Units & Measurements — Page 12 | JEE Physics

Q111

In the equation

\left[X+\dfrac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T, X

is pressure,

Y

is volume,

\mathrm{R}

is universal gas constant and

T

is temperature. The physical quantity equivalent to the ratio

\dfrac{a}{b}

is:

A Impulse

B Energy

C Pressure gradient

D Coefficient of viscosity

Correct Answer

Option B

Solution

Given that,

\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T

$\therefore$ $X$ and $\dfrac{a}{Y^2}$ have the same dimensions and $Y$ and $b$ have the same dimensions, let's analyze the dimensions of $\dfrac{a}{b}$ .

Since $X$ represents pressure, it has dimensions of $[M L^{-1} T^{-2}]$ .

Since $X$ and $\dfrac{a}{Y^2}$ have the same dimensions, we have:

\left[\frac{a}{Y^2}\right] = [M L^{-1} T^{-2}]

Then, the dimensions of $a$ are:

[a] = [M L^{-1} T^{-2}] [Y^2] = [M L^5 T^{-2}]

Now, since $Y$ and $b$ have the same dimensions and $Y$ represents volume, we have:

[b] = [L^3]

Now, let's find the dimensions of the ratio $\dfrac{a}{b}$ :

\frac{[a]}{[b]} = \frac{[M L^5 T^{-2}]}{[L^3]} = [M L^2 T^{-2}]

Indeed, the dimensions of $\dfrac{a}{b}$ are $[M L^2 T^{-2}]$ , which corresponds to the dimensions of energy.

Q112

A body of mass

(5 \pm 0.5) ~\mathrm{kg}

is moving with a velocity of

(20 \pm 0.4) ~\mathrm{m} / \mathrm{s}

. Its kinetic energy will be

A

(1000 \pm 140) ~\mathrm{J}

B

(500 \pm 0.14) ~\mathrm{J}

C

(1000 \pm 0.14) ~\mathrm{J}

D

(500 \pm 140) ~\mathrm{J}

Correct Answer

Option A

Solution

To find the kinetic energy of the body, we can use the formula:

KE = \frac{1}{2}mv^2

Given the mass

m = (5 \pm 0.5) \,\text{kg}

and the velocity

v = (20 \pm 0.4) \,\text{m/s}

, we can find the kinetic energy and its uncertainty by applying the rules for the propagation of errors in multiplication.

For the product of two quantities, the relative error is the sum of the relative errors of the individual quantities:

\frac{\Delta (mv^2)}{mv^2} = \frac{\Delta m}{m} + \frac{\Delta v}{v} + \frac{\Delta v}{v}

Now, substitute the given values:

\frac{\Delta (mv^2)}{mv^2} = \frac{0.5}{5} + \frac{0.4}{20} + \frac{0.4}{20}

\frac{\Delta (mv^2)}{mv^2} = 0.1 + 0.02 + 0.02

\frac{\Delta (mv^2)}{mv^2} = 0.14

Now, calculate the kinetic energy:

KE = \frac{1}{2}(5)(20)^2 = 1000 \,\text{J}

To find the uncertainty in the kinetic energy, multiply the relative error by the calculated kinetic energy:

\Delta KE = 0.14 \times 1000 = 140 \,\text{J}

So, the kinetic energy of the body is

(1000 \pm 140) \,\text{J}

.

Q113

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :

A

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

B

\mathrm{FV}^{4} \mathrm{~T}^{-6}

C

\mathrm{F}^{2} \mathrm{~V}^{-2} \mathrm{~T}^{6}

D

\mathrm{FV}^{-4} \mathrm{~T}^{-2}

Correct Answer

Option D

Solution

We know that force (F) has dimensions of

\mathrm{MLT}^{-2}

, velocity (V) has dimensions of

\mathrm{LT}^{-1}

, and time (T) has dimensions of

\mathrm{T}

. To express density ( $\rho$ ), which has dimensions of

\mathrm{ML}^{-3}

, in terms of F, V, and T, we need to find the exponents of F, V, and T. Let the dimensions of density be expressed as:

\mathrm{[M]}^a \mathrm{[L]}^b \mathrm{[T]}^c

Substituting the dimensions of F, V, and T:

\mathrm{[F]}^a \mathrm{[V]}^b \mathrm{[T]}^c = (\mathrm{MLT}^{-2})^a (\mathrm{LT}^{-1})^b (\mathrm{T})^c

Since density has dimensions of

\mathrm{ML}^{-3}

, we can set up the following equations:

a = 1

(for the mass term M)

a + b = -3

(for the length term L)

-2a - b + c = 0

(for the time term T) Solving these equations, we get:

a = 1

b = -4

c = -2

Thus, the dimensional formula of density in terms of F, V, and T is:

\mathrm{FV}^{-4} \mathrm{T}^{-2}

Q114

Given below are two statements : Statements I : Astronomical unit (Au), Parsec (Pc) and Light year (ly) are units for measuring astronomical distances. Statements II : $$\mathrm{Au} In the light of the above statements, choose the most appropriate answer from the options given below:

A Both Statements I and Statements II are incorrect.

B Both Statements I and Statements II are correct,

C Statements I is incorrect but Statements II is correct.

D Statements I is correct but Statements II is incorrect.

Correct Answer

Option D

Solution

Statement I is correct.

Astronomical unit (AU), parsec (pc), and light year (ly) are indeed units for measuring astronomical distances.

However, Statement II is incorrect.

The correct order of these units is:

\mathrm{AU} < \mathrm{ly} < \mathrm{Parsec} (\mathrm{Pc})

1 AU is the average distance from the Earth to the Sun, which is about 93 million miles or 150 million kilometers. 1 light year is the distance that light travels in one year in a vacuum, which is approximately 9.461 trillion kilometers (5.878 trillion miles). 1 parsec is equal to about 3.26 light years, or approximately 30.9 trillion kilometers (19.2 trillion miles).

Parsec (parallax of one arcsecond) is the distance at which an arc of length $1 \mathrm{Au}$ subtends an angle of one arcsecond $(1 \mathrm{Pc}=3.26 \mathrm{ly})$ .

So, the correct answer is Statement I is correct but Statement II is incorrect.

Q115

In an experiment with vernier callipers of least count

0.1 \mathrm{~mm}

, when two jaws are joined together the zero of vernier scale lies right to the zero of the main scale and 6th division of vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between

3.2 \mathrm{~cm}

and

3.3 \mathrm{~cm}

marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as

A

3 .18 \mathrm{~cm}

B

3.22 \mathrm{~cm}

C

3.26 \mathrm{~cm}

D

3.25 \mathrm{~cm}

Correct Answer

Option A

Solution

In this experiment, the vernier callipers have a least count of 0.1 mm, which is equal to 0.01 cm.

First, let's calculate the zero error since the zero of the vernier scale does not coincide with the zero of the main scale when the two jaws are joined together.

The zero error can be found using the formula: Zero Error = (Number of divisions coinciding) × Least Count Zero Error = 6 × 0.01 cm = 0.06 cm Since the zero of the vernier scale lies to the right of the main scale zero, the zero error is positive.

Now, let's find the main scale reading (MSR) when measuring the diameter of the spherical bob.

The MSR is the value just before the zero of the vernier scale, which is 3.2 cm in this case.

Next, let's find the vernier scale reading (VSR) when measuring the diameter.

The VSR is the product of the coinciding division number and the least count: VSR = (Number of divisions coinciding) × Least Count VSR = 4 × 0.01 cm = 0.04 cm Now, we can find the total reading, which is the sum of the MSR and VSR: Total Reading = MSR + VSR = 3.2 cm + 0.04 cm = 3.24 cm Since there is a positive zero error, we need to subtract it from the total reading to get the corrected diameter: Corrected Diameter = Total Reading - Zero Error = 3.24 cm - 0.06 cm = 3.18 cm So, the diameter of the bob is measured as 3.18 cm.

Q116

A physical quantity P is given as

P = {{{a^2}{b^3}} \over {c\sqrt d }}

The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be

A 12%

B 13%

C 16%

D 14%

Correct Answer

Option B

Solution

The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.

Given the physical quantity P as

P = \frac{a^2b^3}{c\sqrt{d}}

The percentage error in P, denoted as $\Delta P/P$ , is given by:

\frac{\Delta P}{P} = 2 \left(\frac{\Delta a}{a}\right) + 3 \left(\frac{\Delta b}{b}\right) + \left(\frac{\Delta c}{c}\right) + \frac{1}{2} \left(\frac{\Delta d}{d}\right)

Substituting the given percentage errors for a, b, c, and d:

\frac{\Delta P}{P} = 2(0.01) + 3(0.02) + 0.03 + \frac{1}{2}(0.04) = 0.02 + 0.06 + 0.03 + 0.02 = 0.13

Therefore, the percentage error in P is 13%.

Q117

The resistance

R=\dfrac{V}{I}

where

\mathrm{V}=(200 \pm 5) \mathrm{V}

and

I=(20 \pm 0.2) \mathrm{A}

, the percentage error in the measurement of

\mathrm{R}

is :

A 5.5%

B 3%

C 7%

D 3.5%

Correct Answer

Option D

Solution

\mathrm{R}=\frac{\mathrm{V}}{1}

According to error analysis

\begin{aligned} & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{\mathrm{dV}}{\mathrm{V}}+\frac{\mathrm{dI}}{\mathrm{I}} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{5}{200}+\frac{0.2}{20} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{7}{200} \\ & \% \text{ error } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=\frac{7}{200} \times 100=3.5 \% \end{aligned}

Q118

Two resistances are given as

\mathrm{R}_{1}=(10 \pm 0.5) \Omega

and

\mathrm{R}_{2}=(15 \pm 0.5) \Omega

. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

A 2.33

B 5.33

C 4.33

D 6.33

Correct Answer

Option C

Solution

In the problem, we are given two resistances, $R_1$ and $R_2$ , each with a certain measurement error, $\Delta R_1$ and $\Delta R_2$ .

These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.

The formula for the equivalent resistance $R$ of two resistors $R_1$ and $R_2$ in parallel is:

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

We want to find the percentage error in $R$ , which is given by $(\Delta R / R) \times 100\%$ .

In order to find $\Delta R / R$ , we differentiate both sides of the above equation with respect to $R$ , $R_1$ , and $R_2$ .

This gives us:

\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}

We can then solve this equation for $\Delta R / R$ :

\frac{\Delta R}{R} = \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)R

Substituting the given values, $R_1 = 10 \, \Omega$ , $R_2 = 15 \, \Omega$ , $\Delta R_1 = \Delta R_2 = 0.5 \, \Omega$ , and $R = R_1R_2/(R_1+R_2) = 6 \, \Omega$ , we get:

\frac{\Delta R}{R} = \left(\frac{0.5}{100} + \frac{0.5}{225}\right) \times 6 = \frac{13}{300}

Finally, to convert this to a percentage, we multiply by 100, giving:

\frac{\Delta R}{R} \times 100 = \frac{13}{3} = 4.33 \%

This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\%$ .

Q119

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

A

\dfrac{5}{11}

B

\dfrac{10}{11}

C

\dfrac{50}{11}

D

\dfrac{1}{2}

Correct Answer

Option A

Solution

The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is determined by the difference in the measurement of one main scale division and one Vernier scale division.

In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale.

This means that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale.

Since each division on the main scale is 5 units, this is equal to:

10 \text{ divisions on main scale} \times 5 \text{ units per division} = 50 \text{ units}

The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions on the Vernier scale.

This means that:

1 \text{ division on the Vernier scale} = \frac{50 \text{ units}}{11}

Therefore, the least count of the Vernier caliper, which is the difference between one division on the main scale and one division on the Vernier scale, can be calculated as follows:

\text{Least count} = \text{value of one main scale division} - \text{value of one Vernier scale division}

\text{Least count} = 5 \text{ units} - \frac{50 \text{ units}}{11}

\text{Least count} = \frac{55 \text{ units}}{11} - \frac{50 \text{ units}}{11}

\text{Least count} = \frac{5 \text{ units}}{11}

Thus, the least count of the Vernier caliper is

\frac{5 \text{ units}}{11}

Therefore, the correct answer is: Option A:

\frac{5}{11}

Q120

The dimensional formula of angular impulse is :

A

\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]

B

\left[\mathrm{M} \mathrm{L}^{-2} \mathrm{~T}^{-1}\right]

C

\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]

D

\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1}\right]

Correct Answer

Option C

Solution

Angular impulse is given when a torque is applied for a certain amount of time.

The angular impulse changes the angular momentum of an object and has the same dimensions as angular momentum.

The dimensional formula for torque $\tau$ is the same as that for work, since torque is a kind of rotational work, which is given by force times distance (or in rotational terms, it can be considered as force times lever arm).

The dimensional formula for force is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$ , and when multiplied by distance $\left[\mathrm{L}\right]$ , we get: $\left[\mathrm{Torque}\right] = \left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times \left[\mathrm{L}\right] = \left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$ Now, angular impulse is torque times time, so we multiply the dimension of torque by time $\left[\mathrm{T}\right]$ : $\left[\mathrm{Angular\ Impulse}\right] = \left[\mathrm {M} \mathrm{L}^2 \mathrm{T}^{-2}\right] \times \left[\mathrm{T}\right] = \left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}\right]$ So the correct dimensional formula for angular impulse is given by Option C, which is $\left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}\right]$ .