Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 3 Ex 3.1 — all 7 questions solved, covering conversion between degree and radian measure and the arc-length formula l = rθ.
Every question in this exercise comes down to just two tools: the conversion relation π radian = 180°, and the arc-length formula l = rθ (arc length equals radius times the angle in radians). Questions 1–2 are pure conversion drills in both directions, including one part with negative angles and one with minutes. Questions 3, 4, 6 and 7 are real-world applications — a spinning wheel, an arc on a circle, two circles compared, and a swinging pendulum — all solved by rearranging l = rθ. Question 5 adds a small geometric insight: when a chord equals the radius, the triangle it forms with the centre is equilateral, immediately giving a 60° angle.
We use the conversion Radian measure = \dfrac{\pi}{180}\times Degree measure.
(i) 25^\circ=\dfrac{\pi}{180}\times25=\dfrac{25\pi}{180}
(ii) First convert the minutes: 47^\circ30'=47\tfrac{1}{2}^\circ=\dfrac{95}{2}^\circ. So -47^\circ30'=-\dfrac{95}{2}^\circ.
Converting: -\dfrac{95}{2}\times\dfrac{\pi}{180}=-\dfrac{95\pi}{360}
(iii) 240^\circ=\dfrac{\pi}{180}\times240=\dfrac{240\pi}{180}
(iv) 520^\circ=\dfrac{\pi}{180}\times520=\dfrac{520\pi}{180}
We use the conversion Degree measure = \dfrac{180}{\pi}\times Radian measure, with \dfrac{180}{\pi}=\dfrac{180\times7}{22}=\dfrac{630}{11}.
(i) \dfrac{11}{16}\times\dfrac{630}{11}=\dfrac{630}{16}=39.375^\circ
Converting the decimal part: 0.375^\circ\times60=22.5'=22'30''.
(ii) -4\times\dfrac{630}{11}=-\dfrac{2520}{11}=-229.0909\ldots^\circ
Converting the decimal part: 0.0909^\circ\times60\approx5.45', and 0.45'\times60\approx27''.
(iii) \dfrac{5\pi}{3}\times\dfrac{180}{\pi}=\dfrac{5\times180}{3}=300^\circ
(iv) \dfrac{7\pi}{6}\times\dfrac{180}{\pi}=\dfrac{7\times180}{6}=210^\circ
Revolutions per second: \dfrac{360}{60}=6 revolutions per second.
Each revolution turns through 2\pi radians, so in one second the wheel turns through:
6\times2\pi=12\pi radians
Given: r=100 cm, l=22 cm. Using \theta=\dfrac{l}{r}:
\theta=\dfrac{22}{100}=0.22 radian
Converting to degrees: 0.22\times\dfrac{180}{\pi}=0.22\times\dfrac{1260}{22}=12.6^\circ
Converting the decimal part: 0.6^\circ\times60=36'.
Radius r=\dfrac{40}{2}=20 cm. Since the chord (20 cm) equals the radius (20 cm), the triangle formed by the two radii and the chord has all three sides equal — it is equilateral.
Therefore the central angle \theta=60^\circ=\dfrac{\pi}{3} radian.
Using l=r\theta:
l=20\times\dfrac{\pi}{3}=\dfrac{20\pi}{3} cm
Substituting \pi=\tfrac{22}{7}: l=\dfrac{20\times22}{3\times7}=\dfrac{440}{21}\approx20.95 cm
Let r_1,r_2 be the radii, with \theta_1=60^\circ=\dfrac{\pi}{3} and \theta_2=75^\circ=\dfrac{5\pi}{12} radian.
Since the arc length l is the same in both circles, l=r_1\theta_1=r_2\theta_2, so:
\dfrac{r_1}{r_2}=\dfrac{\theta_2}{\theta_1}=\dfrac{5\pi/12}{\pi/3}=\dfrac{5\pi}{12}\times\dfrac{3}{\pi}=\dfrac{15}{12}=\dfrac{5}{4}
The pendulum's length is the radius, r=75 cm. Using \theta=\dfrac{l}{r}:
(i) \theta=\dfrac{10}{75}=\dfrac{2}{15} radian
(ii) \theta=\dfrac{15}{75}=\dfrac{1}{5} radian
(iii) \theta=\dfrac{21}{75}=\dfrac{7}{25} radian
Expert CBSE Coaching · Class 9–12