JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 3)
Sand is being dropped from a stationary dropper at a rate of $$0.5 \,\mathrm{kgs}^{-1}$$ on a conveyor belt moving with a velocity of $$5 \mathrm{~ms}^{-1}$$. The power needed to keep the belt moving with the same velocity will be :
1.25 W
2.5 W
6.25 W
12.5 W
Explanation
$${{dm} \over {dt}} = 0.5$$ kg/s
$$v = 5$$ m/s
$$F = {{vdm} \over {dt}} = 2.5$$ kg m/s2
$$P = \overline F \,.\,\overline v = (2.5)(5)$$ W
$$ = 12.5$$ W
Comments (0)
