Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
a spherical body of r radius and relative density 0.5 is floating in equlibrium in water with half of it immersed in water. the work done in pushing the ball so that whole of it is immersed in water?
work done by gravity(u)+work done by pushing(v)=work done by buoyant force(w)
w=2u, therefore v=u=(4πr3/3)*0.5*1000*gr
please let me know if you want more explanation
my previous solution is WRONG
There are three forces involved: gravity,pushing and buoyancy
gravity(g) + pushing(p)=buoyancy(b)
thus, p=b-g, so work done by p(wp)=work done by buoyancy(wb)-wg
let mass of ball=m, wg=mgr
wb=∫b(x) dx, from x=0 to x=r, where x is the distance by which the centre of the ball is below the surgace of water
The answer for this question is (5pi/12)dgr4 where (d is the density of water and r is the radius of the spherical body) . the answer is completely independent of the relative density of the floating body.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !