Physical World Units And Measurements – NEET UG PhysicsPractice Questions & PYQs

Impulse is defined as the product of an applied force and the duration of its application, expressed by the equation $I = F \times t$. By substituting the dimensions of force $[MLT^{-2}]$ and time $[T]$ into this relationship, we calculate the dimensional formula for impulse to be $[MLT^{-1}]$. Because linear momentum represents the product of mass and velocity, it shares this exact same dimensional representation of $[MLT^{-1}]$, making the two quantities dimensionally equivalent.

Newton's law of viscosity defines the viscous force through the equation $F = -\eta A \frac{\Delta v}{\Delta Z}$, where $\eta$ represents the coefficient of viscosity. By rearranging this formula to isolate $\eta$, we get $\eta = \frac{F}{A (\Delta v / \Delta Z)}$. Substituting the dimensional formulas for force as $[MLT^{-2}]$, area as $[L^2]$, and the velocity gradient as $[T^{-1}]$ into this equation allows for calculation through basic exponent arithmetic. Dividing $[MLT^{-2}]$ by the product of $[L^2]$ and $[T^{-1}]$ simplifies to $[ML^{-1} T^{-1}]$.

The Principle of Homogeneity of Dimensions states that only physical quantities with identical dimensional formulas can be added or subtracted. Since the term $\frac{a}{V^{2}}$ is added to pressure $P$, both expressions must share the same dimensional formula, which for pressure is $[ML^{-1}T^{-2}]$. By setting the dimensions of $\frac{a}{V^{2}}$ equal to those of $P$, we can isolate $a$ using the relation $[a] = [P][V]^{2}$. Substituting the known dimensions where volume is $[L^{3}]$, we calculate $[a] =$[ML^{-1}T^{-2}]$ \times ([L^{3}]$)^{2}$, which simplifies to $[ML^{-1}T^{-2}] × [L^{6}]$ to yield the final dimensions of $[ML^{5}T^{-2}]$.

Torque is fundamentally defined as the product of force and the perpendicular distance from the axis of rotation. Since force possesses the dimensional formula $[\text{MLT}^{-2}]$ and distance corresponds to length represented as $[\text{L}]$, we simply multiply these dimensions to determine the result. Combining these values yields the dimensional formula for torque as $[\text{ML}^{2}\text{T}^{-2}]$.

The least count of a vernier caliper is the smallest measurement the instrument can resolve, determined by the difference between one main scale division and one vernier scale division. This relationship is represented by the formula $LC = 1 \text{ MSD} - 1 \text{ VSD}$. Since 20 vernier divisions coincide with 16 main scale divisions, one vernier division is equivalent to 0.8 main scale divisions. Substituting this value into the equation yields $LC = 1 \text{ MSD} - 0.8 \text{ MSD} = 0.2 \text{ MSD}$. Given that 1 main scale division measures 1 mm, the least count is 0.2 mm, which converts to 0.02 cm.

The maximum uncertainty in a calculated value like density is determined by summing the relative errors of the individual measurements used. Because density is defined by the ratio $\rho = \frac{m}{V}$, the total relative error is calculated using the formula $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}$. This principle ensures that errors from both independent variables are fully accounted for in the final result.

Using the given values, the relative error consists of the ratio of the mass error to the mass, $\frac{0.01}{22.42}$, added to the ratio of the volume error to the volume, $\frac{0.1}{4.7}$. Calculating these fractions gives approximately $0.00045$ and $0.02128$, respectively, which combine to a total relative error of approximately $0.0217$. When converted into a percentage by multiplying by $100\%$, this figure indicates an uncertainty of $2\%$.

The density of a substance is defined as mass divided by volume, and because the volume of a cube is calculated as the side length cubed, density is expressed mathematically as $\rho = \frac{m}{l^3}$. According to the rules of error propagation, the maximum relative error in a quantity is found by adding the relative errors of the constituent measurements, multiplied by their respective powers in the formula. Therefore, the maximum percentage error in density is given by the formula $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta l}{l}$. Substituting the provided values, the calculation involves adding 3% for the mass error to the product of 3 and 2% for the length error, which results in a final maximum error of 9%.

Accurate measurement with a Vernier calliper relies on determining the instrument's least count and accounting for any systemic zero error. The least count is the difference between one main scale division and one vernier scale division; because ten vernier divisions span nine main scale divisions, each vernier division represents 0.9 of a main scale division. With one main scale division measuring 0.1 cm, the calculation $1 - 0.9 = 0.1$ of a main scale division results in a least count of 0.01 cm.

The measurement is found by adding the main scale reading to the vernier scale contribution, which is the product of the coinciding division and the least count. Adding the main scale reading of 5 cm to the vernier reading of $8 \times 0.01 = 0.08$ cm gives an observed diameter of 5.08 cm. Because the instrument exhibits a positive zero error of 0.1 cm, the initial reading is slightly inflated. Subtracting this error of 0.1 cm from the observed value of 5.08 cm yields a final corrected diameter of 4.98 cm.

Dimensional homogeneity dictates that the physical dimensions on both sides of any valid proportionality must be identical, allowing us to equate the powers of the fundamental dimensions of mass $M$, length $L$, and time $T$. Given the proportionality $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$, we define the dimensions of each variable as surface tension $S$ having dimensions $MT^{-2}$, area $A$ having dimensions $L^2$, density $\rho$ having dimensions $ML^{-3}$, and radius $R$ having dimensions $L$. Substituting these definitions into the proportionality gives the dimensional equation $M^0 L^0 T^1 = (MT^{-2})^\alpha (L^2)^\beta (ML^{-3})^\gamma (L)^\delta$.

By grouping the terms based on their dimensions, we obtain the expression $M^0 L^0 T^1 = M^{\alpha+\gamma} L^{2\beta-3\gamma+\delta} T^{-2\alpha}$. Equating the exponents for each dimension produces a system of linear equations: the time dimension requires $-2\alpha = 1$, which yields $\alpha = -1/2$. The mass dimension requires $\alpha + \gamma = 0$, implying that $\gamma = 1/2$. Finally, the length dimension necessitates $2\beta - 3\gamma + \delta = 0$, and by applying the value $\beta = -1$, the equation becomes $2(-1) - 3(1/2) + \delta = 0$, which simplifies to $\delta = 7/2$.

When a physical quantity is determined by a relationship involving products, quotients, and exponents, the total percentage error is found by adding the individual percentage errors of each variable, multiplied by their respective powers. For the expression $P = \frac{a^3 b^2}{c \sqrt{d}}$, the total percentage error is derived by summing three times the percentage error in $a$, two times the percentage error in $b$, the percentage error in $c$, and half the percentage error in $d$, treating the square root as an exponent of $\frac{1}{2}$.

The calculation follows the form $\frac{\Delta P}{P} \times 100 = 3\left(\frac{\Delta a}{a} \times 100\right) + 2\left(\frac{\Delta b}{b} \times 100\right) + \left(\frac{\Delta c}{c} \times 100\right) + \frac{1}{2}\left(\frac{\Delta d}{d} \times 100\right)$. Substituting the values for the percentage errors leads to $3(1) + 2(3) + 2 + \frac{1}{2}(4)$. Performing this addition yields $3 + 6 + 2 + 2$, resulting in a total percentage error of 13 percent.

The Principle of Homogeneity of Dimensions dictates that only physical quantities possessing identical dimensions can be added or subtracted, meaning every term in the expression $F = \alpha t^2 + \beta t$ must share the same dimensional formula as force. Setting the dimensions of these individual terms equal to one another, we find that $[\alpha t^2] = [F]$ implies $[\alpha] = [F]/[t^2]$, while $[\beta t] = [F]$ implies $[\beta] = [F]/[t]$. When we evaluate the ratio $\frac{\alpha t}{\beta}$ by substituting these derived dimensional forms, the expression \frac{([F]/}[t^2]${) \cdot [t]}{[F]/[t]}$ simplifies to unity, confirming that this particular grouping results in a dimensionless quantity.

The total diameter measured by a screw gauge is found by adding the main scale reading to the circular scale reading, which is the product of the coinciding division and the device's least count. The least count is determined by the ratio $\frac{\text{Pitch}}{\text{Number of divisions}}$, which with a pitch of 1 mm and 100 divisions equals 0.01 mm. With the main scale reading at 1 mm and the 63rd division coinciding with the reference line, the calculation for the total reading is:

$\text{Total Reading} = 1 \text{ mm} + (63 \times 0.01 \text{ mm}) = 1.63 \text{ mm}$

Since the required options are in centimeters, dividing this value by 10 converts the measurement to 0.163 cm.

The vernier constant, also referred to as the least count of a vernier callipers, represents the minimum length the instrument can measure and is defined as the difference between the length of one main scale division (MSD) and one vernier scale division (VSD). Since the device is calibrated such that N+1 vernier scale divisions correspond exactly to N main scale divisions, a single vernier scale division can be expressed as $\text{VSD} = \frac{N}{N+1} \text{ MSD}$. Calculating the vernier constant by subtracting the VSD from the MSD results in the equation $\text{V.C.} = \text{MSD} - \frac{N}{N+1} \text{ MSD}$.

Factoring out the MSD term simplifies this to $\text{V.C.} = \text{MSD} \left( 1 - \frac{N}{N+1} \right)$, which reduces to:
$\text{V.C.} = \frac{\text{MSD}}{N+1}$
Because 1 MSD = 0.1 mm, and the requested unit is centimeters, we convert this to 0.01 cm. Substituting this value into the expression produces:
$\text{V.C.} = \frac{0.01}{N+1} \text{ cm} = \frac{1}{100(N+1)} \text{ cm}$

The principle of homogeneity dictates that we can only add or subtract physical quantities of the same dimension, so in the denominator $\sqrt{x}+B$, $B$ must have the same dimension as $\sqrt{x}$. Since $x$ is a length, its dimension is $L$, implying $\sqrt{x}$ is $L^{1/2}$ and therefore the dimension of $B$ is $[L^{1/2}]$.

Potential energy $V$ has the dimensional formula $[M L^2 T^{-2}]$. Equating the total dimensional formula $V = \frac{A x^2}{\sqrt{x}+B}$ allows us to determine the dimension of $A$ by setting $[M L^2 T^{-2}] = \frac{[A] L^2}{L^{1/2}}$. Solving for $A$ gives $[A] = [M L^{1/2} T^{-2}]$.

To find the dimensions of the final expression $\frac{A^2}{B}$, we square the dimension of $A$ to obtain $[M^2 L T^{-4}]$ and divide by the dimension of $B$, which is $[L^{1/2}]$. This calculation results in $[M^2 L^{1/2} T^{-4}]$.

Dimensional analysis identifies quantities as dimensionless when their defining ratios cancel out all base units, leaving them with dimensions of $[M^0 L^0 T^0]$. The solid angle is defined as:

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

Since this calculation yields a ratio of area to the square of distance, it results in a dimensionless value. Because plane angles are defined as the ratio of arc length to radius and strain is defined as the ratio of change in length to initial length, both also function as dimensionless ratios that share the same $[M^0 L^0 T^0]$ dimensions.

According to Ohm's law, resistance is defined by the relation $R = V/I$, which yields a central value of 10 $\Omega$. When determining the uncertainty in a derived quantity like resistance, the relative error is obtained by adding the relative errors of the individual measurements, as expressed by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$. Substituting the provided values leads to a total relative error of $4/200 + 0.2/20 = 6/200$. Multiplying this relative error by the central resistance of 10 gives the absolute uncertainty of 0.3 $\Omega$. Consequently, the resistance is stated as $(10 \pm 0.3) \Omega$.

Random errors are defined by their lack of any identifiable, consistent, or systematic source, which prevents them from being tracked or corrected through standard calibration. When experimental conditions such as voltage or ambient temperature undergo irregular, unpredictable shifts, these variations manifest as random errors. Because these disturbances do not stem from faulty equipment or observer bias, they are inherently categorized as random rather than instrumental or personal errors.

The density of a cylindrical wire is calculated using the formula
$\rho = \frac{M}{\pi r^2 \ell}$
which indicates that the overall uncertainty depends on the experimental measurements of mass, radius, and length. When determining the maximum percentage error for a quantity derived from products or quotients, we sum the relative errors of the individual components, while multiplying the relative error of any term by its exponent in the formula.

The relative error in density is represented by the relationship
$\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + \frac{2 \Delta r}{r} + \frac{\Delta \ell}{\ell}$
Substituting the provided experimental values into this expression gives
$\frac{\Delta \rho}{\rho} = \left( \frac{0.002}{0.4} + \frac{2 \times 0.001}{0.3} + \frac{0.02}{5} \right)$
Evaluating these terms leads to values of $0.005$, $0.0067$, and $0.004$, which sum to $0.0157$. Multiplying this decimal result by $100$ to express it as a percentage yields a maximum error of approximately $1.6\%$.

Dimensional analysis is a powerful tool for identifying physical quantities by examining the exponents of fundamental dimensions like mass, length, and time. To identify the quantity with dimensions of reciprocal mass, we must evaluate the mass exponent in each given option: angular momentum is $[ML^{2}T^{-1}]$, the coefficient of thermal conductivity is $[MLT^{-3}K^{-1}]$, and torque is $[ML^{2}T^{-2}]$. Among these, the gravitational constant is distinct because its dimensional formula is $[M^{-1}L^{3}T^{-2}]$, which confirms it is the only quantity that possesses the required inverse power of mass.

Finding the representative value of a physical quantity involves calculating the arithmetic mean of multiple observations. Because the measurements are limited by the precision of the vernier callipers, the final result must reflect the correct number of significant figures, which corresponds to the precision of the individual readings provided.

The mean diameter is found by summing the five recorded values of 3.33 cm, 3.32 cm, 3.34 cm, 3.33 cm, and 3.32 cm, which gives a total sum of 16.64 cm. Dividing this total by the number of observations yields the following calculation:
$\text{Mean} = \frac{3.33 + 3.32 + 3.34 + 3.33 + 3.32}{5} = \frac{16.64}{5} = 3.328 \text{ cm}$
Since the original measurements are given to three significant figures, the result must be rounded to the same degree of precision, yielding 3.33 cm.