Request a call back

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

JEE Class main Answered

Sir pls solve the following.
question image
Asked by rsudipto | 22 Dec, 2018, 08:07: PM
answered-by-expert Expert Answer
As seen from figure , net force acting on the block = (mg sin30 - μ mg cos30) Newton
Acceleration = g(sin30 - μ cos30) m/s2
 
distance S travelled along inclined plane in 2 s,  starting from rest,   S = (1/2)×a×t2 = (1/2)×g(sin30 - μ cos30)×2×2 = 8
 
after simplifying above equation, we get 9.8×(1- √3μ) = 8  .....(1)

 From (1), we get μ = 0.106
Answered by Thiyagarajan K | 22 Dec, 2018, 10:25: PM
JEE main - Physics
Asked by kmani310507 | 28 Apr, 2024, 04:38: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by arivaryakashyap | 23 Apr, 2024, 10:40: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ksahu8511 | 19 Apr, 2024, 11:55: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by mohammedimroz | 13 Apr, 2024, 09:48: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by medhamahesh007 | 02 Apr, 2024, 11:11: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by gundlasumathi93 | 31 Mar, 2024, 02:13: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by chhayasharma9494 | 31 Mar, 2024, 12:47: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by Machinenineha | 27 Mar, 2024, 05:28: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×