A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/sec, then the rate (in cm/sec) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1m above the ground is?

Correct option is C.
Let x be the distance of the bottom of the ladder from the wall and y be the height of the top of the ladder from the ground.
Then, by Pythagorean theorem, x² + y² = 4 (since the length of the ladder is 2m = 200cm).
Differentiating with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Given that dy/dt = -25 cm/sec (negative because the ladder is sliding down).
When y = 1m = 100cm, x² + 100² = 400²
x² = 400² - 100² = 300²
x = 100√3 cm
Substituting the values in the equation 2x(dx/dt) + 2y(dy/dt) = 0, we have:
2(100√3)(dx/dt) + 2(100)(-25) = 0
200√3 (dx/dt) = 5000
d(x)/dt = 5000 / (200√3) = 25/√3 = 25√3/3 cm/sec
However, this solution seems to have an error in calculation. Let's redo it.
Given x² + y² = 4 (in meters)
x² + y² = 4
Differentiating with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
When y = 1m, x² + 1² = 2² => x = √3 m = 100√3 cm
dy/dt = -0.25 m/s = -25 cm/s
2(100√3)(dx/dt) + 2(100)(-25) = 0
200√3 (dx/dt) = 5000
d(x)/dt = 25/√3 = 25√3/3 ≈ 14.43 cm/s
Let's reconsider the given solution: 25/3 cm/sec. There may be an error in the provided solution or the question's values.