Physics_paper_1_TZ1_HL_

Question

The ratio of the diameter of an atom to the diameter of its nucleus is:
A. $10^1$

B. $10^3$

C. $10^5$

D. $10^7$

▶️Answer/Explanation

Ans:C

The diameter of an atom is much larger than the diameter of its nucleus. The nucleus of an atom is extremely small compared to the size of the entire atom.

The approximate diameter of an atom is on the order of 0.1 nanometers (1 Ångström), while the diameter of the nucleus is on the order of 1 femtometer (1 fm). Therefore, the ratio of the diameter of an atom to the diameter of its nucleus is:

The ratio of the diameter of an atom to the diameter of its nucleus is:

$\frac{\text{Diameter of atom}}{\text{Diameter of nucleus}} = \frac{0.1\, \text{nm}}{1\, \text{fm}} = 10^5$

Question

The kinetic energy of a body is determined from measurements of its momentum p and its mass m.
The percentage uncertainties in the measurements are:

What is the percentage uncertainty in the kinetic energy?

A. 7%

B. 10%

C. 13%

D. 14%

▶️Answer/Explanation

Ans:B

$\begin{aligned} p^2 & =2 m k \\ k & =\frac{p^2}{2 m} \\ \pm \frac{\Delta k}{k} & = \pm\left(2 \times \frac{\Delta p}{p}+1 \times \frac{\Delta m}{m}\right)\end{aligned}$

$\left(\frac{\Delta k}{k} \times 100\right)=2 \times 3+4=10 \%$

Question

A car travels clockwise around a circular track of radius R. What is the magnitude of displacement from X to Y?

A. $ R \frac{3 \pi}{2}$

B. $R \frac{\pi}{2}$

C. $R \sqrt{2}$

D. $R$

▶️Answer/Explanation

Ans:C

The displacement from point X to point Y is a chord of the circle and can be represented by the line connecting points X and Y.

Question

A stone of mass m is projected vertically upwards with speed u from the top of a cliff. The speed of the stone when it is just about to hit the ground is v.

What is the magnitude of the change in momentum of the stone?
A. $m\left(\frac{v+u}{2}\right)$

B. $m\left(\frac{v-u}{2}\right)$

C. $m(v+u)$

D. $m(v-u)$

▶️Answer/Explanation

Ans:C

The initial momentum of the stone, when it is projected vertically upwards with speed $u$, is $+mu$ (upwards is considered positive).

The final momentum of the stone, just before it hits the ground, is $-mv$ (downwards is considered negative).

Now, the change in momentum is:

Change in momentum = Final momentum – Initial momentum
Change in momentum = $-mv – mu$

Change in momentum = $m(-v – u)$

Change in momentum = $m(-(v + u))$

So, the magnitude of the change in momentum is:

$|m(v + u)|$

Question

A car accelerates uniformly. The car passes point $X$ at time $t_1$ with velocity $v_1$ and point $Y$ at time $t_2$ with velocity $v_2$. The distance $X Y$ is $s$.

The following expressions are proposed for the magnitude of its acceleration a:

I. $a=\frac{2 s}{\left(t_2-t_1\right)^2}$

II. $a=\frac{v_2^2-v_1^2}{2 s}$

III. $a=\frac{v_2-v_1}{t_2-t_1}$

Which is correct?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

▶️Answer/Explanation

Ans:C

To determine which expressions for the magnitude of acceleration ($a$) are correct, we can analyze each of the proposed expressions one by one:

I. $a = \frac{2s}{(t_2 – t_1)^2}$

Expression I is incorrect because it assumes that the initial velocity $\left(v_1\right)$ is zero, which may not be the case for the uniformly accelerating car.

II. $a = \frac{v_2^2 – v_1^2}{2s}$

This expression is also correct. It’s derived from the kinematic equation for uniformly accelerated motion: $v^2 = u^2 + 2as$, where $v_2$ is the final velocity, $v_1$ is the initial velocity, $s$ is the displacement, and $a$ is the acceleration. Solving for $a$ results in this expression.

III. $a = \frac{v_2 – v_1}{t_2 – t_1}$

This expression is also correct and is derived from the definition of acceleration as the rate of change of velocity over time.

So, expressions (II, and III) are correct.

Question

A ball attached to a string is made to rotate with constant speed along a horizontal circle. The string is attached to the ceiling and makes an angle of $\theta^{\circ}$ with the vertical. The tension in the string is $T$.

What is correct about the horizontal component and vertical component of the net force on the ball?

▶️Answer/Explanation

Ans:D

From FBD we can say option D is correct net force in vertical must be zero as it is rotating in along a horizontal circle.

Question

A block of mass 2.0kg is placed on a trolley of mass 5.0kg, moving horizontally. A force of 8.0N is applied to the block which slides on the surface of the trolley. The frictional force between the trolley and the ground is zero.

The trolley accelerates at a rate of $1.0 \mathrm{~m} \mathrm{~s}^{-2}$. What is the coefficient of dynamic friction between the block and the trolley?

A. $ 0.05$

B. $ 0.15$

C. $ 0.25$

D. $ 0.35$

▶️Answer/Explanation

Ans:C

friction $= 5\times 1=5~\rm N$

$\mu _k (2 \times 10) =5$

$\mu=\frac{1}{4}$

Question

The input power of an electric motor is $200 \mathrm{~W}$. It is used to raise a mass of $10 \mathrm{~kg}$ at constant speed. If the efficiency of the motor is $40 \%$, through what height will the mass be raised in 1 second?

A. $0.5 \mathrm{~m}$

B. $0.8 \mathrm{~m}$

C. $1.2 \mathrm{~m}$

D. $2.0 \mathrm{~m}$

▶️Answer/Explanation

Ans:B

$\text{Useful power} = \text{Efficiency} \times \text{Input power}$

$\text{Useful power} = 0.40 \times 200 \, \text{W} = 80 \, \text{W}$

$\text{Work (W)} = \text{Power (P)} \times \text{Time (t)}$

In this case, Power (P) is 80 W, and Time (t) is 1 second.

$\text{Work (W)} = 80 \, \text{W} \times 1 \, \text{s} = 80 \, \text{J} \, (\text{joules})$

The work done is equal to the change in potential energy of the mass. Use the formula for gravitational potential energy:

$\text{Gravitational Potential Energy (U)} = mgh$

Rearrange the formula to solve for \(h):

$h = \frac{W}{mg}$

$h = \frac{80 \, \text{J}}{10 \, \text{kg} \times 9.81 \, \text{m/s}^2} \approx 0.815 \, \text{m}$

Rounded to one decimal place, the height to which the mass is raised in 1 second is approximately $0.8 \, \text{m}$.

Question

The temperature of an object is changed from $\theta_1{ }^{\circ} \mathrm{C}$ to $\theta_2{ }^{\circ} \mathrm{C}$. What is the change in temperature measured in kelvin?

A. $\left(\theta_2-\theta_1\right)$

B. $\left(\theta_2-\theta_1\right)+273$

C. $\left(\theta_2-\theta_1\right)-273$

D. $273-\left(\theta_2-\theta_1\right)$

▶️Answer/Explanation

Ans:A

The change in temperature measured in kelvin is equal to the change in temperature in degrees Celsius because the Kelvin scale is based on the Celsius scale, but it starts at absolute zero (0 K = -273.15°C). Therefore, the correct answer is: $\left(\theta_2 – \theta_1\right)$

Question

A metal cube $X$ of length $L$ is heated gaining thermal energy $Q$. Its temperature rises by $\Delta T$. $A$ second cube $Y$, of length $2 L$, made of the same material, gains thermal energy of $2 Q$.

What is the temperature rise of $\mathrm{Y}$ ?

A. $\frac{\Delta T}{8}$

B. $\frac{\Delta T}{4}$

C. $\Delta T$

D. $2 \Delta T$

▶️Answer/Explanation

Ans:B

The change in thermal energy ($Q$) is related to the mass of the material, the specific heat capacity ($c$), and the change in temperature ($\Delta T$) by the equation:

\[Q = mc\Delta T\]

For both cubes $X$ and $Y$, made of the same material, the specific heat capacity is the same ($c$).

Now, let’s consider cube $X$. Its length is $L$, so its volume ($V_X$) is $L^3$. Cube $Y$ has twice the length ($2L$, so its volume ($V_Y$) is $(2L)^3 = 8L^3$.

Since the mass ($m$) is directly proportional to volume and density, and both cubes are made of the same material, we can say that:

\[m_X = k \cdot V_X\]
\[m_Y = k \cdot V_Y\]

Where $k$ is a constant.

So, the ratio of the mass of cube $Y$ to the mass of cube $X$ is:

\[\frac{m_Y}{m_X} = \frac{k \cdot V_Y}{k \cdot V_X} = \frac{8L^3}{L^3} = 8\]

Now, let’s consider the thermal energy gained:

For cube $X$: $Q_X = m_Xc\Delta T$

For cube $Y$: $Q_Y = m_Yc\Delta T_Y$

Given that $Q_Y = 2Q_X$ and $\frac{m_Y}{m_X} = 8$, we can write:

\[2Q_X = 8(m_Xc\Delta T)\]

Now, we can simplify:

\[2Q_X = 8Q_X\]

Divide both sides by 2:

\[Q_X = 4Q_X\]

This implies that $\Delta T_Y$ for cube $Y$ must be four times the change in temperature $\Delta T$ for cube $X$.

So,

\[\Delta T_Y = 4\Delta T\]

The correct answer is B. $\frac{\Delta T}{4}$.

Question

Which graph represents the variation with displacement of the potential energy $P$ and the total energy $T$ of a system undergoing simple harmonic motion (SHM)?

▶️Answer/Explanation

Ans:B

Question

A wave is polarized. What must be correct about the wave?
It is a…

A. transverse wave.

B. longitudinal wave.

C. standing wave.

D. travelling wave.

▶️Answer/Explanation

Ans:A

A polarized wave is a:

A. transverse wave.

Polarization refers to the orientation of the oscillations of the wave. In a transverse wave, the oscillations are perpendicular to the direction of wave propagation, and this is the type of wave that can be polarized. Longitudinal waves, on the other hand, have oscillations parallel to the direction of wave propagation and cannot be polarized in the same way. Standing and traveling waves are characteristics related to the motion and behavior of waves and are not directly related to polarization.

Question

A group of students perform an experiment to find the refractive index of a glass block. They measure various values of the angle of incidence $i$ and angle of refraction $r$ for a ray entering the glass from air. They plot a graph of the $\sin r$ against $\sin i$.

They determine the gradient of the graph to be $m$.
Which of the following gives the critical angle of the glass?

A. $\sin ^{-1}(m)$

B. $\sin ^{-1}\left(\frac{1}{m}\right)$

C. $m$

D. $\frac{1}{m}$

▶️Answer/Explanation

Ans:A

$\begin{aligned} \quad & n_1 \text { sin i }=n_2 \sin r \\ & \text { Snalls Law. } \\ m(slope) = & \frac{\sin r}{\sin i}=\frac{n_1}{n_2}=\frac{1}{n}\end{aligned}$

$n_1=1$ and $n_2 =n$

Now , TIR comes to action from which we have to calculate the critical angle (and happens only when ray goes from denser to rarer medium)

$\begin{aligned}
& n \cdot \sin \theta_c=1 \cdot \sin 90 \\
& n \cdot \sin \theta_c=1 \Rightarrow \theta_c=\sin ^{-1}\left(\frac{1}{n}\right)\\
&\theta_c=sin ^{-1} m
\end{aligned}$

Question

A standing wave is formed in a pipe open at one end and closed at the other. The length of the pipe is L and the speed of sound in the pipe is V.

$n$ is a positive integer.
What expression is correct about the frequencies of the harmonics in the pipe?

A. $\frac{(2 n-1) V}{2 L}$

B. $\frac{(2 n-1) V}{4 L}$

C. $\frac{n V}{2 L}$

D. $\frac{n V}{4 L}$

▶️Answer/Explanation

Ans:B

The general frequency for closed at one end and open to other is ,

$f_n=\frac{n \cdot w}{4 L} \quad n$-odd Number and even Harmonics are not present in this.

But in question $n$ is given positive number which it has even as well as odd to , so we have to exclude the even one , and for that we will replace n to $2n+1/2n-1$

$f_n=\frac{2n+1 \cdot w}{4 L}$ or $f_n=\frac{2n-1 \cdot w}{4 L}$

Correct match option – B

Question

Two positive charges of magnitude $q$ and $2 q$ are fixed as shown. At which position is the electric field, due to these charges, equal to zero?

▶️Answer/Explanation

Ans:B

At position A and D electric can never be zero.

At B distance from 2q is much more then c and magnitude is also 2 times so more possibility of cancel out each other because At C distance from q is lower and magnitude is also smaller in compare to 2q so there is more chance that $E_2q$ will ore domniate over $E_q$

Question

$\mathrm{P}$ and $\mathrm{Q}$ are two conductors of the same material connected in series. $\mathrm{Q}$ has a diameter twice that of $P$.
What is $\frac{\text { drift speed of electrons in } P}{\text { drift speed of electrons in } Q}$ ?

A. 4

B. 2

C. $\frac{1}{2}$

D. $\frac{1}{4}$

▶️Answer/Explanation

Ans:A

The drift speed of electrons in a conductor depends on various factors, including the current passing through the conductor and its properties. In this scenario, since conductors P and Q are connected in series, they carry the same current. The drift speed is primarily determined by the current, cross-sectional area, charge of the electron, and the number density of electrons in the material.

Let’s consider conductor P with diameter $d$ and conductor Q with diameter $2d$. The cross-sectional area ($A$) of the conductor is directly proportional to the square of its diameter. So, for conductor Q, the cross-sectional area is four times that of conductor P.

Given that the current is the same through both conductors and the drift speed is inversely proportional to the cross-sectional area, we can compare the drift speeds of electrons in P and Q as follows:

\[
\frac{\text{Drift speed in P}}{\text{Drift speed in Q}} = \frac{A_Q}{A_P} = \frac{4A_P}{A_P} = 4
\]

So, the drift speed of electrons in conductor P is four times that of conductor Q.

Question

Three lamps (X, Y and Z) are connected as shown in the circuit. The emf of the cell is 20V. The internal resistance of the cell is negligible. The power dissipated by X, Y and Z is 10W, 20W and 20W respectively.

What is the voltage across Lamp X and Lamp Y?

▶️Answer/Explanation

Ans:B

Question

An electron enters a region of uniform magnetic field at a speed v. The direction of the electron is perpendicular to the magnetic field. The path of the electron inside the magnetic field is circular with radius r.

The speed of the electron is varied to obtain different values of r.

Which graph represents the variation of speed v with r?

▶️Answer/Explanation

Ans:A

When an electron enters a region of a uniform magnetic field with a velocity perpendicular to the magnetic field lines, it experiences a magnetic force that acts as a centripetal force, causing the electron to move in a circular path. This is due to the interaction of the magnetic field with the moving charge of the electron.

The magnetic force ($F_{\text{mag}}$) on the electron is given by:

\[F_{\text{mag}} = qvB\]

This magnetic force provides the centripetal force required to keep the electron in a circular path. The centripetal force ($F_{\text{cent}}$) for an object moving in a circle of radius $r$ with speed $v$ is given by:

\[F_{\text{cent}} = \frac{mv^2}{r}\]

Since the magnetic force is acting as the centripetal force, we can equate the two expressions:

\[qvB = \frac{mv^2}{r}\]

Now, we can solve for the radius ($r$) of the circular path:

\[r = \frac{mv}{qB}\]

So, as the speed of the electron ($v$) is varied, the radius of the circular path ($r$) will change accordingly. The greater the speed, the larger the radius of the circular path, and vice versa. This relationship shows that the radius of the circular path is directly proportional to the speed of the electron, provided that the magnetic field strength and the charge of the electron remain constant.

Question

P and R are parallel wires carrying the same current into the plane of the paper. P and R are equidistant from a point Q. The line PQ is perpendicular to the line RQ.

The magnetic field due to $P$ at $Q$ is $X$. What is the magnitude of the resultant magnetic field at $Q$ due to both wires?

A. $\frac{x}{2}$

B. $x$

C. $x \sqrt{2}$

D. $2 x$

▶️Answer/Explanation

Ans:C

$B_{eq}=\sqrt{x^2+x^2}\Rightarrow x\sqrt{x}$

Question

A stuntman rides a motorcycle on the inside surface of a cylinder.

Which is the correct free-body diagram showing all the forces acting on the cyclist at that position?

▶️Answer/Explanation

Ans:C

Question

The energy levels E of an atom are shown.

Which emission spectrum represents the transitions?

▶️Answer/Explanation

Ans:D

Question

Three claims are made about the structure of the atom.

I. Most of the atom is empty space.
II. The positive charge of the atom is concentrated in a small volume.
III. The electrons have discrete energy levels.

Which of these claims can be deduced from the Rutherford-Geiger-Marsden scattering experiment?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

▶️Answer/Explanation

Ans:A

The Rutherford-Geiger-Marsden scattering experiment, commonly known as the Rutherford gold foil experiment, provided important insights into the structure of the atom. Let’s examine each of the claims in light of this experiment:

I. Most of the atom is empty space:

This claim can be deduced from the Rutherford-Geiger-Marsden experiment. Rutherford’s experiment showed that alpha particles, which are positively charged, mostly passed through the gold foil with little to no deflection, indicating that the atom is mostly empty space. However, a small fraction of the alpha particles did experience significant deflection, suggesting that there is a concentrated positive charge in a small volume within the atom.

II. The positive charge of the atom is concentrated in a small volume:

This claim can also be deduced from the Rutherford-Geiger-Marsden experiment. The significant deflection of some alpha particles indicated that the positive charge in the atom is concentrated in a small, dense nucleus at the center of the atom.

III. The electrons have discrete energy levels:

The Rutherford-Geiger-Marsden experiment did not directly provide information about the discrete energy levels of electrons. This aspect of atomic structure is better explained by later developments in atomic theory, such as Bohr’s model of the atom and quantum mechanics.

Based on the information provided by the Rutherford-Geiger-Marsden experiment, claims I and II can be deduced, but claim III cannot. Therefore, the correct answer is:A. I and II only

Question

This interaction between a proton and a pion violates two or more conservation laws.
$
\mathrm{p}+\pi^{-} \rightarrow \mathrm{K}^{-}+\pi^{+}
$
Quark composition of particles:
$
\pi^{-}=\mathrm{du}, \pi^{+}=\mathrm{u} \overline{\mathrm{d}}, \mathrm{K}^{-}=\mathrm{su}, \mathrm{p}=\mathrm{uud}
$

Which laws are violated by this interaction?

I. Conservation of charge

II. Conservation of strangeness

III. Conservation of baryon number

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

▶️Answer/Explanation

Ans:C

In this interaction, the initial and final states involve different particles, and we need to consider the conservation of various quantum numbers. Let’s evaluate the conservation laws:

I. Conservation of charge:

Initial state: p (proton) has a charge of +1, and π⁻ (pion) has a charge of -1. So, the total charge of the initial state is 1 – 1 = 0.
Final state: K⁻ (kaon) has a charge of -1, and π⁺ (pion) has a charge of +1. So, the total charge of the final state is -1 + 1 = 0.
The conservation of charge is not violated in this interaction.

II. Conservation of strangeness:

Strangeness is not conserved in this interaction. The initial state has a strangeness of 0 (since p and π⁻ have no strangeness), while the final state has a strangeness of -1 (K⁻ has strangeness -1, and π⁺ has strangeness 0).

III. Conservation of baryon number:

Baryon number is not conserved in this interaction. The initial state has a baryon number of 1 (since p is a baryon), while the final state has a baryon number of 0 (since there are no baryons in the final state).

So, the correct answer is: C. II and III only

Conservation of charge (I) is not violated, but both the conservation of strangeness (II) and the conservation of baryon number (III) are violated in this interaction.

Question

A student claims that the following three factors may affect the rate of global warming.

I. Increased volcanic activity

II. Increased solubility of carbon dioxide $\left(\mathrm{CO}_2\right)$ in the ocean

III. Increased rate of deforestation

Which factors can increase the rate of global warming?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

▶️Answer/Explanation

Ans:B

The factors that can increase the rate of global warming are:

I. Increased volcanic activity: Volcanic eruptions can release large amounts of greenhouse gases, such as carbon dioxide (CO2) and sulfur dioxide (SO2), into the atmosphere. These gases can contribute to global warming.

III. Increased rate of deforestation: Deforestation leads to a reduction in the number of trees and vegetation that can absorb and store carbon dioxide. When forests are cleared, the stored carbon is released into the atmosphere, increasing the concentration of greenhouse gases, which contributes to global warming.

II. Increased solubility of carbon dioxide (CO2) in the ocean: This factor does not directly increase the rate of global warming. In fact, the increased solubility of CO2 in the ocean can have a cooling effect on the planet by absorbing excess CO2 from the atmosphere. While it may mitigate global warming to some extent, it is not a factor that directly increases the rate of global warming.

So, the correct answer is not B (I and III only) but:

D. I, II, and III

All three factors can have an impact on the rate of global warming, with factors I and III contributing to warming and factor II having a potential mitigating effect.

Question

Two surfaces $X$ and $Y$ emit radiation of the same surface intensity. $X$ emits a radiation of peak wavelength twice that of $Y$.
What is $\frac{\text { emissivity of } X}{\text { emissivity of } Y}$ ?

A. $\frac{1}{16}$

B. $\frac{1}{2}$

C. 2

D. 16

▶️Answer/Explanation

Ans:D

The emissivity of a surface is a measure of how efficiently it emits radiation compared to an ideal blackbody at the same temperature. The emissivity is often denoted by the symbol ε.

According to Wien’s displacement law, the peak wavelength (λ_max) of the radiation emitted by a blackbody is inversely proportional to its temperature (T):

\[λ_{\text{max}} \propto \frac{1}{T}.\]

Now, if surface X emits radiation with a peak wavelength that is twice that of surface Y, we can say:

\[\frac{λ_{\text{max},X}}{λ_{\text{max},Y}} = 2.\]

Now, let’s consider the emissivity of X (ε_X) and Y (ε_Y). The emissivity of a surface is dependent on the surface’s ability to emit radiation efficiently. Given that both surfaces emit radiation of the same surface intensity, we can write:

\[I_X = I_Y.\]

The intensity of radiation emitted by a blackbody is given by the Stefan-Boltzmann law:

\[I = εσT^4,\]

where ε is the emissivity, σ is the Stefan-Boltzmann constant, and T is the absolute temperature.

Since the intensities are the same for X and Y:

\[ε_XσT_X^4 = ε_YσT_Y^4.\]

Now, we can compare the temperatures of X and Y. Using Wien’s displacement law:

\[\frac{λ_{\text{max},X}}{λ_{\text{max},Y}} = \frac{T_Y}{T_X}.\]

We already know that $λ_{\text{max},X}/λ_{\text{max},Y = 2$, so we can rewrite this as:

\[2 = \frac{T_Y}{T_X}.\]

Now, substitute this into the equation for the emissivity:

\[ε_XσT_X^4 = ε_YσT_Y^4.\]

\[ε_XσT_X^4 = ε_Yσ(2T_X)^4.\]

Now, simplify:

\[ε_XσT_X^4 = ε_Yσ16T_X^4.\]

The σ and T_X^4 terms cancel on both sides:

\[ε_X = 16ε_Y.\]

To find the ratio ε_X/ε_Y:

\[\frac{ε_X}{ε_Y} = \frac{16ε_Y}{ε_Y} = 16.\]

Question

A simple pendulum oscillates with frequency $f$. The length of the pendulum is halved. What is the new frequency of the pendulum?

A. $2 f$

B. $\sqrt{2} f$

C. $\frac{f}{\sqrt{2}}$

D. $\frac{f}{2}$

▶️Answer/Explanation

Ans:B

The frequency $f$ of a simple pendulum is given by the formula:

\[f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}\]

If you halve the length of the pendulum ($L \rightarrow \frac{L}{2}$), the new frequency ($f’$) can be calculated as:

\[f’ = \frac{1}{2\pi}\sqrt{\frac{g}{\frac{L}{2}}}\]

Simplify the expression:

\[f’ = \frac{1}{2\pi}\sqrt{\frac{2g}{L}}\]

Now, let’s compare $f$ and $f’$:

\[\frac{f’}{f} = \frac{\frac{1}{2\pi}\sqrt{\frac{2g}{L}}}{\frac{1}{2\pi}\sqrt{\frac{g}{L}}}\]

Now, simplify further:

\[\frac{f’}{f} = \sqrt{\frac{2g}{L} \cdot \frac{L}{g}}\]

\[\frac{f’}{f} = \sqrt{2}\]

So, the new frequency $f’$ is $\sqrt{2}$ times the original frequency $f$. Therefore, the correct answer is:B.

Question

The intensity pattern of monochromatic light of wavelength $\lambda$, is projected onto a screen.

What combination produces this pattern?

▶️Answer/Explanation

Ans:D

Question

What is the pattern observed when white light passes through a diffraction grating?

▶️Answer/Explanation

Ans:C

When white light passes through a diffraction grating, a pattern known as a “spectral or diffraction pattern” is observed. This pattern is a result of the dispersion of light into its component colors due to the phenomenon of diffraction and interference.

Question

Source S produces sound waves of speed $v$ and frequency $f$. S moves with constant velocity $\frac{v}{5}$ away from a stationary observer.

What is the frequency measured by the observer?

A. $\frac{4}{5} f$

B. $\frac{5}{6} f$

C. $\frac{6}{5} f$

D. $\frac{5}{4} f$

▶️Answer/Explanation

Ans:B

The frequency observed by a stationary observer when a source is in motion can be calculated using the Doppler effect formula for sound:

\[f’ = \frac{f \cdot (v + v_o)}{v + v_s},\]

$f$ is the frequency emitted by the source.
$v$ is the speed of sound.
$v_o$ is the velocity of the observer, which is stationary (0).
$v_s$ is the velocity of the source, which is moving away at $\frac{v}{5}$.

Now, substitute these values into the formula:

\[f’ = \frac{f \cdot (v + 0)}{v + \frac{v}{5}}.\]

Simplify the equation:

\[f’ = \frac{f \cdot v}{\frac{6}{5}v} = \frac{5}{6}f.\]).

Question

Two isolated point masses, P of mass m and Q of mass 2m, are separated by a distance 3d. X is a point a distance d from P and 2d from Q.

What is the net gravitational field strength at X and the net gravitational potential at X?

▶️Answer/Explanation

Ans:D

To find the net gravitational field strength and potential at point X due to the two masses, you can use the principles of superposition for gravitational fields and potentials.

Gravitational Field Strength at X (g_X):

The gravitational field strength at a point due to a single point mass is given by:

\[g = \frac{GM}{r^2},\]

Let’s calculate the field strength at X due to each of the masses:

For mass P (m), the distance from P to X is $d$, so the field strength at X due to P is $g_P = \frac{Gm}{d^2}$.
For mass Q (2m), the distance from Q to X is $2d$, so the field strength at X due to Q is $g_Q = \frac{G(2m)}{(2d)^2} = \frac{Gm}{2d^2}$.

Since these field strengths are in opposite directions (toward P and away from Q), you need to subtract them to find the net field strength at X:

\[g_X = g_P – g_Q = \frac{Gm}{d^2} – \frac{Gm}{2d^2} = \frac{Gm}{2d^2}.\]

Gravitational Potential at X (V_X):

The gravitational potential at a point due to a single point mass is given by:

\[V = -\frac{GM}{r},\]

where the negative sign indicates that the potential decreases with increasing distance from the mass.

Let’s calculate the potential at X due to each of the masses:

For mass P (m), the potential at X due to P is $V_P = -\frac{Gm}{d}$.
For mass Q (2m), the potential at X due to Q is $V_Q = -\frac{G(2m)}{2d} = -\frac{Gm}{d}.$

Again, since these potentials are in opposite directions, you need to add them to find the net potential at X:

\[V_X = V_P + V_Q = -\frac{Gm}{d} – \frac{Gm}{d} = -\frac{2Gm}{d}.\]

So, the net gravitational field strength at point X is $\frac{3Gm}{4d^2}$, and the net gravitational potential at point X is $-\frac{2Gm}{d}$.

Question

A negatively charged particle is stationary halfway between two horizontal charged plates. The plates are separated by a distance d with potential difference V between them.

What is the magnitude of the electric field and direction of the electric field at the position of the particle?

▶️Answer/Explanation

Ans:D

$\begin{gathered}E=-\frac{d V}{d r}=\frac{V}{d} \\ E=U / d\end{gathered}$

As negative charge is stationary it means the weight must be balanced by Electric Force.

Which means Electric force will act upward direction as weight act downward.

From this $ \overrightarrow{F_e}=q \vec{E} $ we can conclude Electric field have opposite direction due to negative charge.

Electric field will direction will be down.

Question

The escape speed from the surface of earth is $v_{\mathrm{esc}}$. The radius of earth is $R$. A satellite of mass $m$ is in orbit at a height $\frac{R}{4}$ above the surface of the Earth. What is the energy required to move the satellite to infinity?

A. $\frac{m v_{\text {esc }}^2}{5}$

B. $\frac{2 m v_{\text {esc }}^2}{5}$

C. $m v_{\text {esc }}^2$

D. $2 m v_{\text {esc }}^2$

▶️Answer/Explanation

Ans:A

To calculate the energy required to move the satellite from its orbit at a height of $R/4$ above the surface of the Earth to infinity, you need to consider both the gravitational potential energy and the kinetic energy.

The initial gravitational potential energy of the satellite at a height $R/4$ above the surface of the Earth is given by:

\[U_{\text{initial}} = -\frac{GMm}{R + \frac{R}{4}}\]

The negative sign indicates that this is the gravitational potential energy, and it is negative because it is measured relative to infinity.

The satellite is in orbit, so its kinetic energy is given by:

\[K = \frac{1}{2}mv^2\]

Where $v$ is the orbital velocity at that height. The orbital velocity at a height $h$ above the Earth’s surface is given by:

\[v = \sqrt{\frac{GM}{R + h}}\]

So, the kinetic energy at the initial position is:

\[K_{\text{initial}} = \frac{1}{2}m\left(\sqrt{\frac{GM}{R + \frac{R}{4}}}\right)^2\]

Now, let’s calculate the final potential energy at infinity. As the satellite moves to infinity, the gravitational potential energy approaches zero, and the kinetic energy also becomes zero. Therefore, the final energy at infinity is zero.

So, the energy required to move the satellite to infinity is the difference between the initial energy and the final energy:

\[E_{\text{required}} = U_{\text{initial}} + K_{\text{initial}} – 0\]

Substitute the expressions for $U_{\text{initial}}$ and $K_{\text{initial}}$:

\[E_{\text{required}} = -\frac{GMm}{R + \frac{R}{4}} + \frac{1}{2}m\left(\sqrt{\frac{GM}{R + \frac{R}{4}}}\right)^2\]

Simplify the expression:

\[E_{\text{required}} = -\frac{GMm}{\frac{5R}{4}} + \frac{1}{2}m\left(\frac{\sqrt{\frac{GM}{R + \frac{R}{4}}}}{1}\right)^2\]

Now, calculate the energy required:

\[E_{\text{required}} = -\frac{4GMm}{5R} + \frac{1}{2}m\left(\frac{\sqrt{\frac{GM}{R + \frac{R}{4}}}}{1}\right)^2\]

Simplify further:

\[E_{\text{required}} = -\frac{4GMm}{5R} + \frac{1}{2}m\left(\sqrt{\frac{GM}{R + \frac{R}{4}}}\right)^2\]

\[E_{\text{required}} = -\frac{4GMm}{5R} + \frac{1}{2}m\frac{GM}{R + \frac{R}{4}}\Rightarrow |m\times \frac{2GM}{5R}|\]

\[V_{esc}=\sqrt{\frac{2 G M}{R}}\]

So from this \[E_{\text{required}} =\frac{m v_{\text {esc }}^2}{5}\]

Question

Which law is equivalent to the law of conservation of energy?

A. Coulomb’s law

B. Ohm’s Law

C. Newton’s first law

D. Lenz’s law

▶️Answer/Explanation

Ans:D

Lenz’s law is indeed based on the principle of the conservation of energy. It states that the direction of an induced electromotive force (emf) and the induced current in a closed circuit will always be such that they oppose the change in magnetic flux that produced them. This is in accordance with the law of conservation of energy because it implies that work is done to overcome the opposition, and this work results in the transformation of energy.

When an external force is applied to move a conductor through a changing magnetic field (or vice versa), work is done, and this work is ultimately converted into electrical energy in the form of induced current. The law of conservation of energy dictates that the total energy in the system remains constant. Therefore, the extra effort to overcome the opposition and induce the current is indeed transformed into electrical energy, thus conserving energy in the process.

Question

Wire XY moves perpendicular to a magnetic field in the direction shown.

The graph shows the variation with time of the displacement of XY.

What is the graph of the electromotive force $(\mathrm{emf}) \varepsilon$ induced across $\mathrm{XY}$ ?

▶️Answer/Explanation

Ans:C

Induced Voltage (emf)
An emf is induced in a conductor moving in a magnetic field. A conducting wire of length $L$ moves perpendicularly to a uniform magnetic field $\boldsymbol{B}$ with constant velocity $\boldsymbol{v}$.

Force on electrons in the wire:
\[
\vec{F}=q \vec{v} \times \vec{B}
\]

Since force $\boldsymbol{F}$ on electrons is upward, $I$ is downward in the wire.
\[
V=B v L \text { (Potential difference) }
\]

An emf is induced and a current flows in the wire as long as it moves in the magnetic field. (principle of the electric generator).

So, from this Emf is directly proportional to v(velocity).

So, from $0$ to $t_1$ velocity will increase , and it become constant up to $t_2$ and will start decreasing .

So, graph will like option -C

Question

Three changes are made to a transformer.

I. increasing the thickness of wire in the coils
II. laminating the soft iron core
III. using wire with lower resistivity

Which changes will reduce power losses in the transformer?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

▶️Answer/Explanation

Ans:D

Reducing power losses in a transformer involves minimizing both copper losses and iron losses. Let’s evaluate each change and its impact on power losses:

I. Increasing the thickness of wire in the coils:

Increasing the thickness of wire reduces the resistance of the coil, which decreases copper losses (I²R losses). So, this change reduces copper losses and is beneficial for reducing power losses.

II. Laminating the soft iron core:

Laminating the core reduces eddy current losses, which are a type of iron loss in transformers. By minimizing eddy currents, this change reduces iron losses, making it beneficial for reducing power losses.

III. Using wire with lower resistivity:

Using wire with lower resistivity further decreases the resistance of the coil, reducing copper losses (I²R losses). This change is also beneficial for reducing power losses.

So, all three changes will reduce power losses in the transformer. Therefore, the correct answer is:D

Question

A resistor of resistance $R$ is connected to an alternating current power supply. The peak voltage across the resistor is $V_0$. What is the mean power dissipated by the resistor?

A. $\frac{V_0^2 \sqrt{2}}{R}$

B. $\frac{V_0^2}{R}$

C. $\frac{V_0^2}{R \sqrt{2}}$

D. $\frac{V_0^2}{2 R}$

▶️Answer/Explanation

Ans:D

The mean power $(P)$ dissipated by a resistor in an alternating current (AC) circuit can be calculated using the root mean square (RMS) values of current and voltage. In this case, the peak voltage $\left(V_0\right)$ is given. The RMS voltage $\left(V_{\text {rms }}\right)$ for a sinusoidal AC waveform is related to the peak voltage as follows:
\[
V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}
\]

The mean power dissipated by the resistor is given by:
\[
P=I_{\mathrm{rms}}^2 \cdot R
\]

In an $\mathrm{AC}$ circuit with a purely resistive component, the RMS current is related to the RMS voltage and resistance as follows:
\[
I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{R}=\frac{V_0}{R \sqrt{2}}
\]

Now, we can calculate the mean power:
\[
P=\left(\frac{V_0}{R \sqrt{2}}\right)^2 \cdot R=\frac{V_0^2}{2 R}
\]

Question

A gamma ray can split into an electron and a positron when it passes through certain materials. Which process describes this phenomenon?

A. Pair production

B. Pair annihilation

C. Nuclear fission

D. Radioactive decay

▶️Answer/Explanation

Ans:A

The phenomenon in which a gamma ray can split into an electron and a positron when passing through certain materials is described as “Pair production.” Pair production is a process in which a gamma ray of sufficient energy is converted into an electron and a positron. A fundamental law of mechanics, given by Newton, is that in any process total linear (as well as angular) momentum remains unchanged.

Question

In the Bohr model for hydrogen, the radius of the electron orbit in the $\mathrm{n}=2$ state is four times that of the radius in the $n=1$ state.
What is $\frac{\text { speed of the electron in the } n=2 \text { state }}{\text { speed of the electron in the } n=1 \text { state }}$ ?

A. $\frac{1}{4}$

B. $\frac{1}{2}$

C. 2

D. 4

▶️Answer/Explanation

Ans:B

In the Bohr model for hydrogen, the radius $(r)$ of the electron orbit is inversely proportional to the principal quantum number $(n)$ according to the following relationship:
\[
r \propto \frac{1}{n^2}
\]

So, if the radius of the electron orbit in the $n=2$ state is four times that of the $n=1$ state, we can express it as:
\[
\frac{r_{n=2}}{r_{n=1}}=4
\]

In the Bohr model of the hydrogen atom, the speed of the electron $(v)$ is indeed inversely proportional to the square root of the radius $(r)$ of the electron’s orbit. The correct relationship is:

\[
v \propto \frac{1}{\sqrt{r}}
\]

So, if the radius of the electron orbit in the $n=2$ state is four times that of the $n=1$ state $r_{n=2}=4 r_{n=1}$ ), then the speed of the electron in the $n=2$ state compared to the $n=1$ state is:
\[
\frac{v_{n=2}}{v}=\frac{1}{\sqrt{4}}=\frac{1}{2}
\]

Question

Which statement about atomic nuclei is correct? The density is…

A. directly proportional to mass number.

B. inversely proportional to nuclear radius.

C. inversely proportional to volume.

D. constant for all nuclei.

▶️Answer/Explanation

Ans:D

In atomic nuclei, the density is relatively constant across different nuclei. This is because the density of a nucleus is determined by the ratio of its mass to its volume. While the mass and volume of nuclei can vary, the ratio remains relatively constant for most nuclei. So, the correct answer is D

Question

Radioactive nuclide $X$ decays into a stable nuclide $Y$. The decay constant of $X$ is $\lambda$. The variation with time $t$ of number of nuclei of $X$ and $Y$ are shown on the same axes.

What is the expression for $s$ ?
A. $\frac{\ln 2}{\lambda}$

B. $\frac{1}{\lambda}$

C. $\frac{\lambda}{\ln 2}$

D. $\ln 2$

▶️Answer/Explanation

Ans:A

Certainly! In the context of radioactive decay, the parameter $s$ represents the half-life of a radioactive substance. The half-life is the amount of time it takes for half of a sample of radioactive nuclei to decay into a stable product. It’s a fundamental concept in nuclear physics and is a measure of how quickly a radioactive substance decays.

The half-life ($s$) is related to the decay constant ($\lambda$) as follows:

\[s = \frac{\ln 2}{\lambda}\]

Where:

$s$ is the half-life of the radioactive substance.
$\lambda$ is the decay constant of the substance.
$\ln 2$ is the natural logarithm of 2.

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