Verified Solution • 2025–26 Session moderate

QJEEM22Pha19b: JEE Main 2022 Physics - Class 11 JEEM22Pha19b Solution: Sand is being dropped from a stationary dropper at a rate of on …

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs−10.5 \,\mathrm{kgs}^{-1} on a conveyor belt moving with a velocity of 5 ms−15 \mathrm{~ms}^{-1} . The power needed to keep the belt moving with the same velocity will be :

Multiple Choice Options

  • A 1.25 W
  • B 2.5 W
  • C 6.25 W
  • D 12.5 W

Step-by-Step Verified Solution

dmdt=0.5{{dm} \over {dt}} = 0.5 kg/s

v=5v = 5 m/s

F=vdmdt=2.5F = {{vdm} \over {dt}} = 2.5 kg m/s2

P=F‾ . v‾=(2.5)(5)P = \overline F \,.\,\overline v = (2.5)(5) W

=12.5= 12.5 W

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