Units & Measurements — Page 16 | JEE Physics

Q151

The position of a particle moving on

x

-axis is given by

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

, where

t

is time. The dimension of

\dfrac{A B C}{D}

is

A

\mathrm{L}

B

\mathrm{L}^3 \mathrm{~T}^{-2}

C

\mathrm{L}^2 \mathrm{~T}^{-2}

D

\mathrm{L}^2

Correct Answer

Option C

Solution

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 $\dfrac{A B C}{D}$ is:

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

The correct answer is Option C.

Q152

The energy of a system is given as

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

, where t is the time and

\beta=0.3 \mathrm{~s}^{-1}

. The errors in the measurement of

\alpha

and

t

are

1.2 \%

and

1.6 \%

, respectively. At

t=5 \mathrm{~s}

, maximum percentage error in the energy is :

A

6 \%

B

11.6 \%

C

4 \%

D

8.4 \%

Correct Answer

Option A

Solution

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: $\dfrac{dE}{E} = 3\,\dfrac{d\alpha}{\alpha} - \beta\, dt.$ For maximum error estimation, we consider the absolute values and sum the contributions: $\left|\dfrac{\Delta E}{E}\right| \approx 3\,\left|\dfrac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|.$ 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: $\left|\dfrac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08).$ 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: $\left|\dfrac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06,$ which is $6\%$ .

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is: $\boxed{6\%}$ Answer: Option A (6%).

Q153

For an experimental expression

y=\dfrac{32.3 \times 1125}{27.4}

, where all the digits are significant. Then to report the value of

y

we should write

A

y=1326.186

B

y=1326.2

C

y=1326.19

D

y=1330

Correct Answer

Option D

Solution

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

Q154

A quantity Q is formulated as

X^{-2}Y^{+\dfrac{3}{2}}Z^{-\dfrac{2}{5}}

. X, Y, and Z are independent parameters which have fractional errors of 0.1, 0.2, and 0.5, respectively in measurement. The maximum fractional error of Q is

A 0.6

B 0.8

C 0.7

D 0.1

Correct Answer

Option C

Solution

The quantity $Q$ is expressed as $Q = X^{-2}Y^{+\dfrac{3}{2}}Z^{-\dfrac{2}{5}}$ .

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: $\text{Fractional error in } Q = \left| -2 \right| \dfrac{\Delta X}{X} + \left| \dfrac{3}{2} \right| \dfrac{\Delta Y}{Y} + \left| -\dfrac{2}{5} \right| \dfrac{\Delta Z}{Z}$ Substituting the fractional errors into the equation: $= 2 \times 0.1 + \dfrac{3}{2} \times 0.2 + \dfrac{2}{5} \times 0.5$ 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.

Q155

The dimension of

\sqrt{\dfrac{\mu_0}{\epsilon_0}}

is equal to that of: (

\mu_0

= Vacuum permeability and

\epsilon_0

= Vacuum permittivity)

A Voltage

B Inductance

C Resistance

D Capacitance

Correct Answer

Option C

Solution

The dimension of $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ 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: $L = \dfrac{\mu_0 \cdot \text{Number of turns} \cdot \text{Area}}{\text{Length}}$ Capacitance ( $C$ ) is given by: $C = \dfrac{\text{Area} \cdot \epsilon_0}{\text{Distance}}$ From these, the ratio $\dfrac{L}{C}$ can be expressed as $\dfrac{\mu_0}{\epsilon_0}$ : $\dfrac{L}{C} \propto \dfrac{\mu_0}{\epsilon_0}$ Taking the square root of this ratio, we have: $\sqrt{\dfrac{\mu_0}{\epsilon_0}} \propto \sqrt{\dfrac{L}{C}}$ This simplifies further to considering the relationship between time constant ( $\tau$ ) and resistance ( $R$ ): $\dfrac{L}{C} = \dfrac{\tau R}{(\tau / R)} = R^2$ Thus, by taking the square root: $\sqrt{\dfrac{\mu_0}{\epsilon_0}} = R$ In conclusion, the dimension of $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ is equivalent to the dimension of resistance, $R$ .

Q156

If

\mu_0

and

\epsilon_0

are the permeability and permittivity of free space, respectively, then the dimension of

\left(\dfrac{1}{\mu_0 \epsilon_0}\right)

is :

A

\mathrm{T}^2 / \mathrm{L}

B

\mathrm{L}^2 / \mathrm{T}^2

C

\mathrm{T}^2 / \mathrm{L}^2

D

\mathrm{L} / \mathrm{T}^2

Correct Answer

Option B

Solution

The expression $\dfrac{1}{\mu_0 \epsilon_0}$ is related to the speed of light $c$ , given by the equation: $c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$ Therefore, we have: $\dfrac{1}{\mu_0 \epsilon_0} = c^2$ 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 $\dfrac{1}{\mu_0 \epsilon_0}$ are $\text{L}^2 \text{T}^{-2}$ .

Q157

Given a charge q , current I and permeability of vacuum

\mu_{\mathrm{o}^*}

. Which of the following quantity has the dimension of momentum ?

A

{ }

q I / \mu_o

B

q^2 \mu_o I

C

q \mu_0 / I

D

\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}

Correct Answer

Option D

Solution

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 $Q \mu_0 I$ .

Q158

Match with .

List - I		List - II
(A)	Mass density	(I)	[ML2T−3]
(B)	Impulse	(II)	[MLT−1]
(C)	Power	(III)	[ML2T0]
(D)	Moment of inertia	(IV)	[ML−3T0]

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

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

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

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

Correct Answer

Option D

Solution

(A) Mass density: Mass density is defined as mass per unit volume. The dimension of mass is

[M]

. The dimension of volume is

[L^3]

. Therefore, the dimension of mass density is

\frac{[M]}{[L^3]} = [ML^{-3}T^{0}]

.

Matches with (IV).

(B) Impulse: Impulse is defined as the change in momentum, or force multiplied by time.

Impulse = Force x Time The dimension of force is

[MLT^{-2}]

. The dimension of time is

[T]

. Therefore, the dimension of impulse is

[MLT^{-2}] \cdot [T] = [MLT^{-1}]

.

Matches with (II).

(C) Power: Power is defined as the rate of doing work or energy per unit time.

Power = Work / Time The dimension of work (or energy) is

[ML^2T^{-2}]

. The dimension of time is

[T]

. Therefore, the dimension of power is

\frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]

. Matches with (I). (D) Moment of inertia: Moment of inertia is defined as

I = mr^2

, where m is mass and r is the distance from the axis of rotation. The dimension of mass is

[M]

. The dimension of distance squared is

[L^2]

. Therefore, the dimension of moment of inertia is

[ML^2T^{0}]

.

Matches with (III).

So, the correct matches are: (A) - (IV) (B) - (II) (C) - (I) (D) - (III) Therefore, the correct option is D.

Q159

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm ) :

A 0.002

B 0.0025

C 0.0005

D 0.001

Correct Answer

Option A

Solution

Determine one main scale division (msd): $300 \, \text{msd} = 15 \, \text{cm}$ Therefore, $1 \, \text{msd} = \dfrac{15}{300} \, \text{cm} = 0.05 \, \text{cm}$ Determine one vernier scale division (vsd): $25 \, \text{vsd} = 24 \, \text{msd}$ Therefore, $1 \, \text{vsd} = \dfrac{24}{25} \, \text{msd}$ Calculate the least count (LC): The least count is given by the difference between one main scale division and one vernier scale division: $\text{LC} = 1 \, \text{msd} - 1 \, \text{vsd}$ Substitute the expression for 1 vsd: $\text{LC} = 1 \, \text{msd} - \dfrac{24}{25} \times 1 \, \text{msd}$ This simplifies to: $\text{LC} = \dfrac{1}{25} \times 1 \, \text{msd}$ Calculate the LC in cm: Substituting the value of 1 msd: $\text{LC} = \dfrac{1}{25} \times 0.05 \, \text{cm} = 0.002 \, \text{cm}$ Hence, the least count of the traveling microscope is 0.002 cm.

Q160

In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of

M^P L^Q T^R A^S

, where value of '

Q

' and '

R

' are

A

(3,-5)

B

(-2,1)

C

(-2,2)

D

(1,-1)

Correct Answer

Option D

Solution

To solve this problem, we need to find the dimensions of the electric flux and the magnetic flux, and then take their ratio.

Determine the dimension of electric flux (Φₑ): Electric flux is defined as

\Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A}.

The electric field $\mathbf{E}$ has dimensions:

[E] = \frac{\text{Force}}{\text{Charge}}.

Since force has dimensions

M L T^{-2},

and charge (in SI) has dimensions

A\,T,

we get:

[E] = \frac{M L T^{-2}}{A\,T} = M L T^{-3} A^{-1}.

The area element

dA

has dimensions

L^2.

Thus, the electric flux has dimensions:

[\Phi_E] = [E][A] = M L T^{-3} A^{-1} \times L^2 = M L^3 T^{-3} A^{-1}.

Determine the dimension of magnetic flux (Φ_B): Magnetic flux is defined as

\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A}.

The magnetic field $\mathbf{B}$ has dimensions determined via the Lorentz force law (using

F = q\,v\,B

), giving:

[B] = \frac{F}{q\,v} = \frac{M L T^{-2}}{(A\,T)(L T^{-1})} = M T^{-2} A^{-1}.

Again, the area element has dimensions

L^2.

So the magnetic flux has dimensions:

[\Phi_B] = [B][A] = (M T^{-2} A^{-1})(L^2) = M L^2 T^{-2} A^{-1}.

Find the ratio of electric flux to magnetic flux: Taking the ratio:

\frac{\Phi_E}{\Phi_B} = \frac{M L^3 T^{-3} A^{-1}}{M L^2 T^{-2} A^{-1}}.

Canceling like terms: Mass $(M)$ cancels. Current $(A^{-1})$ cancels. For length,

L^{3-2} = L^1.

For time,

T^{-3 -(-2)} = T^{-1}.

Therefore, the ratio has dimensions:

L^{1}T^{-1}.

This means in the expression

M^P L^Q T^R A^S,

the exponents for length and time are

Q = 1

and

R = -1.

Looking at the options provided, this corresponds to Option D:

(1,-1).

Thus, the answer is Option D.