The minute hand of a watch is $1.5 \,cm$ long. How far does its tip move in $40$ minutes? ( Use $\pi=3.14$ ).
The minute hand of a watch is $1.5 \,cm$ long. How far does its tip move in $40$ minutes? ( Use $\pi=3.14$ ).
In $60$ minutes, the minute hand of a watch completes one revolution. Therefore, in $40$ minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, ${\theta = 23 \times {{360}^\circ }}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by
$l=r \theta=1.5 \times \frac{4 \pi}{3} \,cm =2 \pi \,cm =2 \times 3.14 \,cm =6.28 \,cm$
Similar Questions
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Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
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