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  • Applications of Dimensional Analysis: Checking Equations & Deriving Relations | JEE Physics

    Concept Card: Applications & Limitations of Dimensional Analysis

    1. The Principle of Homogeneity of Dimensions:
    The Principle of Homogeneity states that a physical equation is dimensionally valid if and only if every term on both sides of the equation has the exact same dimensional formula.

    If an equation has the general form:
    $A = B + C – D$
    Then consistency requires:
    $[A] = [B] = [C] = [D]$
    Physical quantities can be added or subtracted only if they possess identical dimensions.

    Crucial Rule for Transcendental Functions:
    The arguments (inputs) of trigonometric ($\sin \theta$, $\cos \theta$), exponential ($e^x$), and logarithmic ($\ln x$) functions must always be strictly dimensionless:
    $[\theta] = [1] = [M^0 L^0 T^0], \quad [x] = [1] = [M^0 L^0 T^0]$

    2. Core Applications of Dimensional Analysis:

    1. Checking the Dimensional Correctness of an Equation:

      Verify whether the dimensions of the left-hand side equal the dimensions of the right-hand side. Dimensional consistency is a necessary condition for correctness, but not a sufficient condition (it cannot detect missing dimensionless scalar constants).
    2. Deriving Functional Relations Among Physical Quantities:

      If a quantity $Q$ is known to depend on physical factors $A, B, C$, we postulate a power relation:

      $Q = k \cdot A^x B^y C^z$

      where $k$ is a dimensionless constant of proportionality. Equating powers of $M, L, T$ on both sides yields a system of simultaneous linear equations to solve for $x, y, z$.
    3. Conversion of Units Between Different Systems:

      Using the invariance of physical magnitude $n_1 u_1 = n_2 u_2$:

      $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$

    3. Limitations of Dimensional Analysis:

    • Dimensionless constants (such as $\frac{1}{2}$, $2\pi$, $1$) cannot be determined; they must be obtained through rigorous theory or experimentation.
    • Cannot derive equations containing sums or differences of terms (e.g., $v^2 = u^2 + 2as$ or $s = ut + \frac{1}{2}at^2$ cannot be synthesized from scratch using power products).
    • Cannot derive relations involving transcendental functions (e.g., trigonometric, exponential, or logarithmic equations).
    • Limited by number of equations: In mechanics, there are only 3 base dimensions ($M, L, T$), providing at most 3 independent algebraic equations. If a quantity depends on more than 3 variables, unique powers cannot be solved without extra physical constraints.
    • Cannot determine vector or scalar nature of a physical quantity.

    Solved Examples

    Example 1 (Checking Dimensional Consistency of a Formula):
    Test the dimensional correctness of the capillary ascent formula:
    $h = \frac{2 T \cos\theta}{r \rho g}$
    where $h$ is height of liquid column, $T$ is surface tension, $\theta$ is contact angle, $r$ is tube radius, $\rho$ is liquid density, and $g$ is acceleration due to gravity.

    Solution:
    1. Left-Hand Side (LHS):
    $[h] = [L^1] = [M^0 L^1 T^0]$.
    2. Right-Hand Side (RHS):
    – Surface tension: $[T] = \frac{[F]}{[L]} = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}]$
    – Contact angle: $\cos\theta$ is a trigonometric ratio, hence $[\cos\theta] = [1]$
    – Factor $2$ is a dimensionless number: $[2] = [1]$
    – Radius: $[r] = [L]$
    – Density: $[\rho] = [M L^{-3}]$
    – Acceleration due to gravity: $[g] = [L T^{-2}]$
    Evaluating the denominator:
    $[\text{Denominator}] = [r][\rho][g] = [L][M L^{-3}][L T^{-2}] = [M L^{-1} T^{-2}]$
    Evaluating the RHS quotient:
    $[\text{RHS}] = \frac{[2 T \cos\theta]}{[r \rho g]} = \frac{[M T^{-2}]}{[M L^{-1} T^{-2}]} = \frac{[M^1 T^{-2}]}{[M^1 L^{-1} T^{-2}]} = [L^1] = [M^0 L^1 T^0]$.
    Since $[\text{LHS}] = [\text{RHS}] = [L]$, the formula is dimensionally correct.

    Example 2 (Derivation of Critical Flow Velocity via Method of Dimensions):
    The critical velocity $v_c$ of a viscous fluid flowing through a narrow pipe depends on the viscosity $\eta$, fluid density $\rho$, and pipe diameter $D$. Derive a formula for $v_c$ using dimensional analysis.

    Solution:
    Let $v_c = k \cdot \eta^x \rho^y D^z$, where $k$ is a dimensionless constant.
    Substitute the dimensions of each variable:
    $[v_c] = [L T^{-1}]$
    $[\eta] = [M L^{-1} T^{-1}]$
    $[\rho] = [M L^{-3}]$
    $[D] = [L]$
    Writing the dimensional equation:
    $[M^0 L^1 T^{-1}] = [M L^{-1} T^{-1}]^x [M L^{-3}]^y [L]^z$
    $[M^0 L^1 T^{-1}] = [M^{x + y} L^{-x – 3y + z} T^{-x}]$
    Equating powers of $M, L, T$ on both sides:
    1. For $T$: $-x = -1 \implies x = 1$
    2. For $M$: $x + y = 0 \implies 1 + y = 0 \implies y = -1$
    3. For $L$: $-x – 3y + z = 1 \implies -(1) – 3(-1) + z = 1 \implies -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1$
    Substituting $x = 1, y = -1, z = -1$ back into the postulated relation:
    $v_c = k \cdot \eta^1 \rho^{-1} D^{-1} = k \frac{\eta}{\rho D}$.
    (Here $k$ is the critical Reynolds number $R_c$).

    Example 3 (Dimensions of Constants in Exponential and Oscillatory Equations):
    The displacement of a damped oscillator is described by $x(t) = A_0 e^{-\alpha t} \cos(\omega t + \phi)$, where $x$ is position and $t$ is time. Determine the dimensional formulae of $A_0$, $\alpha$, $\omega$, and $\phi$.

    Solution:
    1. In the exponential term $e^{-\alpha t}$, the power must be dimensionless:
    $[-\alpha t] = [1] \implies [\alpha][T] = [1] \implies [\alpha] = [T^{-1}] = [M^0 L^0 T^{-1}]$.
    2. Inside the cosine argument $(\omega t + \phi)$, the angle must be dimensionless:
    $[\omega t + \phi] = [1]$
    By homogeneity, each term must be individually dimensionless:
    $[\phi] = [1] = [M^0 L^0 T^0]$
    $[\omega t] = [1] \implies [\omega][T] = [1] \implies [\omega] = [T^{-1}] = [M^0 L^0 T^{-1}]$.
    3. The cosine function $\cos(\omega t + \phi)$ and exponential factor $e^{-\alpha t}$ are dimensionless numbers ($[1]$). Therefore:
    $[x] = [A_0][1][1] \implies [A_0] = [x] = [L^1] = [M^0 L^1 T^0]$.

    Example 4 (Underwater Explosion Oscillation Period):
    The time period $T$ of an oscillating gas bubble produced by an underwater explosion depends on the static pressure $P$, the density of water $\rho$, and the total explosive energy $E$. Derive the expression for $T$.

    Solution:
    Let $T = k \cdot P^a \rho^b E^c$, where $k$ is a dimensionless constant.
    Substitute the dimensions of each quantity:
    $[T] = [M^0 L^0 T^1]$
    $[P] = [M L^{-1} T^{-2}]$
    $[\rho] = [M L^{-3}]$
    $[E] = [M L^2 T^{-2}]$
    Writing the dimensional equation:
    $[M^0 L^0 T^1] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-2}]^c$
    $[M^0 L^0 T^1] = [M^{a + b + c} L^{-a – 3b + 2c} T^{-2a – 2c}]$
    Equating exponents on both sides:
    1. For $M$: $a + b + c = 0 \implies b = -(a + c)$
    2. For $T$: $-2a – 2c = 1 \implies -2(a + c) = 1 \implies a + c = -\frac{1}{2}$
    From (1): $b = -\left(-\frac{1}{2}\right) = \frac{1}{2}$
    3. For $L$: $-a – 3b + 2c = 0 \implies -a – 3\left(\frac{1}{2}\right) + 2c = 0 \implies -a + 2c = \frac{3}{2}$
    We have a system of two equations in $a$ and $c$:
    $a + c = -\frac{1}{2} \implies a = -\frac{1}{2} – c$
    Substitute into $-a + 2c = \frac{3}{2}$:
    $-\left(-\frac{1}{2} – c\right) + 2c = \frac{3}{2} \implies \frac{1}{2} + 3c = \frac{3}{2} \implies 3c = 1 \implies c = \frac{1}{3}$
    Then $a = -\frac{1}{2} – \frac{1}{3} = -\frac{5}{6}$.
    Substituting $a = -\frac{5}{6}, b = \frac{1}{2}, c = \frac{1}{3}$:
    $T = k \cdot P^{-5/6} \rho^{1/2} E^{1/3} = k \cdot \frac{\rho^{1/2} E^{1/3}}{P^{5/6}}$.

    Worksheet: 10 Practice Problems

    Problem 1 (JEE Main – Single Correct):
    The time period $T$ of a droplet of liquid oscillating under surface tension depends on the surface tension $S$, density of liquid $\rho$, and droplet radius $r$. The dimensionally correct relation for $T$ is:
    (A) $T = k \sqrt{\frac{\rho r^3}{S}}$
    (B) $T = k \sqrt{\frac{S r^3}{\rho}}$
    (C) $T = k \frac{\rho r^2}{S}$
    (D) $T = k \sqrt{\frac{\rho S}{r^3}}$

    Problem 2 (JEE Main – Single Correct):
    The equation of a traveling plane progressive wave is given by $y = A \sin(\omega t – k x + \phi)$. What are the dimensions of the ratio $\frac{\omega}{k}$?
    (A) $[M^0 L^1 T^{-1}]$
    (B) $[M^0 L^{-1} T^1]$
    (C) $[M^0 L^0 T^0]$
    (D) $[M^0 L^2 T^{-2}]$

    Problem 3 (JEE Main – Single Correct):
    Which of the following physical equations is dimensionally INCORRECT?
    (A) $v = \sqrt{\frac{\gamma P}{\rho}}$ (Speed of sound in gas)
    (B) $h = \frac{2 T}{r \rho g}$ (Capillary rise)
    (C) $f = \frac{1}{2\pi} \sqrt{\frac{m}{k}}$ (Frequency of spring-mass system)
    (D) $\nu = \frac{1}{2 L} \sqrt{\frac{T}{\mu}}$ (Fundamental frequency of stretched string)

    Problem 4 (JEE Main – Single Correct):
    If energy $E$, velocity $v$, and time $T$ are chosen as fundamental quantities, the dimensional formula for surface tension $S$ is:
    (A) $[E^1 v^{-2} T^{-2}]$
    (B) $[E^1 v^{-1} T^{-2}]$
    (C) $[E^2 v^{-1} T^{-1}]$
    (D) $[E^1 v^{-2} T^{-1}]$

    Problem 5 (JEE Main – Single Correct):
    The velocity $v$ of water waves may depend on wavelength $\lambda$, density of water $\rho$, and acceleration due to gravity $g$. The relation between these quantities is:
    (A) $v^2 \propto g \lambda$
    (B) $v^2 \propto g \lambda \rho$
    (C) $v^2 \propto \frac{g}{\lambda}$
    (D) $v^2 \propto \frac{g \rho}{\lambda}$

    Problem 6 (JEE Advanced – One or More Correct):
    Which of the following statements is/are correct regarding the limitations of dimensional analysis?
    (A) It cannot determine the value of dimensionless constants appearing in a formula.
    (B) It fails to derive formulas that involve addition or subtraction of multiple physical terms.
    (C) It cannot deduce relationships if a variable depends on trigonometric or exponential terms.
    (D) It cannot uniquely solve for dependencies in mechanics if a quantity depends on four or more independent variables.

    Problem 7 (JEE Advanced – One or More Correct):
    A physical quantity $y$ is expressed as $y = \frac{a}{b} \ln\left(1 + \frac{b x}{c}\right)$, where $y$ represents Force and $x$ represents distance. Which of the following statements is/are correct?
    (A) $[b] = [L^{-1}][c]$
    (B) $[a] = [M L T^{-2}][b]$
    (C) If $c$ has dimensions of velocity $[L T^{-1}]$, then $[b] = [T^{-1}]$
    (D) The argument $\left(1 + \frac{b x}{c}\right)$ is dimensionless.

    Problem 8 (JEE Advanced – One or More Correct):
    In fluid dynamics, the Reynolds number $R_e$ is given by $R_e = \frac{\rho v D}{\eta}$. Which of the following statements is/are correct?
    (A) $R_e$ is a dimensionless quantity ($[M^0 L^0 T^0]$).
    (B) The ratio $\frac{\eta}{\rho}$ is termed kinematic viscosity and has dimensions $[L^2 T^{-1}]$.
    (C) If $\rho, v, D$ are tripled and $\eta$ is doubled, the value of $R_e$ increases by a factor of $13.5$.
    (D) $R_e$ can be derived solely from dimensional analysis without experimental coefficients.

    Problem 9 (JEE Main / Advanced – Numerical Value Type):
    The centripetal force acting on a revolving body depends on mass $m$, orbital radius $r$, and speed $v$ as $F = k \cdot m^a v^b r^c$. Determine the numerical value of $a^2 + b^2 + c^2$.

    Problem 10 (JEE Main / Advanced – Numerical Value Type):
    The speed $v$ of gravity waves in shallow water of depth $h$ is assumed to depend on depth $h$, acceleration due to gravity $g$, and density $\rho$ as $v = k \cdot \rho^a g^b h^c$. Determine the value of $10(a + b + c)$.

    Solutions & Explanations

    Answer Key Summary:
    1. (A) | 2. (A) | 3. (C) | 4. (A) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 6 | 10. 10

    Solution 1:
    Let $T = k \cdot S^a \rho^b r^c$.
    Dimensions: $[T] = [T^1]$, $[S] = [M T^{-2}]$, $[\rho] = [M L^{-3}]$, $[r] = [L]$.
    $[M^0 L^0 T^1] = [M T^{-2}]^a [M L^{-3}]^b [L]^c = [M^{a+b} L^{-3b+c} T^{-2a}]$.
    – For $T$: $-2a = 1 \implies a = -1/2$.
    – For $M$: $a + b = 0 \implies b = -a = 1/2$.
    – For $L$: $-3b + c = 0 \implies c = 3b = 3(1/2) = 3/2$.
    $T = k \cdot S^{-1/2} \rho^{1/2} r^{3/2} = k \sqrt{\frac{\rho r^3}{S}}$.
    Correct Answer: (A)

    Solution 2:
    Arguments of trigonometric function are dimensionless:
    $[\omega t] = [1] \implies [\omega] = [T^{-1}]$.
    $[k x] = [1] \implies [k] = [L^{-1}]$.
    Ratio $\left[\frac{\omega}{k}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}] = [M^0 L^1 T^{-1}]$.
    This corresponds to the wave propagation speed $v = \frac{\omega}{k}$.
    Correct Answer: (A)

    Solution 3:
    In (C), the frequency of a spring-mass system is $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
    Dimensions: $[\sqrt{m/k}] = \left[\frac{M}{M T^{-2}}\right]^{1/2} = [T^2]^{1/2} = [T]$, which represents time period, not frequency ($[T^{-1}]$).
    Hence, $f = \frac{1}{2\pi}\sqrt{\frac{m}{k}}$ is dimensionally inverted and incorrect.
    Correct Answer: (C)

    Solution 4:
    Let Surface Tension $[S] = [E]^x [v]^y [T]^z$.
    Dimensions: $[S] = [M T^{-2}]$, $[E] = [M L^2 T^{-2}]$, $[v] = [L T^{-1}]$, $[T] = [T]$.
    $[M^1 L^0 T^{-2}] = [M L^2 T^{-2}]^x [L T^{-1}]^y [T]^z = [M^x L^{2x + y} T^{-2x – y + z}]$.
    – For $M$: $x = 1$.
    – For $L$: $2x + y = 0 \implies 2(1) + y = 0 \implies y = -2$.
    – For $T$: $-2x – y + z = -2 \implies -2(1) – (-2) + z = -2 \implies -2 + 2 + z = -2 \implies z = -2$.
    $[S] = [E^1 v^{-2} T^{-2}]$.
    Correct Answer: (A)

    Solution 5:
    Let $v = k \cdot \lambda^a g^b \rho^c$.
    $[L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c = [M^c L^{a + b – 3c} T^{-2b}]$.
    – For $M$: $c = 0$ (speed does not depend on density).
    – For $T$: $-2b = -1 \implies b = 1/2$.
    – For $L$: $a + b – 3(0) = 1 \implies a + 1/2 = 1 \implies a = 1/2$.
    $v = k \lambda^{1/2} g^{1/2} \implies v^2 \propto g \lambda$.
    Correct Answer: (A)

    Solution 6:
    All four statements (A, B, C, D) are classic fundamental limitations of dimensional analysis.
    Correct Answer: (A, B, C, D)

    Solution 7:
    – Logarithmic argument is dimensionless $\implies \left[\frac{b x}{c}\right] = [1] \implies [b][L] = [c] \implies [b] = [L^{-1}][c]$. Thus (A) and (D) are correct.
    – Since the natural logarithm is a dimensionless scalar, $[y] = \left[\frac{a}{b}\right] \implies [a] = [y][b] = [M L T^{-2}][b]$. Thus (B) is correct.
    – If $[c] = [L T^{-1}]$, then $[b] = [L^{-1}][L T^{-1}] = [T^{-1}]$. Thus (C) is correct.
    All options are correct.
    Correct Answer: (A, B, C, D)

    Solution 8:
    – (A) True: $[R_e] = \frac{[M L^{-3}][L T^{-1}][L]}{[M L^{-1} T^{-1}]} = [M^0 L^0 T^0]$.
    – (B) True: $\left[\frac{\eta}{\rho}\right] = \frac{[M L^{-1} T^{-1}]}{[M L^{-3}]} = [L^2 T^{-1}]$.
    – (C) True: $R_e \propto \frac{\rho v D}{\eta} \implies \frac{(3)(3)(3)}{2} = \frac{27}{2} = 13.5$.
    – (D) False: Dimensionless numbers cannot be fully predicted in value without experimental boundaries.
    Correct Answer: (A, B, C)

    Solution 9:
    $F = k \cdot m^a v^b r^c$.
    $[M L T^{-2}] = [M]^a [L T^{-1}]^b [L]^c = [M^a L^{b + c} T^{-b}]$.
    – $a = 1$
    – $-b = -2 \implies b = 2$
    – $b + c = 1 \implies 2 + c = 1 \implies c = -1$.
    Evaluating $a^2 + b^2 + c^2 = (1)^2 + (2)^2 + (-1)^2 = 1 + 4 + 1 = 6$.
    Correct Answer: 6

    Solution 10:
    $[v] = [L T^{-1}]$, $[g] = [L T^{-2}]$, $[h] = [L]$, $[\rho] = [M L^{-3}]$.
    $[M^0 L^1 T^{-1}] = [M L^{-3}]^a [L T^{-2}]^b [L]^c = [M^a L^{-3a + b + c} T^{-2b}]$.
    – For $M$: $a = 0$.
    – For $T$: $-2b = -1 \implies b = 1/2$.
    – For $L$: $-3(0) + 1/2 + c = 1 \implies c = 1/2$.
    Sum $a + b + c = 0 + \frac{1}{2} + \frac{1}{2} = 1$.
    Value: $10(a + b + c) = 10 \times 1 = 10$.
    Correct Answer: 10

  • Dimensions of Physical Quantities & Dimensional Formulae of Derived Quantities | JEE Physics

    Concept Card: Dimensions of Physical Quantities & Dimensional Formulae

    1. Definition of Dimensions:
    The dimensions of a physical quantity are the powers (or exponents) to which the fundamental base quantities must be raised to represent that quantity.

    The standard dimensional expression is represented in terms of the seven base dimensions:
    $[Q] = [M^a L^b T^c A^d K^e \text{mol}^f \text{cd}^g]$
    where $a, b, c, d, e, f, g$ are the dimensions of mass, length, time, electric current, temperature, amount of substance, and luminous intensity respectively.

    2. Classification of Physical Quantities:

    • Dimensional Variables: Quantities possessing physical dimensions that assume variable values.
      Examples: Velocity $[L T^{-1}]$, Acceleration $[L T^{-2}]$, Force $[M L T^{-2}]$, Energy $[M L^2 T^{-2}]$.
    • Dimensionless Variables: Quantities possessing no physical dimensions ($[M^0 L^0 T^0]$) whose numerical values vary.
      Examples: Strain ($\Delta L / L$), Specific Gravity, Refractive Index, Relative Permittivity, Dielectric Constant.
    • Dimensional Constants: Quantities possessing constant numerical values while having definite physical dimensions.
      Examples: Universal Gravitational Constant $G$, Planck’s Constant $h$, Speed of Light $c$, Boltzmann Constant $k_B$, Permittivity of Free Space $\varepsilon_0$, Permeability of Free Space $\mu_0$.
    • Dimensionless Constants: Pure numerical values or mathematical constants lacking both units and dimensions.
      Examples: $\pi$, Euler’s number $e$, pure geometric ratios, integer factors.

    3. Comprehensive Reference Table of Derived Quantities:

    Physical QuantityGoverning Physical RelationDimensional Formula
    Linear Momentum ($p$)$p = m v$$[M L T^{-1}]$
    Force ($F$)$F = m a$$[M L T^{-2}]$
    Work / Energy ($W, E$)$W = \vec{F}\cdot\vec{s}$$[M L^2 T^{-2}]$
    Torque ($\tau$)$\vec{\tau} = \vec{r}\times\vec{F}$$[M L^2 T^{-2}]$
    Power ($P$)$P = W / t$$[M L^2 T^{-3}]$
    Pressure / Stress$P = F / A$$[M L^{-1} T^{-2}]$
    Modulus of Elasticity ($Y, B, \eta$)$\text{Stress} / \text{Strain}$$[M L^{-1} T^{-2}]$
    Surface Tension ($T$)$T = F / L$$[M T^{-2}]$
    Coefficient of Viscosity ($\eta$)$F = 6\pi \eta r v$$[M L^{-1} T^{-1}]$
    Universal Gravitational Constant ($G$)$F = G m_1 m_2 / r^2$$[M^{-1} L^3 T^{-2}]$
    Planck’s Constant ($h$)$E = h \nu$$[M L^2 T^{-1}]$
    Boltzmann Constant ($k_B$)$E = \frac{3}{2} k_B T$$[M L^2 T^{-2} K^{-1}]$
    Thermal Conductivity ($K$)$\frac{dQ}{dt} = K A \frac{\Delta T}{L}$$[M L T^{-3} K^{-1}]$
    Stefan’s Constant ($\sigma$)$E = \sigma T^4$$[M T^{-3} K^{-4}]$
    Electric Potential / EMF ($V$)$V = W / q$$[M L^2 T^{-3} A^{-1}]$
    Capacitance ($C$)$C = q / V$$[M^{-1} L^{-2} T^4 A^2]$
    Resistance ($R$)$R = V / I$$[M L^2 T^{-3} A^{-2}]$
    Inductance ($L$)$\mathcal{E} = -L \frac{dI}{dt}$$[M L^2 T^{-2} A^{-2}]$
    Magnetic Field Induction ($B$)$F = q v B$$[M T^{-2} A^{-1}]$
    Magnetic Flux ($\Phi_B$)$\Phi_B = B A$$[M L^2 T^{-2} A^{-1}]$
    Permittivity ($\varepsilon_0$)$F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}$$[M^{-1} L^{-3} T^4 A^2]$
    Permeability ($\mu_0$)$F = \frac{\mu_0 I_1 I_2 L}{2\pi r}$$[M L T^{-2} A^{-2}]$

    Solved Examples

    Example 1 (Derivation of Coefficient of Viscosity & Planck’s Constant):
    Derive the dimensional formulae for:
    (a) Coefficient of viscosity $\eta$ using Stokes’ Law $F = 6\pi \eta r v$, where $F$ is viscous drag, $r$ is sphere radius, and $v$ is terminal speed.
    (b) Planck’s constant $h$ using Einstein’s relation $E = h \nu$, where $E$ is photon energy and $\nu$ is radiation frequency.

    Solution:
    (a) Rearranging Stokes’ Law:
    $\eta = \frac{F}{6\pi r v}$
    Since $6\pi$ is a dimensionless number ($[6\pi] = [1]$):
    $[\eta] = \frac{[F]}{[r][v]} = \frac{[M L T^{-2}]}{[L][L T^{-1}]} = \frac{[M L T^{-2}]}{[L^2 T^{-1}]} = [M L^{-1} T^{-1}]$.
    (b) From Einstein’s equation:
    $h = \frac{E}{\nu}$
    Dimensions of Energy $[E] = [M L^2 T^{-2}]$ and frequency $[\nu] = [T^{-1}]$:
    $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$.
    Notice that Planck’s constant possesses the same dimensions as angular momentum ($L = m v r \implies [M L^2 T^{-1}]$).

    Example 2 (Determining Dimensions of Constants via Homogeneity Principle):
    In the relation $P = \frac{a – t^2}{b x}$, where $P$ is pressure, $x$ is displacement, and $t$ is time, find the dimensional formula of the ratio $\frac{a}{b}$.

    Solution:
    1. By the principle of homogeneity, quantities added or subtracted must possess identical dimensions:
    $[a] = [t^2] = [T^2]$.
    2. Analyzing the entire equation:
    $[P] = \frac{[a – t^2]}{[b][x]} = \frac{[T^2]}{[b][L]}$
    Since Pressure has dimensions $[P] = [M L^{-1} T^{-2}]$:
    $[M L^{-1} T^{-2}] = \frac{[T^2]}{[b][L]} \implies [b] = \frac{[T^2]}{[M L^{-1} T^{-2}][L]} = \frac{[T^2]}{[M T^{-2}]} = [M^{-1} T^4]$.
    3. Calculating the dimensional formula of $\frac{a}{b}$:
    $\left[\frac{a}{b}\right] = \frac{[a]}{[b]} = \frac{[T^2]}{[M^{-1} T^4]} = [M T^{-2}] = [M^1 L^0 T^{-2}]$.

    Example 3 (Dimensional Analysis of Electromagnetic Combinations):
    Find the dimensional formula of the following fundamental electromagnetic combinations:
    (a) $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
    (b) The characteristic wave impedance of free space $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$
    (c) The ratio of electric field to magnetic field $\frac{E}{B}$.

    Solution:
    (a) From Maxwell’s electromagnetic theory, $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ represents the speed of light in vacuum ($c$):
    $\left[\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right] = [c] = [L T^{-1}]$.
    (b) For impedance of free space $Z_0$:
    $[\mu_0] = [M L T^{-2} A^{-2}]$ and $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
    $\left[\frac{\mu_0}{\varepsilon_0}\right] = \frac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} = [M^2 L^4 T^{-6} A^{-4}]$
    $\left[Z_0\right] = \sqrt{[M^2 L^4 T^{-6} A^{-4}]} = [M L^2 T^{-3} A^{-2}]$.
    Notice that $[M L^2 T^{-3} A^{-2}]$ is identical to the dimensions of electrical resistance ($[R]$). In vacuum, $Z_0 \approx 377\,\Omega$.
    (c) From the Lorentz force relation $F_E = q E$ and $F_B = q v B$, when electric and magnetic forces balance:
    $q E = q v B \implies \frac{E}{B} = v$
    $\left[\frac{E}{B}\right] = [v] = [L T^{-1}]$.

    Example 4 (Expressing Fundamental Quantities in Terms of Universal Constants):
    If the speed of light $c$, Planck’s constant $h$, and the universal gravitational constant $G$ are taken as fundamental base quantities, express mass ($M$) in terms of $c, h, G$.

    Solution:
    Let $[M] = [c]^x [h]^y [G]^z$.
    Substitute the respective dimensional formulas:
    $[M^1 L^0 T^0] = [L T^{-1}]^x [M L^2 T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$
    $[M^1 L^0 T^0] = [M^{y – z} L^{x + 2y + 3z} T^{-x – y – 2z}]$
    Equating exponents:
    1. For $M$: $y – z = 1 \implies y = z + 1$
    2. For $L$: $x + 2y + 3z = 0$
    3. For $T$: $-x – y – 2z = 0 \implies x = -y – 2z$
    Substitute (3) into (2):
    $(-y – 2z) + 2y + 3z = 0 \implies y + z = 0 \implies y = -z$
    Equating with (1):
    $-z = z + 1 \implies 2z = -1 \implies z = -\frac{1}{2}$
    $y = -z = \frac{1}{2}$
    $x = -y – 2z = -\frac{1}{2} – 2\left(-\frac{1}{2}\right) = -\frac{1}{2} + 1 = \frac{1}{2}$
    Hence, $[M] = [c^{1/2} h^{1/2} G^{-1/2}] = \sqrt{\frac{c h}{G}}$ (known as Planck Mass).

    Worksheet: 10 Practice Problems

    Problem 1 (JEE Main – Single Correct):
    Which of the following pairs of physical quantities have the exact same dimensions?
    (A) Torque and Angular momentum
    (B) Torque and Work
    (C) Linear momentum and Work
    (D) Force and Torque

    Problem 2 (JEE Main – Single Correct):
    The dimensional formula for the ratio of Planck’s constant $h$ to elementary charge $e$ (i.e., $[h/e]$) is:
    (A) $[M L^2 T^{-1} A^{-1}]$
    (B) $[M L^2 T^{-2} A^{-1}]$
    (C) $[M L T^{-1} A^{-2}]$
    (D) $[M L^2 T^{-3} A^{-1}]$

    Problem 3 (JEE Main – Single Correct):
    The velocity $v$ of a particle at time $t$ is given by the relation $v = a t + \frac{b}{t + c}$. The dimensional formulae of $a, b$, and $c$ are respectively:
    (A) $[L T^{-2}], [L], [T]$
    (B) $[L^2], [T], [L T^{-2}]$
    (C) $[L T^2], [L T], [T]$
    (D) $[L], [L T], [T^2]$

    Problem 4 (JEE Main – Single Correct):
    Stefan’s law states that radiant energy emitted per unit area per unit time by a blackbody is $E = \sigma T^4$. The dimensional formula of Stefan’s constant $\sigma$ is:
    (A) $[M L^0 T^{-3} K^{-4}]$
    (B) $[M L^2 T^{-3} K^{-4}]$
    (C) $[M L^{-1} T^{-2} K^{-4}]$
    (D) $[M L T^{-3} K^{-2}]$

    Problem 5 (JEE Main – Single Correct):
    Which of the following physical quantities has the dimensional formula $[M L^{-1} T^{-1}]$?
    (A) Surface Tension
    (B) Modulus of Rigidity
    (C) Coefficient of Viscosity
    (D) Power

    Problem 6 (JEE Advanced – One or More Correct):
    Which of the following groups of physical quantities share identical dimensions?
    (A) Pressure, Stress, Young’s Modulus
    (B) Work, Heat, Energy, Torque
    (C) Electric potential, Electromotive force (EMF), Potential difference
    (D) Surface tension, Spring constant, Force per unit length

    Problem 7 (JEE Advanced – One or More Correct):
    In electrical circuit theory, if $L$ represents inductance, $C$ represents capacitance, and $R$ represents electrical resistance, which of the following combinations have the dimension of time $[T]$?
    (A) $\frac{L}{R}$
    (B) $R C$
    (C) $\sqrt{L C}$
    (D) $\frac{1}{R C}$

    Problem 8 (JEE Advanced – One or More Correct):
    Let $\varepsilon_0$ denote the permittivity of vacuum and $\mu_0$ denote the permeability of vacuum. Which of the following statements is/are correct?
    (A) $[\mu_0 \varepsilon_0] = [L^{-2} T^2]$
    (B) $\left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right] = [M L^2 T^{-3} A^{-2}]$
    (C) $\left[\frac{1}{2} \varepsilon_0 E^2\right] = [M L^{-1} T^{-2}]$ (energy density)
    (D) $\left[\frac{B^2}{2\mu_0}\right] = [M L^{-1} T^{-2}]$ (magnetic energy density)

    Problem 9 (JEE Main / Advanced – Numerical Value Type):
    The dimensional formula of the magnetic permeability of free space $\mu_0$ is expressed as $[M^a L^b T^c A^d]$. Determine the numerical value of $|a| + |b| + |c| + |d|$.

    Problem 10 (JEE Main / Advanced – Numerical Value Type):
    The capacitance $C$ of a parallel plate capacitor has the dimensional formula $[M^{-1} L^{-2} T^x A^y]$. Determine the integer value of $x + y$.

    Solutions & Explanations

    Answer Key Summary:
    1. (B) | 2. (B) | 3. (A) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 6 | 10. 6

    Solution 1:
    – Torque $\vec{\tau} = \vec{r} \times \vec{F} \implies [\tau] = [L][M L T^{-2}] = [M L^2 T^{-2}]$.
    – Work $W = \vec{F} \cdot \vec{d} \implies [W] = [M L T^{-2}][L] = [M L^2 T^{-2}]$.
    Both torque and work have identical dimensions $[M L^2 T^{-2}]$.
    Correct Answer: (B)

    Solution 2:
    Planck’s constant: $[h] = [M L^2 T^{-1}]$.
    Elementary charge: $[e] = [q] = [A T]$.
    Ratio: $\left[\frac{h}{e}\right] = \frac{[M L^2 T^{-1}]}{[A T]} = [M L^2 T^{-2} A^{-1}]$.
    Notice this corresponds to magnetic flux ($\Phi_B = L I \implies [M L^2 T^{-2} A^{-1}]$).
    Correct Answer: (B)

    Solution 3:
    From the principle of homogeneity:
    1. In the denominator $(t + c)$, quantities added must have identical dimensions $\implies [c] = [t] = [T]$.
    2. In $v = a t$, $[a t] = [v] \implies [a][T] = [L T^{-1}] \implies [a] = [L T^{-2}]$.
    3. In $\frac{b}{t + c} = v$, $\frac{[b]}{[T]} = [L T^{-1}] \implies [b] = [L]$.
    Thus, $[a] = [L T^{-2}]$, $[b] = [L]$, $[c] = [T]$.
    Correct Answer: (A)

    Solution 4:
    Emissive power $E = \frac{\text{Energy}}{\text{Area} \times \text{Time}} \implies [E] = \frac{[M L^2 T^{-2}]}{[L^2][T]} = [M T^{-3}]$.
    Stefan’s Law: $E = \sigma T^4 \implies \sigma = \frac{E}{T^4}$.
    $[\sigma] = \frac{[M T^{-3}]}{[K^4]} = [M L^0 T^{-3} K^{-4}]$.
    Correct Answer: (A)

    Solution 5:
    From Stokes’ Law, $F = 6\pi \eta r v \implies \eta = \frac{F}{6\pi r v}$.
    $[\eta] = \frac{[M L T^{-2}]}{[L][L T^{-1}]} = [M L^{-1} T^{-1}]$.
    – Surface tension: $[M T^{-2}]$
    – Modulus of rigidity: $[M L^{-1} T^{-2}]$
    – Power: $[M L^2 T^{-3}]$.
    Correct Answer: (C)

    Solution 6:
    – (A) All represent force per unit area: $[M L^{-1} T^{-2}]$.
    – (B) Work, Heat, Energy, Torque all share $[M L^2 T^{-2}]$.
    – (C) Electric potential, EMF, and potential difference all represent work per unit charge: $[M L^2 T^{-3} A^{-1}]$.
    – (D) Surface tension ($F/L$) and Spring constant ($F/x$) both have $[M T^{-2}]$.
    All sets are correct.
    Correct Answer: (A, B, C, D)

    Solution 7:
    – Inductive time constant: $\tau_L = \frac{L}{R} \implies [T]$.
    – Capacitive time constant: $\tau_C = R C \implies [T]$.
    – Resonant oscillation frequency $\omega = \frac{1}{\sqrt{L C}} \implies [\sqrt{L C}] = \left[\frac{1}{\omega}\right] = [T]$.
    – $\frac{1}{R C}$ has dimensions of frequency $[T^{-1}]$.
    Hence, (A), (B), and (C) have dimensions of time.
    Correct Answer: (A, B, C)

    Solution 8:
    – (A) $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies \mu_0 \varepsilon_0 = \frac{1}{c^2} = [L^{-2} T^2]$. True.
    – (B) $\sqrt{\frac{\mu_0}{\varepsilon_0}}$ is wave impedance of vacuum ($377\,\Omega$), which has dimensions of resistance: $[M L^2 T^{-3} A^{-2}]$. True.
    – (C) $\frac{1}{2}\varepsilon_0 E^2$ is electrostatic energy density ($\text{Energy}/\text{Volume} = [M L^2 T^{-2}] / [L^3] = [M L^{-1} T^{-2}]$). True.
    – (D) $\frac{B^2}{2\mu_0}$ is magnetic energy density ($\text{Energy}/\text{Volume} = [M L^{-1} T^{-2}]$). True.
    All statements are correct.
    Correct Answer: (A, B, C, D)

    Solution 9:
    From Biot-Savart Law / Ampere’s Force Law: $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} \implies \mu_0 = \frac{2\pi r F}{I_1 I_2 L}$.
    $[\mu_0] = \frac{[L][M L T^{-2}]}{[A^2][L]} = [M^1 L^1 T^{-2} A^{-2}]$.
    Here $a = 1, b = 1, c = -2, d = -2$.
    Sum of absolute values: $|a| + |b| + |c| + |d| = 1 + 1 + 2 + 2 = 6$.
    Correct Answer: 6

    Solution 10:
    Capacitance $C = \frac{q}{V} = \frac{q^2}{W}$.
    $[q] = [A T] \implies [q^2] = [A^2 T^2]$.
    $[W] = [M L^2 T^{-2}]$.
    $[C] = \frac{[A^2 T^2]}{[M L^2 T^{-2}]} = [M^{-1} L^{-2} T^4 A^2]$.
    Comparing with $[M^{-1} L^{-2} T^x A^y]$:
    $x = 4$ and $y = 2$.
    $x + y = 4 + 2 = 6$.
    Correct Answer: 6

  • Errors in Measurement & Error Propagation: Absolute, Relative & Percentage Error | JEE Physics

    Concept Card: Errors in Measurement & Propagation of Errors

    1. Classification of Experimental Errors:
    Error is the difference between the true value and the measured value of a physical quantity:

    • Systematic Errors: Errors that tend to occur consistently in one direction (either consistently positive or consistently negative).
      • Instrumental errors: Imperfect calibration or zero error of the measuring instrument.
      • Imperfection in technique/procedure: Heat loss in calorimetry, air resistance effects, buoyancy errors.
      • Personal errors: Parallax errors and bias in taking readings.
      • Remedy: Identified and eliminated by calibration, zero correction, and refined procedures. Cannot be minimized by simple averaging.
    • Random Errors: Unpredictable, irregular variations occurring due to random fluctuations in experimental conditions (e.g., temperature changes, line voltage fluctuations, mechanical vibrations).
      • Follow a Gaussian normal distribution.
      • Remedy: Minimized by repeating the measurement $n$ times. If $n$ observations are taken, the random error is reduced by a factor of $\frac{1}{\sqrt{n}}$.
    • Least Count Error: The smallest division on the measuring scale determines the resolution limit and represents the absolute uncertainty of a single observation.

    2. Mathematical Definitions of Errors:
    For a set of $n$ measurements $a_1, a_2, \dots, a_n$ of a physical quantity:

    • Arithmetic Mean (Best Estimate of True Value):

      $a_{\text{mean}} = \bar{a} = \frac{1}{n}\sum_{i=1}^{n} a_i$
    • Absolute Error ($\Delta a_i$):

      $\Delta a_i = |a_{\text{mean}} – a_i|$ (always reported as a positive magnitude)
    • Mean Absolute Error ($\Delta a_{\text{mean}}$):

      $\Delta a_{\text{mean}} = \frac{1}{n}\sum_{i=1}^{n} \Delta a_i$

      Final reported measurement: $a = a_{\text{mean}} \pm \Delta a_{\text{mean}}$
    • Relative / Fractional Error:

      $\text{Relative Error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}$
    • Percentage Error ($\%E$):

      $\%E = \left(\frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}\right) \times 100\%$

    3. Rules for Combination and Propagation of Errors:

    • Sum and Difference: If $Z = A + B$ or $Z = A – B$, absolute errors always add:

      $\Delta Z = \Delta A + \Delta B$

      Maximum relative error: $\frac{\Delta Z}{Z} = \frac{\Delta A + \Delta B}{A \pm B}$
    • Product and Quotient: If $Z = A \cdot B$ or $Z = \frac{A}{B}$, relative errors add:

      $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$

      $\%Z = \%A + \%B$
    • Generalized Power Rule: If $Z = \frac{A^p B^q}{C^r}$:

      $\frac{\Delta Z}{Z} = p\left(\frac{\Delta A}{A}\right) + q\left(\frac{\Delta B}{B}\right) + r\left(\frac{\Delta C}{C}\right)$

      $\%Z = p(\%A) + q(\%B) + r(\%C)$
    • Calculus Method for Non-Linear Equations: If $Z = f(x, y)$:

      $\Delta Z_{\text{max}} = \left|\frac{\partial f}{\partial x}\right|\Delta x + \left|\frac{\partial f}{\partial y}\right|\Delta y$

    Solved Examples

    Example 1 (Mean Absolute Error & Percentage Error Calculation):
    The time period of oscillation of a simple pendulum is recorded in 5 successive measurements as: $2.63\text{ s}$, $2.56\text{ s}$, $2.42\text{ s}$, $2.71\text{ s}$, and $2.80\text{ s}$. Determine the mean period, mean absolute error, relative error, and percentage error.

    Solution:
    1. Mean time period:
    $T_{\text{mean}} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} = \frac{13.12}{5} = 2.624\text{ s} \approx 2.62\text{ s}$
    2. Absolute errors in each reading:
    $|\Delta T_1| = |2.62 – 2.63| = 0.01\text{ s}$
    $|\Delta T_2| = |2.62 – 2.56| = 0.06\text{ s}$
    $|\Delta T_3| = |2.62 – 2.42| = 0.20\text{ s}$
    $|\Delta T_4| = |2.62 – 2.71| = 0.09\text{ s}$
    $|\Delta T_5| = |2.62 – 2.80| = 0.18\text{ s}$
    3. Mean absolute error:
    $\Delta T_{\text{mean}} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} = \frac{0.54}{5} = 0.108\text{ s} \approx 0.11\text{ s}$
    Reported period: $T = (2.62 \pm 0.11)\text{ s}$.
    4. Relative error and percentage error:
    $\text{Relative Error} = \frac{0.11}{2.62} = 0.04198 \approx 0.042$
    $\text{Percentage Error} = 0.04198 \times 100\% = 4.2\%$.

    Example 2 (Errors in Parallel Combination of Resistors):
    Two resistors $R_1 = (100 \pm 3)\,\Omega$ and $R_2 = (200 \pm 4)\,\Omega$ are connected in parallel. Determine the equivalent resistance and the maximum percentage error in the equivalent resistance.

    Solution:
    1. Equivalent resistance $R_p$:
    $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} = 66.7\,\Omega$
    2. Error formula using differentiation:
    Differentiating $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$:
    $-\frac{\Delta R_p}{R_p^2} = -\frac{\Delta R_1}{R_1^2} – \frac{\Delta R_2}{R_2^2} \implies \frac{\Delta R_p}{R_p} = R_p \left[\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right]$
    Substitute the numerical values:
    $\frac{\Delta R_p}{R_p} = 66.7 \times \left[\frac{3}{(100)^2} + \frac{4}{(200)^2}\right] = 66.7 \times \left[\frac{3}{10000} + \frac{4}{40000}\right] = 66.7 \times \left[3 \times 10^{-4} + 1 \times 10^{-4}\right]$
    $\frac{\Delta R_p}{R_p} = 66.7 \times 4 \times 10^{-4} = 0.02668$
    3. Maximum percentage error:
    $\%R_p = 0.02668 \times 100\% \approx 2.7\%$.
    Absolute error: $\Delta R_p = 0.02668 \times 66.7\,\Omega \approx 1.8\,\Omega$.
    Reported value: $R_p = (66.7 \pm 1.8)\,\Omega$.

    Example 3 (Density Calculation with Propagation of Errors):
    A rectangular metal block has mass $m = (50.0 \pm 0.5)\text{ g}$, length $l = (10.0 \pm 0.1)\text{ cm}$, breadth $b = (5.0 \pm 0.1)\text{ cm}$, and thickness $h = (2.0 \pm 0.1)\text{ cm}$. Find the maximum percentage error in the density of the metal block.

    Solution:
    Density is given by $\rho = \frac{m}{V} = \frac{m}{l \cdot b \cdot h}$.
    Using the product and quotient error rule:
    $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + \frac{\Delta b}{b} + \frac{\Delta h}{h}$
    Calculate individual fractional errors:
    $\frac{\Delta m}{m} = \frac{0.5}{50.0} = 0.01 = 1.0\%$
    $\frac{\Delta l}{l} = \frac{0.1}{10.0} = 0.01 = 1.0\%$
    $\frac{\Delta b}{b} = \frac{0.1}{5.0} = 0.02 = 2.0\%$
    $\frac{\Delta h}{h} = \frac{0.1}{2.0} = 0.05 = 5.0\%$
    Total percentage error:
    $\%\rho = 1.0\% + 1.0\% + 2.0\% + 5.0\% = 9.0\%$.

    Example 4 (Determination of $g$ using Simple Pendulum):
    In an experiment to determine acceleration due to gravity $g$ using a simple pendulum, $T = 2\pi \sqrt{\frac{L}{g}}$. The measured length is $L = (100.0 \pm 0.1)\text{ cm}$, and the time for $20$ oscillations is measured as $t = (40.0 \pm 0.2)\text{ s}$ using a stopwatch. Find the percentage error in the measured value of $g$.

    Solution:
    From $T = 2\pi \sqrt{\frac{L}{g}} \implies g = 4\pi^2 \frac{L}{T^2}$.
    Since $T = \frac{t}{n}$ where $n = 20$ is an exact count, $\frac{\Delta T}{T} = \frac{\Delta t}{t}$.
    Thus, the fractional error formula is:
    $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\left(\frac{\Delta t}{t}\right)$
    Substitute the values:
    $\frac{\Delta L}{L} = \frac{0.1\text{ cm}}{100.0\text{ cm}} = 0.001 = 0.1\%$
    $\frac{\Delta t}{t} = \frac{0.2\text{ s}}{40.0\text{ s}} = 0.005 = 0.5\%$
    $\frac{\Delta g}{g} = 0.1\% + 2(0.5\%) = 0.1\% + 1.0\% = 1.1\%$.
    The maximum percentage error in $g$ is $1.1\%$.

    Worksheet: 10 Practice Problems

    Problem 1 (JEE Main – Single Correct):
    A physical quantity $P$ is determined from the relation $P = \frac{A^3 B^2}{\sqrt{C} \cdot D}$. The percentage errors in measuring $A, B, C$, and $D$ are $1\%$, $2\%$, $4\%$, and $3\%$ respectively. The maximum percentage error in $P$ is:
    (A) $12\%$
    (B) $14\%$
    (C) $10\%$
    (D) $8\%$

    Problem 2 (JEE Main – Single Correct):
    A student measures the diameter of a wire using a screw gauge with least count $0.001\text{ cm}$ and lists measurements as $0.050\text{ cm}$, $0.052\text{ cm}$, $0.051\text{ cm}$, and $0.051\text{ cm}$. The relative error in the diameter is approximately:
    (A) $0.01$
    (B) $0.02$
    (C) $0.05$
    (D) $0.005$

    Problem 3 (JEE Main – Single Correct):
    The resistance of a wire is given by $R = \frac{V}{I}$. If $V = (50 \pm 2)\text{ V}$ and $I = (5 \pm 0.2)\text{ A}$, the percentage error in $R$ is:
    (A) $4\%$
    (B) $8\%$
    (C) $6\%$
    (D) $2\%$

    Problem 4 (JEE Main – Single Correct):
    The error in the measurement of the radius of a solid sphere is $0.8\%$. The maximum percentage error in the calculated volume of the sphere is:
    (A) $0.8\%$
    (B) $1.6\%$
    (C) $2.4\%$
    (D) $3.2\%$

    Problem 5 (JEE Main – Single Correct):
    Two temperatures measured by a thermometer are $T_1 = (20.0 \pm 0.4)^\circ\text{C}$ and $T_2 = (45.0 \pm 0.3)^\circ\text{C}$. The temperature difference $\Delta T = T_2 – T_1$ along with its absolute error should be stated as:
    (A) $(25.0 \pm 0.1)^\circ\text{C}$
    (B) $(25.0 \pm 0.7)^\circ\text{C}$
    (C) $(25.0 \pm 0.35)^\circ\text{C}$
    (D) $(25.0 \pm 0.5)^\circ\text{C}$

    Problem 6 (JEE Advanced – One or More Correct):
    Which of the following statements is/are TRUE regarding errors in physical measurements?
    (A) Systematic errors cannot be minimized by simply taking a large number of readings.
    (B) Random errors are caused by unknown and unpredictable causes.
    (C) The fractional error is a dimensionless quantity.
    (D) If $Z = A – B$, the relative error in $Z$ is always smaller than the relative error in $A$.

    Problem 7 (JEE Advanced – One or More Correct):
    A physical quantity $Q$ is given by $Q = \frac{A – B}{A + B}$, where $A$ and $B$ are independently measured positive quantities with absolute errors $\Delta A$ and $\Delta B$. Which of the following expressions represent the maximum fractional error $\frac{\Delta Q}{Q}$?
    (A) $\frac{\Delta A + \Delta B}{A – B} + \frac{\Delta A + \Delta B}{A + B}$
    (B) $\frac{2(B \Delta A + A \Delta B)}{A^2 – B^2}$
    (C) $\frac{\Delta A}{A} + \frac{\Delta B}{B}$
    (D) $\frac{\Delta A – \Delta B}{A + B}$

    Problem 8 (JEE Advanced – One or More Correct):
    The heat generated in a resistor is given by $H = I^2 R t$. The percentage errors in the measurement of electric current $I$, resistance $R$, and time $t$ are $2\%$, $1\%$, and $1\%$ respectively. Which of the following statements is/are correct?
    (A) The maximum percentage error in $H$ is $6\%$.
    (B) The measurement of current $I$ contributes the maximum percentage uncertainty to $H$.
    (C) If the percentage error in $I$ is reduced to $1\%$, the error in $H$ becomes $4\%$.
    (D) The percentage error in $H$ is independent of the value of resistance $R$.

    Problem 9 (JEE Main / Advanced – Numerical Value Type):
    The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is the linear mass density. If the percentage error in tension is $3\%$ and the percentage error in linear mass density is $1\%$, the maximum percentage error in the wave speed is $x\%$. Find the value of $10 \times x$.

    Problem 10 (JEE Main / Advanced – Numerical Value Type):
    The focal length of a thin convex lens is determined using the lens formula $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$. The object distance is measured as $u = -(20 \pm 0.2)\text{ cm}$ and the real image distance is measured as $v = +(30 \pm 0.3)\text{ cm}$. The percentage error in the focal length $f$ is $p\%$. Find the value of $10 \times p$.

    Solutions & Explanations

    Answer Key Summary:
    1. (A) | 2. (A) | 3. (B) | 4. (C) | 5. (B) | 6. (A, B, C) | 7. (A, B) | 8. (A, B, C, D) | 9. 20 | 10. 10

    Solution 1:
    $P = \frac{A^3 B^2}{C^{1/2} D}$.
    $\frac{\Delta P}{P} = 3\left(\frac{\Delta A}{A}\right) + 2\left(\frac{\Delta B}{B}\right) + \frac{1}{2}\left(\frac{\Delta C}{C}\right) + \left(\frac{\Delta D}{D}\right)$.
    $\%P = 3(1\%) + 2(2\%) + \frac{1}{2}(4\%) + 3\% = 3\% + 4\% + 2\% + 3\% = 12\%$.
    Correct Answer: (A)

    Solution 2:
    Mean diameter $d_{\text{mean}} = \frac{0.050 + 0.052 + 0.051 + 0.051}{4} = \frac{0.204}{4} = 0.051\text{ cm}$.
    Absolute errors:
    $|\Delta d_1| = |0.051 – 0.050| = 0.001\text{ cm}$
    $|\Delta d_2| = |0.051 – 0.052| = 0.001\text{ cm}$
    $|\Delta d_3| = |0.051 – 0.051| = 0.000\text{ cm}$
    $|\Delta d_4| = |0.051 – 0.051| = 0.000\text{ cm}$
    Mean absolute error $\Delta d_{\text{mean}} = \frac{0.001 + 0.001 + 0 + 0}{4} = 0.0005\text{ cm}$.
    $\text{Relative error} = \frac{0.0005}{0.051} \approx 0.0098 \approx 0.01$.
    Correct Answer: (A)

    Solution 3:
    $R = \frac{V}{I} \implies \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
    $\frac{\Delta V}{V} = \frac{2}{50} = 4\%$
    $\frac{\Delta I}{I} = \frac{0.2}{5} = 4\%$
    $\%R = 4\% + 4\% = 8\%$.
    Correct Answer: (B)

    Solution 4:
    Volume of sphere $V = \frac{4}{3}\pi r^3 \implies \frac{\Delta V}{V} = 3\left(\frac{\Delta r}{r}\right)$.
    $\%V = 3 \times 0.8\% = 2.4\%$.
    Correct Answer: (C)

    Solution 5:
    For subtraction $\Delta T = T_2 – T_1 = 45.0 – 20.0 = 25.0^\circ\text{C}$.
    Absolute errors in difference always add:
    $\Delta(\Delta T) = \Delta T_1 + \Delta T_2 = 0.4 + 0.3 = 0.7^\circ\text{C}$.
    Reported value: $(25.0 \pm 0.7)^\circ\text{C}$.
    Correct Answer: (B)

    Solution 6:
    – (A) True: Systematic errors are unidirectional and constant; averaging does not eliminate them.
    – (B) True: Random errors fluctuate symmetrically and follow Gaussian distribution.
    – (C) True: Fractional error is a ratio of identical quantities $\frac{\Delta x}{x}$, hence dimensionless.
    – (D) False: Since denominator $A – B$ can be very small, relative error $\frac{\Delta A + \Delta B}{A – B}$ can be significantly larger.
    Correct Answer: (A, B, C)

    Solution 7:
    Taking natural logarithm of $Q = \frac{A – B}{A + B}$:
    $\ln Q = \ln(A – B) – \ln(A + B)$
    Differentiating:
    $\frac{dQ}{Q} = \frac{dA – dB}{A – B} – \frac{dA + dB}{A + B}$
    For maximum error, add the absolute contributions:
    $\frac{\Delta Q}{Q} = \frac{\Delta A + \Delta B}{A – B} + \frac{\Delta A + \Delta B}{A + B}$ → (A) is correct.
    Combining the terms over a common denominator:
    $\frac{\Delta Q}{Q} = \frac{(\Delta A + \Delta B)(A + B) + (\Delta A + \Delta B)(A – B)}{A^2 – B^2} = \frac{2 A \Delta A + 2 A \Delta B \dots}{A^2 – B^2}$
    Alternatively, differentiating directly:
    $dQ = \frac{(A + B)(dA – dB) – (A – B)(dA + dB)}{(A + B)^2} = \frac{2 B \, dA – 2 A \, dB}{(A + B)^2}$
    Dividing by $Q = \frac{A – B}{A + B}$:
    $\frac{dQ}{Q} = \frac{2(B \, dA – A \, dB)}{(A – B)(A + B)} \implies \frac{\Delta Q_{\text{max}}}{Q} = \frac{2(B \Delta A + A \Delta B)}{A^2 – B^2}$ → (B) is also correct.
    Correct Answer: (A, B)

    Solution 8:
    $H = I^2 R t \implies \frac{\Delta H}{H} = 2\left(\frac{\Delta I}{I}\right) + \frac{\Delta R}{R} + \frac{\Delta t}{t}$.
    $\%H = 2(2\%) + 1\% + 1\% = 4\% + 1\% + 1\% = 6\%$.
    – Current has power $2$, contributing $4\%$ out of $6\%$ total (maximum contributor).
    – If $\%I = 1\%$, $\%H = 2(1\%) + 1\% + 1\% = 4\%$.
    – Percentage error depends only on relative errors, not the absolute magnitude of $R$.
    All options are correct.
    Correct Answer: (A, B, C, D)

    Solution 9:
    $v = T^{1/2} \mu^{-1/2}$.
    $\frac{\Delta v}{v} = \frac{1}{2}\left(\frac{\Delta T}{T}\right) + \frac{1}{2}\left(\frac{\Delta \mu}{\mu}\right)$.
    $\%v = \frac{1}{2}(3\%) + \frac{1}{2}(1\%) = 1.5\% + 0.5\% = 2.0\%$.
    So $x = 2.0 \implies 10 \times x = 10 \times 2 = 20$.
    Correct Answer: 20

    Solution 10:
    For lens formula $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$:
    With $u = -20\text{ cm}$ and $v = +30\text{ cm}$:
    $\frac{1}{f} = \frac{1}{30} – \left(-\frac{1}{20}\right) = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12} \implies f = 12\text{ cm}$.
    Differentiating $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$:
    $-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} \implies \frac{\Delta f}{f} = f \left[\frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}\right]$
    Substitute $|u| = 20$, $\Delta u = 0.2$, $v = 30$, $\Delta v = 0.3$, $f = 12$:
    $\frac{\Delta f}{f} = 12 \left[\frac{0.3}{900} + \frac{0.2}{400}\right] = 12 \left[\frac{1}{3000} + \frac{1}{2000}\right] = 12 \left[\frac{2 + 3}{6000}\right] = 12 \times \frac{5}{6000} = \frac{60}{6000} = 0.01 = 1.0\%$.
    Thus, percentage error $p = 1.0\% \implies 10 \times p = 10 \times 1.0 = 10$.
    Correct Answer: 10

  • Significant Figures & Least Count of Measuring Instruments: Rules & Applications | JEE Physics

    Concept Card: Least Count of Measuring Instruments & Significant Figures

    1. Least Count (LC) of Measuring Instruments:
    The least count is the smallest value that can be directly measured by an instrument. It defines the precision and instrumental resolution limit.

    A. Vernier Calipers:
    $\text{Least Count (LC)} = 1\text{ Main Scale Division (MSD)} – 1\text{ Vernier Scale Division (VSD)}$
    If $n$ divisions of the Vernier scale coincide with $(n – 1)$ divisions of the main scale:
    $n\text{ VSD} = (n – 1)\text{ MSD} \implies 1\text{ VSD} = \frac{n – 1}{n}\text{ MSD}$
    $\text{LC} = 1\text{ MSD} – \frac{n – 1}{n}\text{ MSD} = \frac{1\text{ MSD}}{n}$
    Total Reading:
    $\text{Total Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) – (\text{Zero Error})$

    B. Screw Gauge (Micrometer):
    $\text{Pitch} = \frac{\text{Distance advanced on linear main scale}}{\text{Number of full rotations completed}}$
    $\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of circular scale divisions (CSD)}}$
    Total Reading:
    $\text{Total Reading} = \text{Linear Scale Reading (LSR)} + (\text{CSR} \times \text{LC}) – (\text{Zero Error})$

    C. Zero Error Types:

    • Positive Zero Error: The zero mark of the vernier or circular scale lies to the right (or above) the main scale reference line when jaws/anvils are closed. The measured value is larger than actual. $\text{Correction} = -|\text{Zero Error}|$.
    • Negative Zero Error: The zero mark lies to the left (or below) the main scale reference line. The measured value is smaller than actual. $\text{Correction} = +|\text{Zero Error}|$.

    2. Rules for Determining Significant Figures (SF):

    1. All non-zero digits are significant (e.g., $18.45$ has 4 SF).
    2. All zeros occurring between two non-zero digits are significant (e.g., $2005$ has 4 SF, $1.0004$ has 5 SF).
    3. Leading zeros (to the left of the first non-zero digit) are never significant; they merely locate the decimal point (e.g., $0.0034$ has 2 SF).
    4. Trailing zeros after a decimal point are significant (e.g., $5.200$ has 4 SF, $0.0700$ has 3 SF).
    5. Trailing zeros in an integer without a decimal point are generally non-significant unless given in scientific notation (e.g., $4300$ has 2 SF; $4.30 \times 10^3$ has 3 SF).
    6. Exact numbers or pure geometric factors have infinite significant figures (e.g., $\pi$, $2$ in $2\pi r$).

    3. Arithmetic Operations with Significant Figures:

    • Addition & Subtraction: The final calculated result must have no more decimal places than the measurement with the fewest decimal places.
    • Multiplication & Division: The final calculated result must contain no more significant figures than the quantity with the least number of significant figures.
    • Rounding Off Rule: If the dropped digit is $5$ followed only by zeros, the preceding digit is left unchanged if it is even, and increased by $1$ if it is odd (e.g., $4.750 \to 4.8$, while $4.650 \to 4.6$).

    Solved Examples

    Example 1 (Vernier Calipers with Positive Zero Error):
    A vernier calipers has $1\text{ mm}$ main scale divisions. $10$ divisions of the vernier scale coincide with $9$ main scale divisions. Before measuring, with jaws closed, the zero of the vernier scale is to the right of the main scale zero and its $3\text{rd}$ division coincides with a main scale mark. While measuring the diameter of a cylinder, the main scale reads $4.2\text{ cm}$ and the $7\text{th}$ vernier division coincides. Find the true diameter.

    Solution:
    1. Least count of instrument:
    $\text{LC} = \frac{1\text{ MSD}}{10} = \frac{1\text{ mm}}{10} = 0.1\text{ mm} = 0.01\text{ cm}$.
    2. Zero error:
    Since zero lies to the right, it is a positive zero error:
    $\text{Zero Error} = +3 \times \text{LC} = +3 \times 0.01\text{ cm} = +0.03\text{ cm}$.
    3. Observed reading:
    $\text{Observed Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) = 4.2\text{ cm} + (7 \times 0.01\text{ cm}) = 4.27\text{ cm}$.
    4. Corrected reading:
    $\text{True Reading} = \text{Observed Reading} – \text{Zero Error} = 4.27\text{ cm} – (+0.03\text{ cm}) = 4.24\text{ cm}$.

    Example 2 (Screw Gauge with Negative Zero Error):
    A screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular scale. When the anvils are in contact, the circular scale zero is $3$ divisions below the baseline. When a wire is clamped, the linear scale shows $2.5\text{ mm}$ and the $28\text{th}$ circular division aligns with the baseline. Determine the actual diameter.

    Solution:
    1. Least count:
    $\text{LC} = \frac{\text{Pitch}}{\text{Number of CSD}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$.
    2. Zero error:
    Since the zero mark has not yet reached the baseline (lies $3$ divisions below), the zero error is negative:
    $\text{Zero Error} = -3 \times \text{LC} = -0.03\text{ mm}$.
    3. Observed reading:
    $\text{Observed Reading} = \text{LSR} + (\text{CSR} \times \text{LC}) = 2.5\text{ mm} + (28 \times 0.01\text{ mm}) = 2.78\text{ mm}$.
    4. True diameter:
    $\text{True Reading} = \text{Observed Reading} – (\text{Zero Error}) = 2.78\text{ mm} – (-0.03\text{ mm}) = 2.81\text{ mm}$.

    Example 3 (Significant Figures in Multiplication & Addition):
    A thin metal plate has length $l = 3.124\text{ m}$, breadth $b = 1.05\text{ m}$, and thickness $t = 2.5\text{ cm}$. Calculate the total volume of the plate expressed to the correct number of significant figures.

    Solution:
    1. Convert all dimensions to meters:
    $l = 3.124\text{ m}$ (4 significant figures)
    $b = 1.05\text{ m}$ (3 significant figures)
    $t = 2.5\text{ cm} = 0.025\text{ m}$ (2 significant figures)
    2. Compute volume:
    $V = l \times b \times t = 3.124 \times 1.05 \times 0.025\text{ m}^3 = 0.082005\text{ m}^3$
    3. Apply significant figure rule for multiplication:
    The factor with the fewest significant figures is $t$ (which has 2 SF). The final result must therefore be rounded off to 2 significant figures:
    $V = 0.082\text{ m}^3 = 8.2 \times 10^{-2}\text{ m}^3$.

    Example 4 (Density Calculation with Appropriate Rounding):
    The mass of a solid block is measured as $m = 48.20\text{ g}$ and its volume is measured as $V = 6.4\text{ cm}^3$. Compute the density of the block to the appropriate number of significant figures.

    Solution:
    1. Mass has 4 significant figures ($48.20\text{ g}$).
    2. Volume has 2 significant figures ($6.4\text{ cm}^3$).
    3. Density calculation:
    $\rho = \frac{m}{V} = \frac{48.20}{6.4} = 7.53125\text{ g/cm}^3$
    Since volume limits the precision to 2 significant figures, we round $7.53125$ to 2 significant figures:
    $\rho = 7.5\text{ g/cm}^3$.

    Worksheet: 10 Practice Problems

    Problem 1 (JEE Main – Single Correct):
    $1\text{ cm}$ of the main scale of a vernier calipers is divided into $10$ equal parts. If $20$ divisions of the vernier scale coincide with $19$ small main scale divisions, the least count of the calipers is:
    (A) $0.05\text{ mm}$
    (B) $0.01\text{ mm}$
    (C) $0.005\text{ mm}$
    (D) $0.02\text{ mm}$

    Problem 2 (JEE Main – Single Correct):
    The number of significant figures in the numbers $0.0050$, $2.040$, and $400.0$ are respectively:
    (A) 2, 4, 4
    (B) 1, 3, 1
    (C) 2, 3, 4
    (D) 2, 4, 1

    Problem 3 (JEE Main – Single Correct):
    Evaluate the arithmetic expression $(15.82 – 2.5) \times 0.20$ and report the result to the correct number of significant figures:
    (A) $2.664$
    (B) $2.66$
    (C) $2.7$
    (D) $3$

    Problem 4 (JEE Main – Single Correct):
    A screw gauge has pitch $0.5\text{ mm}$ and $100$ divisions on its circular scale. When measured with jaws closed, the $4\text{th}$ division of the circular scale lies below the reference line. While measuring a wire’s thickness, the linear scale reads $2\text{ mm}$ and the circular scale reads $46$. The true thickness of the wire is:
    (A) $2.25\text{ mm}$
    (B) $2.23\text{ mm}$
    (C) $2.21\text{ mm}$
    (D) $2.50\text{ mm}$

    Problem 5 (JEE Main – Single Correct):
    When $3.924\text{ g}$ is subtracted from $8.7\text{ g}$, the result expressed to the correct significant figures is:
    (A) $4.776\text{ g}$
    (B) $4.78\text{ g}$
    (C) $4.8\text{ g}$
    (D) $5\text{ g}$

    Problem 6 (JEE Advanced – One or More Correct):
    Which of the following statements is/are TRUE regarding significant figures and measurements?
    (A) Change of units does not change the number of significant figures in a measurement.
    (B) The least count of an instrument determines the absolute uncertainty of a single reading.
    (C) The number $3.2000 \times 10^3$ has 5 significant figures.
    (D) A measured value with more significant figures is inherently more precise than one with fewer significant figures.

    Problem 7 (JEE Advanced – One or More Correct):
    In a specially designed vernier calipers, $1\text{ MSD} = 1\text{ mm}$, and $(N + 1)$ divisions of the vernier scale coincide with $N$ main scale divisions (retrograde vernier). Which of the following statements is/are correct?
    (A) $1\text{ VSD} = \frac{N}{N + 1}\text{ mm}$
    (B) $\text{Least count} = \frac{1}{N + 1}\text{ mm}$
    (C) The vernier scale divisions are smaller than the main scale divisions.
    (D) The vernier scale reading is added to the main scale reading when taking measurements.

    Problem 8 (JEE Advanced – One or More Correct):
    A screw gauge has a pitch of $1\text{ mm}$ and $100$ circular scale divisions. Which of the following readings can be measured with this instrument without estimation?
    (A) $5.42\text{ mm}$
    (B) $5.425\text{ mm}$
    (C) $0.07\text{ mm}$
    (D) $3.00\text{ mm}$

    Problem 9 (JEE Main / Advanced – Numerical Value Type):
    A vernier calipers has $20$ divisions on its vernier scale which coincide with $19$ divisions on the main scale. The smallest division on the main scale is $0.5\text{ mm}$. When measuring a spherical bob, the main scale reads $2.5\text{ cm}$ and the $14\text{th}$ vernier division coincides with a main scale line. If the instrument has a positive zero error of $+0.05\text{ mm}$, the corrected diameter of the bob is $D\text{ mm}$. Find the value of $D \times 10$.

    Problem 10 (JEE Main / Advanced – Numerical Value Type):
    The mass of a cube is measured as $m = 10.00\text{ g}$ and its edge length is measured as $l = 2.0\text{ cm}$. The density of the cube reported to the correct number of significant figures is $d\text{ g/cm}^3$. Calculate the numerical value of $100 \times d$.

    Solutions & Explanations

    Answer Key Summary:
    1. (A) | 2. (A) | 3. (C) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, C, D) | 9. 253 | 10. 130

    Solution 1:
    $1\text{ MSD} = \frac{1\text{ cm}}{10} = 1\text{ mm} = 0.1\text{ cm}$.
    $20\text{ VSD} = 19\text{ MSD} \implies 1\text{ VSD} = \frac{19}{20}\text{ MSD}$.
    $\text{LC} = 1\text{ MSD} – 1\text{ VSD} = 1\text{ MSD} – \frac{19}{20}\text{ MSD} = \frac{1}{20}\text{ MSD} = \frac{1\text{ mm}}{20} = 0.05\text{ mm}$.
    Correct Answer: (A)

    Solution 2:
    – $0.0050$: Leading zeros are not significant; trailing zero after decimal is significant $\implies 2$ SF.
    – $2.040$: Embedded zero and trailing zero after decimal are significant $\implies 4$ SF.
    – $400.0$: Decimal is present, so trailing zeros are significant $\implies 4$ SF.
    Correct Answer: (A)

    Solution 3:
    First step (subtraction): $15.82 – 2.5 = 13.32$.
    Since $2.5$ has $1$ decimal place, the result of subtraction is limited to $1$ decimal place ($13.3$).
    Second step (multiplication): $13.3 \times 0.20$.
    $13.3$ has $3$ SF and $0.20$ has $2$ SF. The final product must have $2$ significant figures:
    $13.3 \times 0.20 = 2.66 \implies \text{rounded to 2 SF} = 2.7$.
    Correct Answer: (C)

    Solution 4:
    $\text{LC} = \frac{\text{Pitch}}{\text{CSD}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}$.
    Zero error: The $4\text{th}$ division lies below the reference line, so the error is negative:
    $\text{Zero Error} = -4 \times \text{LC} = -4 \times 0.005\text{ mm} = -0.020\text{ mm}$.
    Observed reading:
    $\text{Observed} = \text{LSR} + (\text{CSR} \times \text{LC}) = 2\text{ mm} + (46 \times 0.005\text{ mm}) = 2\text{ mm} + 0.230\text{ mm} = 2.230\text{ mm}$.
    True thickness:
    $\text{True Reading} = 2.230\text{ mm} – (-0.020\text{ mm}) = 2.250\text{ mm} = 2.25\text{ mm}$.
    Correct Answer: (A)

    Solution 5:
    $8.7\text{ g} – 3.924\text{ g} = 4.776\text{ g}$.
    In subtraction, the result can only have as many decimal places as the number with the least decimal places ($8.7$ has 1 decimal place).
    Rounding $4.776$ to 1 decimal place gives $4.8\text{ g}$.
    Correct Answer: (C)

    Solution 6:
    (A) True: Converting $2.4\text{ m}$ to $240\text{ cm}$ retains 2 SF ($2.4 \times 10^2\text{ cm}$).
    (B) True: Absolute resolution limit is bounded by instrument least count.
    (C) True: In $3.2000 \times 10^3$, all digits before the exponent are significant (5 SF).
    (D) True: More significant figures reflect smaller relative uncertainty and higher precision.
    Correct Answer: (A, B, C, D)

    Solution 7:
    $(N + 1)\text{ VSD} = N\text{ MSD} \implies 1\text{ VSD} = \frac{N}{N + 1}\text{ MSD} = \frac{N}{N + 1}\text{ mm}$.
    $\text{LC} = 1\text{ MSD} – 1\text{ VSD} = 1 – \frac{N}{N + 1} = \frac{1}{N + 1}\text{ mm}$.
    Since $\frac{N}{N + 1} < 1$, $1\text{ VSD} < 1\text{ MSD}$.
    Total reading is still formed by $\text{MSR} + (\text{VSR} \times \text{LC})$.
    All statements (A, B, C, D) are correct.
    Correct Answer: (A, B, C, D)

    Solution 8:
    $\text{LC} = \frac{1\text{ mm}}{100} = 0.01\text{ mm}$.
    An instrument with $\text{LC} = 0.01\text{ mm}$ can directly measure any value that is an integer multiple of $0.01\text{ mm}$ up to two decimal places in mm.
    – $5.42\text{ mm}$ is $542 \times 0.01\text{ mm}$ → Valid.
    – $0.07\text{ mm}$ is $7 \times 0.01\text{ mm}$ → Valid.
    – $3.00\text{ mm}$ is $300 \times 0.01\text{ mm}$ → Valid.
    – $5.425\text{ mm}$ requires estimation of half a division (third decimal place) → Not directly measurable.
    Correct Answer: (A, C, D)

    Solution 9:
    $\text{LC} = \frac{1\text{ MSD}}{20} = \frac{0.5\text{ mm}}{20} = 0.025\text{ mm}$.
    Main scale reading: $\text{MSR} = 2.5\text{ cm} = 25.0\text{ mm}$.
    Vernier scale reading: $\text{VSR} = 14 \times 0.025\text{ mm} = 0.35\text{ mm}$.
    Observed reading: $25.0 + 0.35 = 25.35\text{ mm}$.
    Zero error is $+0.05\text{ mm}$.
    Corrected diameter: $D = 25.35 – 0.05 = 25.30\text{ mm} = 25.3\text{ mm}$.
    Therefore, $D \times 10 = 25.3 \times 10 = 253$.
    Correct Answer: 253

    Solution 10:
    Volume of cube $V = l^3 = (2.0\text{ cm})^3 = 8.0\text{ cm}^3$ (2 significant figures).
    Mass $m = 10.00\text{ g}$ (4 significant figures).
    Density $d = \frac{m}{V} = \frac{10.00}{8.0} = 1.25\text{ g/cm}^3$.
    Rounding to 2 significant figures (since $V$ has 2 SF):
    Following rounding off rules for $5$ preceded by an even number ($2$ is even), $1.25 \to 1.2$ or rounding normally to nearest even gives $1.2\text{ g/cm}^3$ (or standard rounding $1.3\text{ g/cm}^3$). With standard nearest rounding: $d = 1.3\text{ g/cm}^3$.
    Thus, $100 \times d = 100 \times 1.3 = 130$.
    Correct Answer: 130

  • Units of Measurement & Systems of Units (SI, CGS, MKS): Fundamental and Derived Units | JEE Physics

    Concept Card: Systems of Units & Physical Quantities

    1. Physical Quantities & Invariance:
    A physical quantity $Q$ is completely expressed by a numerical value $n$ and a unit $u$:
    $Q = n \times u$
    Since the physical magnitude remains constant regardless of the measurement scale:
    $n_1 u_1 = n_2 u_2 = \text{constant} \implies n \propto \frac{1}{u}$
    The larger the chosen unit, the smaller the numerical magnitude.

    2. Historical Systems of Units:

    • CGS System: Centimeter ($\text{cm}$), Gram ($\text{g}$), Second ($\text{s}$).
    • FPS System: Foot ($\text{ft}$), Pound ($\text{lb}$), Second ($\text{s}$).
    • MKS System: Meter ($\text{m}$), Kilogram ($\text{kg}$), Second ($\text{s}$).
    • SI System (Système International d’Unités): Internationally accepted metric system containing 7 base units and 2 supplementary units.

    3. The 7 SI Base Quantities:

    1. Length: meter ($\text{m}$)
    2. Mass: kilogram ($\text{kg}$)
    3. Time: second ($\text{s}$)
    4. Electric Current: ampere ($\text{A}$)
    5. Thermodynamic Temperature: kelvin ($\text{K}$)
    6. Amount of Substance: mole ($\text{mol}$)
    7. Luminous Intensity: candela ($\text{cd}$)

    4. Supplementary Units:

    • Plane Angle ($d\theta = \frac{ds}{r}$): radian ($\text{rad}$) — Dimensionless $[M^0 L^0 T^0]$.
    • Solid Angle ($d\Omega = \frac{dA}{r^2}$): steradian ($\text{sr}$) — Dimensionless $[M^0 L^0 T^0]$.

    5. Derived Quantities & Important Standard Conversions:

    • Force: $1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2} = 10^5\text{ dynes}$
    • Work / Energy: $1\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2} = 10^7\text{ ergs}$
    • Pressure: $1\text{ Pa} = 1\text{ N}\cdot\text{m}^{-2} = 10\text{ dyne}\cdot\text{cm}^{-2}$
    • Universal Gravitational Constant: $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2\cdot\text{kg}^{-2} = 6.67 \times 10^{-8}\text{ dyne}\cdot\text{cm}^2\cdot\text{g}^{-2}$
    • Astronomical Unit ($1\text{ AU}$): $1.496 \times 10^{11}\text{ m}$
    • Light Year ($1\text{ ly}$): $9.46 \times 10^{15}\text{ m}$
    • Parsec ($1\text{ pc}$): $3.08 \times 10^{16}\text{ m} \approx 3.26\text{ ly}$

    Solved Examples

    Example 1 (Direct Unit Conversion via $n_1 u_1 = n_2 u_2$):
    The density of mercury is $13.6\text{ g/cm}^3$ in CGS units. Determine its numerical value in SI units ($\text{kg/m}^3$).

    Solution:
    Density has the dimension $[\rho] = [M L^{-3}]$.
    Using $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^{-3}$:
    $n_2 = 13.6 \times \left[\frac{1\text{ g}}{1\text{ kg}}\right]^1 \times \left[\frac{1\text{ cm}}{1\text{ m}}\right]^{-3}$
    $n_2 = 13.6 \times \left[10^{-3}\right] \times \left[10^{-2}\right]^{-3} = 13.6 \times 10^{-3} \times 10^6 = 13.6 \times 10^3 = 13600$
    Thus, the density in SI units is $1.36 \times 10^4\text{ kg/m}^3$.

    Example 2 (Finding Units of Constants from Physical Laws):
    In the Van der Waals equation of state for a real gas:
    $\left(P + \frac{a}{V^2}\right)(V – b) = R T$
    where $P$ is pressure and $V$ is volume, determine the SI units of the constants $a$ and $b$.

    Solution:
    By the principle of homogeneity, only physical quantities having identical dimensions can be added or subtracted.
    1. For constant $b$:
    $[b] = [V] \implies \text{Unit of } b = \text{m}^3$.
    2. For constant $a$:
    $\left[\frac{a}{V^2}\right] = [P] \implies [a] = [P][V]^2$
    Since the SI unit of $P$ is $\text{N/m}^2$ (or $\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$) and $V$ is $\text{m}^3$:
    $\text{Unit of } a = (\text{N/m}^2) \times (\text{m}^3)^2 = \text{N}\cdot\text{m}^4 = \text{kg}\cdot\text{m}^5\cdot\text{s}^{-2}$.

    Example 3 (Base Quantity Representation in Alternative Fundamental Systems):
    If force ($F$), velocity ($v$), and time ($t$) are chosen as the fundamental base quantities, express mass ($M$) in terms of these quantities.

    Solution:
    Let $[M] = [F]^a [v]^b [t]^c$.
    Substituting standard dimensions in terms of $M, L, T$:
    $[M^1 L^0 T^0] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c = [M^a L^{a+b} T^{-2a-b+c}]$
    Equating exponents on both sides:
    For $M$: $a = 1$
    For $L$: $a + b = 0 \implies b = -a = -1$
    For $T$: $-2a – b + c = 0 \implies -2(1) – (-1) + c = 0 \implies -2 + 1 + c = 0 \implies c = 1$
    Hence, $[M] = [F^1 v^{-1} t^1] = F v^{-1} t$.

    Example 4 (Energy Conversion to a Custom Arbitrary System):
    A heat engine absorbs $4.2 \times 10^6\text{ J}$ of energy. Find the numerical magnitude of this energy in a new system where the unit of mass is $10\text{ kg}$, the unit of length is $100\text{ m}$, and the unit of time is $1\text{ minute}$.

    Solution:
    Energy has dimensions $[E] = [M L^2 T^{-2}]$.
    Given: $n_1 = 4.2 \times 10^6$, $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
    New system: $M_2 = 10\text{ kg}$, $L_2 = 100\text{ m}$, $T_2 = 1\text{ min} = 60\text{ s}$.
    $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$
    $n_2 = (4.2 \times 10^6) \times \left(\frac{1}{10}\right)^1 \times \left(\frac{1}{100}\right)^2 \times \left(\frac{1}{60}\right)^{-2}$
    $n_2 = (4.2 \times 10^6) \times 10^{-1} \times 10^{-4} \times (3600) = (4.2 \times 10^1) \times 3600 = 42 \times 3600 = 151200$
    The numerical value in the new system is $1.512 \times 10^5$.

    Worksheet: 10 Practice Problems

    Problem 1 (JEE Main – Single Correct):
    The base SI units of the universal gravitational constant $G$ are:
    (A) $\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$
    (B) $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$
    (C) $\text{kg}^{-2}\cdot\text{m}^3\cdot\text{s}^{-1}$
    (D) $\text{kg}^{-1}\cdot\text{m}^2\cdot\text{s}^{-2}$

    Problem 2 (JEE Main – Single Correct):
    If the unit of length and the unit of force are both increased by a factor of 4, the unit of energy will increase by a factor of:
    (A) 4
    (B) 8
    (C) 16
    (D) 32

    Problem 3 (JEE Main – Single Correct):
    In a new system of units, the unit of mass is $\alpha\text{ kg}$, the unit of length is $\beta\text{ m}$, and the unit of time is $\gamma\text{ s}$. The magnitude of $1\text{ calorie} = 4.2\text{ J}$ in this new system is:
    (A) $4.2 \, \alpha^{-1} \beta^{-2} \gamma^2$
    (B) $4.2 \, \alpha \beta^2 \gamma^{-2}$
    (C) $4.2 \, \alpha^{-1} \beta^{-1} \gamma^2$
    (D) $4.2 \, \alpha^2 \beta^{-2} \gamma$

    Problem 4 (JEE Main – Single Correct):
    The ratio of the SI unit of universal gas constant $R$ to its CGS unit is:
    (A) $10^7$
    (B) $10^5$
    (C) $10^{-7}$
    (D) $10^{-5}$

    Problem 5 (JEE Main – Single Correct):
    In the relation $X = 3 Y Z^2$, $X$ has dimensions of capacitance ($[M^{-1} L^{-2} T^4 A^2]$) and $Z$ has dimensions of magnetic induction ($[M T^{-2} A^{-1}]$). What are the SI units of $Y$?
    (A) $\text{kg}^{-3}\cdot\text{m}^{-2}\cdot\text{s}^8\cdot\text{A}^4$
    (B) $\text{kg}^{-2}\cdot\text{m}^{-1}\cdot\text{s}^6\cdot\text{A}^3$
    (C) $\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2$
    (D) $\text{kg}^{-3}\cdot\text{m}^{-1}\cdot\text{s}^7\cdot\text{A}^4$

    Problem 6 (JEE Advanced – One or More Correct):
    Which of the following statements is/are TRUE concerning physical quantities and units?
    (A) A quantity can have a unit while being dimensionless.
    (B) A quantity can have dimensions while having no unit.
    (C) All fundamental quantities in SI are scalar quantities.
    (D) The numerical value of a physical quantity varies inversely with the magnitude of the unit chosen.

    Problem 7 (JEE Advanced – One or More Correct):
    If the speed of light $c$, Planck’s constant $h$, and the universal gravitational constant $G$ are chosen as fundamental quantities, which of the following expressions represent a fundamental base quantity?
    (A) Planck length $l_p = \sqrt{\frac{G h}{c^3}}$
    (B) Planck mass $m_p = \sqrt{\frac{h c}{G}}$
    (C) Planck time $t_p = \sqrt{\frac{G h}{c^5}}$
    (D) Planck energy $E_p = \sqrt{\frac{h c^5}{G}}$

    Problem 8 (JEE Advanced – One or More Correct):
    Select the pair(s) of physical quantities that share the exact same base SI units:
    (A) Torque and Work
    (B) Angular momentum and Planck’s constant
    (C) Surface tension and Spring constant
    (D) Stress and Modulus of Elasticity

    Problem 9 (JEE Main / Advanced – Numerical Value Type):
    A force of $10\text{ N}$ acts on an object. In a hypothetical system where the unit of mass is $2\text{ kg}$, the unit of length is $0.5\text{ m}$, and the unit of time is $2\text{ s}$, the numerical value of this force is $F_{\text{new}}$. Determine the value of $F_{\text{new}}$.

    Problem 10 (JEE Main / Advanced – Numerical Value Type):
    An electric heater delivers a power of $1000\text{ W}$. In a new system of units where the unit of mass is $100\text{ g}$, the unit of length is $10\text{ cm}$, and the unit of time is $0.1\text{ s}$, this power is equal to $k \times 10^4$ new units. Find the integer value of $k$.

    Solutions & Explanations

    Answer Key Summary:
    1. (A) | 2. (C) | 3. (A) | 4. (A) | 5. (A) | 6. (A, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 40 | 10. 100

    Solution 1:
    From Newton’s law of gravitation, $F = \frac{G m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2}$.
    In base SI units:
    $[G] = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot(\text{m}^2)}{\text{kg}^2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$.
    Correct Answer: (A)

    Solution 2:
    Energy = $\text{Force} \times \text{Length}$.
    If unit of force becomes $4 F_0$ and unit of length becomes $4 L_0$, then:
    $\text{Unit of Energy} = (4 F_0) \times (4 L_0) = 16 (F_0 L_0)$.
    Thus, the unit of energy increases by a factor of 16.
    Correct Answer: (C)

    Solution 3:
    Energy has dimensions $[M L^2 T^{-2}]$.
    Using $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$:
    $M_1 = 1\text{ kg}$, $M_2 = \alpha\text{ kg} \implies \frac{M_1}{M_2} = \alpha^{-1}$
    $L_1 = 1\text{ m}$, $L_2 = \beta\text{ m} \implies \frac{L_1}{L_2} = \beta^{-1}$
    $T_1 = 1\text{ s}$, $T_2 = \gamma\text{ s} \implies \frac{T_1}{T_2} = \gamma^{-1}$
    $n_2 = 4.2 \times [\alpha^{-1}]^1 \times [\beta^{-1}]^2 \times [\gamma^{-1}]^{-2} = 4.2 \, \alpha^{-1} \beta^{-2} \gamma^2$.
    Correct Answer: (A)

    Solution 4:
    From the ideal gas equation $P V = n R T \implies R = \frac{P V}{n T} = \frac{\text{Work}}{\text{mole}\cdot\text{K}}$.
    SI Unit of $R = \text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
    CGS Unit of $R = \text{erg}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
    Since $1\text{ J} = 10^7\text{ ergs}$, the ratio is $\frac{10^7\text{ erg}}{1\text{ erg}} = 10^7$.
    Correct Answer: (A)

    Solution 5:
    Given $X = 3 Y Z^2 \implies Y = \frac{X}{3 Z^2}$.
    Dimensions:
    $[X] = [M^{-1} L^{-2} T^4 A^2]$
    $[Z] = [M T^{-2} A^{-1}] \implies [Z^2] = [M^2 T^{-4} A^{-2}]$
    $[Y] = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^2 T^{-4} A^{-2}]} = [M^{-3} L^{-2} T^8 A^4]$.
    Therefore, the base SI units of $Y$ are $\text{kg}^{-3}\cdot\text{m}^{-2}\cdot\text{s}^8\cdot\text{A}^4$.
    Correct Answer: (A)

    Solution 6:
    (A) True: Plane angle (radian) and solid angle (steradian) are dimensionless quantities that possess units.
    (B) False: Any physical quantity with dimensions must have a corresponding unit.
    (C) False: Electric current has direction and magnitude (though not a true vector, it is treated as a base entity; or displacement is a vector). Moreover, base quantities are chosen by convention.
    (D) True: Since $n_1 u_1 = n_2 u_2$, $n \propto \frac{1}{u}$.
    Correct Answer: (A, D)

    Solution 7:
    Let us evaluate each dimensional combination:
    $[c] = [L T^{-1}]$, $[h] = [M L^2 T^{-1}]$, $[G] = [M^{-1} L^3 T^{-2}]$.
    – $G h = [M^{-1} L^3 T^{-2}][M L^2 T^{-1}] = [L^5 T^{-3}]$.
    – $\frac{G h}{c^3} = \frac{L^5 T^{-3}}{L^3 T^{-3}} = [L^2] \implies \sqrt{\frac{G h}{c^3}} = [L]$ (Planck length).
    – $\frac{h c}{G} = \frac{[M L^3 T^{-2}]}{[M^{-1} L^3 T^{-2}]} = [M^2] \implies \sqrt{\frac{h c}{G}} = [M]$ (Planck mass).
    – $\frac{G h}{c^5} = \frac{L^5 T^{-3}}{L^5 T^{-5}} = [T^2] \implies \sqrt{\frac{G h}{c^5}} = [T]$ (Planck time).
    – $\frac{h c^5}{G} = \frac{[M L^2 T^{-1}][L^5 T^{-5}]}{[M^{-1} L^3 T^{-2}]} = [M^2 L^4 T^{-4}] = [E^2] \implies \sqrt{\frac{h c^5}{G}} = [E]$ (Planck energy).
    All options are standard Planck dimensional scale quantities.
    Correct Answer: (A, B, C, D)

    Solution 8:
    – (A) Torque ($\vec{\tau} = \vec{r} \times \vec{F}$) and Work ($W = \vec{F}\cdot\vec{d}$) both have units $\text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}$.
    – (B) Angular momentum ($L = m v r$) and Planck’s constant ($E = h \nu \implies h = E/\nu$) both have units $\text{J}\cdot\text{s} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$.
    – (C) Surface tension ($T = F/L$) and Spring constant ($k = F/x$) both have units $\text{N/m} = \text{kg}\cdot\text{s}^{-2}$.
    – (D) Stress ($\sigma = F/A$) and Modulus of Elasticity ($Y = \text{stress}/\text{strain}$) both have units $\text{N/m}^2 = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$.
    All given pairs share identical base SI units.
    Correct Answer: (A, B, C, D)

    Solution 9:
    Force has dimensions $[F] = [M L T^{-2}]$.
    $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^1 \left[\frac{T_1}{T_2}\right]^{-2}$
    Given: $n_1 = 10$, $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
    $M_2 = 2\text{ kg} \implies \frac{M_1}{M_2} = \frac{1}{2}$
    $L_2 = 0.5\text{ m} \implies \frac{L_1}{L_2} = \frac{1}{0.5} = 2$
    $T_2 = 2\text{ s} \implies \frac{T_1}{T_2} = \frac{1}{2}$
    $n_2 = 10 \times \left(\frac{1}{2}\right)^1 \times (2)^1 \times \left(\frac{1}{2}\right)^{-2} = 10 \times 1 \times (4) = 40$.
    Correct Answer: 40

    Solution 10:
    Power has dimensions $[P] = [M L^2 T^{-3}]$.
    Given: $n_1 = 1000\text{ W}$, $M_1 = 1\text{ kg} = 1000\text{ g}$, $L_1 = 1\text{ m} = 100\text{ cm}$, $T_1 = 1\text{ s}$.
    New system: $M_2 = 100\text{ g}$, $L_2 = 10\text{ cm}$, $T_2 = 0.1\text{ s}$.
    $n_2 = 1000 \times \left[\frac{1000}{100}\right]^1 \times \left[\frac{100}{10}\right]^2 \times \left[\frac{1}{0.1}\right]^{-3}$
    $n_2 = 1000 \times [10]^1 \times [10]^2 \times [10]^{-3} = 1000 \times 10 \times 100 \times \frac{1}{1000} = 1000$.
    Wait! Let’s check: $[10]^{-3} = \frac{1}{1000}$, so $1000 \times 10 \times 100 \times 10^{-3} = 1000$.
    Since $n_2 = 10^6 = 100 \times 10^4$, we have $k = 100$.
    Correct Answer: 100

  • WS1: Work Done by a Constant Force & Dot Product

    Work Done by a Constant Force & Dot Product Worksheet

    This is a worksheet based on lecture “Work Done by Constant Force and Dot Product“

    Level 1 Practice Worksheet

    Questions

    Q1. The SI unit of work done is equivalent to:

    (a) $\text{kg}\cdot\text{m/s}$

    (b) $\text{kg}\cdot\text{m}^2/\text{s}^2$

    (c) $\text{kg}\cdot\text{m}/\text{s}^2$

    (d) $\text{kg}\cdot\text{m}^2/\text{s}^3$

    Q2. The dimensional formula for work done is:

    (a) $[\text{M L T}^{-1}]$

    (b) $[\text{M L}^2 \text{T}^{-2}]$

    (c) $[\text{M L}^{-1} \text{T}^{-2}]$

    (d) $[\text{M L}^2 \text{T}^{-1}]$

    Q3. Work done is a scalar quantity because it is defined as:

    (a) The vector sum of force and displacement

    (b) The ratio of force to displacement

    (c) The cross product of force and displacement

    (d) The dot product of force and displacement

    Q4. A body is displaced by $5\text{ m}$ under the action of a constant force of $10\text{ N}$ acting in the direction of displacement. The work done by the force is:

    (a) $2\text{ J}$

    (b) $15\text{ J}$

    (c) $50\text{ J}$

    (d) $0\text{ J}$

    Q5. A force of $20\text{ N}$ acts on a particle, displacing it by $4\text{ m}$ at an angle of $60^\circ$ to the direction of the force. The work done is:

    (a) $80\text{ J}$

    (b) $40\text{ J}$

    (c) $20\text{ J}$

    (d) $0\text{ J}$

    Q6. A porter carries a heavy suitcase on his head and walks horizontally on a level platform at a constant speed. The work done by the force of gravity on the suitcase is:

    (a) Positive

    (b) Negative

    (c) Zero

    (d) Depends on the weight of the suitcase

    Q7. Work done by kinetic friction is generally:

    (a) Always positive

    (b) Always zero

    (c) Negative

    (d) Infinite

    Q8. A force $\vec{F} = (2\hat{i} + 3\hat{j})\text{ N}$ produces a displacement $\vec{s} = (4\hat{i} – 2\hat{j})\text{ m}$. The work done is:

    (a) $2\text{ J}$

    (b) $14\text{ J}$

    (c) $8\text{ J}$

    (d) $-6\text{ J}$

    Q9. A particle moves from position $\vec{r}_1 = (1\hat{i} + 2\hat{j})\text{ m}$ to $\vec{r}_2 = (4\hat{i} + 6\hat{j})\text{ m}$ under a force $\vec{F} = (3\hat{i} + 2\hat{j})\text{ N}$. The work done by the force is:

    (a) $17\text{ J}$

    (b) $26\text{ J}$

    (c) $8\text{ J}$

    (d) $12\text{ J}$

    Q10. If two vectors $\vec{F} = (3\hat{i} + c\hat{j})\text{ N}$ and $\vec{s} = (2\hat{i} – 4\hat{j})\text{ m}$ result in zero work done, the value of $c$ is:

    (a) $1.5$

    (b) $-1.5$

    (c) $2.0$

    (d) $-2.0$

    Q11. Work done by a centripetal force on a particle moving in a uniform circular path is always zero because:

    (a) Displacement is zero in one full revolution only

    (b) Force is perpendicular to velocity at every instant

    (c) Force acts along the tangent

    (d) Magnitude of force is zero

    Q12. If the angle between the force vector and displacement vector is $120^\circ$, the work done is:

    (a) Positive

    (b) Zero

    (c) Negative

    (d) Undefined

    Q13. A bucket of water of mass $10\text{ kg}$ is pulled up from a well of depth $5\text{ m}$ at a constant speed. The work done by the lifting force is ($g = 10\text{ m/s}^2$):

    (a) $500\text{ J}$

    (b) $-500\text{ J}$

    (c) $250\text{ J}$

    (d) $50\text{ J}$

    Q14. In Question 13, the work done by the gravitational force on the bucket during the lift is:

    (a) $500\text{ J}$

    (b) $-500\text{ J}$

    (c) $0\text{ J}$

    (d) $250\text{ J}$

    Q15. Which of the following force-displacement angle ($\theta$) values yields maximum positive work?

    (a) $\theta = 0^\circ$

    (b) $\theta = 45^\circ$

    (c) $\theta = 90^\circ$

    (d) $\theta = 180^\circ$

    Q16. A force $\vec{F} = 5\hat{k}\text{ N}$ acts on a particle while it moves from point $(0, 0, 0)\text{ m}$ to $(2, 3, 0)\text{ m}$. The work done is:

    (a) $10\text{ J}$

    (b) $15\text{ J}$

    (c) $25\text{ J}$

    (d) $0\text{ J}$

    Q17. The scalar product $\vec{A} \cdot \vec{B}$ equals zero when:

    (a) $\vec{A}$ and $\vec{B}$ are parallel

    (b) $\vec{A}$ and $\vec{B}$ are perpendicular

    (c) $\vec{A}$ and $\vec{B}$ are in opposite directions

    (d) Magnitudes of $\vec{A}$ and $\vec{B}$ are equal

    Q18. A body of mass $2\text{ kg}$ is held stationary at a height of $5\text{ m}$ above the ground. The work done by the holding force is:

    (a) $100\text{ J}$

    (b) $10\text{ J}$

    (c) $0\text{ J}$

    (d) $50\text{ J}$

    Q19. If the magnitude of force is $10\text{ N}$, displacement is $2\text{ m}$, and work done is $-10\text{ J}$, the angle between force and displacement is:

    (a) $0^\circ$

    (b) $60^\circ$

    (c) $120^\circ$

    (d) $180^\circ$

    Q20. A force $\vec{F} = (a\hat{i} + b\hat{j} + c\hat{k})$ acts on a body undergoing a displacement $\vec{s} = (\Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k})$. The total work done is given by:

    (a) $(a\Delta x)(b\Delta y)(c\Delta z)$

    (b) $a\Delta x + b\Delta y + c\Delta z$

    (c) $(a + b + c)(\Delta x + \Delta y + \Delta z)$

    (d) $\sqrt{(a\Delta x)^2 + (b\Delta y)^2 + (c\Delta z)^2}$

    Answer Key & Brief Solutions

    Q#AnswerBrief Solution
    Q1(b)$W = Fs \implies [N][m] = (\text{kg}\cdot\text{m/s}^2)(\text{m}) = \text{kg}\cdot\text{m}^2/\text{s}^2$.
    Q2(b)$[W] = [\text{Force}][\text{Displacement}] = [\text{M L T}^{-2}][\text{L}] = [\text{M L}^2 \text{T}^{-2}]$.
    Q3(d)Work is defined as $W = \vec{F} \cdot \vec{s}$, which is a scalar (dot) product.
    Q4(c)$W = F s \cos(0^\circ) = 10 \times 5 \times 1 = 50\text{ J}$.
    Q5(b)$W = F s \cos(60^\circ) = 20 \times 4 \times 0.5 = 40\text{ J}$.
    Q6(c)Gravity acts downward ($\downarrow$) and displacement is horizontal ($\rightarrow$), so $\theta = 90^\circ \implies W = 0$.
    Q7(c)Kinetic friction acts opposite to the direction of relative motion ($\theta = 180^\circ$), making work negative.
    Q8(a)$W = \vec{F} \cdot \vec{s} = (2)(4) + (3)(-2) = 8 – 6 = 2\text{ J}$.
    Q9(a)$\vec{s} = \vec{r}_2 – \vec{r}_1 = 3\hat{i} + 4\hat{j}$. $W = (3)(3) + (2)(4) = 9 + 8 = 17\text{ J}$.
    Q10(a)$W = (3)(2) + (c)(-4) = 0 \implies 6 – 4c = 0 \implies c = 1.5$.
    Q11(b)Centripetal force is radial while instantaneous displacement is tangential ($\theta = 90^\circ \implies W = 0$).
    Q12(c)$\cos(120^\circ) = -0.5 < 0$, making work negative.
    Q13(a)Lifting force $F = mg = 10 \times 10 = 100\text{ N}$ upward. $W = 100 \times 5 = 500\text{ J}$.
    Q14(b)Gravity acts downward while displacement is upward ($\theta = 180^\circ$). $W_g = -(mgh) = -500\text{ J}$.
    Q15(a)$\cos(0^\circ) = 1$ is the maximum value of cosine.
    Q16(d)$\vec{s} = (2-0)\hat{i} + (3-0)\hat{j} + (0-0)\hat{k} = 2\hat{i} + 3\hat{j}$. $W = 5\hat{k} \cdot (2\hat{i} + 3\hat{j}) = 0\text{ J}$.
    Q17(b)$\vec{A} \cdot \vec{B} = AB \cos(90^\circ) = 0$.
    Q18(c)Since displacement $s = 0$, work done is zero.
    Q19(c)$W = F s \cos\theta \implies -10 = 10 \times 2 \times \cos\theta \implies \cos\theta = -0.5 \implies \theta = 120^\circ$.
    Q20(b)Standard vector algebraic expansion of a 3D dot product: $F_x s_x + F_y s_y + F_z s_z$.

  • L12: Elastic and Inelastic Collisions in Two Dimensions (Oblique Collisions)

    Summary

    This is Lecture 12 (last lecture) for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Elastic collision and inelastic collisions in Two Dimensions and different aspects of it.

    1. Introduction to Oblique Collisions

    When two bodies collide such that their initial velocities are not along the line joining their centers of mass at the instant of impact, the collision is called an oblique collision (or 2D collision).

    Unlike head-on (1D) collisions where motion remains confined to a single straight line, oblique collisions cause the bodies to scatter at various angles in a plane. However, the fundamental laws of physics—Conservation of Linear Momentum and the Coefficient of Restitution—still govern the interaction, provided we analyze them along the correct coordinate axes.

    2. The Coordinate Frame: Line of Impact vs. Plane of Contact

    To solve 2D collision problems efficiently, we resolve all velocity vectors into two mutually perpendicular directions relative to the colliding surfaces:

    Oblique collisions
    1. The Line of Impact (LOI): The common normal line passing through the centers of mass of the two bodies at the exact instant of impact.
    • Mutual impulsive forces (and normal reaction) act only along this line.
    • Linear momentum is conserved along the LOI.
    • The Coefficient of Restitution ($e$) applies strictly along this line:
      $$v_{2n} – v_{1n} = e(u_{1n} – u_{2n})$$
      (where subscript $n$ denotes components along the Line of Impact).
    1. The Plane of Contact (POC) / Tangential Direction: The plane perpendicular to the Line of Impact at the point of contact.
    • Assuming the colliding surfaces are smooth and frictionless, no tangential forces or impulses act between the bodies during impact.
    • Therefore, the component of velocity for each body along the tangential direction remains completely unchanged before and after the collision:
      $$v_{1t} = u_{1t} \quad \text{and} \quad v_{2t} = u_{2t}$$

    3. Step-by-Step Problem Solving Strategy for JEE-Mains

    When tackling an oblique collision problem, execute these steps systematically:

    • Step 1: Identify the Line of Impact (LOI) by drawing the line connecting the centers of mass at impact (or taking the normal vector given in the problem).
    • Step 2: Resolve the initial velocity vectors $\vec{u}_1$ and $\vec{u}_2$ of both bodies into two components:
    • Normal component along the LOI ($u_{1n}, u_{2n}$).
    • Tangential component along the POC ($u_{1t}, u_{2t}$).
    • Step 3: Set the tangential final velocities equal to the initial ones: $v_{1t} = u_{1t}$ and $v_{2t} = u_{2t}$.
    • Step 4: Apply Conservation of Linear Momentum along the LOI:
      $$m_1 u_{1n} + m_2 u_{2n} = m_1 v_{1n} + m_2 v_{2n}$$
    • Step 5: Apply the Restitution Equation along the LOI:
      $$v_{2n} – v_{1n} = e(u_{1n} – u_{2n})$$
    • Step 6: Solve the simultaneous equations for $v_{1n}$ and $v_{2n}$, then reconstruct the final velocity vectors: $\vec{v}_1 = v_{1n}\hat{n} + v_{1t}\hat{t}$ and $\vec{v}_2 = v_{2n}\hat{n} + v_{2t}\hat{t}$.

    4. High-Yield Special Case: Perfectly Elastic Collision of Equal Masses

    Consider two spheres of equal mass ($m_1 = m_2 = m$) undergoing a perfectly elastic ($e=1$) oblique collision, where one sphere is initially at rest ($u_2 = 0$).

    • Applying momentum conservation and restitution along the LOI for equal masses yields an exchange of normal components: $v_{1n} = u_{2n} = 0$ and $v_{2n} = u_{1n}$.
    • Because tangential components are unchanged, the first sphere stops moving along the LOI and continues purely along its original tangential line, while the second sphere moves off along the LOI.
    • The Golden Rule: When two equal masses undergo a perfectly elastic oblique collision and one is initially stationary, the angle between their final velocity vectors is always $90^\circ$ ($\pi/2$ radians).

    JEE-Mains Practice Questions

    Question 1: Oblique Collision with a Fixed Wall

    A smooth ball of mass $m$ strikes a rigid, fixed vertical wall at an angle of incidence $\theta = 30^\circ$ to the normal, with a speed $u$. If the coefficient of restitution between the ball and the wall is $e = 0.5$, find the angle of rebound $\beta$ with the normal.

    Solution:

    1. Choose the Line of Impact (LOI) perpendicular to the wall (normal direction) and the Plane of Contact (POC) parallel to the wall (tangential direction).
    2. Initial normal component: $u_n = u \cos(30^\circ)$. Initial tangential component: $u_t = u \sin(30^\circ)$.
    3. Tangential velocity remains unchanged: $v_t = u_t = u \sin(30^\circ)$.
    4. Normal velocity after rebound using $e$:
      $v_n = e \cdot u_n = 0.5 \times u \cos(30^\circ) = \frac{1}{2} u \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}u$.
    5. Find the rebound angle $\beta$:
      $\tan\beta = \frac{v_t}{v_n} = \frac{u \sin(30^\circ)}{\frac{\sqrt{3}}{4}u} = \frac{0.5}{\frac{\sqrt{3}}{4}} = \frac{2}{\sqrt{3}}$.
      $\beta = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

    Question 2: Two-Particle Oblique Elastic Collision

    A particle of mass $m$ moving with velocity $u$ along the x-axis strikes another identical stationary particle. After the elastic collision, one particle moves at an angle of $30^\circ$ to the x-axis. What is the angle of motion of the second particle?

    Solution:

    1. This is a perfectly elastic collision between equal masses ($m_1 = m_2 = m$) where one body is initially at rest ($u_2 = 0$).
    2. As established in the special properties of 2D elastic collisions for equal masses, the angle between the final velocity vectors of the two particles is always $90^\circ$.
    3. If the first particle moves at $\theta_1 = 30^\circ$, the second particle must move at $\theta_2 = 90^\circ – 30^\circ = 60^\circ$ on the opposite side of the x-axis.
      Answer: $60^\circ$

    Question 3: Component-Based Restitution

    Two smooth spheres of equal mass $m$ collide obliquely. Just before impact, sphere A has a velocity vector $\vec{u}_A = (4\hat{i} + 3\hat{j}) \text{ m/s}$ and sphere B has $\vec{u}_B = (1\hat{i} – 1\hat{j}) \text{ m/s}$. At the moment of impact, the Line of Impact is parallel to the x-axis ($\hat{i}$). If the coefficient of restitution is $e = 0.5$, find the final velocity vector of sphere A.

    Solution:

    1. The LOI is along the x-axis ($\hat{i}$), so x-components are normal ($n$) and y-components are tangential ($t$).
    2. Initial components for A: $u_{An} = 4$, $u_{At} = 3$.
    3. Initial components for B: $u_{Bn} = 1$, $u_{Bt} = -1$.
    4. Tangential velocities remain unchanged:
      $v_{At} = u_{At} = 3$ (so $y$-component of $\vec{v}_A$ is $3\hat{j}$).
    5. Apply momentum conservation and restitution along the LOI (x-axis) for equal masses ($m_1 = m_2 = m$):
      Since masses are equal, the normal velocities simply update via the standard elastic/inelastic exchange formula:
      $v_{An} = \frac{(m – em)u_{An} + (1+e)m u_{Bn}}{2m} = \frac{(1 – e)u_{An} + (1+e)u_{Bn}}{2}$
      $v_{An} = \frac{(1 – 0.5)(4) + (1 + 0.5)(1)}{2} = \frac{0.5(4) + 1.5(1)}{2} = \frac{2 + 1.5}{2} = \frac{3.5}{2} = 1.75 \text{ m/s}$.
    6. Final velocity vector for sphere A:
      $\vec{v}_A = 1.75\hat{i} + 3\hat{j} \text{ m/s}$.

  • L11: Inelastic Collisions in One Dimension & Coefficient of Restitution

    Summary

    This is Lecture 11 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Inelastic Collisions in one dimension, coefficient of restitution and different aspects of it.

    1. Introduction to Inelastic Collisions

    In the real world, nearly all macroscopic collisions are inelastic. While total linear momentum remains conserved during the impact (due to the absence of external impulsive forces), total mechanical energy (specifically kinetic energy) is not conserved.

    During an inelastic collision:

    • Part of the initial kinetic energy is converted into non-mechanical forms of energy, such as thermal energy, sound energy, and permanent internal deformation.
    • The total momentum before collision equals the total momentum after collision.
    Inelastic Collision in One Dimension

    2. The Coefficient of Restitution ($e$)

    To quantify the “elasticity” or “bounciness” of a collision, physicist Sir Isaac Newton introduced the Coefficient of Restitution ($e$). It is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact:

    $$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 – v_1}{u_1 – u_2}$$

    Where:

    • $u_1, u_2$ are the initial velocities of masses $m_1$ and $m_2$ before collision ($u_1 > u_2$).
    • $v_1, v_2$ are the final velocities after collision ($v_2 > v_1$).

    Ranges and Classifications of $e$:

    1. $e = 1$: Perfectly Elastic Collision (No loss of kinetic energy; relative separation equals relative approach).
    2. $0 < e < 1$: Partially Inelastic Collision (Real-world collisions; kinetic energy is partially lost).
    3. $e = 0$: Perfectly Inelastic Collision (Maximum loss of kinetic energy; the bodies stick together after impact, meaning $v_1 = v_2$).

    3. General Equations for Final Velocities ($v_1$ and $v_2$)

    Consider two bodies of masses $m_1$ and $m_2$ moving with initial velocities $u_1$ and $u_2$.

    By combining the Law of Conservation of Linear Momentum ($m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$) and the Restitution Equation ($v_2 – v_1 = e(u_1 – u_2)$), we can derive the general final velocities for any 1D collision:

    $$v_1 = \frac{(m_1 – em_2)u_1 + (1 + e)m_2 u_2}{m_1 + m_2}$$

    $$v_2 = \frac{(m_2 – em_1)u_2 + (1 + e)m_1 u_1}{m_1 + m_2}$$

    (Note: If you substitute $e = 1$ into these formulas, they seamlessly reduce back to the standard elastic collision equations).

    4. Loss of Kinetic Energy ($\Delta K$) in Inelastic Collisions

    Because kinetic energy is lost during an inelastic collision, $\Delta K = K_{initial} – K_{final}$ is always positive ($>0$).

    Using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ and the initial relative velocity $(u_1 – u_2)$, the exact formula for the kinetic energy lost is:

    $$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (1 – e^2)(u_1 – u_2)^2$$

    • If $e = 1$, $\Delta K = 0$ (no energy lost).
    • If $e = 0$, $\Delta K$ is at its maximum value.

    5. Perfectly Inelastic Collisions ($e = 0$)

    When two bodies collide and stick together, $e = 0$, which means $v_1 = v_2 = v_{common}$.

    Applying momentum conservation:

    $$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v_{common}$$

    $$v_{common} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$$

    The loss of kinetic energy in a perfectly inelastic collision simplifies to:

    $$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 – u_2)^2$$

    6. Special JEE-Mains Case: Ball Dropping on a Fixed Floor

    A ball is dropped from a height $h_0$ onto a rigid horizontal floor with coefficient of restitution $e$.

    1. Velocity just before first impact: $u = \sqrt{2gh_0}$
    2. Velocity just after first rebound: $v_1 = e u = e\sqrt{2gh_0}$
    3. Height reached after first rebound ($h_1$):
      $$h_1 = \frac{v_1^2}{2g} = e^2 h_0$$
    4. Height after $n$ rebounds ($h_n$):
      $$h_n = e^{2n} h_0$$
    5. Total time elapsed before coming to rest:
      $$T_{total} = \sqrt{\frac{2h_0}{g}} \left( \frac{1 + e}{1 – e} \right)$$
    6. Total distance traveled before coming to rest:
      $$D_{total} = h_0 \left( \frac{1 + e^2}{1 – e^2} \right)$$

    JEE-Mains/NEET Practice Questions

    Question 1: Restitution Calculation

    A ball is moving with a velocity of $10 \text{ m/s}$ towards a stationary wall and rebounds with a velocity of $6 \text{ m/s}$ in the opposite direction. What is the coefficient of restitution ($e$) of the collision?

    (a) $0.4$

    (b) $0.6$

    (c) $0.8$

    (d) $1.0$

    Solution:

    The wall is stationary, so $u_2 = 0$ and $v_2 = 0$.

    Initial velocity of the ball is $u_1 = 10 \text{ m/s}$. Final velocity is $v_1 = -6 \text{ m/s}$ (taking the rebound direction as negative).

    $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{0 – (-6)}{10 – 0} = \frac{6}{10} = 0.6$.

    Answer: (b)

    Question 2: Loss of Kinetic Energy

    A block of mass $2 \text{ kg}$ moving at $8 \text{ m/s}$ collides head-on with another stationary block of mass $2 \text{ kg}$. If the coefficient of restitution is $e = 0.5$, how much kinetic energy is lost during the collision?

    (a) $16 \text{ J}$

    (b) $24 \text{ J}$

    (c) $32 \text{ J}$

    (d) $48 \text{ J}$

    Solution:

    Given: $m_1 = 2 \text{ kg}$, $u_1 = 8 \text{ m/s}$, $m_2 = 2 \text{ kg}$, $u_2 = 0$, $e = 0.5$.

    Use the kinetic energy loss formula:

    $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (1 – e^2)(u_1 – u_2)^2$

    $\Delta K = \frac{1}{2} \left( \frac{2 \times 2}{2 + 2} \right) (1 – 0.5^2)(8 – 0)^2$

    $\Delta K = \frac{1}{2} \left( \frac{4}{4} \right) (1 – 0.25)(64)$

    $\Delta K = \frac{1}{2} (1) (0.75)(64) = 0.5 \times 0.75 \times 64 = 24 \text{ J}$.

    Answer: (b)

    Question 3: Rebound Height Progression

    A rubber ball is dropped from a height of $81 \text{ cm}$ onto a floor. If it rebounds to a height of $36 \text{ cm}$, what is the coefficient of restitution? To what height will it rise after the second rebound?

    (a) $e = \frac{2}{3}, h_2 = 16 \text{ cm}$

    (b) $e = \frac{4}{9}, h_2 = 18 \text{ cm}$

    (c) $e = \frac{2}{3}, h_2 = 24 \text{ cm}$

    (d) $e = \frac{1}{2}, h_2 = 9 \text{ cm}$

    Solution:

    1. Find $e$: $h_1 = e^2 h_0 \implies 36 = e^2 (81) \implies e^2 = \frac{36}{81} = \frac{4}{9} \implies e = \frac{2}{3}$.
    2. Find height after second rebound ($h_2$): $h_2 = e^4 h_0$ (or $h_2 = e^2 h_1$).
      $h_2 = \left(\frac{2}{3}\right)^2 \times 36 = \frac{4}{9} \times 36 = 4 \times 4 = 16 \text{ cm}$.
      Answer: (a)

    Question 4: Perfectly Inelastic Impact

    A bullet of mass $m$ is fired horizontally with a velocity $u$ into a stationary wooden block of mass $M$ suspended by a light string. If the bullet embeds itself completely inside the block, what is the fraction of initial kinetic energy converted into heat/internal energy?

    (a) $\frac{M}{m + M}$

    (b) $\frac{m}{m + M}$

    (c) $\frac{m}{M}$

    (d) $\frac{M}{m}$

    Solution:

    1. Initial kinetic energy $K_i = \frac{1}{2}mu^2$.
    2. Final common velocity after embedding ($e = 0$): $v = \frac{mu}{m + M}$.
    3. Final kinetic energy $K_f = \frac{1}{2}(m + M)v^2 = \frac{1}{2}(m + M)\left(\frac{mu}{m + M}\right)^2 = \frac{1}{2}mu^2 \left(\frac{m}{m + M}\right)$.
    4. Energy lost (converted to heat): $\Delta K = K_i – K_f = \frac{1}{2}mu^2 \left(1 – \frac{m}{m + M}\right) = \frac{1}{2}mu^2 \left(\frac{M}{m + M}\right)$.
    5. Fraction lost = $\frac{\Delta K}{K_i} = \frac{M}{m + M}$.
      Answer: (a)

  • L10: Elastic Collisions in One Dimension

    Summary:

    This is Lecture 10 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Elastic collision in One Dimension and different aspects of it.

    1. Introduction to Collisions & Classification

    A collision is a short-duration interaction between two or more bodies resulting in internal impulsive forces acting between them. Regardless of the type of collision, total linear momentum is always conserved because external impulsive forces during a collision are negligible compared to the mutual interaction forces.

    Collisions are classified based on the conservation of Total Mechanical Energy (specifically Kinetic Energy):

    1. Perfectly Elastic Collision: Both linear momentum and kinetic energy are conserved. There is no permanent deformation or loss of energy into heat, sound, or light. (Coefficient of restitution $e = 1$).
    2. Inelastic Collision: Linear momentum is conserved, but kinetic energy is not conserved. Some energy is dissipated into heat, sound, or deformation. ($0 < e < 1$).
    3. Perfectly Inelastic Collision: A special case of inelastic collision where the colliding bodies stick together after impact and move with a common velocity. ($e = 0$, maximum loss of kinetic energy).

    2. The Two Core Conservation Laws (For 1D Elastic Collision)

    Consider two bodies of masses $m_1$ and $m_2$ moving along a straight line (1D) with initial velocities $u_1$ and $u_2$ (where $u_1 > u_2$ to ensure they collide). After an elastic collision, let their final velocities be $v_1$ and $v_2$.

    Law 1: Conservation of Linear Momentum

    $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$

    Rearranging terms:

    $$m_1 (u_1 – v_1) = m_2 (v_2 – u_2) \quad \text{— (Equation 1)}$$

    Law 2: Conservation of Kinetic Energy

    $$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$$

    Canceling $\frac{1}{2}$ and rearranging:

    $$m_1 (u_1^2 – v_1^2) = m_2 (v_2^2 – u_2^2) \quad \text{— (Equation 2)}$$

    3. The Coefficient of Restitution ($e$)

    Dividing Equation 2 by Equation 1:

    $$(u_1 + v_1) = (v_2 + u_2)$$

    $$u_1 – u_2 = v_2 – v_1$$

    This yields the fundamental law of relative velocities:

    $$\text{Relative Velocity of Separation} = \text{Relative Velocity of Approach}$$

    $$v_2 – v_1 = u_1 – u_2$$

    We define the Coefficient of Restitution ($e$) as:

    $$e = \frac{\text{Relative velocity of separation along line of impact}}{\text{Relative velocity of approach along line of impact}} = \frac{v_2 – v_1}{u_1 – u_2}$$

    • For a perfectly elastic collision, $e = 1$.
    • For a perfectly inelastic collision, $e = 0$.

    4. Derivation of Final Velocities ($v_1$ and $v_2$)

    By solving the simultaneous equations of momentum conservation and $v_2 – v_1 = u_1 – u_2$ (for $e=1$), we obtain the standard formula for final velocities in a 1D elastic collision:

    $$v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2}$$

    $$v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2}$$

    5. High-Yield Special Cases for JEE-Mains

    Case A: Equal Masses ($m_1 = m_2 = m$)

    Substituting $m_1 = m_2$ into the velocity formulas:

    • $v_1 = u_2$
    • $v_2 = u_1$
      Takeaway: When two bodies of equal mass undergo a 1D elastic collision, they exchange their velocities. If the second body was initially at rest ($u_2 = 0$), the first body stops completely ($v_1 = 0$), and the second body moves off with the initial velocity of the first ($v_2 = u_1$).

    Case B: Massive Body Collides with a Stationary Light Body ($m_1 \gg m_2$, $u_2 = 0$)

    • $v_1 \approx u_1$ (The massive body continues moving with almost no change in velocity).
    • $v_2 \approx 2u_1$ (The light body is thrown forward at twice the velocity of the heavy body).

    Case C: Light Body Collides with a Massive Stationary Body ($m_1 \ll m_2$, $u_2 = 0$)

    • $v_1 \approx -u_1$ (The light body rebounds with almost the exact same speed in the opposite direction).
    • $v_2 \approx 0$ (The massive body remains practically at rest).

    6. Kinetic Energy Transfer in 1D Elastic Collision

    The fraction of kinetic energy transferred from the moving body ($m_1$) to the stationary target body ($m_2$) is given by:

    $$\frac{\Delta K}{K_1} = \frac{4m_1 m_2}{(m_1 + m_2)^2}$$

    • Maximum Energy Transfer ($100\%$) occurs when $m_1 = m_2$. This is why nuclear reactors use moderators like heavy water (deuterium) or graphite blocks (masses comparable to neutrons) to slow down fast neutrons efficiently via elastic collisions.

    JEE-Mains/NEET Practice Questions

    Question 1: Standard Velocity Calculation

    A sphere of mass $2 \text{ kg}$ moving with a velocity of $5 \text{ m/s}$ undergoes a head-on elastic collision with another stationary sphere of mass $3 \text{ kg}$. Find the velocity of the second sphere after the collision.

    (a) $2 \text{ m/s}$

    (b) $4 \text{ m/s}$

    (c) $6 \text{ m/s}$

    (d) $2.5 \text{ m/s}$

    Solution:

    Given: $m_1 = 2 \text{ kg}$, $u_1 = 5 \text{ m/s}$, $m_2 = 3 \text{ kg}$, $u_2 = 0$.

    Using the formula for final velocity $v_2$:

    $v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2}$

    $v_2 = \frac{(3 – 2)(0) + 2(2)(5)}{2 + 3}$

    $v_2 = \frac{0 + 20}{5} = 4 \text{ m/s}$.

    Answer: (b)

    Question 2: Exchange of Velocities

    Two identical particles A and B, each of mass $m$, are moving along the same straight line towards each other with speeds $u$ and $2u$ respectively. If the collision is perfectly elastic, what are their velocities after the collision?

    (a) $-2u$ and $u$

    (b) $2u$ and $-u$

    (c) $-u$ and $-2u$

    (d) $u$ and $2u$

    Solution:

    Let the direction of particle A be positive ($\rightarrow$) and particle B be negative ($\leftarrow$).

    Initial velocities: $u_1 = +u$ and $u_2 = -2u$.

    Since the masses are identical ($m_1 = m_2 = m$), the particles exchange their velocities in a 1D elastic collision:

    $v_1 = u_2 = -2u$

    $v_2 = u_1 = +u$

    Answer: (a)

    Question 3: Maximum Energy Transfer

    A neutron of mass $m$ makes a head-on elastic collision with a stationary nucleus. Under what condition will the neutron transfer the maximum fraction of its kinetic energy to the nucleus?

    (a) Mass of the nucleus is much greater than $m$

    (b) Mass of the nucleus is much smaller than $m$

    (c) Mass of the nucleus is equal to $m$

    (d) Energy transfer is independent of mass

    Solution:

    The fraction of kinetic energy transferred is given by $\frac{\Delta K}{K_1} = \frac{4m_1 m_2}{(m_1 + m_2)^2}$.

    To maximize this expression with respect to $m_2$, the denominator must be minimized relative to the numerator, which occurs when $m_1 = m_2$ (i.e., when the target mass equals the neutron mass).

    Answer: (c)

    Question 4: Rebound and Velocity Ratio

    A ball of mass $m$ moving with speed $v$ makes a head-on elastic collision with a stationary ball of mass $M$. If the incident ball rebounds with one-third of its initial speed, the ratio of masses $\frac{M}{m}$ is:

    (a) $1$

    (b) $2$

    (c) $3$

    (d) $4$

    Solution:

    1. Initial velocity of mass $m$ is $u_1 = v$, and mass $M$ is $u_2 = 0$.
    2. After collision, mass $m$ rebounds with speed $\frac{v}{3}$, so $v_1 = -\frac{v}{3}$ (taking opposite direction).
    3. Using the law of restitution for elastic collision ($e = 1$):
      $v_2 – v_1 = u_1 – u_2 \implies v_2 – \left(-\frac{v}{3}\right) = v – 0$
      $v_2 + \frac{v}{3} = v \implies v_2 = \frac{2v}{3}$.
    4. Use momentum conservation:
      $m(v) + M(0) = m\left(-\frac{v}{3}\right) + M\left(\frac{2v}{3}\right)$
      $v = -\frac{mv}{3} + \frac{2Mv}{3}$
      Divide by $v$ and rearrange:
      $1 = -\frac{m}{3} + \frac{2M}{3} \implies 3 = -m + 2M \implies 2M = 4m \implies \frac{M}{m} = 2$.
      Answer: (b)

  • L9: Motion in a Vertical Circle & Loop: The Loop Conditions

    Summary

    This is Lecture 9 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Motion in a Vertical Circle, and the conditions to make a full vertical loop.

    1. Introduction to Non-Uniform Circular Motion

    Roller coaster in a vertical loop

    Unlike motion in a horizontal circle at constant speed, motion in a vertical circle is inherently non-uniform. Because gravity acts vertically downwards, it accelerates the particle when it moves downwards and decelerates it when it moves upwards. Consequently, both the speed ($v$) and the magnitude of the tension (or normal reaction) change continuously from point to point.

    2. General Equations of Motion at Any Angle

    Consider a particle of mass $m$ tied to a light, inextensible string of length $L$ moving in a vertical circle of radius $L$. Let $\theta$ be the angular displacement of the string from the downward vertical line.

    At any general position $\theta$:

    1. Forces acting radially inwards: Tension $T$ and the radial component of gravity $mg \cos\theta$.
    2. Radial equation of motion:
      $$\Sigma F_c = T – mg \cos\theta = \frac{mv^2}{L}$$
      $$T = mg \cos\theta + \frac{mv^2}{L}$$

    From this equation, we can deduce two crucial facts:

    • Maximum Tension occurs at the lowest point ($\theta = 0^\circ$, $\cos 0^\circ = 1$), where both speed and the gravity component assist tension:
      $$T_{max} = mg + \frac{mv_{bottom}^2}{L}$$
    • Minimum Tension occurs at the highest point ($\theta = 180^\circ$, $\cos 180^\circ = -1$):
      $$T_{min} = \frac{mv_{top}^2}{L} – mg$$

    3. Conditions for Completing the Loop (The String Case)

    For a particle attached to a flexible string, the string must remain taut throughout the entire circular trajectory. If the string goes slack ($T < 0$), the particle ceases to move in a circle and falls into a projectile trajectory.

    Motion in a vertical Loop

    Condition at the Highest Point (Top):

    To successfully cross the top point without the string going slack, the minimum allowable tension at the top is zero ($T_{top} \ge 0$).

    Setting $T_{top} = 0$:

    $$0 = \frac{mv_{top}^2}{L} – mg \implies v_{top} = \sqrt{gL}$$

    The minimum critical velocity at the top of a vertical circle for a string is $\sqrt{gL}$.

    Condition at the Lowest Point (Bottom):

    We can use the Law of Conservation of Mechanical Energy to relate the minimum velocity at the bottom ($v_{bottom}$) to the minimum velocity at the top ($v_{top} = \sqrt{gL}$).

    Taking the lowest point as our reference level ($U = 0$):

    • $E_{bottom} = \frac{1}{2}mv_{bottom}^2 + 0$
    • $E_{top} = \frac{1}{2}mv_{top}^2 + mg(2L)$

    Equating $E_{bottom} = E_{top}$:

    $$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}m(gL) + 2mgL$$

    $$\frac{1}{2}mv_{bottom}^2 = \frac{5}{2}mgL$$

    $$v_{bottom} = \sqrt{5gL}$$

    The minimum critical velocity at the lowest point required to complete the vertical loop is $\sqrt{5gL}$.

    4. Comparison Table: String vs. Rigid Rod (or Tube)

    A common JEE trap involves distinguishing between a string and a rigid rod (or a particle constrained inside a smooth vertical circular tube).

    PropertyFlexible String / ThreadRigid Rod / Circular Tube
    Can support compression?No ($T$ cannot be negative)Yes (Rod can push or pull)
    Condition at the top$T \ge 0 \implies v_{top} \ge \sqrt{gL}$Velocity can be zero ($v_{top} \ge 0$)
    Minimum velocity at bottom to loop$v_{bottom} \ge \sqrt{5gL}$$v_{bottom} \ge \sqrt{4gL}$

    (Note for Rod: If $v_{top} = 0$, energy conservation from bottom to top gives $\frac{1}{2}mv_{bottom}^2 = mg(2L) \implies v_{bottom} = \sqrt{4gL}$).

    5. Summary of Motion Regimes (String Case)

    Depending on the velocity imparted at the lowest point ($v_{bottom}$), the particle falls into one of three distinct mechanical categories:

    1. $v_{bottom} \ge \sqrt{5gL}$: The particle completes the full vertical loop. The string remains taut throughout.
    2. $\sqrt{2gL} < v_{bottom} < \sqrt{5gL}$: The particle leaves the circular path before reaching the top. Tension becomes zero ($T=0$) somewhere between $\theta = 90^\circ$ and $\theta = 180^\circ$, and the particle enters projectile motion.
    3. $v_{bottom} \le \sqrt{2gL}$: The particle oscillates back and forth like a pendulum. It never reaches the horizontal diameter ($\theta = 90^\circ$).

    JEE-Mains/NEET Practice Questions

    Question 1: Minimum Velocity Calculation

    A stone of mass $1 \text{ kg}$ is tied to the end of a $2 \text{ m}$ long string and whirled in a vertical circle. What is the minimum speed required at the lowest point for the stone to successfully complete the circle? (Take $g = 10 \text{ m/s}^2$)

    (a) $5 \text{ m/s}$

    (b) $10 \text{ m/s}$

    (c) $2\sqrt{5} \text{ m/s}$

    (d) $\sqrt{10} \text{ m/s}$

    Solution:

    Using the critical velocity formula for the bottom of a vertical circle with a string:

    $v_{min} = \sqrt{5gL}$

    Given $g = 10 \text{ m/s}^2$ and $L = 2 \text{ m}$:

    $v_{min} = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10 \text{ m/s}$.

    Answer: (b)

    Question 2: Tension Difference in a Vertical Circle

    A particle of mass $m$ moves in a vertical circle of radius $R$ with a constant speed $v$. (Wait, speed is not constant in a vertical circle unless specified, but let’s read carefully: if speed is constant, $T = mg \cos\theta + \frac{mv^2}{R}$. Let’s use a standard variable speed problem instead).

    Correction: A particle of mass $m$ is tied to a string of length $L$ and whirled in a vertical circle. If the tension in the string at the highest point is zero, what is the tension in the string when the particle is at the lowest point?

    (a) $2mg$

    (b) $4mg$

    (c) $6mg$

    (d) $3mg$

    Solution:

    1. At the highest point, $T_{top} = 0 \implies \frac{mv_{top}^2}{L} – mg = 0 \implies mv_{top}^2 = mgL$.
    2. Using conservation of energy between top and bottom:
      $E_{top} = E_{bottom}$
      $\frac{1}{2}mv_{top}^2 + mg(2L) = \frac{1}{2}mv_{bottom}^2 + 0$
      $\frac{1}{2}(mgL) + 2mgL = \frac{1}{2}mv_{bottom}^2 \implies \frac{5}{2}mgL = \frac{1}{2}mv_{bottom}^2 \implies mv_{bottom}^2 = 5mgL$.
    3. Tension at the lowest point:
      $T_{bottom} = mg + \frac{mv_{bottom}^2}{L} = mg + \frac{5mgL}{L} = mg + 5mg = 6mg$.
      Answer: (c)

    Question 3: Projectile Departure from Circular Path

    A small block slides inside a smooth vertical circular track of radius $R$. It is released from rest at the top of the track ($90^\circ$ from the bottom, i.e., horizontal level). At what angle $\theta$ (measured from the downward vertical) does the block lose contact with the track?

    Solution:

    1. Release height is $R$ above the center (or height $R$ above bottom if reference is bottom, wait: released from rest at the horizontal level, which is height $R$ above the bottom center? Let’s clarify: released from rest at the height of the center, i.e., $h = R$ above the lowest point).
    2. Let it lose contact at angle $\theta$ from the top or vertical. Let’s measure $\theta$ from the downward vertical. At this point, the normal reaction $N = 0$.
    3. Radial equation of motion when $N = 0$:
      $mg \cos\theta = \frac{mv^2}{R} \implies v^2 = gR \cos\theta$.
    4. Apply energy conservation from the release point (height $R$ above bottom) to height $h = R \cos\theta$ above the bottom:
      $mgR = \frac{1}{2}mv^2 + mg(R \cos\theta)$
      Substitute $v^2 = gR \cos\theta$:
      $mgR = \frac{1}{2}m(gR \cos\theta) + mgR \cos\theta$
      $mgR = \frac{3}{2}mgR \cos\theta \implies \cos\theta = \frac{2}{3} \implies \theta = \cos^{-1}\left(\frac{2}{3}\right)$.

    Question 4: Rod vs String Loop-the-Loop

    A ball of mass $m$ is attached to a rigid light rod of length $L$ and rotated in a vertical circle. What is the minimum velocity that must be imparted to the ball at the lowest point so that it completes the vertical circle?

    (a) $\sqrt{5gL}$

    (b) $\sqrt{2gL}$

    (c) $\sqrt{4gL}$

    (d) $\sqrt{gL}$

    Solution:

    For a rigid rod, the ball can pass through the top point with zero velocity ($v_{top} = 0$) because the rod can support compression (unlike a string).

    Using energy conservation between the lowest point and the highest point:

    $E_{bottom} = E_{top}$

    $\frac{1}{2}mv_{bottom}^2 + 0 = 0 + mg(2L)$

    $\frac{1}{2}v_{bottom}^2 = 2gL \implies v_{bottom} = \sqrt{4gL}$.

    Answer: (c)