Units & Measurements — Page 13 | JEE Physics

Q121

A physical quantity

Q

is found to depend on quantities

a, b, c

by the relation

Q=\dfrac{a^4 b^3}{c^2}

. The percentage error in

a, b

and

c

are

3 \%, 4 \%

and

5 \%

respectively. Then, the percentage error in

Q

is :

A 43%

B 34%

C 66%

D 14%

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} \\ & \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{b}}{\mathrm{b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} \end{aligned}

\frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right)

\begin{aligned} \% \text{ error in } \mathrm{Q} & =4 \times 3 \%+3 \times 4 \%+2 \times 5 \% \\ & =12 \%+12 \%+10 \% \\ & =34 \% \end{aligned}

Q122

The radius

(\mathrm{r})

, length

(l)

and resistance

(\mathrm{R})

of a metal wire was measured in the laboratory as

\begin{aligned} & \mathrm{r}=(0.35 \pm 0.05) ~\mathrm{cm} \\\\ & \mathrm{R}=(100 \pm 10) ~\mathrm{ohm} \\\\ & l=(15 \pm 0.2)~ \mathrm{cm} \end{aligned}

The percentage error in resistivity of the material of the wire is :

A

37.3 \%

B

25.6 \%

C

35.6 \%

D

39.9 \%

Correct Answer

Option D

Solution

To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity.

The resistivity $\rho$ of a wire is given by: $\rho = \dfrac{RA}{l}$ where:

R

is the resistance

A

is the cross-sectional area of the wire

l

is the length of the wire The cross-sectional area

A

of the wire with radius

r

is: $A = \pi r^2$ We can plug this into the equation for resistivity to get: $\rho = \dfrac{R \pi r^2}{l}$ Now, to find the percentage error in resistivity, we need to find the percentage errors in

R

,

r

, and

l

and then use the following rule for combining errors: For a given function,

f = f(x,y,z,...)

, where

x, y, z,...

are the measured quantities with possible errors, the percentage error in

f

, denoted as

(\delta f)_{\%}

, can be approximated by adding the relative percentage errors of the input quantities. If

f

has the form of a product and quotient of the measured quantities as in our case (

\rho = \frac{R \pi r^2}{l}

), the percentage error in

f

is given by: $(\delta f)_{\%} = (\delta x)_{\%} + (\delta y)_{\%} + (\delta z)_{\%} + ...$ Where

(\delta x)_{\%}

,

(\delta y)_{\%}

, and

(\delta z)_{\%}

are the percentage errors in each measured quantity

x, y, z, ...

respectively. For our case: The percentage error in radius

(\delta r)_{\%}

is given by the error in

r

divided by the average radius and then multiplied by 100: $(\delta r)_{\%} = \left(\dfrac{0.05}{0.35}\right) \times 100$ The percentage error in resistance

(\delta R)_{\%}

is: $(\delta R)_{\%} = \left(\dfrac{10}{100}\right) \times 100$ The percentage error in length

(\delta l)_{\%}

is: $(\delta l)_{\%} = \left(\dfrac{0.2}{15}\right) \times 100$ Now let's calculate each: $(\delta r)_{\%} = \left(\dfrac{0.05}{0.35}\right) \times 100 \approx 14.29\%$ $(\delta R)_{\%} = \left(\dfrac{10}{100}\right) \times 100 = 10\%$ $(\delta l)_{\%} = \left(\dfrac{0.2}{15}\right) \times 100 \approx 1.33\%$ However, since the area

A

is proportional to

r^2

, the percentage error in

A

will be twice the percentage error in

r

.

Thus: $(\delta A)_{\%} = 2 \times (\delta r)_{\%} = 2 \times 14.29\% \approx 28.58\%$ Finally, we add the percentage errors to find the percentage error in resistivity: $(\delta \rho)_{\%} = (\delta R)_{\%} + (\delta A)_{\%} + (\delta l)_{\%}$ $(\delta \rho)_{\%} = 10\% + 28.58\% + 1.33\% \approx 39.91\%$ This calculation gives us a value close to 39.91%, which means the correct option is closest to this value.

Thus, the best answer is: Option D $39.9\%$

Q123

Given below are two statements : Statement (I) : Planck's constant and angular momentum have same dimensions. Statement (II) : Linear momentum and moment of force have same dimensions. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Both Statement I and Statement II are false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option A

Solution

\begin{aligned} & {[\mathrm{h}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{~L}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{P}]=\mathrm{MLT}^{-1}} \\ & {[\tau]=\mathrm{ML}^2 \mathrm{~T}^{-2}} \end{aligned}

(Here

\mathrm{h}

is Planck's constant,

\mathrm{L}

is angular momentum,

\mathrm{P}

is linear momentum and $\tau$ is moment of force)

Q124

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. Reason (R) : The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. In the light of the above statements, choose the correct answer from the options given below :

A Both (A) and (R) are correct and (R) is the correct explanation of (A)

B Both (A) and (R) are correct but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Correct Answer

Option B

Solution

The provided statements (A) Assertion and (R) Reason have to be analyzed to determine their correctness and also whether Reason (R) correctly explains Assertion (A).

Assertion (A) is addressing a scenario where a positive zero error occurs in a Vernier caliper.

Positive zero error refers to the condition when the zero mark on the Vernier scale is to the right of the zero mark on the main scale when the jaws of the caliper are completely closed.

In this case, even when measuring a zero length, the Vernier caliper will show a positive reading, indicating an error.

This error is constant and will get added to actual measurements, causing the instrument to give readings that are more than the actual measurement.

So, Assertion (A) is true.

Reason (R) explains the possible causes of zero error in Vernier calipers.

Zero errors can indeed occur due to imperfect manufacturing processes, where the scales are not perfectly aligned.

They can also happen due to rough handling, for instance, if the instrument is dropped, which might cause a permanent deformation leading to a continuous zero error.

Hence, Reason (R) is true as well.

Finally, we need to analyze whether Reason (R) is the correct explanation of Assertion (A).

Although both statements are correct, the Reason (R) does not explain why a positive zero error would lead to the reading being more than the actual reading, it merely states the possible causes of zero errors.

Therefore, the correct relationship between the statements is that they are both true, but (R) does not provide the correct explanation for (A).

The correct option is : Option B Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Q125

If the percentage errors in measuring the length and the diameter of a wire are

0.1 \%

each. The percentage error in measuring its resistance will be:

A 0.144%

B 0.2%

C 0.1%

D 0.3%

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \\ & \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% \text{ and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \% \end{aligned}

Q126

A force is represented by

F=a x^2+b t^{\dfrac{1}{2}}

where

x=

distance and

t=

time. The dimensions of

b^2 / a

are:

A

\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]

B

\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]

C

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

D

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

Correct Answer

Option B

Solution

To determine the dimensions of

\frac{b^2}{a}

, let's start by identifying the dimensions of each term in the equation

F=a x^2+b t^{\frac{1}{2}}

, where

F

represents force,

x

represents distance, and

t

represents time. The dimension of force (

F

) is given by [MLT-2], where

M

is mass,

L

is length, and

T

is time. The term

ax^2

has the same dimension as force, so:

[a] = \frac{[F]}{[x]^2} = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2}

The term

bt^{\frac{1}{2}}

also has the same dimension as force, which gives:

[b] = \frac{[F]}{[t]^{\frac{1}{2}}} = \frac{MLT^{-2}}{T^{\frac{1}{2}}} = M L T^{-\frac{5}{2}}

Now, to find

\frac{b^2}{a}

, we substitute the dimensions of

b

and

a

:

\left[\frac{b^2}{a}\right] = \frac{\left(M^2 L^2 T^{-5}\right)}{\left(M L^{-1} T^{-2}\right)} = [M^1 L^3 T^{-3}]

Therefore, the dimensions of

\frac{b^2}{a}

are

\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]

, which corresponds to mass times length cubed per time cubed.

Q127

The de-Broglie wavelength associated with a particle of mass

m

and energy

E

is

h / \sqrt{2 m E}

. The dimensional formula for Planck's constant is :

A

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

B

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

C

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

D

\left[\mathrm{MLT}^{-2}\right]

Correct Answer

Option C

Solution

To determine the dimensional formula for Planck's constant, we will start by analyzing the given de-Broglie wavelength equation:

\lambda = \frac{h}{\sqrt{2 m E}}

Here, $\lambda$ is the wavelength,

h

is the Planck's constant,

m

is the mass of the particle, and

E

is the energy of the particle.

First, let's derive the dimensional formula for each term involved: 1.

Wavelength $\lambda$ has the dimensional formula of length

[L]

. 2. Mass

m

has the dimensional formula

[\text{M}]

. 3. Energy

E

has the dimensional formula of work, which is force times distance:

[E] = [F][L] = [\text{MLT}^{-2}][L] = [\text{ML}^{2}\text{T}^{-2}]

. Now, let's rewrite the equation in terms of the dimensions:

[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{ML}^{2}\text{T}^{-2}]}}

Simplifying inside the square root:

[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{M}] [\text{L}^{2}\text{T}^{-2}]}}

[L] = \frac{[h]}{\sqrt{2 [\text{M}^{2}] [\text{L}^{2}\text{T}^{-2}]}}

Since the constants like 2 do not affect the dimensional formula, we can simplify further:

[L] = \frac{[h]}{[\text{M L T}^{-1}]}

Cross multiplying to solve for the dimensional formula of

h

:

[h] = [L] [\text{M L T}^{-1}]

[h] = [\text{M L}^{2} \text{T}^{-1}]

Therefore, the dimensional formula for Planck's constant

h

is:

\left[\text{ML}^{2}\text{T}^{-1}\right]

Hence, the correct option is: Option C

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

Q128

The dimensional formula of latent heat is :

A

\left[\mathrm{M}^{\mathrm{0}} \mathrm{LT}^{-2}\right]

B

\left[\mathrm{MLT}^{-2}\right]

C

\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]

D

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

Correct Answer

Option C

Solution

To derive the dimensional formula of latent heat, we need to understand what latent heat actually refers to.

Latent heat is the amount of heat absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

The formula for latent heat ( $L$ ) is given by: $Q = mL$ where: $Q$ = Heat absorbed or released (with the dimension of energy $[\mathrm{ML}^2\mathrm{T}^{-2}]$ ), $m$ = Mass of the substance ( $[\mathrm{M}]$ ), $L$ = Latent heat.

To find the dimensions of latent heat, we rearrange the formula to solve for $L$ : $L = \dfrac{Q}{m}$ Knowing the dimensions of $Q$ (energy, which is equivalent to work done, with dimensions $[\mathrm{ML}^2\mathrm{T}^{-2}]$ ) and $m$ (mass, with dimensions $[\mathrm{M}]$ ), we can substitute these into the equation to find $L$ 's dimensions: $L = \dfrac{[\mathrm{ML}^2\mathrm{T}^{-2}]}{[\mathrm{M}]}$ This simplifies to: $L = [\mathrm{L}^2\mathrm{T}^{-2}]$ Therefore, the dimensional formula of latent heat is: $L = [\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$ So, the correct option is: Option C $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Q129

One main scale division of a vernier caliper is equal to

\mathrm{m}

units. If

\mathrm{n}^{\text{th }}

division of main scale coincides with

(n+1)^{\text{th }}

division of vernier scale, the least count of the vernier caliper is :

A

\dfrac{1}{(\mathrm{n}+1)}

B

\dfrac{m}{(n+1)}

C

\dfrac{n}{(n+1)}

D

\dfrac{\mathrm{m}}{\mathrm{n}(\mathrm{n}+1)}

Correct Answer

Option B

Solution

The least count of a vernier caliper is defined as the smallest distance that it can measure and is calculated by the difference in length between one main scale division and one vernier scale division.

It can be represented as:

\text{Least Count} = \text{Main scale division} - \text{Vernier scale division}

Given that one main scale division is equal to

\mathrm{m}

units and the

\mathrm{n}^{\text{th }}

division of main scale coincides with the

(n+1)^{\text{th }}

division of the vernier scale, this means that

n

divisions on the main scale is equal to

(n+1)

divisions on the vernier scale. Since one main scale division is

\mathrm{m}

units,

n

divisions on the main scale would be

n \times \mathrm{m}

units. If

n

divisions on the main scale are equal to

(n+1)

divisions on the vernier scale, we can determine the length of one vernier scale division as

\text{Length of one vernier scale division} = \frac{n \times \mathrm{m}}{n+1}

Thus, the least count, which is the difference between one main scale division and one vernier scale division, is:

\text{Least Count} = \mathrm{m} - \frac{n \times \mathrm{m}}{n+1} = \mathrm{m} \left(1 - \frac{n}{n+1} \right) = \mathrm{m} \left( \frac{n+1-n}{n+1} \right) = \mathrm{m} \left( \frac{1}{n+1} \right)

This simplifies to:

\text{Least Count} = \frac{\mathrm{m}}{\mathrm{n + 1}}

Therefore, the correct option is: Option B:

\frac{\mathrm{m}}{\mathrm{n+1}}

Q130

The equation of stationary wave is :

y=2 \mathrm{a} \sin \left(\dfrac{2 \pi \mathrm{nt}}{\lambda}\right) \cos \left(\dfrac{2 \pi x}{\lambda}\right) \text{. }

Which of the following is NOT correct :

A The dimensions of

\mathrm{nt}

is [L]

B The dimensions of

n

is

[\mathrm{LT}^{-1}]

C The dimensions of

x

is [L]

D The dimensions of

n / \lambda

is [T]

Correct Answer

Option D

Solution

To determine which of the options is NOT correct, we need to analyze the dimensional consistency of each term in the given equation of the stationary wave:

y = 2a \sin \left(\frac{2\pi nt}{\lambda}\right) \cos \left(\frac{2\pi x}{\lambda}\right).

Let's break down the dimensions for each relevant term: 1. Analyzing

\mathrm{nt}

: The argument of the sine function

\frac{2\pi nt}{\lambda}

must be dimensionless. Therefore, the dimensions of

\mathrm{nt}

should be the same as the dimensions of $\lambda$ (wavelength), which is [L]. Thus, the dimensions of

nt

should be [L]. Hence, Option A is correct. 2. Analyzing

n

: From the above analysis, since

\mathrm{nt}

has the dimension [L] and

t

(time) has the dimension [T], it follows that:

n = \frac{\text{Dimension of } nt}{\text{Dimension of } t} = \frac{[L]}{[T]} = [\mathrm{LT}^{-1}].

Hence, Option B is also correct. 3. Analyzing

x

: In the argument of the cosine function

\frac{2\pi x}{\lambda}

, since it must be dimensionless, the dimensions of

x

should be the same as the dimensions of $\lambda$ (wavelength), which is [L]. Hence, the dimensions of

x

should be [L]. Therefore, Option C is correct. 4. Analyzing

\frac{n}{\lambda}

: From the dimensions we have determined for

n

and $\lambda$ :

\frac{n}{\lambda} = \frac{[\mathrm{LT}^{-1}]}{[L]} = [\mathrm{T}^{-1}].

Thus, the dimensions of

\frac{n}{\lambda}

should be [T-1], not [T].

This indicates that Option D is NOT correct.

Conclusion: Option D is the correct answer since it is NOT dimensionally correct.