Units & Measurements — Page 15 | JEE Physics

Q141

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of

\left[\mathrm{M}^{\mathrm{P}} \mathrm{L}^{\mathrm{Q}} \mathrm{T}^R A^{\mathrm{S}}\right]

. The value of P and Q are :

A

-1,0

B

0,-1

C

-1,1

D

1,-1

Correct Answer

Option B

Solution

Electric dipole moment $(\overrightarrow{\mathrm{P}})=\mathrm{q} \times 2 \ell$ Magnetic dipole moment $(\overrightarrow{\mathrm{M}})=\mathrm{IA}$

\left[\frac{\mathrm{P}}{\mathrm{M}}\right]=\left[\frac{\mathrm{LTA}}{\mathrm{~L}^2 \mathrm{~A}}\right]=\mathrm{L}^{-1} \mathrm{~T}=\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^1 \mathrm{~A}^0

After compering values of $\mathrm{P} \& \mathrm{Q}$ are $0,-1$ Correct Answer : Option 4

Q142

In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division

=49 \mathrm{~MSD} ; 20

divisions on main scale in each

\mathrm{cm}

For mark on paper

\text{ MSR }=8.45 \mathrm{~cm}, \mathrm{VC}=26

For mark on paper seen through slab

\mathrm{MSR}=7.12 \mathrm{~cm}, \mathrm{VC}=41

For powder particle on the top surface of the glass slab

\text{ MSR }=4.05 \mathrm{~cm}, \mathrm{VC}=1

(MSR

=

Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :

A 1.52

B 1.35

C 1.24

D 1.42

Correct Answer

Option D

Solution

To find the refractive index of the glass slab, we first need to calculate the actual readings using the Main Scale Reading (MSR) and Vernier Coincidence (VC), and understand how to translate these values into a measurement of the refractive index.

We will start by calculating the Least Count (LC) of the vernier calipers.

The least count (LC) of the traveling microscope is given by the formula: $\text{LC} = \text{MSD} - \text{VSD}$ Where MSD is the value of one main scale division and VSD is the value of one vernier scale division in terms of the main scale.

We are told 50 vernier scale divisions equal 49 main scale divisions (MSD), so 1 VSD is $\dfrac{49}{50}$ of an MSD.

Given that 20 divisions on the main scale represent 1 cm, each main scale division (MSD) represents: $1 \, \text{MSD} = \dfrac{1 \, \text{cm}}{20} = 0.05 \, \text{cm}$ Therefore, the least count (LC) of the microscope is: $\text{LC} = 0.05 \, \text{cm} - \left( \dfrac{49}{50} \times 0.05 \, \text{cm} \right) = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm}$ Now, using the least count to find the total reading (TR) from both supplied observations: For the mark on paper, the total reading (TR) is: $\text{TR} = \text{MSR} + (\text{VC} \times \text{LC})$ $\text{TR}_{\text{paper}} = 8.45 \, \text{cm} + (26 \times 0.001 \, \text{cm}) = 8.45 \, \text{cm} + 0.026 \, \text{cm} = 8.476 \, \text{cm}$ For the mark on the paper seen through the slab, the total reading is: $\text{TR}_{\text{seen through slab}} = 7.12 \, \text{cm} + (41 \times 0.001 \, \text{cm}) = 7.12 \, \text{cm} + 0.041 \, \text{cm} = 7.161 \, \text{cm}$ For the powder particle on the top surface of the glass slab: $\text{TR}_{\text{top surface}} = 4.05 \, \text{cm} + (1 \times 0.001 \, \text{cm}) = 4.05 \, \text{cm} + 0.001 \, \text{cm} = 4.051 \, \text{cm}$ The real depth (RD) observed directly is the difference between the first and third observations (mark on paper and powder particle on top surface): $\text{RD} = \text{TR}_{\text{paper}} - \text{TR}_{\text{top surface}} = 8.476 \, \text{cm} - 4.051 \, \text{cm} = 4.425 \, \text{cm}$ The apparent depth (AD) when viewed through the slab is the difference between the second and third observations: $\text{AD} = \text{TR}_{\text{seen through slab}} - \text{TR}_{\text{top surface}} = 7.161 \, \text{cm} - 4.051 \, \text{cm} = 3.11 \, \text{cm}$ The refractive index ( $n$ ) of the glass slab can be found using the formula: $n = \dfrac{\text{Real Depth (RD)}}{\text{Apparent Depth (AD)}} = \dfrac{4.425 \, \text{cm}}{3.11 \, \text{cm}}$ Calculating this gives: $n = \dfrac{4.425}{3.11} \approx 1.42$ Therefore, the refractive index of the glass slab is approximately 1.42, which corresponds to Option D.

Q143

To find the spring constant

(k)

of a spring experimentally, a student commits

2 \%

positive error in the measurement of time and

1 \%

negative error in measurement of mass. The percentage error in determining value of

k

is :

A 5%

B 3%

C 1%

D 4%

Correct Answer

Option A

Solution

To determine the spring constant

k

of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

T = 2 \pi \sqrt{\frac{m}{k}}

Here,

T

is the period of oscillation,

m

is the mass, and

k

is the spring constant. Rearranging the formula to solve for

k

, we get:

k = \frac{4 \pi^2 m}{T^2}

To find the error in

k

, we have to consider the errors in both the measurements of

T

and

m

. Let's denote the percentage errors as follows:

\Delta T / T \cdot 100\% = 2\%

(positive error)

\Delta m / m \cdot 100\% = -1\%

(negative error) According to the rules of error propagation, the relative error in

k

can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect

k

. Since

T

is squared in the denominator and

m

is linear in the numerator, the calculation is as follows:

\frac{\Delta k}{k} = \left| -2 \cdot \frac{\Delta T}{T} \right| + \left| 1 \cdot \frac{\Delta m}{m} \right|

Substituting the percentage errors:

\frac{\Delta k}{k} = \left| -2 \cdot 0.02 \right| + \left| 1 \cdot (-0.01) \right|

\frac{\Delta k}{k} = 0.04 + 0.01

\frac{\Delta k}{k} = 0.05

Thus, the percentage error in determining the value of

k

is:

\frac{\Delta k}{k} \cdot 100\% = 5\%

The correct answer is Option A: 5%

Q144

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is

1 \mathrm{~mm}

and circular scale reading is equal to 42 divisions. Pitch of screw gauge is

1 \mathrm{~mm}

and it has 100 divisions on circular scale. The diameter of the wire is

\dfrac{x}{50} \mathrm{~mm}

. The value of

x

is :

A 42

B 71

C 21

D 142

Correct Answer

Option B

Solution

To determine the diameter of the wire using a screw gauge, we employ the formula: $\text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC})$ where: MSR (Main Scale Reading) is the value read directly from the main scale in mm.

CSR (Circular Scale Reading) is the number of divisions observed on the circular scale.

LC (Least Count) is the value of one division on the circular scale, calculated as $\dfrac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$ .

Given: Main Scale Reading (MSR) = $1 \mathrm{~mm}$ Circular Scale Reading (CSR) = 42 divisions Pitch of screw gauge = $1 \mathrm{~mm}$ Number of divisions on circular scale = 100 First, we find the Least Count (LC): $LC = \dfrac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \dfrac{1 \mathrm{~mm}}{100} = 0.01 \mathrm{~mm}$ Then we calculate the total measurement of the diameter of the wire: $\text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 1 \mathrm{~mm} + (42 \times 0.01 \mathrm{~mm}) = 1 \mathrm{~mm} + 0.42 \mathrm{~mm} = 1.42 \mathrm{~mm}$ Given that the diameter of the wire is also represented as

\frac{x}{50} \mathrm{~mm}

, we can equate this to our found total reading: $1.42 \mathrm{~mm} = \dfrac{x}{50} \mathrm{~mm}$ Solving for

x

: $1.42 = \dfrac{x}{50}$ $x = 1.42 \times 50$ $x = 71$ Therefore, the value of

x

is 71, which corresponds to Option B: 71.

Q145

Match with .

List - I		List - II
(A)	Young’s Modulus	(I)	M L-1 T-1
(B)	Torque	(II)	M L-1 T-2
(C)	Coefficient of Viscosity	(III)	M-1 L3 T-2
(D)	Gravitational Constant	(IV)	M L2 T-2

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

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

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

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

Correct Answer

Option D

Solution

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 $\dfrac{u}{y}$ = 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] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}

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

Hence, option D is correct.

Q146

Match List I with List II. .tg .tg List I List II A. Torque I. Nms

^{ - 1}

B. Stress II. J kg

^{ - 1}

C. Latent Heat III. Nm D. Power IV. Nm

^{ - 2}

Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

Torque $\rightarrow \mathrm{Nm}$ Stress $\rightarrow N / \mathrm{m}^{2}$ Latent heat $\rightarrow \mathrm{J} / \mathrm{kg}$ Power $\rightarrow \mathrm{N} \mathrm{m} / \mathrm{s}$ A-III, B-IV, C-II, D-I

Q147

If

B

is magnetic field and

\mu_0

is permeability of free space, then the dimensions of

\left(B / \mu_0\right)

is

A

\mathrm{MT}^{-2} \mathrm{~A}^{-1}

B

\mathrm{L}^{-1} \mathrm{~A}

C

\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}

D

\mathrm{LT}^{-2} \mathrm{~A}^{-1}

Correct Answer

Option B

Solution

To determine the dimensions of $\left(\dfrac{B}{\mu_0}\right)$ , we start with the formula for the magnetic field $B$ inside a solenoid: $B = \mu_0 n I$ where $n = \dfrac{N}{L}$ 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: $\dfrac{B}{\mu_0} = nI = \dfrac{N}{L}I$ From the above, the dimensions of $\left(\dfrac{B}{\mu_0}\right)$ are: $\left[\dfrac{B}{\mu_0}\right] = \left[L^{-1} A\right]$ Thus, the dimensional formula of $\left(\dfrac{B}{\mu_0}\right)$ corresponds to $L^{-1} \text{A}$ .

Q148

Given below are two statements : Statement I: In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

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.

Q149

The maximum percentage error in the measurment of density of a wire is [Given, mass of wire

=(0.60 \pm 0.003) \mathrm{g}

radius of wire

=(0.50 \pm 0.01) \mathrm{cm}

length of wire

=(10.00 \pm 0.05) \mathrm{cm}]

A 7

B 8

C 5

D 4

Correct Answer

Option C

Solution

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.

Q150

Match with .

List - I		List - II
(A)	Angular Impulse	(I)	M0 L2 T-2
(B)	Latent Heat	(II)	M L2 T-3 A-1
(C)	Electrical resistivity	(III)	M L2 T-1
(D)	Electromotive force	(IV)	M L3 T-3 A-2

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

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

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

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

Correct Answer

Option C

Solution

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)