Blog

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
September 6, 2026
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
September 5, 2026
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
September 5, 2026

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

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

(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

(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) $15\text{ J}$

(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) $40\text{ J}$

(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

(d) Depends on the weight of the suitcase

Q7. Work done by kinetic friction is generally:

(a) Always positive

(b) Always zero

(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) $14\text{ J}$

(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) $26\text{ J}$

(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) $-1.5$

(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

(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

(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) $-500\text{ J}$

(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) $-500\text{ J}$

(d) $250\text{ J}$

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

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

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

(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) $15\text{ J}$

(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

(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) $10\text{ J}$

(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) $60^\circ$

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

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

Answer Key & Brief Solutions

Q#	Answer	Brief 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: $F_x s_x + F_y s_y + F_z s_z$.

August 16, 2026

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:
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:
  $\vec{v}_A = 1.75\hat{i} + 3\hat{j} \text{ m/s}$.
August 16, 2026
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.
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)
August 16, 2026
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)
August 16, 2026

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

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 $\theta$:

Forces acting radially inwards: Tension $T$ and the radial component of gravity $mg \cos\theta$.
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.

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).

Property	Flexible String / Thread	Rigid 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:

$v_{bottom} \ge \sqrt{5gL}$: The particle completes the full vertical loop. The string remains taut throughout.
$\sqrt{2gL} < v_{bottom} < \sqrt{5gL}$: 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.
$v_{bottom} \le \sqrt{2gL}$: 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) $10 \text{ m/s}$

(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) $4mg$

(d) $3mg$

Solution:

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

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).
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 $N = 0$.
Radial equation of motion when $N = 0$:
$mg \cos\theta = \frac{mv^2}{R} \implies v^2 = gR \cos\theta$.
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) $\sqrt{2gL}$

(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)

August 16, 2026

L8: Conservative and Non-Conservative Forces

Summary

This is Lecture 8 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 Conservative and Non Conservative Forces

1. Introduction to Force Fields

In mechanics, when a particle moves through space, it may experience forces that depend on its position, velocity, or time. The nature of these forces dictates whether mechanical energy is conserved or dissipated. Forces are broadly categorized into two major classes: Conservative and Non-Conservative.

2. Conservative Forces

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path. It depends only on the initial and final positions.

Key Properties of Conservative Forces:

Path Independence: The work done by a conservative force along path 1 from point $A$ to point $B$ is identical to the work done along path 2 between the same two points:
$$W_{AB} (\text{Path 1}) = W_{AB} (\text{Path 2})$$
Work in a Closed Loop is Zero: If a particle starts at point $A$, moves along any arbitrary closed loop, and returns to $A$, the net work done by a conservative force is zero:
$$\oint \vec{F}_c \cdot d\vec{r} = 0$$
Association with Potential Energy: A conservative force can always be expressed as the negative gradient of a scalar potential energy function $U$:
$$\vec{F} = -\nabla U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right)$$
Conservation of Mechanical Energy: When only conservative forces do work, the total mechanical energy ($K + U$) remains constant.

The Mathematical Test for Conservative Forces (JEE Favorite):

For a two-dimensional force field $\vec{F} = F_x \hat{i} + F_y \hat{j}$, the force is conservative if and only if it satisfies the exactness condition (derived from curl $\nabla \times \vec{F} = 0$):

$$\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$$

Common Examples:

Gravitational force, ideal spring force ($F = -kx$),

electrostatic force ($F = \frac{kq_1q_2}{r^2}$).

3. Non-Conservative Forces

A non-conservative force is a force for which the work done depends explicitly on the path taken between two points.

Key Properties of Non-Conservative Forces:

Path Dependence: If you take a longer, winding path against a non-conservative force (like friction), more energy is dissipated than if you take a straight path.
Work in a Closed Loop is Non-Zero: Moving an object in a closed loop under a non-conservative force results in net non-zero work (usually negative, as energy is continuously drained):
$$\oint \vec{F}_{nc} \cdot d\vec{r} \neq 0$$
Dissipative Nature: Non-conservative forces typically convert macroscopic mechanical energy into microscopic thermal energy, sound, or internal deformation energy. Mechanical energy is not conserved ($W_{nc} = \Delta E$).
No Potential Energy Function: You cannot define a potential energy field $U$ for a non-conservative force.

Common Examples: Kinetic friction, static friction (when slipping occurs), air resistance, fluid viscosity, and tension from an active non-ideal string/motor.

4. Distinguishing Features Summary Table

Feature	Conservative Forces	Non-Conservative Forces
Path Dependence	Independent of path (depends only on endpoints)	Dependent on the specific path taken
Closed Loop Work	$\oint \vec{F} \cdot d\vec{r} = 0$	$\oint \vec{F} \cdot d\vec{r} \neq 0$
Energy Conversion	Converts between Kinetic and Potential energy	Dissipates Mechanical energy into Thermal/Internal energy
Potential Energy	Can be defined ($U$ exists)	Cannot be defined
Mathematical Test	$\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$ (in 2D)	Fails the exactness derivative test
Examples	Gravity, Spring Force, Electrostatic Force	Friction, Air Drag, Viscous Force

JEE-Mains/NEET Practice Questions

Question 1: Mathematical Test for Conservative Force

Which of the following force fields is conservative?

(a) $\vec{F} = (xy\hat{i} + x^2y\hat{j}) \text{ N}$

(b) $\vec{F} = (y^2\hat{i} + 2xy\hat{j}) \text{ N}$

(d) Both (b) and (c)

Solution: (d)

Test each option using the exactness condition $\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$:

For (a): $F_x = xy \implies \frac{\partial F_x}{\partial y} = x$. $F_y = x^2y \implies \frac{\partial F_y}{\partial x} = 2xy$. Since $x \neq 2xy$, not conservative.
For (b): $F_x = y^2 \implies \frac{\partial F_x}{\partial y} = 2y$. $F_y = 2xy \implies \frac{\partial F_y}{\partial x} = 2y$. Since $2y = 2y$, conservative.
For (c): $F_x = x \implies \frac{\partial F_x}{\partial y} = 0$. $F_y = -y \implies \frac{\partial F_y}{\partial x} = 0$. Since $0 = 0$, conservative.
Therefore, both (b) and (c) are conservative force fields.

Question 2: Work in a Closed Loop

A block is slid along a rough horizontal floor in a complete circular path of radius $R$ and returned to its exact starting point under the action of kinetic friction. The total work done by kinetic friction is:

(a) Zero

(b) $-\mu mg (2\pi R)$

(d) Dependent on the speed of the block

Solution: (b)

Kinetic friction is a non-conservative, dissipative force that always opposes motion. In a closed loop, friction never cancels out; it always performs negative work over every differential segment of the path ($d\vec{r}$).

$W = \oint (-f_k) ds = -f_k \times (\text{Circumference}) = -\mu mg (2\pi R)$.

Question 3: Conceptual Identification

Assertion (A): The work done by the gravitational force in moving a particle around any closed path is zero.

Reason (R): Gravitational force is a conservative force.

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

(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(d) (A) is false, but (R) is true.

Solution: (a)

By definition, one of the primary defining characteristics of a conservative force (like gravity) is that the net work done in any closed path is strictly zero. Thus, Reason (R) directly explains Assertion (A).

Question 4: Potential Energy Existence

A force is given by $\vec{F} = (3x^2 y^2 \hat{i} + 2x^3 y \hat{j}) \text{ N}$. Find the change in potential energy ($\Delta U$) of the system when a particle moves from the origin $(0,0)$ to the point $(1, 2)$.

Solution:

First, check if it’s conservative:

$F_x = 3x^2y^2 \implies \frac{\partial F_x}{\partial y} = 6x^2y$

$F_y = 2x^3y \implies \frac{\partial F_y}{\partial x} = 6x^2y$

Since partial derivatives match, the force is conservative, meaning potential energy exists.

$\Delta U = -W_c = – \int_{(0,0)}^{(1,2)} (F_x dx + F_y dy)$

Notice that $F_x dx + F_y dy = 3x^2y^2 dx + 2x^3y dy$ is the exact differential of $U(x,y) = x^3 y^2$.

Integrating from $(0,0)$ to $(1,2)$:

$U(1,2) = (1)^3 (2)^2 = 4 \text{ J}$.

$U(0,0) = (0)^3 (0)^2 = 0 \text{ J}$.

Change in potential energy $\Delta U = U_f – U_i = 4 – 0 = 4 \text{ J}$. (Thus, work done by the conservative force is $-4 \text{ J}$).

August 16, 2026

L7: Conservation of Mechanical Energy
Summary:

This is Lecture 7 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link

1. Total Mechanical Energy ($E$)

Mechanical energy is the macroscopic energy associated with the motion and configuration of a system. The Total Mechanical Energy ($E$) of a system is defined as the sum of its Kinetic Energy ($K$) and Potential Energy ($U$):

$$E = K + U = \frac{1}{2}mv^2 + U(x, y, z)$$
- Scalar Quantity: Mechanical energy is a scalar, measured in Joules (J).
- Frame Dependent: Because kinetic energy depends on velocity, mechanical energy values are frame-of-reference dependent (though energy conservation laws hold in any inertial frame).
2. Derivation of the Conservation Law

To understand when mechanical energy is conserved, we begin with the fundamental Work-Energy Theorem:

$$W_{net} = \Delta K$$

In any mechanical system, the total net force can be divided into conservative forces (like gravity, spring force, electrostatic force) and non-conservative forces (like friction, air drag, tension from an active motor, or external applied forces).

$$W_{net} = W_c + W_{nc} = \Delta K$$

By definition, the work done by a conservative force is equal to the negative change in potential energy ($W_c = -\Delta U$). Substituting this into our equation:

$$-\Delta U + W_{nc} = \Delta K$$

Rearranging the terms yields:

$$W_{nc} = \Delta K + \Delta U = \Delta E$$

The General Principle:

The work done by non-conservative forces is equal to the total change in mechanical energy of the system.

The Law of Conservation of Mechanical Energy:

If all internal and external forces acting on a system are conservative—meaning no non-conservative forces do work ($W_{nc} = 0$)—then the total mechanical energy remains strictly constant:

$$\Delta E = 0 \implies K_i + U_i = K_f + U_f$$

3. Classic JEE-Mains/NEET Applications

Application A: The Simple Pendulum

A bob of mass $m$ is suspended by a string of length $L$. It is pulled to an angle $\theta_0$ and released from rest. What is its speed when it passes through the lowest point?
- Analysis: Only gravity and string tension act on the bob. Tension is always perpendicular to instantaneous velocity, so it does zero work ($W_T = 0$). Gravity is conservative. Thus, mechanical energy is conserved.
- Calculation:
- Choose reference $U=0$ at the lowest point.
- Initial state (at release angle $\theta_0$): Height above lowest point is $h = L(1 – \cos\theta_0)$. Velocity $v_i = 0$.
- $E_i = K_i + U_i = 0 + mgL(1 – \cos\theta_0)$
- Final state (at lowest point): Height $h = 0$. Velocity $v_f = v$.
- $E_f = K_f + U_f = \frac{1}{2}mv^2 + 0$
- Equating $E_i = E_f$:
  $$\frac{1}{2}mv^2 = mgL(1 – \cos\theta_0) \implies v = \sqrt{2gL(1 – \cos\theta_0)}$$
Application B: Loop-the-Loop (Minimum Velocity at Top)

A small block slides down a frictionless track and enters a vertical circular loop of radius $R$. What is the minimum release height $H$ from rest so that the block successfully completes the loop without leaving the track?
- Condition at the top of the loop: For the block to stay on the track, the normal reaction must be at least zero ($N \ge 0$). At the exact threshold, $N = 0$, meaning gravity provides the necessary centripetal force:
  $$\frac{mv_{top}^2}{R} = mg \implies v_{top}^2 = gR$$
- Applying Energy Conservation between release point and top of loop:
- $E_{release} = 0 + mgH = mgH$
- $E_{top} = \frac{1}{2}mv_{top}^2 + mg(2R)$
- Equating $E_{release} = E_{top}$:
  $$mgH = \frac{1}{2}m(gR) + 2mgR = \frac{5}{2}mgR \implies H = \frac{5}{2}R$$
JEE-Mains/NEET Practice Questions

Question 1: Pendulum and String Obstacle

A simple pendulum of length $L$ with a bob of mass $m$ is released from a horizontal position ($\theta = 90^\circ$). As it swings down, the string hits a small fixed peg located directly below the point of suspension at a distance $\frac{3L}{4}$ from the top. What is the maximum height reached by the bob on the other side?

Solution:
1. When the bob reaches the lowest point, it has dropped a vertical height $L$. By energy conservation, its velocity $v$ satisfies $\frac{1}{2}mv^2 = mgL \implies v^2 = 2gL$.
2. Below the peg, the radius of the circular swing changes to $r = L – \frac{3L}{4} = \frac{L}{4}$.
3. As it swings up around the peg, energy is still conserved because tension does no work. Let it rise to a height $h’$ above the lowest point.
4. $\frac{1}{2}mv^2 = mgh’ \implies mgL = mgh’ \implies h’ = L$.
5. Since the peg is at depth $\frac{3L}{4}$ below the pivot, the total height reached above the lowest point is $L$, which means it rises to the exact same horizontal level it started from! (Answer: $L$ above lowest point).
Question 2: Spring and Incline Combination

A block of mass $m = 1 \text{ kg}$ is released from rest down a smooth inclined plane of inclination $30^\circ$. After sliding down a distance $d = 2 \text{ m}$, it hits an unstretched ideal spring of constant $k = 200 \text{ N/m}$ aligned parallel to the incline. What is the maximum compression of the spring? (Take $g = 10 \text{ m/s}^2$)

Solution:
1. Let the maximum compression of the spring be $x_m$.
2. Total vertical descent from the initial release point to the point of maximum compression is $(d + x_m) \sin 30^\circ$.
3. Apply conservation of mechanical energy between the initial position and the point of maximum compression (where velocity $v = 0$ momentarily):
  $K_i + U_{gravity, i} + U_{spring, i} = K_f + U_{gravity, f} + U_{spring, f}$
  $0 + 0 + 0 = 0 – mg(d + x_m)\sin 30^\circ + \frac{1}{2}kx_m^2$
4. Substitute values:
  $\frac{1}{2}(200)x_m^2 = (1)(10)(2 + x_m)\left(\frac{1}{2}\right)$
  $100x_m^2 = 5(2 + x_m) \implies 100x_m^2 – 5x_m – 10 = 0$
  Divide by $5$: $20x_m^2 – x_m – 2 = 0$
  Solve via quadratic formula: $x_m = \frac{1 \pm \sqrt{1 – 4(20)(-2)}}{40} = \frac{1 \pm \sqrt{161}}{40} \approx \frac{1 + 12.68}{40} \approx 0.34 \text{ m}$ (or $34 \text{ cm}$).
Question 3: Chain Sliding Off a Table

A uniform chain of mass $M$ and length $L$ is placed on a smooth horizontal table such that a fraction $\frac{1}{n}$ of its length hangs vertically over the edge. If the chain is released from rest, find its velocity when the entire chain just leaves the table.

Solution:
1. Let $y = 0$ be the level of the table surface. Initially, the center of mass of the hanging portion $\frac{L}{n}$ is at a depth of $\frac{L}{2n}$ below the table.
2. Initial potential energy of the system:
  $U_i = -\left(\frac{M}{n}\right)g\left(\frac{L}{2n}\right) = -\frac{MgL}{2n^2}$
  (The portion on the table has $U = 0$).
3. Final state: The entire chain hangs vertically. The center of mass of the whole chain is at a depth of $\frac{L}{2}$ below the table.
  $U_f = -Mg\left(\frac{L}{2}\right) = -\frac{MgL}{2}$
4. Apply conservation of mechanical energy ($K_i + U_i = K_f + U_f$):
  $0 – \frac{MgL}{2n^2} = \frac{1}{2}Mv^2 – \frac{MgL}{2}$
  $\frac{1}{2}v^2 = \frac{gL}{2} \left(1 – \frac{1}{n^2}\right) \implies v = \sqrt{gL \left(1 – \frac{1}{n^2}\right)}$
Question 4: Work-Energy Theorem with Friction (Non-Conservative Force)

A block of mass $2 \text{ kg}$ slides down a rough curved track from a height of $5 \text{ m}$ and reaches the bottom with a speed of $8 \text{ m/s}$. How much mechanical energy was lost due to friction? (Take $g = 10 \text{ m/s}^2$)

Solution:
1. Initial mechanical energy at height $h = 5 \text{ m}$ (with $v_i = 0$):
  $E_i = mgh = (2)(10)(5) = 100 \text{ J}$
2. Final mechanical energy at the bottom ($h = 0$, $v = 8 \text{ m/s}$):
  $E_f = \frac{1}{2}mv^2 + 0 = \frac{1}{2}(2)(8^2) = 64 \text{ J}$
3. The work done by non-conservative forces (friction) equals the change in mechanical energy:
  $W_{nc} = E_f – E_i = 64 – 100 = -36 \text{ J}$
4. Therefore, the mechanical energy lost to friction is $36 \text{ J}$.
August 16, 2026

Blog

Concept Card: Errors in Measurement & Propagation of Errors

Solved Examples

Worksheet: 10 Practice Problems

Solutions & Explanations

Concept Card: Least Count of Measuring Instruments & Significant Figures

Solved Examples

Worksheet: 10 Practice Problems

Solutions & Explanations

Concept Card: Systems of Units & Physical Quantities

Solved Examples

Worksheet: 10 Practice Problems

Solutions & Explanations

Work Done by a Constant Force & Dot Product Worksheet

Questions

Answer Key & Brief Solutions

Summary

1. Introduction to Oblique Collisions

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

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

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

JEE-Mains Practice Questions

Summary

1. Introduction to Inelastic Collisions

2. The Coefficient of Restitution ($e$)

Ranges and Classifications of $e$:

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

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

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

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

JEE-Mains/NEET Practice Questions

Summary:

1. Introduction to Collisions & Classification

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

Law 1: Conservation of Linear Momentum

Law 2: Conservation of Kinetic Energy

3. The Coefficient of Restitution ($e$)

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

5. High-Yield Special Cases for JEE-Mains

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

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

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

6. Kinetic Energy Transfer in 1D Elastic Collision

JEE-Mains/NEET Practice Questions

Summary

1. Introduction to Non-Uniform Circular Motion

2. General Equations of Motion at Any Angle

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

Condition at the Highest Point (Top):

Condition at the Lowest Point (Bottom):

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

5. Summary of Motion Regimes (String Case)

JEE-Mains/NEET Practice Questions

Summary

1. Introduction to Force Fields

2. Conservative Forces

Key Properties of Conservative Forces:

The Mathematical Test for Conservative Forces (JEE Favorite):

3. Non-Conservative Forces

Key Properties of Non-Conservative Forces:

4. Distinguishing Features Summary Table

JEE-Mains/NEET Practice Questions

Summary:

1. Total Mechanical Energy ($E$)

2. Derivation of the Conservation Law

3. Classic JEE-Mains/NEET Applications

Application A: The Simple Pendulum

Application B: Loop-the-Loop (Minimum Velocity at Top)

JEE-Mains/NEET Practice Questions