JEE Main 1-Sep-2021 Shift 2 Solved Paper

Let’s draw the free body diagram when body slides down on smooth surface
For smooth surface,
ma = mg sin30º
a = g sin30º
a = g / 2
Distance covered by the block on the smooth surface in time T,
s=ut+

1

2

at2
s=0+

1

2

(

g

2

)T2
⇒s=(

g

4

)T2 ...(i)
Now, let’s draw the free body diagram when body slides down on rough surface
For rough surface,
ma = mg sin30º - μ mg cos30º
a = g sin30º - μ g
Distance covered by the block on the rough surface in time α T,
s=ut+

1

2

at2
s=0+

1

2

(g‌sin‌30‌º−μg‌sin‌30‌μ)t2
s=

g

4

(1−√3μ)(αT)2 ...(ii)
Distance covered by the block is same for both the case,
⇒

g

4

(1−√3μ)(αT)2=

g

4

T2[from Eq. (i) & Eq. (ii)]
⇒1−√3μ=

1

α2

⇒μ=(

α2−1

α2

)

1

√3

Comparing with (

α2−1

α2

)

1

√x

The value of the x = 3.