Physics

We are given two conditions:

The particles are at the same distance from the origin. This means that the magnitudes of their position vectors are equal.

The vectors are perpendicular (at right angles) to each other. This means their dot product is zero.

Let’s work through these step-by-step.

● The vectors are given by:

$\vec{A} = 2\hat{i} + 3n \hat{j} + 2\hat{k},$

$\vec{B} = 2\hat{i} -2\hat{j} + 4p \hat{k}.$

● Since the vectors are perpendicular, their dot product must be zero:

$\vec{A} \cdot \vec{B} = (2)(2) + (3n)(-2) + (2)(4p) = 4 - 6n + 8p = 0.$

This can be rearranged to:

$8p = 6n - 4 \quad \Longrightarrow \quad p = \frac{3n - 2}{4} \quad \text{(Equation 1)}.$

● Since the distances from the origin are equal, the magnitudes of the vectors must be equal. Compute the squares of the magnitudes:

For $\vec{A}: \quad |\vec{A}|^2 = 2^2 + (3n)^2 + 2^2 = 4 + 9n^2 + 4 = 8 + 9n^2.$

For $\vec{B}: \quad |\vec{B}|^2 = 2^2 + (-2)^2 + (4p)^2 = 4 + 4 + 16p^2 = 8 + 16p^2.$

Setting them equal:

$8 + 9n^2 = 8 + 16p^2 \quad \Longrightarrow \quad 9n^2 = 16p^2.$

This simplifies to:

$p^2 = \frac{9n^2}{16} \quad \Longrightarrow \quad p = \pm \frac{3n}{4} \quad \text{(Equation 2)}.$

● Now equate the two expressions for $p$ from Equation 1 and Equation 2.

Case 1: Assume $p = \frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = \frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = 3n \quad \Longrightarrow \quad -2 = 0,$

which is a contradiction.

Case 2: Assume $p = -\frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = -\frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = -3n.$

Simplify by adding $3n$ to both sides:

$6n - 2 = 0 \quad \Longrightarrow \quad 6n = 2 \quad \Longrightarrow \quad n = \frac{1}{3}.$

● Since we found $n = \frac{1}{3},$ its reciprocal is:

$n^{-1} = 3.$

Thus, the value of $n^{-1}$ is $3.$

$\begin{aligned} & {[\tau]=[\mathrm{E}]} \\ & {[\sigma] \neq[\mathrm{I}]} \\ & {[\mathrm{L}]=[\mathrm{h}]} \\ & {[\mathrm{P}]=[\mathrm{Y}]} \end{aligned}$

$\begin{aligned} & {\left[\mathrm{LT}^{-1}\right]=[\mathrm{A}]\left[\mathrm{T}^2\right]=\frac{[\mathrm{B}][\mathrm{T}]}{[\mathrm{C}]+[\mathrm{T}]}} \\ & {[\mathrm{C}]=[\mathrm{T}]} \\ & {[\mathrm{A}]=\left[\mathrm{LT}^{-3}\right]} \\ & {[\mathrm{B}]=\left[\mathrm{LT}^{-1}\right]} \\ & {[\mathrm{ABC}]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right]} \end{aligned}$

We know, Young's modulus, $E = {\sigma \over \varepsilon }$

$ \Rightarrow E = {{{F \over A}} \over {{{\Delta L} \over L}}} = {F \over A}\left( {{L \over {\Delta L}}} \right)$

$ \Rightarrow [E] = {{[F]} \over {[A]}} = {{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$

(A) $ \Rightarrow [E] = [{M^1}{L^{ - 1}}{T^{ - 2}}]$

(B) we know, Torque = $r\times F$

$[\tau ] = [r][F]$

$ = [L][{M^1}{L^1}{T^{ - 2}}]$

$[\tau ] = [{M^1}{L^2}{T^{ - 2}}]$

(C) We know, $F = \mu A{u \over y}$

where, F = force, $\mu$ = coefficient of viscosity

A = area $\frac{u}{y}$ = rate of shear deformation

$ \Rightarrow \mu = {{Fy} \over {Au}}$

$ \Rightarrow [\mu ] = {{[{M^1}{L^1}{T^{ - 2}}][L]} \over {[{L^2}][L{T^{ - 1}}]}}$

$ \Rightarrow [\mu ] = [{M^1}{L^{ - 1}}{T^{ - 1}}]$

(D) We know, $F = {{G{m_1}{m_2}} \over {{r^2}}}$ (Newton's law of gravitation)

$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$<$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$

$ \Rightarrow [G] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}$

$ \Rightarrow [G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Hence, option D is correct.

Given, $Q = {{a{b^4}} \over {cd}}$

$a = (60 \pm 3)\,Pa \Rightarrow a = 60\,Pa,\,\Delta a = 3Pa$

$b = (20 \pm 0.1)\,m \Rightarrow b = 20m,\,\Delta b = 0.1\,m$

$c = (40 \pm 0.2)\,Ns{m^{ - 2}} \Rightarrow c = 40\,Ns{m^{ - 2}},\,\Delta c = 0.2\,Ns{m^{ - 2}}$

$d = (50 \pm 0.1)\,m \Rightarrow \,d = 50\,m,\,\Delta d = 0.1m$

As, $Q = {{a{b^4}} \over {cd}}$

by taking ln on both sides,

$\ln Q = \ln a + u\ln b - \ln c - \ln d$

Now, by differentiating,

${{dQ} \over Q} = {{da} \over a} + 4{{db} \over b} - {{dc} \over c} - {{dd} \over d}$

So, maximum fractional error in Q is given by,

${{\Delta Q} \over Q} = {{\Delta a} \over a} + 4{{\Delta b} \over b} + {{\Delta c} \over c} + {{\Delta d} \over d}$

$ \Rightarrow {{\Delta Q} \over Q} = {3 \over {60}} + 4\left( {{{0.1} \over {20}}} \right) + {{0.2} \over {40}} + {{0.1} \over {40}}$

$ = {1 \over {20}} + {1 \over {50}} + {1 \over {200}} + {1 \over {500}}$

$ \Rightarrow {{\Delta Q} \over Q} = {{50 + 20 + 5 + 2} \over {1000}} = {{77} \over {1000}}$

Hence the % error in $Q = {{\Delta Q} \over Q} \times 100\% $

$ = {{7700} \over {1000}}\% = {x \over {1000}}$ (given)

So, $x = 7700$

To determine the dimensions of $\left(\frac{B}{\mu_0}\right)$, we start with the formula for the magnetic field $B$ inside a solenoid:

$ B = \mu_0 n I $

where $n = \frac{N}{L}$ is the number of turns per unit length, $I$ is the current, and $\mu_0$ is the permeability of free space.

Rearranging the formula gives:

$ \frac{B}{\mu_0} = nI = \frac{N}{L}I $

From the above, the dimensions of $\left(\frac{B}{\mu_0}\right)$ are:

$ \left[\frac{B}{\mu_0}\right] = \left[L^{-1} A\right] $

Thus, the dimensional formula of $\left(\frac{B}{\mu_0}\right)$ corresponds to $L^{-1} \text{A}$.

In general, one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct. Also least count is given by one main scale division divided by number of vernier scale division for normal vernier calliper.

Hence, option 2 is correct.

To determine the maximum percentage error in the density of the wire, follow these steps:

The density of the wire is given by the formula for a cylinder:

$\rho = \frac{m}{\pi r^2 l}$

When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$

Here:

$\frac{\Delta m}{m}$ is the relative error in the mass.

$2\frac{\Delta r}{r}$ is due to the radius being squared.

$\frac{\Delta l}{l}$ is the relative error in the length.

Now plug in the given values:

Mass: $m = 0.60 \, \text{g}$ with an error $\Delta m = 0.003 \, \text{g}$

$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$ or 0.5%.

Radius: $r = 0.50 \, \text{cm}$ with an error $\Delta r = 0.01 \, \text{cm}$

$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$ or 2%. Since the radius is squared in the formula, multiply by 2:

$2\frac{\Delta r}{r} = 2 \times 0.02 = 0.04$ or 4%.

Length: $l = 10.00 \, \text{cm}$ with an error $\Delta l = 0.05 \, \text{cm}$

$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$ or 0.5%.

Sum these contributions to find the overall maximum relative error:

$\frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05$

Converting this relative error into a percentage gives:

$0.05 \times 100\% = 5\%$

Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.

Matching the quantities from List - I with their correct dimensions from List - II is crucial to understanding their physical significance. Here is the correct matching:

(A) Angular Impulse: The dimensions of angular impulse are equivalent to those of angular momentum, represented as $M L^2 T^{-1}$. Hence, (A) corresponds to (III).

(B) Latent Heat: Latent heat has dimensions of energy per unit mass, which simplifies to $M^0 L^2 T^{-2}$. Hence, (B) corresponds to (I).

(C) Electrical resistivity: Electrical resistivity has the dimensions of resistance multiplied by length, which can be represented as $M L^3 T^{-3} A^{-2}$. Hence, (C) corresponds to (IV).

(D) Electromotive force: Electromotive force (emf) is similar to electric potential difference and has dimensions $M L^2 T^{-3} A^{-1}$. Hence, (D) corresponds to (II).

Therefore, the correct answer is Option C:

(A)-(III)

(B)-(I)

(C)-(IV)

(D)-(II)

Let's analyze the position function step by step.

The particle's position is given by:

$x(t) = A \sin t + B \cos^2 t + C t^2 + D,$

where $ t $ represents time, which has the dimension $ T $.

Since the overall dimensions of position $ x(t) $ must be length $ (L) $, each term in the expression must also have dimensions of length.

Term $ A \sin t $:

The sine function, $ \sin t $, is dimensionless (its argument must be dimensionless, and by convention, if $ t $ is in an appropriate unit where the argument is dimensionless, the function itself is dimensionless).

Therefore, $ A $ must carry the dimension of length:

$ [A] = L. $

Term $ B \cos^2 t $:

Similarly, the cosine function is dimensionless, and so is its square.

Thus, $ B $ must have the dimension of length:

$ [B] = L. $

Term $ C t^2 $:

Here, $ t^2 $ has the dimension $ T^2 $.

To have the overall term with the dimension $ L $, we require:

$ [C] \times [T^2] = L. $

So, the dimension of $ C $ must be:

$ [C] = L \, T^{-2}. $

Term $ D $:

This is a constant term added to position, so it must also have the dimension of length:

$ [D] = L. $

Now, we are asked to find the dimension of:

$ \frac{ABC}{D}. $

The dimensions of $ A $, $ B $, and $ C $ are:

$ [A] = L, \quad [B] = L, \quad [C] = L \, T^{-2}. $

Multiplying them together:

$ [ABC] = L \times L \times (L \, T^{-2}) = L^3 \, T^{-2}. $

Since $ [D] = L $, dividing by $ D $ gives:

$ \frac{[ABC]}{[D]} = \frac{L^3 \, T^{-2}}{L} = L^2 \, T^{-2}. $

Thus, the dimension of $ \frac{A B C}{D} $ is:

$ \boxed{L^2 T^{-2}}. $

The correct answer is Option C.

We start with the energy function:

$ E(t)= \alpha^3 e^{-\beta t}, $

where $\beta=0.3\, \text{s}^{-1}$.

Step 1. Take the Logarithm

Taking the logarithm of both sides:

$ \ln E = 3 \ln \alpha - \beta t. $

Step 2. Differentiate to Find the Relative Error

Differentiate both sides:

$ \frac{dE}{E} = 3\,\frac{d\alpha}{\alpha} - \beta\, dt. $

For maximum error estimation, we consider the absolute values and sum the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|. $

Step 3. Plug in the Given Errors

The percentage error in $\alpha$ is $1.2\%$ (i.e., $\Delta \alpha/\alpha = 0.012$).

The percentage error in $t$ is $1.6\%$, so for $t=5\,\text{s}$:

$ \Delta t = 0.016 \times 5 = 0.08\, \text{s}. $

Now substitute these values:

$ \left|\frac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08). $

Calculating each term:

$3(0.012) = 0.036$ (or $3.6\%$),

$0.3(0.08) = 0.024$ (or $2.4\%$).

Step 4. Compute the Total Maximum Percentage Error

Add the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06, $

which is $6\%$.

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is:

$ \boxed{6\%} $

Answer: Option A (6%).

The correct matching is:

(A) Permeability of free space has dimensions

$[MLT^{-2}A^{-2}]$

which corresponds to entry (III).

(B) Magnetic field has dimensions

$[MT^{-2}A^{-1}]$

which corresponds to entry (II).

(C) Magnetic moment in SI units is expressed in Ampere-square meters, i.e.,

$[L^2A]$

which corresponds to entry (IV).

(D) Torsional constant (torque per unit angular displacement) has dimensions

$[ML^2T^{-2}]$

which corresponds to entry (I).

Thus, the correct answer is:

Option A: (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

$ 32.3 \quad (3\ \text{sig figs}), \quad 1125 \quad (4\ \text{sig figs}), \quad 27.4 \quad (3\ \text{sig figs}) $

In multiplication and division, the final answer must be reported with the same number of significant figures as the factor with the fewest significant figures, which here is 3.

First, calculate the unrounded result:

$ y = \frac{32.3 \times 1125}{27.4} \approx \frac{36337.5}{27.4} \approx 1326.186 $

Expressing $1326.186$ in scientific notation:

$ 1326.186 \approx 1.326186 \times 10^3 $

Rounding to 3 significant figures:

The first three significant digits are 1, 3, and 2.

Considering the fourth digit (6) causes the last significant digit to round up.

Thus,

$ 1.326186 \times 10^3 \approx 1.33 \times 10^3 $

In standard decimal form, this is written as:

$ 1330 $

Hence, the final reported value is:

$ y = 1330 $

The least count (L.C.) of a screw gauge is given by:

$ \text{L.C.} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} $

Let the original pitch be $p$ and the original number of divisions be $n$. Then, the original least count is:

$ \frac{p}{n} = 0.01 \, \text{mm} $

After modifications:

The pitch is increased by 75%, so the new pitch is:

$ p_{\text{new}} = p + 0.75p = 1.75p $

The number of divisions is reduced by 50%, so the new number of divisions is:

$ n_{\text{new}} = 0.5n $

The new least count is then:

$ \text{L.C.}_{\text{new}} = \frac{p_{\text{new}}}{n_{\text{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} \times \frac{p}{n} = 3.5 \times \frac{p}{n} $

Substitute the original least count:

$ \text{L.C.}_{\text{new}} = 3.5 \times 0.01 \, \text{mm} = 0.035 \, \text{mm} $

Expressed in the form $\times 10^{-3} \, \text{mm}$, this becomes:

$ 0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm} $

Thus, the new least count is:

$ 35 \times 10^{-3} \, \text{mm} $

Given, L = 5 mm, B = 2.5 mm

We know, least count of a screw gauge,

$L.C. = {\text{Pitch length} \over \text{No. of division on circular scale}}$

$ \Rightarrow L.C. = {{0.75} \over {15}} = 0.05\,mm$

We know, $A = LB$

By taking $ln$ on both sides,

$ \Rightarrow \ln A = \ln L + \ln B$

by differentiating both sides,

$ \Rightarrow {{dA} \over A} = {{dL} \over L} + {{dB} \over B}$

Here, $dL = dB = 0.05\,mm$ (L.C.)

So fractional error,

${{dA} \over A} = {{0.05} \over 5} + {{0.05} \over {2.5}}$

$ = {1 \over {100}} + {2 \over {100}} = {3 \over {100}} = {x \over {100}}$ (given)

Hence, $x = 3$.

The quantity $ Q $ is expressed as $ Q = X^{-2}Y^{+\frac{3}{2}}Z^{-\frac{2}{5}} $. The variables $ X $, $ Y $, and $ Z $ are independent parameters with fractional errors of 0.1, 0.2, and 0.5, respectively. To determine the maximum fractional error in $ Q $, we can use the following method:

$ \text{Fractional error in } Q = \left| -2 \right| \frac{\Delta X}{X} + \left| \frac{3}{2} \right| \frac{\Delta Y}{Y} + \left| -\frac{2}{5} \right| \frac{\Delta Z}{Z} $

Substituting the fractional errors into the equation:

$ = 2 \times 0.1 + \frac{3}{2} \times 0.2 + \frac{2}{5} \times 0.5 $

Calculating each term gives:

$ = 0.2 + 0.3 + 0.2 $

Summing these values results in:

$ = 0.7 $

Thus, the maximum fractional error in $ Q $ is 0.7.

Here are the dimensional formulas:

Coefficient of viscosity, μ

$[\mu]=\frac{\text{pressure}\times\text{time}}{} =\frac{[M\,L^{-1}T^{-2}]\,[T]}{} =[M\,L^{-1}T^{-1}]\quad\longrightarrow\text{(IV)}$

Intensity of a wave, I

$[I]=\frac{\text{power}}{\text{area}} =\frac{[M\,L^2T^{-3}]}{[L^2]} =[M\,L^0T^{-3}]\quad\longrightarrow\text{(I)}$

Pressure gradient, dP/dx

$\left[\frac{dP}{dx}\right] =\frac{[M\,L^{-1}T^{-2}]}{[L]} =[M\,L^{-2}T^{-2}]\quad\longrightarrow\text{(II)}$

Compressibility, κ

$[\kappa]=\frac1{[\text{pressure}]} =[M^{-1}L\,T^2]\quad\longrightarrow\text{(III)}$

Matching gives

(A)–(IV), (B)–(I), (C)–(II), (D)–(III), i.e. Option C.

The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ can be understood as follows:

Vacuum permeability ($ \mu_0 $) and vacuum permittivity ($ \epsilon_0 $) relate to the properties of inductance and capacitance, respectively, in a vacuum. We start by considering the formulas for inductance ($ L $) and capacitance ($ C $):

Inductance ($ L $) is given by:

$ L = \frac{\mu_0 \cdot \text{Number of turns} \cdot \text{Area}}{\text{Length}} $

Capacitance ($ C $) is given by:

$ C = \frac{\text{Area} \cdot \epsilon_0}{\text{Distance}} $

From these, the ratio $\frac{L}{C}$ can be expressed as $\frac{\mu_0}{\epsilon_0}$:

$ \frac{L}{C} \propto \frac{\mu_0}{\epsilon_0} $

Taking the square root of this ratio, we have:

$ \sqrt{\frac{\mu_0}{\epsilon_0}} \propto \sqrt{\frac{L}{C}} $

This simplifies further to considering the relationship between time constant ($\tau$) and resistance ($R$):

$ \frac{L}{C} = \frac{\tau R}{(\tau / R)} = R^2 $

Thus, by taking the square root:

$ \sqrt{\frac{\mu_0}{\epsilon_0}} = R $

In conclusion, the dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equivalent to the dimension of resistance, $ R $.

The expression $\frac{1}{\mu_0 \epsilon_0}$ is related to the speed of light $ c $, given by the equation:

$ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} $

Therefore, we have:

$ \frac{1}{\mu_0 \epsilon_0} = c^2 $

The speed of light $ c $ has the dimensions of $\text{L T}^{-1}$, where $\text{L}$ is the dimension of length and $\text{T}$ is the dimension of time. Thus, when squared, the dimensions become:

$ c^2 = (\text{L T}^{-1})^2 = \text{L}^2 \text{T}^{-2} $

Hence, the dimensions of $\frac{1}{\mu_0 \epsilon_0}$ are $\text{L}^2 \text{T}^{-2}$.

Let's look at the units (dimensions) of the given quantities:

Charge $ Q $ has units of ampere × time $ (A \cdot T) $

Current $ I $ has units of ampere $ (A) $

Permeability of vacuum $ \mu_0 $ has units of mass × length × time$^{-2} $ × ampere$^{-2} $ $ (MLT^{-2}A^{-2}) $

Momentum has units of mass × length × time$^{-1} $ $ (MLT^{-1}) $

We want to combine $ Q $, $ \mu_0 $, and $ I $ so that the result has the same units as momentum $ (MLT^{-1}) $.

Suppose the answer is $ Q^x\, \mu_0^y\, I^z $. Let's write out the units for this:

$ Q^x\, \mu_0^y\, I^z = (A T)^x \, (MLT^{-2}A^{-2})^y \, (A)^z $

Combine the powers of each unit:

Mass (M): exponent = $ y $

Length (L): exponent = $ y $

Time (T): exponent = $ x - 2y $

Ampere (A): exponent = $ x - 2y + z $

We want these to match the units of momentum, $ M^1L^1T^{-1} $:

So we set up:

$ y = 1 $ (so mass and length exponents match momentum)

$ x - 2y = -1 $ (time exponent)

$ x - 2y + z = 0 $ (ampere exponent cancels out)

Solve these equations:

First, $ y = 1 $.

Second, $ x - 2(1) = -1 \implies x = 1 $.

Third, $ x - 2(1) + z = 0 \implies 1 - 2 + z = 0 \implies z = 1 $.

The required combination is $ x = 1 $, $ y = 1 $, $ z = 1 $: So, the quantity with the dimension of momentum is $ Q \mu_0 I $.