Home
Class 12
MATHS
A spherical iron ball of radius 10cm, co...

A spherical iron ball of radius 10cm, coated with a layer of ice of uniform thickness, melts at a rate of `100 pi cm^(3) //` min.The rate at which the thickness decreases when the thickness of ice is 5 cm, is

A

`1cm//min`

B

`2cm//min`

C

`(1)/(376)cm//min`

D

`5cm//min`

Text Solution

Verified by Experts

The correct Answer is:
A

`"Given, "(dV)/(dt)=100pi cm^(3)//min`
`therefore" "(d)/(dt)((4)/(3)pir^(3))=100pi rArr 3r^(2)(dr)/(dt)=(300pi)/(4pi)" "rArr" "((dr)/(dt))_("(r = 5)")=(300)/(4xx3xx25)="1cmmin"`
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST 13

    VMC MODULES ENGLISH|Exercise MATHEMATICS( SECTION-2)|5 Videos
  • MOCK TEST 12

    VMC MODULES ENGLISH|Exercise MATHEMATICS (SECTION 2)|5 Videos
  • MOCK TEST 2

    VMC MODULES ENGLISH|Exercise PART III : MATHEMATICS (SECTION - 2)|10 Videos

Similar Questions

Explore conceptually related problems

A spherical iron ball 10cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50c m^3//m in . When the thickness of ice is 5cm, then find the rate at which the thickness of ice decreases.

A spherical ice ball is melting at the rate of 100 pi cm^(3) /min. The rate at which its radius is decreasing, when its radius is 15 cm, is

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Radius of a circle is increasing at rate of 3 cm//sec Find the rate at which the area of circle is increasing at the instant when radius is 10 cm.

A sphere of 10cm radius has a uniform thickness of ice around it. Ice is melting at rate 50 cm^3//min when thickness is 5cm then rate of change of thickness

A sphere of 10cm radius has a uniform thickness of ice around it. Ice is melting at rate 50 cm^3//min when thickness is 5cm then rate of change of thickness

The volume of a spherical balloon is increasing at a rate of 25 cm^(3)//sec . Find the rate of increase of its curved surface when the radius of balloon is 5 cm.

If the volume of a sphere increase at the rate of , 2pi cm^(3)//sec , then the rate of increase of its radius (in cm/sec), when the volume is 288pi cm^(3)" is :"

Suppose that water is emptied from a spherical tank of radius 10 cm. If the depth of the water in the tank is 4 cm and is decreasing at the rate of 2 cm/sec, then the radius of the top surface of water is decreasing at the rate of

The volume of a spherical balloon is increasing at the rate of 20 cm^3 / sec . Find the rate of change of its surface area at the instant when radius is 5 cm.