Question

100.
96. A man wants to reach from A to
the opposite corner of the square
C (Figure). The sides of the square
are 100 m each. A central square of
50 m x 50 m is filled with sand.
Outside this square, he can walk at a speedl ms!
In the central square, he can walk only at a speed of
vms-! (v< 1). What is smallest value of v for which
he can reach faster via a straight path through the
sand than any path in the square outside the sand?
(a) 0.18 m s-
(b) 0.81 m s-
(c) 0.5 m s-1
(d) 0.95 m s-​

Answer 1

Answer:

Hey mate here is your answer.

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Explanation:

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Option B 0.81 m/s is the answer.

Time taken to go from A to C via straight line path APQC through the S and T sand,

=AP+QC/1+PQ/v=25√2+25√2/1+50√2/v,

=50√2+50(√2/v)=50√2(1/v+1),

Clearly from figure the shortest path outside the sand will be ARC.

Time taken to go from A to C via this path:

T outside=AR+RC/1(s),

Clearly AR=√75²+25²=√75×75+25×25,

=5×5√9+1=25√10 m,

RC=AR,√75²+25²=25√10,

⇒ T outside=2AR=2×25√10(s)=50√10(s),

For T sand< T outside,

⇒50√2(1/v+1)<2×25√10,

⇒2√2/2(1/v+1)<√10,

⇒1/v+1<2√10/2√2=√5/2×2=√5,

⇒1/v<√5/2×2-1⇒1/v<√5-1,

⇒v<1√5-1=0.81m/s,

⇒v<o.81m/s.

Hope it helps.✔✔✔

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