Question

The minimum value of coefficient of friction (u) such that block of mass '5 kg' remains at rest is [NCERT Pg. 102] 5 kg 3. 3 kg (1) 0.3 (2) 0.5 (3) 0.6 (4) 0.4




Explanation:
Here you want to check all the options by applying ú value
If ú=0.6 then,
F=mg=5×10=50kgm/s^2
Ff=ú×F=0.6×50=30N=3kg
so 0.6 Ff balance the block at rest​

Answer 1

Answer:

``to ´`(and any subsequent words) was ignored because we limit queries to 32 words.

So minimum coefficient of friction should be6.0 so that 5 kg block remains at rest.