Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201920 Apr 2019Evening ShiftMathematicsApplication of DerivativesActual

The radius of a sphere increases at the rate of 0.04 ~cm / sec . The rate of increase in the volume of that sphere with respect to its surface area, when its radius is 10 ~cm is

Options

  1. A16
  2. B25
  3. C20
  4. D5

Correct answer

D. 5

Step-by-step solution

Let r be the radius of the sphere. Given, rate of change in radius d r d t =0.04 ~cm / sec Volume of sphere (V)= 4 3 r^3 Differentiating w.r.t, t , we get d V d t = 4 3 (3 r^2 ) d r d t Surface area of sphere (S)=4 r^2 d S d t =4 (2 r) d r d t Eq. (i) divided by Eq. (ii), we get array ll d V d t d S d t = 4 r^2 d r d t 8 r d r d t & d V d S = r 2 = 10 2 & [ r=10 ~cm ] d V d S =5 ~cm & array Hence, option (d) is correct.

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025If f(x)=x e^ x(1-x) , x R , then f(x) is 2025The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All AP EAMCET PYQs