A) 190 m
B) 850 m
C) 591 m
D) 510 m
Correct Answer: B
Solution :
Let the person is at point P. Time taken in first echo, \[{{t}_{1}}=1.5\,s\] \[\therefore \] \[x=\frac{V\times {{t}_{1}}}{2}\] \[=\frac{340\times 7.5}{2}=255\,m\] Similarly, \[y=\frac{V\times {{t}_{2}}}{2}\] \[=\frac{340\times 3.5}{2}\] \[=595\,m\] Distance between two cliffs \[=\text{ }x\text{ }+\text{ }y\] = 255 + 595 = 850 mYou need to login to perform this action.
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