MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec. Find the rate at which the volume of the bubble is increasing at the instant if its radius is 6 cm.
one year ago

Answers : (1)

jitender
114 Points
							£=PIEdA/dt= 4£r^2/dt=2 cm^2/s ... 1st eqn dV/dt=?In 1st multiply and divide by dr/dr.(4£r^2/dr)(dr/dt)=2 8£r dr/dt= 2 dr/dt= 1/4£r....2nd eqndV/dt= 4/3 £r^3/dt=( 4/3 £r^3/dr )dr/dt= (4£r^2)1/4£r=rHence dV/dr. = 6 cm^3/s
						
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details