A man runs round a circular field of radius 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The circumference of the circular field is given by $C = 2 \pi r = 2 \pi 	imes 50 = 100 \pi\, m$.For $20$ rounds, the total distance to be covere

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The circumference of the circular field is given by $C = 2 \pi r = 2 \pi 	imes 50 = 100 \pi\, m$.For $20$ rounds, the total distance to be covered is $D = 20 	imes 100 \pi = 2000 \pi\, m$.The speed of the man is $12\, km/hr$. Converting this to $m/s$:$v = 12 	imes \frac{5}{18} = \frac{60}{18} = \frac{10}{3}\, m/s$.The time taken in seconds is $T = \frac{D}{v} = \frac{2000 \pi}{10/3} = 2000 \pi 	imes \frac{3}{10} = 600 \pi\, s$.Converting the time into minutes:$T_{min} = \frac{600 \pi}{60} = 10 \pi\, minutes$.Using $\pi \approx 3.14$, $T_{min} = 10 	imes 3.14 = 31.4\, minutes$.Rounding to the nearest integer, the time taken is approximately $32\, minutes$.

$A$ man runs round a circular field of radius $50\, m$ at the speed of $12\, km/hr$. What is the time taken by the man to take twenty rounds of the field? (in $minutes$)

Similar Questions

In the given figure,if $O$ is the centre of the circle,then find the value of $y$?

The number of marble slabs of size $20\, cm \times 30\, cm$ required to pave the floor of a square room of side $3\, m$ is:

The area of a sector of a circle of radius $5 \, cm$,formed by an arc of length $3.5 \, cm$,is (in $cm^2$)

$ABCD$ is a trapezium in which $AB \parallel CD$ and $AB = 2CD$. If its diagonals intersect each other at $O$,the ratio of the area of triangle $AOB$ and $COD$ is

$2$ poles of heights $15\, m$ and $30\, m$ stand upright on a playground. If their feet are $36\, m$ apart,find the distance between their tops in meters.

Vedclass Test Series

Exam Paper Generator

Online Exam Module