Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

A block of ice of area A and thickness 0.5 m is floating in the fresh water. In order to just support a man of 100kg, the area of A should be (specefic gravity of ice 0.917 and the density of water = 1000 kg/m^3)  
Asked by jmorakhia | 24 Apr, 2019, 19:08: PM
answered-by-expert Expert Answer
weight of man and weight of ice is balanced by upward thrust
 
Let M be the mass of man, A is the required area of ice block of thickness d, ρi and ρw are the densities of ice and water respectively.
 
Then we have,  Mg + (A×0.5)ρi g = (A×0.5)ρw g       or A =   begin mathsize 12px style fraction numerator M over denominator 0.5 cross times open parentheses rho subscript w space minus space rho subscript i close parentheses end fraction space equals space fraction numerator 100 over denominator 0.5 cross times left parenthesis 1000 minus 917 right parenthesis end fraction space equals space 2.41 space m squared end style
Answered by Thiyagarajan K | 24 Apr, 2019, 22:10: PM
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT