Home Class 11 Applied Maths Chapter 2: Numbers in Day-to-Day Life Exercise 2.1: Clocks
Chapter 2 • Unit I • Section 2.1

Exercise 2.1: Clocks

Comprehensive theory, angular velocities, collision kinematics, and authentic CBSE model solutions (2026-27).

Introductory Discussion: Time & Distance

Lalita Babar is a three-time winner of the Mumbai Marathon. She won a Bronze medal in the 3000 m steeplechase at the 2014 Asian Games and broke the National record with a time of 9 hours 35 minutes and 37 seconds.

Time To Work:
1. Write the length of steeplechase race in decimeters:
Since $1\text{ meter} = 10\text{ decimeters}$,
$\text{Length} = 3000\text{ m} \times 10 = \mathbf{30,000\text{ decimeters}}$.

2. Convert the time taken into seconds:
$\text{Time} = 9\text{ hours } 35\text{ minutes } 37\text{ seconds}$
$= (9 \times 3600) + (35 \times 60) + 37 = 32400 + 2100 + 37 = \mathbf{34,537\text{ seconds}}$.

For thousands of years, humans measured time by observing the passage of day and night, the position of stars, and the cyclical change of seasons. To achieve higher precision, tools such as Sundials, hourglasses, and water clocks were invented over centuries. The most accurate clock in the world today is the Cesium Fountain Atomic Clock developed at NIST Laboratories in Colorado, USA.

Units of Time Measurement

Time is measured in seconds, minutes, hours, and days, derived from astronomical cycles: the rotation of the Earth on its axis and its revolution around the Sun.

Common UnitEquivalent Measurement
1 Minute60 seconds
1 Hour60 minutes = 3600 seconds
1 Day24 hours = 1440 min = 86,400 sec
1 Week7 days
1 Year12 months = $365\frac{1}{4}$ days
Large & Sub-Second UnitEquivalent Measurement
1 Decade10 years
1 Century100 years
1 Millennium1000 years
1 Millisecond$\frac{1}{1000}\text{ sec} = 10^{-3}\text{ second}$
1 Microsecond$\frac{1}{1,000,000}\text{ sec} = 10^{-6}\text{ second}$
Example 1 (CBSE Support Material): What would be the time four and a half hours before 2:15 pm?
Show Step-by-Step Solution
$\text{Target Time} = 2:15\text{ pm} - 4\frac{1}{2}\text{ hours}$
$= 2:15\text{ pm} - (4\text{ hours} + 30\text{ minutes})$
$= 2:15\text{ pm} - 4\text{ hours} - 30\text{ minutes}$
$= 10:15\text{ am} - 30\text{ minutes} = \mathbf{9:45\text{ am}}$.

Structure of a Clock & Angular Kinematics

HOROLOGY: Study and measurement of time & clock-making

A typical analog clock face represents a circular dial of $360^\circ$, divided into 12 major hour intervals (each $30^\circ$) and 60 minute intervals (each $6^\circ$). It is equipped with three concentric hands: the Hour hand, Minute hand, and Seconds hand.

Angular Velocity Derivation:
  • • A complete circular revolution $= 360^\circ$.
  • • Hour Hand: Completes $360^\circ$ in 12 hours $\implies \frac{360^\circ}{12} = \mathbf{30^\circ\text{ per hour}} = \frac{30^\circ}{60} = \mathbf{0.5^\circ\text{ per minute}}$.
  • • Minute Hand: Completes $360^\circ$ in 60 minutes $\implies \frac{360^\circ}{60} = \mathbf{6^\circ\text{ per minute}}$.
  • • Seconds Hand: Completes $360^\circ$ in 60 seconds $\implies \frac{360^\circ}{60} = \mathbf{6^\circ\text{ per second}} = 360^\circ\text{ per minute}$.
Investigation 1:

Calculate the speed of the seconds hand?
Speed $= \frac{360^\circ}{60\text{ s}} = \mathbf{6^\circ/\text{second}}$ (or $360^\circ/\text{minute}$).

Investigation 2:

Find the difference in speed between the minute hand and hour hand?
$\text{Relative Speed} = 6^\circ/\text{min} - 0.5^\circ/\text{min} = \mathbf{5.5^\circ/\text{min}} = \mathbf{\left(\frac{11}{2}\right)^\circ/\text{min}}$.

Collisions (Coincidence) of Hands of a Clock

Two hands collide when the angle between them is $0^\circ$. How many times in a day do the minute and hour hands coincide?

Why Hands Coincide 22 Times in 24 Hours (Not 24 Times):
In a 24-hour day, the hour hand completes two full revolutions. One might intuitively expect $24$ collisions, but between 11:00 and 1:00, the hands collide only once at exactly 12:00 (they do not collide between 11:00 and 12:00). This occurs once in the daytime and once at night.
$$\text{Total Collisions in a Day} = 24 - 2 = \mathbf{22\text{ times}}.$$

Time Between Collisions:
Since 22 collisions occur uniformly across 24 hours:
$$\text{Time of one collision} = \frac{24}{22}\text{ hours} = \frac{12}{11}\text{ hours} = \frac{12 \times 60}{11}\text{ minutes} = \frac{720}{11}\text{ minutes} = \mathbf{65\frac{5}{11}\text{ minutes}} \approx 65.45\text{ minutes}.$$

Stretch Your Brain: Exact Collision Schedule in a 12-Hour Cycle
Collision Number Time in Fractions Exact Time (hh:mm:ss) Apparent Clock Time
1st12:00:0012:00:0012:00
2nd$1:05\frac{5}{11}$1:05:271:05
3rd$2:10\frac{10}{11}$2:10:542:10
4th$3:16\frac{4}{11}$3:16:223:16
5th$4:21\frac{9}{11}$4:21:164:20
6th$5:27\frac{3}{11}$5:27:165:27
7th$6:32\frac{8}{11}$6:32:436:32
8th$7:38\frac{2}{11}$7:38:117:38
9th$8:43\frac{7}{11}$8:43:388:43
10th$9:49\frac{1}{11}$9:49:059:49
11th$11:59\frac{11}{11} = 12:00:00$12:00:0012:00

Time and Angle Formulas

Let $H$ be the running hour and $M$ be the minutes. The angular displacement of the hour hand is $(30H + 0.5M)^\circ$, and that of the minute hand is $6M^\circ$. The angle $\theta$ (or $A$) between them is:

$$\mathbf{A = \left|30H - \frac{11}{2}M\right|}$$ $$\mathbf{T = \frac{2}{11}(30H \pm A)}$$

Note: The angle $A$ is taken as positive for hand movements in the first half (12 to 6) and negative in the second half (6 to 12). For absolute geometric angle, the smaller interior angle is $\le 180^\circ$.

Example 2 (CBSE Support Material): At what time between 4 o'clock and 5 o'clock will the hands of a clock make an angle of $20^\circ$?
Show Step-by-Step Solution
Given: Running hour $H = 4$, target angle $A = 20^\circ$.
Using $T = \frac{2}{11}(H \times 30^\circ \pm A)$:
In the first half (12 to 6), $A$ is taken positive:
$T = \frac{2}{11}[4 \times 30^\circ + 20^\circ] = \frac{2}{11}[120^\circ + 20^\circ] = \frac{2}{11}[140^\circ] = \frac{280^\circ}{11} = 25\frac{5}{11}\text{ minutes}$.
Convert $\frac{5}{11}\text{ minutes}$ to seconds: $\frac{5}{11} \times 60 = \frac{300}{11} \approx 27\text{ seconds}$.
Answer: The hands will make an angle of $20^\circ$ at 4:25:27 (or $25\frac{5}{11}\text{ min past } 4$).
Example 3 (CBSE Support Material): At what time between 3 o'clock and 4 o'clock will the hands of a clock collide?
Show Step-by-Step Solution
At 3 o'clock, the hour hand is at 3 and the minute hand is at 12, so the minute hand is 15 minute spaces behind the hour hand.
Since the minute hand gains 55 minutes in 60 minutes (or $\frac{12}{11}$ min per minute space):
$T = 15 \times \frac{12}{11} = \frac{180}{11} = \mathbf{16\frac{4}{11}\text{ minutes}}$.
Alternatively, using formula with $A = 0^\circ$:
$T = \frac{2}{11}(3 \times 30^\circ) = \frac{180}{11} = 16\frac{4}{11}\text{ minutes}$.
$\frac{4}{11} \times 60 = \frac{240}{11} \approx 22\text{ seconds}$.
Answer: The hands collide at $16\frac{4}{11}$ minutes past 3 (exact time 3:16:22).
Example 4 (CBSE Support Material): Find the angle between the hands of a clock when it struck 15 minutes past 7.
Show Step-by-Step Solution
Given: $H = 7$, $M = 15$.
$A = \left|30H - \frac{11}{2}M\right|$
$= \left|30(7) - \frac{11}{2}(15)\right|$
$= |210^\circ - 82.5^\circ| = \mathbf{127.5^\circ}$.
Answer: The angle formed is $127.5^\circ$ (or $127^\circ 30'$).
Chapter 2 Hub Next: 2.2 Calendar