Units and Measurements - NEET Physics MCQs

Units and Measurements

1.1 Units

1. The unit of thermal conductivity is:

ताप चालकता का मात्रक है:

J m⁻¹ K⁻¹
W m K⁻¹
W m⁻¹ K⁻¹
J m K⁻¹

Exam: NEET (UG) — 05.05.2019

Answer: (c)

Thermal conductivity is the amount of heat that flows per unit time through a unit area with a temperature gradient of one Kelvin per unit length.

The relation for heat conduction is:

\[ \frac{Q}{t} = kA\frac{\Delta T}{\ell} \]

Where:

\(k\) = thermal conductivity
\(Q\) = heat transferred
\(A\) = area of cross-section
\(\Delta T\) = temperature difference
\(\ell\) = length
\(t\) = time

Rearranging the equation:

\[ k = \frac{Q\ell}{A t \Delta T} \]

Therefore, the SI unit of \(k\) is:

\[ \frac{\text{J}\cdot\text{m}} {\text{m}^2\cdot\text{s}\cdot\text{K}} = \frac{\text{J}} {\text{m}\cdot\text{s}\cdot\text{K}} \]

Since:

\[ 1\,\text{W} = 1\,\text{J s}^{-1} \]

we get:

\[ [k] = \text{W m}^{-1}\text{K}^{-1} \]

Hence, the correct answer is (c) W m⁻¹ K⁻¹.

2. The density of a material in CGS system of units is \(4\,\text{g/cm}^3\). In a system of units in which the unit of length is 10 cm and the unit of mass is 100 g, the value of density of the material will be:

मात्रकों की CGS पद्धति में किसी पदार्थ का घनत्व \(4\,\text{g/cm}^3\) है। मात्रकों की एक ऐसी पद्धति में जिसमें लंबाई की इकाई 10 cm तथा द्रव्यमान की इकाई 100 g है, पदार्थ के घनत्व का मान होगा:

0.04
0.4
40
400

Exam: AIPMT (Mains) — 2011

Answer: (c)

The density of the material in the CGS system is:

\[ \rho = 4\,\text{g/cm}^3 \]

In the new system:

\[ 1\text{ unit of mass} = 100\,\text{g} \]

Therefore:

\[ 1\,\text{g} = \frac{1}{100} \text{ unit of mass} \]

Similarly:

\[ 1\text{ unit of length} = 10\,\text{cm} \]

Therefore:

\[ 1\,\text{cm} = \frac{1}{10} \text{ unit of length} \]

Since density is mass per unit volume:

\[ \rho = \frac{4\,\text{g}}{\text{cm}^3} \]

Substituting the new units:

\[ \rho = \frac{ 4\left(\frac{1}{100}\right) }{ \left(\frac{1}{10}\right)^3 } \]

\[ = \frac{4/100}{1/1000} \]

\[ = \frac{4\times1000}{100} = 40 \]

Hence, the density in the new system is 40 units.

Therefore, the correct answer is (c) 40.

3. The unit of permittivity of free space \(\varepsilon_0\) is:

निर्वात की परावैद्युतता \(\varepsilon_0\) का मात्रक होगा:

Newton metre² / Coulomb²
Coulomb² / Newton metre²
Coulomb² / (Newton metre)²
Coulomb / Newton metre

Exam: AIPMT — 2004

Answer: (b)

According to Coulomb's law:

\[ F = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2} \]

Rearranging for \(\varepsilon_0\):

\[ \varepsilon_0 = \frac{q_1q_2} {4\pi F r^2} \]

Ignoring the dimensionless constant \(4\pi\), the unit is:

\[ [\varepsilon_0] = \frac{\text{C}^2} {\text{N}\cdot\text{m}^2} \]

Therefore, the unit of permittivity of free space is:

\[ \boxed{ \text{C}^2\text{N}^{-1}\text{m}^{-2} } \]

Hence, the correct answer is (b) Coulomb² / (Newton metre²).

4. Tesla is the unit of:

टेस्ला किसका मात्रक है?

Electric field — विद्युत क्षेत्र
Magnetic field — चुंबकीय क्षेत्र
Electric flux — विद्युत फ्लक्स
Magnetic flux — चुंबकीय फ्लक्स

Exam: AIPMT — 1997, 1998

Answer: (b)

Tesla (T) is the SI unit of magnetic field or magnetic flux density.

One tesla is defined as the magnetic field that produces a force of one newton on a conductor carrying a current of one ampere when the conductor is perpendicular to the magnetic field and has a length of one metre.

The force on a current-carrying conductor is:

\[ F = BIL\sin\theta \]

For \(\theta=90^\circ\):

\[ F = BIL \]

Therefore:

\[ B = \frac{F}{IL} \]

Hence, the SI unit of magnetic field is:

\[ 1\,\text{T} = 1\,\frac{\text{N}}{\text{A}\cdot\text{m}} \]

Therefore:

Magnetic field → Tesla (T)
Electric field → V/m or N/C
Electric flux → N m²/C
Magnetic flux → Weber (Wb)

Hence, the correct answer is (b) Magnetic field.