- Home
- Standard 11
- Physics
3-2.Motion in Plane
hard
A clock has $75 \mathrm{~cm}, 60 \mathrm{~cm}$ long second hand and minute hand respectively. In $30$ minutes duration the tip of second hand will travel $x$ distance more than the tip of minute hand. The value of $x$ in meter is nearly (Take $\pi=3.14$ ) :
A$220.0$
B$140.5$
C$139.4$
D$118.9$
(JEE MAIN-2024)
Solution
$\mathrm{x}_{\text {min }}=\pi \times \mathrm{r}_{\text {min }}$
$=\pi \times \frac{60}{100} \mathrm{~m}.$
$\mathrm{x}_{\text {second }}=30 \times 2 \pi \times \mathrm{r}_{\text {second }}$
$=30 \times 2 \pi \times \frac{75}{100}$
$\mathrm{x}=\mathrm{x}_{\mathrm{scond}}-\mathrm{x}_{\mathrm{min}}$
$=139.4 \mathrm{~m}$
$=\pi \times \frac{60}{100} \mathrm{~m}.$
$\mathrm{x}_{\text {second }}=30 \times 2 \pi \times \mathrm{r}_{\text {second }}$
$=30 \times 2 \pi \times \frac{75}{100}$
$\mathrm{x}=\mathrm{x}_{\mathrm{scond}}-\mathrm{x}_{\mathrm{min}}$
$=139.4 \mathrm{~m}$
Standard 11
Physics
