KCET 2017 Question Paper with Solutions

The coil has an inductive reactance ${𝑋}_{𝐿}=\frac{1}{\sqrt{\surd 3}}\Omega$ and resistance $𝑅=1\Omega$. We need to determine the time lag between the maximum voltage and current when connected to a $200V,50\mathrm{Hz}$ AC supply.

To find this, we start by calculating the phase angle $𝜙$ using the tangent of the phase angle:

$\tan 𝜙=\frac{{𝑋}_{𝐿}}{𝑅}=\frac{1/\sqrt{\surd 3}}{1}=\frac{1}{\sqrt{\surd 3}}$

Then, the phase angle $𝜙$ is:

$𝜙={\tan }^{-1}\left(\frac{1}{\sqrt{\surd 3}}\right)={30}^{\circ }$

Now, we convert the phase angle to radians to find the time lag:

$𝜔𝑡=\frac{𝜋}{6}$

Solving for $𝑡$:

$𝑡=\frac{𝜋}{6𝜔}=\frac{𝜋}{6\times 2𝜋𝑓}$

Simplifying further:

$𝑡=\frac{1}{12𝑓}$

Given the frequency $𝑓=50\mathrm{Hz}$:

$𝑡=\frac{1}{12\times 50}=\frac{1}{600}s$

Therefore, the time lag between maximum voltage and current is $\frac{1}{600}s$.

Given Initial Conditions:

Initial velocity, $𝑢=20m/s$

Stopping distance, $𝑠=40m$

Determine Retardation (Deceleration):

We use the equation from Newton's second law of motion:

${𝑣}^2={𝑢}^2+2𝑎𝑠$

Since the car stops, its final velocity $𝑣=0$. Therefore, we have:

$0^2-{𝑢}^2=2\times 𝑎\times 𝑠$

Substituting the given values:

$0-{20}^2=2\times 𝑎\times 40$

$-400=80𝑎$

Solving for $𝑎$:

$𝑎=\frac{-400}{80}=-5m/s^2$

Increased Velocity Condition:

When the velocity is doubled, the new initial velocity is ${𝑢}^{\prime }=2𝑢=40m/s$.

Calculate New Stopping Distance:

Using the same formula with the new initial velocity:

$0^2-(40{)}^2=2\times (-5)\times 𝑠$

$-1600=-10𝑠$

Solving for $𝑠$:

$𝑠=\frac{1600}{10}=160m$

Thus, the stopping distance when the car's velocity is doubled is $160m$.

To understand the transformation from ${}_{238}^{92}\text{U}$ to ${}_{234}^{92}\text{U}$, let’s break it down step by step:

Alpha Decay:

An alpha particle is the same as a helium nucleus, denoted by ${}_4^2𝛼$.

When ${}_{238}^{92}\text{U}$ emits an alpha particle, it loses 4 units from its mass number and 2 units from its atomic number.

This turns uranium into thorium:

${}_{238}^{92}\text{U}{\to }_{234}^{90}\text{Th}{+}_4^2𝛼$

Beta Decay:

Beta decay (specifically, beta minus decay) converts a neutron into a proton while emitting an electron (${𝛽}^{-}$).

The atomic number increases by 1, but the mass number remains unchanged.

Since ${}_{234}^{90}\text{Th}$ has an atomic number of 90 and we want to reach ${}_{234}^{92}\text{U}$, we need to increase the atomic number by 2.

This happens via two successive beta decays:

First beta decay:

${}_{234}^{90}\text{Th}{\to }_{234}^{91}\text{Pa}+{𝛽}^{-}$

Second beta decay:

${}_{234}^{91}\text{Pa}{\to }_{234}^{92}\text{U}+{𝛽}^{-}$

Net Effect:

The overall process involves the emission of:

1 alpha particle

2 beta particles

Therefore, the correct option is:

Option B: $1𝛼$ and $2𝛽$.

According to Ohm's Law, the current flowing through a conductor is directly proportional to the potential difference $\left(𝑉\right)$ across the ends of the conductor, provided physical conduction of the conductor such as temperature, mechanical strain etc, are kept constant.

$\begin{array}{cc}& 𝑖𝑒.𝑉\propto 𝐼\\ & 𝑉=𝐼𝑅\\ & \frac{𝑉}{𝐼}=𝑅\text{ (constant) }\end{array}$

Ferromagnetic substances have a very strong response to an applied magnetic field. The magnetic susceptibility (χ) is defined by

$𝜒=\frac{𝑀}{𝐻},$

where $𝑀$ is the magnetization and $𝐻$ is the applied magnetic field. In ferromagnets, even a small $𝐻$ can produce a very large $𝑀$ due to the alignment of magnetic domains. This is why the susceptibility is often much greater than 1.

To summarize the options:

Zero: Incorrect, because the material responds strongly.

$>1$: Although technically true, it doesn’t emphasize how large the value can be.

$<1$: Incorrect, as this would be more typical for diamagnets.

$\gg 1$: Correct, because it indicates the susceptibility is many times greater than 1.

Thus, the correct answer is:

Option D: $\gg 1$.

Magnetic braking in trains works on the principle of eddy currents. Here’s a quick breakdown:

When a magnet moves relative to a conductive material (like aluminum), it induces circular currents (eddy currents) in that conductor.

According to Lenz's law, these eddy currents create their own magnetic fields that oppose the motion causing them.

This opposition produces a braking force, which helps in slowing down the train without physical friction.

So, the correct answer is:

$\text{Option A: eddy current.}$

To find the density, we use the formula:

$𝜌=\frac{𝑚}{𝑉}$

where

• $𝑚=49.53\text{g}$

• $𝑉=1.5{\text{cm}}^3$

Step 1: Calculate the density

$𝜌=\frac{49.53\text{g}}{1.5{\text{cm}}^3}\approx 33.02{\text{g/cm}}^3$

Step 2: Consider Significant Figures

The mass $49.53\text{g}$ has 4 significant figures.

The volume $1.5{\text{cm}}^3$ has 2 significant figures.

When dividing, the final answer must have the same number of significant figures as the value with the fewest significant figures. Here, that is 2 significant figures.

Step 3: Round to 2 Significant Figures

Rounding $33.02$ to 2 significant figures gives:

$33{\text{g/cm}}^3$

Among the options provided, the value that closely represents this magnitude is:

Option D: 33.0

Thus, the correct answer is Option D.

The value of photo current depends on the intensity of incident light. Hence, when the intensity of incident light increases, then the value of photo current is increases. i.e. intensity of light $\left(𝐼\right)\propto$ photo current

$\begin{array}{cc}& \text{ From graph }\\ & {𝐼}_1<{𝐼}_2\end{array}$

To find the acceleration of the center of mass of two balls thrown simultaneously in the air, we can use the formula for the acceleration of the center of mass:

${𝐚}_{\mathrm{cm}}=\frac{{𝑚}_1{𝐚}_1+{𝑚}_2{𝐚}_2}{{𝑚}_1+{𝑚}_2}$

Here, both balls are subject to the same downward gravitational acceleration, so ${𝑎}_1={𝑎}_2=𝑔$, where $𝑔$ is the acceleration due to gravity.

Substituting these values into the formula, we get:

$\begin{array}{cc}{𝐚}_{\mathrm{cm}}& =\frac{{𝑚}_1𝑔+{𝑚}_2𝑔}{{𝑚}_1+{𝑚}_2}\\ {𝐚}_{\mathrm{cm}}& =\frac{𝑔({𝑚}_1+{𝑚}_2)}{{𝑚}_1+{𝑚}_2}\\ {𝐚}_{\mathrm{cm}}& =𝑔\end{array}$

Thus, the acceleration of the center of mass of the two balls is equal to $𝑔$.

According to electromagnetic wave property, the electric field is perpendicular to the magnetic field. The direction of electromagnetic wave propagation will perpendicular to $𝐄$ and $𝐁$. then $𝐄\times 𝐁=𝐂$

Given:

Mass of the body, $𝑚=50\text{kg}$

When the lift is moving downward, the scale reading is calculated as:

$\text{Scale reading}=\frac{𝑚(𝑔-𝑎)}{𝑔}$

If the lift starts falling freely, then the acceleration $𝑎$ equals gravitational acceleration $𝑔$. So, we have:

$𝑎=𝑔$

Substituting $𝑎=𝑔$ into the scale reading formula, we get:

$\text{Scale reading}=\frac{𝑚(𝑔-𝑔)}{𝑔}=0$

Thus, when the lift is in free fall, the spring balance will read $0$.

Given the problem:

Charge $𝐴=+3\text{nC}=3\times {10}^{-9}\text{C}$

Charge $𝐵=+1\text{nC}=1\times {10}^{-9}\text{C}$

Distance between the two charges:

Initial distance, ${𝑟}_1=5\text{cm}=0.05\text{m}=5\times {10}^{-2}\text{m}$

After moving charge $𝐵$ towards $𝐴$ by 1 cm, the new distance is ${𝑟}_2=4\text{cm}=0.04\text{m}=4\times {10}^{-2}\text{m}$

The formula for the work done $\left(𝑊\right)$ in moving a charge in an electric field is:

$𝑊={𝑈}_{𝐵}-{𝑈}_{𝐴}=\frac{𝑘{𝑞}_1{𝑞}_2}{{𝑟}_2}-\frac{𝑘{𝑞}_1{𝑞}_2}{{𝑟}_1}$

Substituting the values:

$\begin{array}{cc}𝑊& =𝑘{𝑞}_1{𝑞}_2[\frac{1}{{𝑟}_2}-\frac{1}{{𝑟}_1}]\\ & =9\times {10}^9\times 3\times {10}^{-9}\times 1\times {10}^{-9}[\frac{1}{4\times {10}^{-2}}-\frac{1}{5\times {10}^{-2}}]\\ & =9\times {10}^9\times 3\times {10}^{-9}\times 1\times {10}^{-9}\left[\frac{5-4}{4\times 5\times {10}^{-2}}\right]\\ & =27\times {10}^{-9}\times \frac{1}{20\times {10}^{-2}}\\ & =\frac{27}{20}\times {10}^{-7}\\ & =1.35\times {10}^{-7}\text{J}\end{array}$

Thus, the work done to move charge $𝐵$ towards $𝐴$ by 1 cm is $1.35\times {10}^{-7}\text{J}$.

To solve this problem, we need to determine the ratio of the radii of the circular paths described by a proton, a deuteron, and an alpha particle when projected perpendicular to a uniform magnetic field with the same kinetic energy.

Let’s start with the relationship for the radius of circular motion in a magnetic field:

$𝑟=\frac{𝑚𝑣}{𝑞𝐵}$

Given that the kinetic energy (K.E) is the same for each particle, we can use the relationship for kinetic energy:

$\text{K.E}=\frac{1}{2}𝑚{𝑣}^2\Rightarrow 𝑣=\sqrt{\surd \frac{2\times \text{K.E}}{𝑚}}$

Substituting this into the radius formula gives us:

$𝑟=\frac{\sqrt{\surd 2𝑚\times \text{K.E}}}{𝑞𝐵}$

Since the kinetic energy is the same for all particles, $𝑟\propto \frac{\sqrt{\surd 𝑚}}{𝑞}$. Thus, comparing their radii, we have:

For a proton ($𝑝$): ${𝑟}_{𝑝}\propto \sqrt{\surd \frac{{𝑚}_{𝑝}}{{𝑞}_{𝑝}}}$

For a deuteron ($𝑑$): ${𝑟}_{𝑑}\propto \sqrt{\surd \frac{2{𝑚}_{𝑝}}{{𝑞}_{𝑝}}}$

For an alpha particle ($𝛼$): ${𝑟}_{𝛼}\propto \sqrt{\surd \frac{4{𝑚}_{𝑝}}{2{𝑞}_{𝑝}}}=\sqrt{\surd \frac{2{𝑚}_{𝑝}}{{𝑞}_{𝑝}}}$

Therefore, the ratio of their radii is:

${𝑟}_{𝑝}:{𝑟}_{𝑑}:{𝑟}_{𝛼}=1:\sqrt{\surd 2}:1$

Thus, the correct answer for the ratio of the radii is $1:\sqrt{\surd 2}:1$.

The specific heat capacity is defined by the formula

$𝑐=\frac{𝑄}{𝑚\Delta 𝑇}$

where:

$𝑄$ is the heat energy (in Joules),

$𝑚$ is the mass (in kilograms),

$\Delta 𝑇$ is the temperature change (in Kelvin).

From the formula, you can see that the specific heat capacity has units of energy per unit mass per unit temperature change. This results in units of Joules per kilogram Kelvin, or

$J{\mathrm{kg}}^{-1}{K}^{-1}.$

Thus, the SI unit of specific heat capacity is given by Option D.

The emitter of a transistor is of moderate size and heavily doped. Its primary role is to provide a substantial amount of majority charge carriers, which facilitates the current flow through the transistor.

When the magnet is allowed to fall vertically along the axis of loop with its north pole towards the coil. The upper face of the coil will become north pole in an attempt to oppose the approaching north pole of the magnet. Therefore, the acceleration in the magnet is less than $𝑔$.

The efficiency of a Carnot engine is determined by the formula:

$𝜂=1-\frac{{𝑇}_{𝐶}}{{𝑇}_{𝐻}}$

where:

${𝑇}_{𝐻}$ is the temperature of the hot reservoir.

${𝑇}_{𝐶}$ is the temperature of the cold reservoir.

To find the combination with the least efficiency, we need to calculate $𝜂$ for each option.

Option A: ${𝑇}_{𝐻}=100𝐾$, ${𝑇}_{𝐶}=80𝐾$

${𝜂}_{𝐴}=1-\frac{80}{100}=1-0.8=0.2\text{ or }20\%$

Option B: ${𝑇}_{𝐻}=40𝐾$, ${𝑇}_{𝐶}=20𝐾$

${𝜂}_{𝐵}=1-\frac{20}{40}=1-0.5=0.5\text{ or }50\%$

Option C: ${𝑇}_{𝐻}=80𝐾$, ${𝑇}_{𝐶}=60𝐾$

${𝜂}_{𝐶}=1-\frac{60}{80}=1-0.75=0.25\text{ or }25\%$

Option D: ${𝑇}_{𝐻}=60𝐾$, ${𝑇}_{𝐶}=40𝐾$

${𝜂}_{𝐷}=1-\frac{40}{60}\approx 1-0.667\approx 0.333\text{ or }33.3\%$

Comparing the efficiencies, the lowest efficiency is 20% in Option A.

Thus, the combination with the least efficiency is:

Option A: $100𝐾,80𝐾$.

When light travels from air into a denser medium, two important things happen:

The speed of light decreases because the medium's refractive index is higher.

The wavelength of the light decreases, as the frequency remains constant and wavelength is given by

$𝜆=\frac{𝑣}{𝑓}.$

So, by Huygens' principle and the wave nature of light, both the speed and the wavelength decrease.

Therefore, the correct answer is:

Option D: Wavelength and speed decreases.

$\begin{array}{cc}& \text{ Current flowing in galvanometer, }\\ & \begin{array}{cc}& 𝐼=\frac{3}{(50+2950)}\\ & 𝐼={10}^{-3}A\end{array}\end{array}$

Current for 30 division $={10}^{-3}A$

Current for 20 division $=\frac{{10}^{-3}}{30}\times 20=\frac{2}{3}\times {10}^{-3}A$

Let the series resistance $=𝑅$

$\begin{array}{cc}\therefore \frac{2}{3}\times {10}^{-3}& =\frac{3}{(50+𝑅)}\\ 𝑅& =4450\Omega \end{array}$

Option (c) is nearest option in the given options.

When a rod is inserted into the coil, than its inductance increases. Therefore, current through the bulb will decreases and also brightness decreases.

Given:

Magnetic moment, $𝑀=6\times {10}^{-2}{\text{A-m}}^2$

Moment of inertia, $𝐼=12\times {10}^{-6}{\text{kg-m}}^2$

Magnetic field, $𝐵=2\times {10}^{-2}\text{T}$

To find the time period for oscillations, we use the formula:

$𝑡=2𝜋\sqrt{\surd \frac{𝐼}{𝑀𝐵}}$

Calculating:

$𝑡=2𝜋\sqrt{\surd \frac{12\times {10}^{-6}}{6\times {10}^{-2}\times 2\times {10}^{-2}}}$

$=2𝜋\sqrt{\surd \frac{1.2\times {10}^{-5}}{12\times {10}^{-4}}}$

$=2𝜋\times 0.1=0.2𝜋$

For 20 oscillations, the total time taken is:

$\text{Time (T)}=20\times 0.2𝜋$

$=4𝜋\approx 12\text{s}$

Thus, the time taken for the dipole to complete 20 oscillations is approximately 12 seconds.

Given the following parameters:

Number of electrons removed $\left(𝑛\right)$: $4\times {10}^{10}$

**Diameter of the sphere $\left(𝐷\right):\ast \ast \text{(}20\text{cm}$

**Radius of the sphere $\left(𝑟\right):\ast \ast \text{(}\frac{20}{2}=10\text{cm}=0.1\text{m}$

We need to calculate the magnitude of the electric field at a distance of $20\text{cm}$ from the center of the sphere.

The formula for the electric field outside the sphere is:

$𝐸=\frac{𝑞}{4𝜋{𝜀}_0{𝑟}^2}=\frac{𝑛𝑒}{4𝜋{𝜀}_0{𝑟}^2}$

Where:

$𝑒=1.6\times {10}^{-19}\text{C}$ (charge of an electron)

${𝜀}_0=8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ (permittivity of free space)

Since $𝑟=20\text{cm}=0.2\text{m}$, substituting the values:

$𝐸=\frac{4\times {10}^{10}\times 1.6\times {10}^{-19}\times 9\times {10}^9}{(0.2{)}^2}$

Breaking down the calculation:

$𝐸=\frac{4\times 1.6\times 9\times {10}^4}{0.04}=\frac{4\times 1.6\times 9\times {10}^4}{20\times 20\times {10}^{-4}}$

Simplifying further:

$𝐸=1.6\times 9\times {10}^2=1.44\times {10}^2$

Thus, the magnitude of the electric field at a distance of $20\text{cm}$ from the center is:

$𝐸=1440{\text{NC}}^{-1}$

The mass defect of ${}_4^2\mathrm{He}$ is 0.03 u. Given the mass number is 4, the binding energy is calculated using the formula:

$\text{Binding energy}=0.03\times 931=27.93\text{MeV}$

The binding energy per nucleon is then determined by dividing the total binding energy by the number of nucleons:

$\text{Per nucleon binding energy}=\frac{27.93}{4}=6.9825\text{MeV}$

Given:

Diameter of the cylindrical conductor (D) = 0.1 mm

Radius (r) = $\frac{𝐷}{2}=\frac{0.1}{2}$ mm

Current (I) = 90 mA = $90\times {10}^{-3}$ A

To find the current density (J), we use the formula:

$\text{Current density (J)}=\frac{𝐼}{𝐴}=\frac{𝐼}{𝜋{𝑟}^2}$

Substituting the known values:

$𝐽=\frac{90\times {10}^{-3}}{\frac{22}{7}\times {\left(\frac{0.1\times {10}^{-3}}{2}\right)}^2}$

Approximating $𝜋\approx 3.14$:

$𝐽=\frac{90\times {10}^{-3}}{3.14\times {\left(\frac{0.1\times {10}^{-3}}{2}\right)}^2}$

$𝐽=12000\times {10}^3$

Thus, the current density:

$𝐽=12\times {10}^7{A/m}^2$

The false property is Option A.

Here’s why:

A bar magnet produces its own magnetic field. In fact, you can observe this magnetic field by the attraction and repulsion it exerts on other magnetic objects.

Option B is true since like poles repel and unlike poles attract.

Option C is true because if you break a bar magnet, each piece becomes a smaller magnet with its own north and south poles; you cannot isolate a single magnetic pole.

Option D is true since a freely suspended bar magnet will align itself along the North-South direction due to Earth’s magnetic field.

Thus, the statement "It doesn't produce magnetic field" (Option A) is false.

According to cartesian sign convention, in ray optics all distance in the direction of incident ray are taken positive.

Given:

Height of object, ${ℎ}_0=10\text{cm}$

Radius of curvature, $𝑅=15\text{cm}$

Distance of object, $𝑢=-10\text{cm}$

First, calculate the focal length of the concave mirror using:

$𝑓=\frac{𝑅}{2}=\frac{15}{2}\text{cm}$

According to the mirror formula:

$\frac{1}{𝑓}=\frac{1}{𝑢}+\frac{1}{𝑣}$

Substituting the known values:

$\frac{1}{-\frac{15}{2}}=\frac{1}{-10}+\frac{1}{𝑣}$

Solving for $\frac{1}{𝑣}$:

$\frac{1}{𝑣}=\frac{-2}{15}+\frac{1}{10}$

$\Rightarrow \frac{1}{𝑣}=\frac{-4+3}{30}$

$\Rightarrow \frac{1}{𝑣}=\frac{-1}{30}$

$\Rightarrow 𝑣=-30\text{cm}$

Now, calculate the magnification ($𝑚$) using the formula:

$𝑚=\frac{{ℎ}_{𝑖}}{{ℎ}_{𝑜}}=\frac{-𝑣}{𝑢}$

Thus:

$\frac{{ℎ}_{𝑖}}{10}=\frac{-(-30)}{-10}$

$\Rightarrow {ℎ}_{𝑖}=-3$

Therefore, the image will be magnified and inverted.

The energy gap (or band gap) is the energy difference between the valence band and the conduction band in a material.

Semiconductors have a finite band gap that allows electrons to jump from the valence band to the conduction band with the absorption of energy. Typically, the band gap in semiconductors is less than 3 eV. For example, Germanium has a band gap of around $0.66\text{ eV}$.

Metals, on the other hand, have overlapping conduction and valence bands, meaning they do not have a meaningful band gap (effectively, the gap is 0 eV).

Given the options:

Option A: Germanium (a semiconductor with $\sim 0.66\text{ eV}$ band gap, which is less than 3 eV)

Options B (Iron), C (Copper), and D (Aluminium) are metals and do not exhibit a band gap in the context of semiconductor physics.

Thus, the correct answer is:

Option A

Germanium

The minimum value of effective capacitance is achieved when the capacitors are connected in series.

To find the equivalent capacitance for capacitors in series, use the formula:

$\frac{1}{{𝐶}_{\mathrm{eq}}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}+\frac{1}{{𝐶}_3}$

For capacitors with capacitances $1\text{pF}$, $2\text{pF}$, and $4\text{pF}$, the equivalent capacitance calculation is:

$\frac{1}{{𝐶}_{\mathrm{eq}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}$

This combines to:

$\frac{1}{{𝐶}_{\mathrm{eq}}}=\frac{4+2+1}{4}=\frac{7}{4}$

Solving for ${𝐶}_{\mathrm{eq}}$ gives:

${𝐶}_{\mathrm{eq}}=\frac{4}{7}\mathrm{pF}$

Thus, the minimum effective capacitance obtainable by connecting the three capacitors in series is $\frac{4}{7}\mathrm{pF}$.

In the initial scenario, two spheres have electric charges ${𝑞}_1=+2\text{nC}=2\times {10}^{-9}\text{C}$ and ${𝑞}_2=-8\text{nC}=-8\times {10}^{-9}\text{C}$. The force between them is given by:

$𝐹=\frac{𝑘\cdot {𝑞}_1\cdot {𝑞}_2}{{𝑑}^2}$

Substituting the values:

$𝐹=\frac{𝑘\cdot (2\times {10}^{-9})\cdot (-8\times {10}^{-9})}{{𝑑}^2}=\frac{𝑘\cdot (-16\times {10}^{-18})}{{𝑑}^2}$

This is our equation (i).

After the spheres are allowed to touch each other, the total charge becomes:

$2\text{nC}-8\text{nC}=-6\text{nC}$

Thus, each sphere carries a charge of:

$\frac{-6\text{nC}}{2}=-3\text{nC}=-3\times {10}^{-9}\text{C}$

The new force ${𝐹}_1$ after separation is:

${𝐹}_1=\frac{𝑘\cdot (3\times {10}^{-9})\cdot (3\times {10}^{-9})}{{𝑑}^{\prime 2}}=\frac{𝑘\cdot (9\times {10}^{-18})}{{𝑑}^{\prime 2}}$

This is equation (ii).

To maintain the same magnitude of force as before:

$\frac{𝑘\cdot (16\times {10}^{-18})}{{𝑑}^2}=\frac{𝑘\cdot (9\times {10}^{-18})}{{𝑑}^{\prime 2}}$

Solving for the new distance ${𝑑}^{\prime }$:

$\frac{{𝑑}^2}{{𝑑}^{\prime 2}}=\frac{16\times {10}^{-18}}{9\times {10}^{-18}}=\frac{16}{9}$

Taking the square root:

$\frac{𝑑}{{𝑑}^{\prime }}=\sqrt{\surd \frac{16}{9}}=\frac{4}{3}$

Therefore:

$𝑑=\frac{4}{3}{𝑑}^{\prime }$

Solving for ${𝑑}^{\prime }$:

${𝑑}^{\prime }=\frac{3}{4}𝑑$

So, the new distance between the spheres to achieve the same magnitude of repulsive force is $\frac{3}{4}𝑑$.

To find the value of acceleration due to gravity at a depth of 1600 km, we can use the formula:

${𝑔}_{𝑑}=𝑔(1-\frac{𝑑}{𝑅})$

Where:

$𝑔$ is the acceleration due to gravity on the surface, approximately $9.8{\mathrm{ms}}^{-2}$.

$𝑑$ is the depth, given as $1600\mathrm{km}$.

$𝑅$ is the radius of the Earth, approximately $6400\mathrm{km}$.

Substituting the values into the formula:

${𝑔}_{𝑑}=9.8(1-\frac{1600}{6400})$

Simplifying the expression inside the parentheses:

${𝑔}_{𝑑}=9.8(1-\frac{1}{4})$

Continue simplifying:

${𝑔}_{𝑑}=9.8\left(\frac{4-1}{4}\right)=9.8\times \frac{3}{4}$

Finally, calculating the result:

${𝑔}_{𝑑}=7.35{\mathrm{ms}}^{-2}$

Thus, the value of acceleration due to gravity at a depth of 1600 km is $7.35{\mathrm{ms}}^{-2}$.

To determine the ratio of the lengths of pendulums $𝐴$ and $𝐵$, we start with the given information:

For pendulum $𝐴$:

It completes 10 oscillations in 20 seconds.

The time period ${𝑇}_{𝐴}$ is the total time divided by the number of oscillations:

${𝑇}_{𝐴}=\frac{20\text{seconds}}{10}=2\text{seconds}$

The formula for the period of a pendulum is given by:

${𝑇}_{𝐴}=2𝜋\sqrt{\surd \frac{{𝐿}_{𝐴}}{𝑔}}$

where ${𝐿}_{𝐴}$ is the length of pendulum $𝐴$ and $𝑔$ is the acceleration due to gravity.

For pendulum $𝐵$:

It completes 8 oscillations in 10 seconds.

The time period ${𝑇}_{𝐵}$ is:

${𝑇}_{𝐵}=\frac{10\text{seconds}}{8}=1.25\text{seconds}$

The formula for the period of pendulum $𝐵$ is:

${𝑇}_{𝐵}=2𝜋\sqrt{\surd \frac{{𝐿}_{𝐵}}{𝑔}}$

where ${𝐿}_{𝐵}$ is the length of pendulum $𝐵$.

To find the ratio of their lengths, we need to compare the periods using the formula for the period:

Step 1: Express the square of the periods based on their formula:

${𝑇}_2^{𝐴}={\left(2𝜋\sqrt{\surd \frac{{𝐿}_{𝐴}}{𝑔}}\right)}^2\Rightarrow \left(\frac{{𝐿}_{𝐴}}{𝑔}\right)={\left(\frac{{𝑇}_{𝐴}}{2𝜋}\right)}^2$

${𝑇}_2^{𝐵}={\left(2𝜋\sqrt{\surd \frac{{𝐿}_{𝐵}}{𝑔}}\right)}^2\Rightarrow \left(\frac{{𝐿}_{𝐵}}{𝑔}\right)={\left(\frac{{𝑇}_{𝐵}}{2𝜋}\right)}^2$

Step 2: Divide the two expressions to find the ratio $\frac{{𝐿}_{𝐴}}{{𝐿}_{𝐵}}$:

$\frac{{𝐿}_{𝐴}}{{𝐿}_{𝐵}}={\left(\frac{{𝑇}_{𝐴}}{{𝑇}_{𝐵}}\right)}^2$

Substitute the values of ${𝑇}_{𝐴}$ and ${𝑇}_{𝐵}$:

$\frac{{𝐿}_{𝐴}}{{𝐿}_{𝐵}}={\left(\frac{2}{1.25}\right)}^2={\left(\frac{160}{100}\right)}^2=\frac{64}{25}$

Thus, the ratio of the lengths of pendulum $𝐴$ to $𝐵$ is $\frac{64}{25}$.

Given:

Water lifted: 6 tonnes (6000 kg)

Depth of well: 25 m

Height of first floor: 35 m

Time taken: 20 minutes (or 1200 seconds)

Gravitational acceleration: $𝑔=10{\text{m/s}}^2$

To find the power of the pump:

Calculate the work done:

The work done in lifting the water is given by the formula:

$\text{Work}=𝑚\times 𝑔\times ℎ$

Here, $ℎ$ is the height difference from the well to the first floor, which is $35\text{ m}-25\text{ m}=10\text{ m}$.

Substitute the values:

$\text{Work}=6000\times 10\times 10=600,000\text{Joules}$

Calculate the power:

Power is the rate at which work is done and is given by the formula:

$𝑃=\frac{\text{Work}}{\text{Time}}$

Substituting the work and time:

$𝑃=\frac{600,000}{1200}=500\text{W}$

$𝑃=3,000\text{W}=3\text{kW}$

Therefore, the power of the pump is $3\text{kW}$.

In a closed pipe, the wave pattern that forms is a standing wave (also called a stationary wave). Here’s why:

Type of Wave:

Sound in air is a longitudinal wave. This means that the particles in the medium (air molecules) oscillate in the same direction as the wave propagation.

Nature of the Wave in a Closed Pipe:

In a closed pipe—where one end is closed and the other end is open—the wave is reflected at the closed end, interfering with the incoming wave to produce a standing (stationary) wave pattern. In this standing wave, certain points (nodes) remain fixed, while other points (antinodes) reach maximum displacement.

Why Not the Other Options:

"Longitudinal and progressive" (Option A) describes sound waves in an open medium where the wave travels continuously. In a closed pipe, reflections lead to a standing wave.

"Transverse and progressive" (Option B) and "transverse and stationary" (Option C) are incorrect because sound waves in air are not transverse—they involve compressions and rarefactions along the direction of propagation.

Thus, the correct answer is:

$\text{Option D: Longitudinal and stationary}.$

This means that the wave pattern in a closed pipe is made up of longitudinal sound waves that are standing rather than continuously traveling along the pipe.

The voltage drop across $5k\Omega$ resistance is given by

$\begin{array}{cc}& 𝑉=02\times {10}^{-3}\times 5\times {10}^3\\ & 𝑉=1.0V\end{array}$

Applying the KVL (Kirchoff's Voltage Law)

${𝑉}_{𝐴𝐵}=1+02+1$

$\begin{array}{cc}& [\because \text{ Voltage drop acros diode }=02V]\\ {𝑉}_{𝐴𝐵}=22V\end{array}$

Given the vectors:

$𝐀=2\widehat{𝐢}+3\widehat{𝐣}+8\widehat{𝐤}$

$𝐁=4\widehat{𝐣}-4\widehat{𝐢}+𝛼\widehat{𝐤}$

$𝐁=-4\widehat{𝐢}+4\widehat{𝐣}+𝛼\widehat{𝐤}$

Since $𝐀$ is perpendicular to $𝐁$, their dot product is zero:

$𝐀\cdot 𝐁=0$

This translates to:

$(2\widehat{𝐢}+3\widehat{𝐣}+8\widehat{𝐤})\cdot (-4\widehat{𝐢}+4\widehat{𝐣}+𝛼\widehat{𝐤})=0$

Calculating the dot product gives:

$2(-4)+3\left(4\right)+8\left(𝛼\right)=0$

Simplifying the equation:

$-8+12+8𝛼=0$

$8𝛼+4=0$

$8𝛼=-4$

$𝛼=-\frac{4}{8}=-\frac{1}{2}$

Therefore, the value of $𝛼$ is $-\frac{1}{2}$.

When an object follows a circular path at a constant speed, the motion of object is called uniform circular motion.

Although the speed does not vary the particle is accelerating because the velocity changes its direction at every point on circulat track.

Given, length of wire $\left(𝑙\right)=50\mathrm{cm}=50\times {10}^{-2}m$ current $\left(𝐼\right)=2.5A$

magnetic field $\left(𝐵\right)=0.5T$

$𝑔=10{\mathrm{ms}}^{-2}$

For the balance condition,

$\begin{array}{cc}{𝐹}_{𝐵}& =𝑚𝑔\\ 𝐵𝐼𝐼& =𝑚𝑔\\ 𝑚& =\frac{𝐵𝐼𝐼}{𝑔}\\ & =\frac{0.5\times 2.5\times 50\times {10}^{-2}}{10}=62.5g\end{array}$

Young's modulus is defined as the ratio of tensile stress to longitudinal strain. This relationship can be expressed in the following formula:

$\text{Young’s modulus}=\frac{\text{tensile stress}}{\text{longitudinal strain}}$

The correct answer is Option B: Zener diode.

Here’s why:

A voltage regulator maintains a constant voltage level regardless of changes in the load current or supply voltage.

A Zener diode, when reverse-biased, enters a breakdown region at its specified Zener voltage. In this region, it maintains approximately that voltage across its terminals even if the current varies.

This characteristic makes the Zener diode ideal for use in simple voltage regulation circuits.

The other options do not serve this function:

A laser diode is primarily used for emitting focused light in optical applications.

A solar cell converts light energy into electrical energy.

A photodiode detects light and converts it into a current but does not regulate voltage.

Thus, the Zener diode is the semiconductor device used as a voltage regulator.

$\text{ NAND gate is an universal gate. }$

The de-Broglie wavelength $𝜆$ of a particle is given by the formula:

$𝜆=\frac{ℎ}{𝑚𝑣}$

Where:

$ℎ$ is the Planck's constant,

$𝑚$ is the mass of the particle,

$𝑣$ is the velocity of the particle.

When a particle is dropped from a height $𝐻$, it accelerates due to gravity, and its velocity $𝑣$ becomes:

$𝑣=\sqrt{\surd 2𝑔𝐻}$

Substituting this expression for velocity into the de-Broglie wavelength formula, we get:

$𝜆=\frac{ℎ}{𝑚\sqrt{\surd 2𝑔𝐻}}$

This shows that the de-Broglie wavelength $𝜆$ is proportional to ${𝐻}^{-1/2}$, meaning that as the height $𝐻$ increases, the wavelength decreases according to this relationship:

$𝜆\propto {𝐻}^{-1/2}$