_1}{K_2}=\frac{R\sqrt{\frac{2}{5}}}{R\sqrt{\frac{2}{3}}}$$

The R terms cancel out, and we are left with :

$$\frac{K_1}{K_2}=\frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}}$$

Simplify by taking the square root of the fraction :

$$\frac{K_1}{K_2}=\frac{\sqrt{2}\sqrt{3}}{\sqrt{5}\sqrt{2}}$$

Now, the square root of 2 terms cancel out, giving us :

$$\frac{K_1}{K_2}=\frac{\sqrt{3}}{\sqrt{5}}$$

Along the axis of rotation.

Position of Neutral point (Zero Gravitational Field)

$$r_1=\frac{\sqrt{m_1}R}{\sqrt{m_1}+\sqrt{m_2}}=\frac{\sqrt{m}R}{\sqrt{m}+\sqrt{9m}}=\frac{R}{4}$$

$${\mathrm{r}}_2=\mathrm{R}-\mathrm{R}/4=3\mathrm{R}/4$$

Now Gravitational potential at point $\mathrm{P}$

$$\begin{array}{rl}& V_P=-\frac{GM}{R/4}-\frac{9(\mathrm{GM})}{3\mathrm{R}/4}\\ & =\frac{-16\mathrm{GM}}{\mathrm{R}}\end{array}$$

For a satellite orbiting just above the surface of the Earth, we can use the formula for the period T of the satellite in terms of the Earth's density d and the gravitational constant G:

$$T=2\pi \sqrt{\frac{a^3}{GM}}$$

where a is the semi-major axis of the orbit (which is approximately equal to the Earth's radius R for a satellite orbiting just above the surface) and M is the mass of the Earth.

We can express the mass of the Earth M in terms of its density d and volume:

$$M=dV=d\times \frac{4}{3}\pi R^3$$

Now, substitute this expression for M into the equation for T:

$$T=2\pi \sqrt{\frac{a^3}{Gd\times \frac{4}{3}\pi R^3}}$$

Since the satellite is orbiting just above the Earth's surface, we can approximate a ≈ R:

$$T=2\pi \sqrt{\frac{R^3}{Gd\times \frac{4}{3}\pi R^3}}$$

Simplify the equation:

$$T=2\pi \sqrt{\frac{1}{Gd\times \frac{4}{3}\pi }}$$

Now, square both sides of the equation:

$$T^2=4{\pi }^2\frac{1}{Gd\times \frac{4}{3}\pi }$$

Simplify further:

$$T^2=\frac{3\pi }{Gd}$$

Thus, the quantity $\frac{3\pi }{Gd}$ represents $T^2$.

$$E=2T\left(4\pi {\mathrm{R}}^2\right)$$

$$=2(0.03)(4)(3.14){\left(2\times 10^{-2}\right)}^2$$

$$=3.01\times 10^{-4}\mathrm{J}$$

When a wire is suspended from the ceiling and stretched by a weight W attached at its free end, the force acting on the wire is equal to the weight W. This force causes longitudinal stress in the wire.

Longitudinal stress is defined as the force acting on an object divided by its cross-sectional area. In this case, the force acting on the wire is W, and the cross-sectional area of the wire is A.

Thus, the longitudinal stress at any point of the wire with cross-sectional area A is given by :

$$\text{Longitudinal Stress}=\frac{\text{Force}}{\text{Area}}=\frac{W}{A}$$

$$v_{\mathrm{rms}}\propto \sqrt{T}$$

$$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}}$$

$=$ let initial speed is $\mathrm{v}$

As speed is increased by $3$ times so final speed become $4v$

$$\Rightarrow \frac{\mathrm{v}}{4\mathrm{v}}=\sqrt{\frac{223}{\mathrm{T}}}$$

$$\mathrm{T}=3568\mathrm{K}$$

So temp. in ${}^{\circ }\mathrm{C}=3568-273=3295^{\circ }\mathrm{C}$

Efficiency of carnot engine

$$\begin{array}{rl}& \%\eta =\left(1-\frac{T_{\text{sink }}}{{\mathrm{T}}_{\text{source }}}\right)\times 100\\ & T_{\text{source }}=327^{\circ }\mathrm{C}=600\mathrm{K}\\ & 50=\left(1-\frac{{\mathrm{T}}_{\text{sink }}}{600}\right)\times 100\\ & \frac{1}{2}=1-\frac{{\mathrm{T}}_{\text{sink }}}{600}\\ & {\mathrm{T}}_{\text{Sink }}=300\mathrm{K}\end{array}$$

So temp. of sink is ${}^{\circ }\mathrm{C}=300-2763=27^{\circ }\mathrm{C}$

$$\begin{array}{rl}& \mathrm{x}=\mathrm{A}\sin (\omega \mathrm{t})\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\mathrm{A}\omega \cos (\omega \mathrm{t})\\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}=-{\omega }^2\mathrm{A}\sin (\omega \mathrm{t})\\ & \mathrm{a}=-{\left(\frac{2\pi }{8}\right)}^2\times 1\sin \left(\frac{2\pi }{8}\times 2\right)\\ & \Rightarrow \mathrm{a}=-\frac{{\pi }^2}{16}\times \sin \left(\frac{\pi }{2}\right)\\ & \therefore \mathrm{a}=\frac{-{\pi }^2}{16}\mathrm{m}/{\mathrm{s}}^2\end{array}$$

The fundamental frequency of an open pipe, n_oop, is given by V / (2L), where V is the speed of sound and L is the length of the pipe.

For a closed pipe of the same length, the fundamental frequency, n_cop, is V / (4L) since only half as many wavelengths fit into the same length due to the closed end.

Therefore, the ratio of frequencies of an open pipe to a closed pipe is:

$$\frac{n_{\text{oop}}}{n_{\text{cop}}}=\frac{V/(2L)}{V/(4L)}=\frac{4L}{2L}=2:1$$

This shows that the fundamental frequency of an open pipe is twice that of a closed pipe of the same length.

The torque τ experienced by an electric dipole in an electric field is given by the formula:

$$\tau =pE\sin \theta$$

where p is the electric dipole moment, E is the electric field intensity, and θ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:

$$p=qd$$

where q is the charge on the dipole, and d is the dipole length.

We are given the following values:

• Torque τ = 4 N·m
• Electric field intensity E = $2\times 10^5{\mathrm{NC}}^{-1}$
• Angle θ = $30^{\circ }$
• Dipole length d = 2 cm = 0.02 m

We need to find the charge q on the dipole. Let's first solve for the electric dipole moment p:

$\tau =pE\sin \theta$

$\Rightarrow$ $p=\frac{\tau }{E\sin \theta }$

Substituting the given values:

$$p=\frac{4}{(2\times 10^5)\sin 30^{\circ }}=\frac{4}{(2\times 10^5)(0.5)}=\frac{4}{10^5}=4\times 10^{-5}\mathrm{C}\mathrm{m}$$

Now, let's solve for the charge q using the formula:

$\Rightarrow$ $p=qd$

$q=\frac{p}{d}$

Substituting the values for p and d:

$$q=\frac{4\times 10^{-5}}{0.02}=2\times 10^{-3}\mathrm{C}=2\mathrm{mC}$$

So, the magnitude of the charge on the dipole is 2 mC.

$${\varphi }_{\mathrm{closed}}=0$$

So, ${\varphi }_{\mathrm{in}}={\varphi }_{\mathrm{out}}$

Number of field lines entering is equal number of field lines leaving.

$$\mathrm{v}=\frac{\mathrm{Kq}}{2\times 10^{-2}}-\frac{\mathrm{Kq}}{8\times 10^{-2}}$$

$$=\mathrm{Kq}\left[\frac{3}{8}\right]\times 10^2$$

$$\mathrm{i}=\frac{10-5}{10}=\frac{5}{10}\mathrm{A}$$

$$=0.5\mathrm{A}$$

from $A$ to $B$ through $E$.

For no reading galvanometer. Potential across it is same.

$${\mathrm{i}}_{400\Omega }\Rightarrow \frac{10-2}{400}=\frac{8}{400}=\frac{1}{50}={\mathrm{i}}_{\mathrm{R}}$$

$${\mathrm{i}}_{\mathrm{R}}\Rightarrow \frac{{\mathrm{V}}_{\mathrm{R}}}{\mathrm{R}}\Rightarrow \frac{2}{\mathrm{R}}=\frac{1}{50}\Rightarrow \mathrm{R}=100\Omega$$

$$R=\left[22\times 10^3\pm 5\%\right]\Omega$$

Acc. to color code

Third Band $\to$ Orange

(color code for digit 3 is orange)

$${\mathrm{R}}_{\mathrm{T}}={\mathrm{R}}_0\left[1+\alpha \left(\mathrm{T}-{\mathrm{T}}_0\right)\right]$$

$$6.8=2[1+\alpha (80-\alpha )]$$

$$\alpha =\frac{2.4}{80}=0.03/{}^{\circ }\mathrm{C}=3\times 10^{-2}/{}^{\circ }\mathrm{C}$$

${\mathrm{I}}_{\mathrm{S}}=\frac{E}{10R}$ ..... (1)

${\mathrm{I}}_{\mathrm{P}}=\frac{\mathrm{E}}{\mathrm{R}/10}=\frac{10\mathrm{E}}{\mathrm{R}}$ .... (2)

$$\mathrm{n}=\frac{{\mathrm{I}}_{\mathrm{P}}}{{\mathrm{I}}_{\mathrm{S}}}=100\Rightarrow \mathrm{n}=100$$

$${\mathrm{C}}_{\mathrm{AB}}=\frac{3\times 6}{3+6}=2\mu \mathrm{F}$$

$$\vec{F}=I(\vec{\ell }\times \vec{B})$$

$$\begin{array}{rl}& =I[(L\hat{i})\times (2\hat{i}+3\hat{j}-4\hat{k})]\\ & \\ & =I(4L\hat{j}+3L\hat{k})\end{array}$$

$$|\vec{F}|=5\text{ IL }$$

$$\mathrm{B}=\frac{{\mu }_0}{4\pi }\frac{\mathrm{I}}{\mathrm{R}}(\pi )-\frac{{\mu }_0}{4\pi }\frac{2I}{\mathrm{R}}$$

$=\frac{{\mu }_0\mathrm{I}}{4\mathrm{R}}\left[1-\frac{2}{\pi }\right]$ outward i.e. away from page.

Magnetic field exist in

Closed Loops (Monopoles do not exist)

$$\varphi \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{A}}=0$$

(Gauss law for magnetism)

The goal is to find the frequency at which resonance occurs in a series LCR circuit. To achieve resonance, the inductive reactance ($X_L$) should equal the capacitive reactance ($X_C$).

Given are:

• Inductance, $L=10\mathrm{mH}=10\times 10^{-3}\mathrm{H}$
• Capacitance, $C=1\mu \mathrm{F}=1\times 10^{-6}\mathrm{F}$
• Resistance, $R=100\Omega$ (which does not directly impact the resonance frequency)

Resonance occurs when the angular frequency ($\omega$) meets the condition $\omega L=\frac{1}{\omega C}$, leading to the formula for calculating the resonance frequency ($f$) as $f=\frac{\omega }{2\pi }=\frac{1}{2\pi \sqrt{LC}}$.

Substituting the given values:

$$f=\frac{1}{2\pi \sqrt{10\times 10^{-3}\cdot 1\times 10^{-6}}}=1.59\mathrm{kHz}$$

Thus, the frequency at which resonance occurs in this LCR circuit is 1.59 kHz.

To find the current in the primary winding of an ideal step-down transformer, we first understand that an ideal transformer's power input (primary side) equals its power output (secondary side). This is because an ideal transformer is assumed to have 100% efficiency, meaning there are no losses in the transformation process.

The power of the lamp (which is connected to the secondary side of the transformer) is given as $60\mathrm{W}$, and it operates at $12\mathrm{V}$. The voltage on the primary side of the transformer is $220\mathrm{V}$.

Using the power formula $P=IV$, where $P$ is the power in watts, $I$ is the current in amperes, and $V$ is the voltage in volts, we equate the power on both sides of the transformer because of its ideal nature, i.e., $P_{\text{primary}}=P_{\text{secondary}}$.

Hence, $60\mathrm{W}=220\mathrm{V}\times I_{\text{primary}}$.

Solving for $I_{\text{primary}}$ gives:

$I_{\text{primary}}=\frac{60\mathrm{W}}{220\mathrm{V}}=\mathrm{0.2727...}\approx 0.27\mathrm{A}$.

Therefore, the current in the primary winding of the transformer is approximately $0.27\mathrm{A}$.

When an AC source is connected to a capacitor, the current (I) depends on the capacitive reactance (X_C), where $X_C=\frac{1}{\omega C}$, with $\omega$ representing the angular frequency $\omega =2\pi f$, and C denoting the capacitance.

The formula for capacitive reactance shows that it is inversely proportional to the frequency (f) of the operation.

Thus, if the frequency decreases, the capacitive reactance increases because $X_C=\frac{1}{\omega C}$ and $\omega$ is proportional to frequency.

As capacitive reactance increases, the displacement current (I), which flows through the capacitor, decreases according to the relationship $I=\frac{V}{X_C}$, with V being the voltage across the capacitor.

So, among the given choices, the correct interpretation is that the displacement current decreases with a decrease in operating frequency.

The formula for calculating the magnetic energy stored in an inductor is given by:

$$E=\frac{1}{2}L{I}^2$$

where:

• $E$ is the energy stored (in joules, J),
• $L$ is the inductance of the inductor (in henrys, H),
• $I$ is the current flowing through the inductor (in amperes, A).

Plugging the given values into the formula:

$$E=\frac{1}{2}\times 4\mu H\times (2A{)}^2$$

$$E=\frac{1}{2}\times 4\times 10^{-6}H\times 4A^2$$

$$E=2\times 10^{-6}H\cdot 4$$

$$E=8\times 10^{-6}J$$

Hence, the energy stored in the inductor is $8\mu J$.

To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit. Since R is given as 10 $\Omega$, we calculate the reactances as follows:

Inductive reactance, $X_L=2\pi fL$, where $f$ is the frequency and $L$ is the inductance. With $f=50$ Hz and $L$ given as $\frac{50}{\pi }$ mH, or $\frac{50\times 10^{-3}}{\pi }$ H, we find $X_L=2\pi \times 50\times \frac{50\times 10^{-3}}{\pi }=5$ $\Omega$.

Capacitive reactance, $X_C=\frac{1}{2\pi fC}$, where $C$ is the capacitance. Given $C=\frac{10^3}{\pi }$ μF, or $\frac{10^{-3}}{\pi }$ F, results in $X_C=\frac{1}{2\pi \times 50\times \frac{10^{-3}}{\pi }}=10\Omega$.

With these values, the net impedance $Z$ of the circuit can be calculated using the formula:

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

Substituting $R=10\Omega$, $X_L=5\Omega$, and $X_C=10\Omega$, we find:

$Z=\sqrt{10^2+(5-10{)}^2}=\sqrt{100+25}=\sqrt{125}=5\sqrt{5}\Omega$

Therefore, the correct option is $5\sqrt{5}\Omega$.

$$C=\frac{E_0}{B_0}$$

$${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{C}}$$

$$=\frac{48}{3\times 10^8}$$

$$=1.6\times 10^{-7}\mathrm{T}$$

The critical angle, denoted as ${\theta }_c$, is the angle of incidence beyond which light is totally internally reflected within a denser medium when it hits the boundary with a less dense medium. To find the critical angle for the medium in question, first, we need to understand the relationship between the speed of light in different media and their refractive indices.

Let's denote the speed of light in air as $V_1$ and in the denser medium as $V_2$. From the given information:

$V_1=\frac{x}{t_1}$ (speed of light in air)

$V_2=\frac{10x}{t_2}$ (speed of light in the denser medium)

The refractive index of a medium (n) is inversely proportional to the speed of light in that medium ($n=\frac{c}{V}$, where c is the speed of light in vacuum). Thus, the refractive index of the denser medium ($n_2$) relative to air ($n_1$) can be obtained by taking the ratio of the speed of light in air to that in the denser medium:

$$\frac{n_2}{n_1}=\frac{V_1}{V_2}$$

Substituting the expressions for $V_1$ and $V_2$:

$$\frac{n_2}{n_1}=\frac{\frac{x}{t_1}}{\frac{10x}{t_2}}=\frac{t_2}{10t_1}$$

For the critical angle ${\theta }_c$, the light in the denser medium (index $n_2$) strikes the boundary with the less dense medium (index $n_1$) such that it refracts at 90 degrees (escapes along the boundary). Snell's Law at this boundary is:

$$n_2\sin {\theta }_c=n_1$$

Given that $n_1=1$ (approximating the refractive index of air), we simplify to:

$$\sin {\theta }_c=\frac{1}{n_2}$$

Substituting $n_2$ from the earlier relation and simplifying:

$$\sin {\theta }_c=\frac{10t_1}{t_2}$$

Hence, the critical angle ${\theta }_c$ is given by:

$${\theta }_c={\sin }^{-1}\left(\frac{10t_1}{t_2}\right)$$

Use $\frac{1}{\mathrm{f}}=[\mu -1]\left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$

$$\frac{1}{{\mathrm{f}}_1}=[1.6-1]\left[\frac{1}{\infty }-\frac{1}{20}\right]=\frac{-3}{100}$$

$$\frac{1}{{\mathrm{f}}_2}=[1.5-1]\left[\frac{1}{20}+\frac{1}{20}\right]=\frac{1}{20}$$

$$\frac{1}{{\mathrm{f}}_3}=\frac{-3}{100}$$

$$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$$

$$\frac{1}{f_{eq}}=-\frac{3}{100}+\frac{1}{20}-\frac{3}{100}=\frac{-1}{100}$$

$$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$$

$$\frac{1}{f_{eq}}=\frac{1}{f}-\frac{1}{f}$$

$${\mathrm{f}}_{\mathrm{eq}}=\infty$$

In Young's double slit experiment, the angular separation of the fringes (θ) is given by :

$$\theta =\frac{m\lambda }{d}$$

where m is the fringe order, λ is the wavelength of the light, and d is the distance between the slits.

Now, let's analyze both statements:

Statement I: If the screen is moved away from the plane of the slits, the angular separation of the fringes remains constant.

This statement is true because the angular separation of the fringes (θ) does not depend on the distance between the screen and the slits. It only depends on the fringe order (m), the wavelength of the light (λ), and the distance between the slits (d).

Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.

This statement is false because if the wavelength (λ) increases, the angular separation of the fringes (θ) also increases according to the formula :

$$\theta =\frac{m\lambda }{d}$$

So, the correct answer is :

Option B: Statement I is true but Statement II is false.

Half life $T=20\min$

Left fraction of activity $\frac{1}{16}$

$$\begin{array}{rl}\because \frac{\mathrm{R}}{{\mathrm{R}}_0}& ={\left(\frac{1}{2}\right)}^{\mathrm{t}/\mathrm{T}}\\ \frac{1}{16}& ={\left(\frac{1}{2}\right)}^{\mathrm{t}/20}\\ {\left(\frac{1}{2}\right)}^4& ={\left(\frac{1}{2}\right)}^{\mathrm{t}/20}\\ 4& =\frac{\mathrm{t}}{20}\\ \mathrm{t}& =80\min \end{array}$$

Shortest wavelength in Balmer series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=2$

$$\because \frac{1}{\lambda }={\mathrm{Rz}}^2\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]$$

$\frac{1}{\lambda }=\frac{R}{4}$ .... (1)

Shortest wavelength is Bracket series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=4$

$\frac{1}{{\lambda }^{\prime }}=\mathrm{R}(1{)}^2\left[\frac{1}{4^2}-\frac{1}{{\infty }^2}\right]\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{\mathrm{R}}{16}$ ..... (2)

Eq. (1) / Eq. (2)

$$\frac{{\lambda }^{\prime }}{\lambda }=\frac{\mathrm{R}}{4}\times \frac{16}{\mathrm{R}}\Rightarrow {\lambda }^{\prime }=4\lambda$$

The formula to calculate the radius of the nth orbit in a hydrogen atom is given by the equation:

$$r_n=r_0\times n^2$$

where,
$r_n$ is the radius of the nth orbit,
$r_0$ is the radius of the innermost orbit (known as the Bohr radius), and
$n$ is the orbit number.

Given that the radius of the innermost orbit ($r_0$) is $5.3\times 10^{-11}\mathrm{m}$ or equivalently $0.53\stackrel{o}{A}$, and we're interested in finding the radius of the third orbit ($n=3$).

So,

$r_3=0.53\times (3{)}^2$ $\stackrel{o}{A}$

$r_3=0.53\times 9$ $\stackrel{o}{A}$

$r_3=4.77$ $\stackrel{o}{A}$

Therefore, the radius of the third allowed orbit of the hydrogen atom is 4.77 $\stackrel{o}{A}$.

Given energy of photon $E=2.20\mathrm{eV}$

Work function of $\mathrm{Cs}{\varphi }_0=2.14\mathrm{eV},\mathrm{K}{\varphi }_0=2.30\mathrm{eV},\mathrm{Na}{\varphi }_0=2.75\mathrm{eV}$

We know that $e^{-}$ emitts when $hv>{\varphi }_0$

here it is clear that energy of photon is more than the work function of $\mathrm{Cs}$ [Caesium] only so Ans. only (Cs).

The minimum wavelength, ${\lambda }_{\min }$, of X-rays produced by an electron can be determined using the equation derived from the energy of a photon and the energy given to an electron by an accelerating voltage, $V$.

The energy of a photon is given by $E=hc/\lambda$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the photon.

When an electron is accelerated through a potential difference of $V$ volts, it gains kinetic energy equal to $eV$, where $e$ is the electron charge.

This energy is then converted into a photon's energy when the electron collides with a target in an X-ray tube, resulting in X-rays of wavelength $\lambda$. Setting the kinetic energy equal to the photon energy gives $eV=hc/\lambda$. Solving for $\lambda$ gives $\lambda =hc/(eV)$.

Therefore, the minimum wavelength (corresponding to the maximum energy photon produced when all the kinetic energy is converted into photon energy) is inversely proportional to the voltage, $V$.

Thus, ${\lambda }_{\min }\propto 1/V$.

Statement I : Photocell/solar cell convert light energy into electric energy/current.

Statement II : We use zener diode in reverse biased condition, when reverse biased voltage more than break down voltage than it act as stablizer.

Capacitor used to remove AC ripples from Rectifier output.

$$y=\overline{\overline{\mathrm{A}}\cdot \overline{\mathrm{B}}}=\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}$$

$=(\mathrm{A}+\mathrm{B})\mathrm{OR}$ Gate

Weight of impure limestone $=20\mathrm{g}$

Weight of pure limestone $\left({\mathrm{CaCO}}_3\right)=20\%$ of $20\mathrm{g}$

$$=\frac{20}{100}\times 20$$

$$=4\mathrm{g}$$

$${\mathrm{n}}_{{\mathrm{CaCO}}_3}=\frac{4}{100}=0.04$$

Image

$${\mathrm{n}}_{{\mathrm{CO}}_2}=0.04$$

$${\mathrm{W}}_{{\mathrm{CO}}_2}=0.04\times 44$$

$$=1.76\mathrm{g}$$

It is statement based question.

Statements B, C & E are correct.

(B) Mass of the electron is $9.10939\times 10^{-31}\mathrm{kg}$

(C) All the isotopes of given elements show same chemical properties.

(E) Dalton's atomic theory, regarded the atom as an ultimate particle of matter.