Units & Measurements — Page 11 | NEET Physics

Q101

Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm . Keeping the significant figures in view, the density of the material of the cube can be best expressed as

X \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}

where the value of

X

is:

A 7.654

B 7.6

C 7.65

D 7.7

Correct Answer

Option B

Solution

Mass of cube: $m = 5.580 \text{ kg}$ (4 significant figures) Side of cube: $a = 9.0 \text{ cm} = 9.0 \times 10^{-2} \text{ m}$ (2 significant figures)

V = a^3 = (9.0 \times 10^{-2})^3 \text{ m}^3

V = 729 \times 10^{-6} \text{ m}^3 = 7.29 \times 10^{-4} \text{ m}^3

\rho = \frac{m}{V} = \frac{5.580}{7.29 \times 10^{-4}} \text{ kg/m}^3

\rho = 7654.3 \text{ kg/m}^3 = 7.6543 \times 10^3 \text{ kg/m}^3

According to NCERT rules, in multiplication or division, the final result should have the same number of significant figures as the quantity with the least number of significant figures.

Mass has 4 significant figures.

Side has only 2 significant figures.

So, the answer must be rounded off to 2 significant figures .

\rho = 7.6 \times 10^3 \text{ kg/m}^3

Final Answer

\boxed{X = 7.6}

The correct option is (B) 7.6

Q102

In a vernier calliper, 20 VSD coincide with 16 MSD (each division of length 1 mm ). The least count of the vernier callipers is:

A 0.2 cm

B 0.01 cm

C 0.02 cm

D 0.1 cm

Correct Answer

Option C

Solution

Least count of a vernier calliper The least count (L.C.) is the smallest measurement a vernier calliper can read.

It is given by:

\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD} \quad ...(i)

It is given that 20 vernier scale divisions (VSD) coincide with 16 main scale divisions (MSD).

Each MSD is 1 mm long.

So,

20\,\text{VSD} = 16\,\text{MSD}

Dividing both sides by 20, we get

1\,\text{VSD} = \frac{16}{20}\,\text{MSD}

Substitute this value in equation (i):

\text{L.C.} = 1\,\text{MSD} - \frac{16}{20}\,\text{MSD}

Simplifying,

\text{L.C.} = \frac{4}{20}\,\text{MSD}

Since 1 MSD = 1 mm,

\text{L.C.} = \frac{1}{5}\,\text{mm} = 0.2\,\text{mm}

In centimetre, the least count is:

\text{L.C.} = 0.02\,\text{cm}

Q103

A balloon is made of a material of surface tension

S

and its inflation outlet (from where gas is filled in it) has small area

A

. It is filled with a gas of density

\rho

and takes a spherical shape of radius

R

. When the gas is allowed to flow freely out of it, its radius

r

changes from

R

to 0 (zero) in time

T

. If the speed

v(r)

of gas coming out of the balloon depends on

r

as

r^\alpha

and

T \propto S^\alpha A^\beta \rho^\gamma R^\delta

then

A

a=-\dfrac{1}{2}, \alpha=-\dfrac{1}{2}, \beta=-1, \gamma=\dfrac{1}{2}, \delta=\dfrac{7}{2}

B

a=\dfrac{1}{2}, \alpha=\dfrac{1}{2}, \beta=-\dfrac{1}{2}, \gamma=\dfrac{1}{2}, \delta=\dfrac{7}{2}

C

a=\dfrac{1}{2}, \alpha=\dfrac{1}{2}, \beta=-1, \gamma=+1, \delta=\dfrac{3}{2}

D

a=-\dfrac{1}{2}, \alpha=-\dfrac{1}{2}, \beta=-1, \gamma=-\dfrac{1}{2}, \delta=\dfrac{5}{2}

Correct Answer

Option A

Solution

The relationship for $T$ is given by: $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$ To find constants $\alpha$ , $\beta$ , $\gamma$ , and $\delta$ , analyze the dimensional formulae: Dimension analysis for Time $T$ : $\text{M}^0 \text{L}^0 \text{T}^1 = \left(\text{M} \text{T}^{-2}\right)^\alpha \left(\text{L}^2\right)^\beta \left(\text{M} \text{L}^{-3}\right)^\gamma \text{L}^\delta$ Equate dimensions: $\text{M}^0 \text{L}^0 \text{T}^1 = \text{M}^{\alpha + \gamma} \text{L}^{2\beta - 3\gamma + \delta} \text{T}^{-2\alpha}$ From comparing dimensions: For mass (M): $\alpha + \gamma = 0$ For length (L): $2\beta - 3\gamma + \delta = 0$ For time (T): $-2\alpha = 1 \Rightarrow \alpha = -\dfrac{1}{2}$ Solving equations: Using $\alpha = -\dfrac{1}{2}$ , $\alpha + \gamma = 0 \Rightarrow -\dfrac{1}{2} + \gamma = 0 \Rightarrow \gamma = \dfrac{1}{2}$ Substituting these into the equation for length: $2\beta - 3\left(\dfrac{1}{2}\right) + \delta = 0$ Assume $\beta = -1$ to solve for $\delta$ : $2(-1) - \dfrac{3}{2} + \delta = 0 \quad \Rightarrow \quad \delta = \dfrac{7}{2}$ By this analysis, the values are $\alpha = -\dfrac{1}{2}$ , $\beta = -1$ , $\gamma = \dfrac{1}{2}$ , and $\delta = \dfrac{7}{2}$ .

Q104

A physical quantity

P

is related to four observations

a, b, c

and

d

as follows:

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

The percentage errors of measurement in

a, b, c

and

d

are

1 \%, 3 \%, 2 \%

, and

4 \%

respectively. The percentage error in the quantity

P

is

A

13 \%

B

15 \%

C

10 \%

D

2 \%

Correct Answer

Option A

Solution

The physical quantity $P$ is defined in terms of four observations $a, b, c,$ and $d$ by the formula: $P = \dfrac{a^3 b^2}{c \sqrt{d}}$ To find the percentage error in $P$ , we use the formula for maximum percentage error.

The general rule for percentage errors when a quantity is a product or a quotient is to add the percentage errors, while for powers, multiply the error by the power.

The percentage error for $P$ is calculated using: $\text{Maximum } \% \text{ error in } P = 3 \left( \dfrac{\Delta a}{a} \times 100 \right) + 2 \left( \dfrac{\Delta b}{b} \times 100 \right) + \left( \dfrac{\Delta c}{c} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta d}{d} \times 100 \right)$ Given the percentage errors: $a$ has a percentage error of $1\%$ $b$ has a percentage error of $3\%$ $c$ has a percentage error of $2\%$ $d$ has a percentage error of $4\%$ Substituting these into the formula: $= 3 \times (1) + 2 \times (3) + (2) + \dfrac{1}{2} \times (4)$ $= 3 + 6 + 2 + 2$ $= 13\%$ Thus, the percentage error in the quantity $P$ is $13\%$ .

Q105

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at

x=0.1 \mathrm{~cm}

when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is

M=5 \mathrm{~cm}

and the number of coinciding vernier division is 8 , the measured diameter after zero error correction, is

A 4.98 cm

B 5.00 cm

C 5.18 cm

D 5.08 cm

Correct Answer

Option A

Solution

To measure the diameter of a spherical object with Vernier calipers, consider the following details: Vernier Scale and Main Scale Relationship : 10 Vernier Scale Divisions (V.S.D.) are equivalent to 9 Main Scale Divisions (M.S.D.).

Least Value on Main Scale : 0.1 cm Zero Error : The zero marking on the Vernier Scale (V.S.) is at 0.1 cm when the caliper jaws are closed.

Given measurements: Main Scale Reading : $M = 5 \, \text{cm}$ Coinciding Vernier Division : 8 Calculations : Determining the Least Count : $\text{Least Count} = \text{1 MSD} - \text{1 VSD}$ $= 1 \times 0.1 \, \text{cm} - \dfrac{9}{10} \times 0.1 \, \text{cm}$ $= 0.1 \, \text{cm} - 0.09 \, \text{cm} = 0.01 \, \text{cm}$ Accounting for Zero Error : Zero Error = +0.1 cm Vernier Scale Reading Calculation : $= 8 \times 0.01 \, \text{cm} = 0.08 \, \text{cm}$ Final Diameter Measurement After Zero Error Correction : $\text{Measured Diameter} = \text{Main Scale Reading} + \text{Vernier Reading} - \text{Zero Error}$ $= 5 \, \text{cm} + 0.08 \, \text{cm} - 0.1 \, \text{cm}$ $= 4.98 \, \text{cm}$ Thus, the corrected measurement for the diameter is 4.98 cm.

Q106

In an electrical circuit, the voltage is measured as

V=(200 \pm 4)

volt and the current is measured as

I=(20 \pm 0.2)

A. The value of the resistance is:

A

(10 \pm 4.2) \Omega

B

(10 \pm 0.3) \Omega

C

(10 \pm 0.1) \Omega

D

(10 \pm 0.8) \Omega

Correct Answer

Option B

Solution

To determine the resistance

R

using Ohm's law, we use the formula:

R = \frac{V}{I}

Given:

V = (200 \pm 4)

volts

I = (20 \pm 0.2)

amperes First, calculate the nominal value of the resistance:

R = \frac{200}{20} = 10 \, \Omega

Next, we need to calculate the uncertainty in the resistance.

For division, the relative uncertainties add up.

The relative uncertainty for voltage is:

\frac{\Delta V}{V} = \frac{4}{200} = 0.02

and for current:

\frac{\Delta I}{I} = \frac{0.2}{20} = 0.01

Adding these relative uncertainties gives us the relative uncertainty for the resistance:

\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = 0.02 + 0.01 = 0.03

Now, we calculate the absolute uncertainty for the resistance:

\Delta R = R \times 0.03 = 10 \times 0.03 = 0.3 \, \Omega

Therefore, the resistance with its uncertainty is:

R = 10 \pm 0.3 \, \Omega

Thus, the correct answer is: Option B

(10 \pm 0.3) \Omega

Q107

The pitch of an error free screw gauge is

1 \mathrm{~mm}

and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads

1 \mathrm{~mm}

and

63^{\text{rd }}

division on the circular scale coincides with the reference line. The diameter of the wire is:

A

1.63 \mathrm{~cm}

B

0.163 \mathrm{~cm}

C

0.163 \mathrm{~m}

D

1.63 \mathrm{~m}

Correct Answer

Option B

Solution

To determine the diameter of the wire using the screw gauge, we need to consider both the pitch scale reading and the circular scale reading.

The formula to calculate the diameter is:

D = P + (C \times L)

Where:

D

is the diameter of the wire.

P

is the pitch scale reading.

C

is the circular scale reading.

L

is the least count of the screw gauge. Given: Pitch of the screw gauge,

P = 1 \mathrm{~mm}

Number of divisions on the circular scale = 100 Pitch scale reading,

P = 1 \mathrm{~mm}

The circular scale reads

63^{\text{rd}}

division. First, we calculate the least count of the screw gauge:

L = \frac{ \text{Pitch} }{ \text{Number of divisions} } = \frac{ 1 \mathrm{~mm} }{ 100 } = 0.01 \mathrm{~mm}

Next, we substitute the values into the formula for the diameter:

D = 1 \mathrm{~mm} + (63 \times 0.01 \mathrm{~mm})

Calculate the product:

63 \times 0.01 \mathrm{~mm} = 0.63 \mathrm{~mm}

Now, add the pitch scale reading and the circular scale reading:

D = 1 \mathrm{~mm} + 0.63 \mathrm{~mm} = 1.63 \mathrm{~mm}

Convert

1.63 \mathrm{~mm}

to centimeters:

1.63 \mathrm{~mm} = 0.163 \mathrm{~cm}

Therefore, the diameter of the wire is: Option B:

0.163 \mathrm{~cm}

Q108

The potential energy of a particle moving along

x

-direction varies as

V=\dfrac{A x^2}{\sqrt{x}+B}

. The dimensions of

\dfrac{A^2}{B}

are:

A

\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{1 / 2} \mathrm{~T}^{-3}\right]

B

\left[\mathrm{M}^{1 / 2} \mathrm{LT}^{-3}\right]

C

\left[M^2 L^{1 / 2} T^{-4}\right]

D

\left[\mathrm{ML}^2 \mathrm{~T}^{-4}\right]

Correct Answer

Option C

Solution

To determine the dimensions of

\frac{A^2}{B}

, we start with the given potential energy expression:

V = \frac{A x^2}{\sqrt{x} + B}

The dimensions of potential energy (

V

) are

[ML^2T^{-2}]

.

So, we need to consider the dimensions of the other terms in the expression to match these dimensions.

Let's break it down step-by-step.

For the term

\sqrt{x}

, where

x

represents distance:

\sqrt{x}

has dimensions

[L^{1/2}]

Since both terms in the denominator

\sqrt{x} + B

have to be of the same dimensions for them to be added together, the dimensions of

B

must also be

[L^{1/2}]

. Now the expression becomes:

V = \frac{A x^2}{L^{1/2}}

Thus, the dimensions must be:

[V] = \left[\frac{A x^2}{L^{1/2}}\right]

Substitute the dimensions of

V

,

x

, and

L

:

[ML^2T^{-2}] = \left[\frac{A L^2}{L^{1/2}}\right]

Simplifying the right-hand side:

[ML^2T^{-2}] = [A L^{3/2}]

Therefore, the dimensions of

A

must be:

[A] = [ML^2T^{-2}] \cdot [L^{-3/2}]

[A] = [ML^{1/2} T^{-2}]

We have dimensions of

A

and

B

. Now we need to find the dimensions of

\frac{A^2}{B}

:

[A^2] = [ML^{1/2} T^{-2}]^2

[A^2] = [M^2 L T^{-4}]

Divide by the dimensions of

B

:

\left[\frac{A^2}{B}\right] = \left[\frac{M^2 L T^{-4}}{L^{1/2}}\right]

\left[\frac{A^2}{B}\right] = [M^2 L^{1/2} T^{-4}]

Thus, the dimensions of

\frac{A^2}{B}

are: Option C:

\left[M^2 L^{1/2} T^{-4}\right]

Q109

In a vernier callipers,

(N+1)

divisions of vernier scale coincide with

N

divisions of main scale. If

1 \mathrm{~MSD}

represents

0.1 \mathrm{~mm}

, the vernier constant (in

\mathrm{cm}

) is:

A

\dfrac{1}{10 \mathrm{~N}}

B

\dfrac{1}{100(N+1)}

C

00 \mathrm{~N}

D

10(N+1)

Correct Answer

Option B

Solution

V.C = MSD $-$ VSD ..... (1) given :

(N+1) \mathrm{VSD}=N

MSD

\mathrm{VSD}=\left(\frac{N}{N+1}\right) \mathrm{MSD}

..... (2) From (1) and (2)

\begin{aligned} & V . C=(M S D)-\frac{N}{N+1}(M S D) \\ & =M S D\left(1-\frac{N}{N+1}\right)=\frac{M S D}{N+1} \\ & =\frac{0.01}{N+1}=\frac{1}{100(N+1)} \end{aligned}

Q110

The quantities which have the same dimensions as those of solid angle are:

A strain and angle

B stress and angle

C strain and arc

D angular speed and stress

Correct Answer

Option A

Solution

Solid angle

d \Omega=\frac{d A}{r^2}

has dimensions

[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0]

Strain

=\frac{\Delta l}{l}

has dimensions

[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0]

Angle measured in radians is also dimensionless

[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0]

\theta=\frac{I}{r}