Exams › JEE Main › Maths
A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness which melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases is
- 1/(36π) cm/min.
- 1/(18π) cm/min.
- 1/(54π) cm/min.
- 5/(6π) cm/min.
Correct answer: 1/(18π) cm/min.
Solution
Outer radius R = 10 + 5 = 15. From V = (4/3)*pi*R^3, dV/dt = 4*pi*R^2*dR/dt. So -50 = 4*pi*(225)*dR/dt, giving |dR/dt| = 50/(900*pi) = 1/(18*pi) cm/min.
Related JEE Main Maths questions
- For the curve defined by x² + y² = a², if k = 1/a, then k can be written as
- For the curve y² = 2x³, identify the point P at which the tangent is perpendicular to the line 4x - 3y + 2 = 0.
- For the function f(x) = 2x² - log|x|, with x ≠ 0, on which interval is the function increasing throughout?
- Two upright poles AP and BQ stand at points A and B respectively. If AP = 16 m, BQ = 22 m, and the distance AB = 20 m, then for a point R on AB, the value of AR for which RP² + RQ² is least is
- For the curve x⁴ + y⁴ = a⁴, a tangent drawn at an arbitrary point meets the coordinate axes at intercepts p and q. Then the quantity p^(-4/3) + q^(-4/3) equals
- A particle travels on the curve y = x³ + 2. Find the point(s) P on the curve where the y-coordinate changes at a rate 8 times the rate of change of the x-coordinate. The points are (4, 11) and (-4, -31/3).
⚔️ Practice JEE Main Maths free + battle 1v1 →