\displaystyle \textbf{Question 1. }\text{If }D,G\text{ and }R\text{ denote respectively the number of degrees, grades} \\ \text{and radians in an angle, then}
\displaystyle \text{(a) }\frac{D}{100}=\frac{G}{90}=\frac{2R}{\pi}\qquad \text{(b) }\frac{D}{90}=\frac{G}{100}=\frac{R}{\pi}
\displaystyle \text{(c) }\frac{D}{90}=\frac{G}{100}=\frac{2R}{\pi}\qquad \text{(d) }\frac{D}{90}=\frac{G}{100}=\frac{R}{2\pi}
\displaystyle \text{Answer:}
\displaystyle 90^\circ=100\text{ grades}=\frac{\pi}{2}\text{ radians}
\displaystyle \therefore \frac{D}{90}=\frac{G}{100}=\frac{R}{\pi/2}=\frac{2R}{\pi}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 2. }\text{If the angles of a triangle are in A.P., then the measures of one} \\ \text{of the angles in radians is}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad \text{(b) }\frac{\pi}{3}\qquad \text{(c) }\frac{\pi}{2}\qquad \text{(d) }\frac{2\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Let the angles be }a-d,\ a,\ a+d
\displaystyle (a-d)+a+(a+d)=\pi
\displaystyle 3a=\pi
\displaystyle a=\frac{\pi}{3}
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 3. }\text{The angle between the minute and hour hands of a clock at }8:30\text{ is}
\displaystyle \text{(a) }80^\circ\qquad \text{(b) }75^\circ\qquad \text{(c) }60^\circ\qquad \text{(d) }105^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Angle made by hour hand}=8\times30^\circ+30\times\frac{1}{2}^\circ=255^\circ
\displaystyle \text{Angle made by minute hand}=30\times6^\circ=180^\circ
\displaystyle \therefore \text{Required angle}=255^\circ-180^\circ=75^\circ
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 4. }\text{At }3:40,\text{ the hour and minute hands of a clock are inclined at}
\displaystyle \text{(a) }\frac{2\pi}{3}\qquad \text{(b) }\frac{7\pi}{12}\qquad \text{(c) }\frac{13\pi}{18}\qquad \text{(d) }\frac{3\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Angle made by hour hand}=3\times30^\circ+40\times\frac{1}{2}^\circ=110^\circ
\displaystyle \text{Angle made by minute hand}=40\times6^\circ=240^\circ
\displaystyle \therefore \text{Required angle}=240^\circ-110^\circ=130^\circ
\displaystyle =130^\circ\times\frac{\pi}{180^\circ}=\frac{13\pi}{18}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 5. }\text{If the arcs of the same length in two circles subtend angles }65^\circ\text{ and } \\ 110^\circ\text{ at the centre, then the ratio of the radii of the circles is}
\displaystyle \text{(a) }22:13\qquad \text{(b) }11:13\qquad \text{(c) }22:15\qquad \text{(d) }21:13
\displaystyle \text{Answer:}
\displaystyle l=r\theta
\displaystyle \text{For same arc length, }r_1\theta_1=r_2\theta_2
\displaystyle \therefore r_1:r_2=\theta_2:\theta_1=110:65=22:13
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 6. }\text{If }OP\text{ makes }4\text{ revolutions in one second, the angular velocity in} \\ \text{radians per second is}
\displaystyle \text{(a) }\pi\qquad \text{(b) }2\pi\qquad \text{(c) }4\pi\qquad \text{(d) }8\pi
\displaystyle \text{Answer:}
\displaystyle 1\text{ revolution}=2\pi\text{ radians}
\displaystyle 4\text{ revolutions}=8\pi\text{ radians}
\displaystyle \therefore \text{Angular velocity}=8\pi\text{ radians per second}
\displaystyle \therefore \text{Correct option is (d).}
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\displaystyle \textbf{Question 7. }\text{A circular wire of radius }7\text{ cm is cut and bent again into an arc of} \\ \text{a circle of radius }12\text{ cm. The angle subtended by the arc at the centre is}
\displaystyle \text{(a) }50^\circ\qquad \text{(b) }210^\circ\qquad \text{(c) }100^\circ\qquad \text{(d) }60^\circ\qquad \text{(e) }195^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Length of wire}=2\pi(7)=14\pi\text{ cm}
\displaystyle l=r\theta
\displaystyle 14\pi=12\theta
\displaystyle \theta=\frac{7\pi}{6}=210^\circ
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 8. }\text{The radius of the circle whose arc of length }15\pi\text{ cm makes an angle of } \\ \frac{3\pi}{4}\text{ radian at the centre is}
\displaystyle \text{(a) }10\text{ cm}\qquad \text{(b) }20\text{ cm}\qquad \text{(c) }11\frac{1}{4}\text{ cm}\qquad \text{(d) }22\frac{1}{2}\text{ cm}
\displaystyle \text{Answer:}
\displaystyle l=r\theta
\displaystyle 15\pi=r\cdot\frac{3\pi}{4}
\displaystyle r=15\pi\cdot\frac{4}{3\pi}=20\text{ cm}
\displaystyle \therefore \text{Correct option is (b).}
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