Back to RATE MEASURE

JEE MATHEMATICS MCQ: RATE MEASURE

1 Easy

1 of 1

Application +4 | -1

The rate of change of the surface area of a sphere of a sphere of radiusr,when the radius is increasing at the rate of 2 cm/sis proportional to

View answer and explanation

Correct option: C

r

Option C (

r

) is correct because the rate of change of the surface area is directly proportional to the radius, as

\frac{{dS}}{{dt}} = 16\pi r

.

More Questions from RATE MEASURE

Practice All Chapter MCQs →

The surface area of a sphere when its volume is increasing at the same rate as its radius, is A particle is moving in a straight line has velocity and displacement equation as

v = 4\sqrt {1 + s}

v

velocity in m/s and

s

m

. The initial velocity of the particle is The diagonal of a square is changing at the rate of 0.5 cm

{s^{ - 1}}

. Then, the rate of change of area , when the area is 400

c{m^2}

is equal to A spherical balloon is filled with

4500 \pi

cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of

72 \pi

cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is: A solid box is formed by placing a cylinder, having equal height and radius on top of a cube such that, the circular base of cylinder is the inscribed circle for square top of the cube. If the radius of cylinder is changing at the rate

\frac{1}{2 \pi+16}

\mathrm{cm} / \mathrm{sec}

, then the rate of change of volume of the box when radius is

2 \mathrm{~cm}

, is (Assuming that box always remain in the given shape) If the area of an expanding circular region increases at a constant rate with respect to time, then the rate of marease of the perimeter with respect to the time