हिंदी

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

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प्रश्न

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

योग
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उत्तर

Given side of square field = 10m

Then the perimeter of that square field = 4 × 10

= 40m

Total time = 2 minute 20 sec

= (2 × 60 + 20)

= 140 sec

Hence, the distance covered in going around the square field = perimeter of the field

Total distance covered in 40 sec = 40m

Total distance covered in 140 sec = `(40/40) xx140` m

= 140 m

This number of rounds = `140/40`

= 3.5

In 140 sec, that farmer will take 3.5 rounds.

Displacement = Diagonal of square = `(102+102) 1/2`

= `(200)1/2`

= `10sqrt2` m

Hence, magnitude of displacement = `10sqrt2` m

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अध्याय 7: Motion - Intext question [पृष्ठ ७४]

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एनसीईआरटी Science [English] Class 9
अध्याय 7 Motion
Intext question | Q 2. | पृष्ठ ७४

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