JEE Main 9 Jan 2019 Shift 2 Solved Paper

F =

GM2

R2

⇒ G = [M−1L3T−2]
E = hv ⇒ h = [ML2T−1]
C = [LT−1]
t α GxhyCz
[T] = [M−1L3T−2]x [ML2T−1]y [LT−1]z
[M0L0T1] = [M−x+

y

‌

L3x+2y +

z

‌

T−2x−2x−y−z]
on comparing the powers of M, L, T
– x + y = 0 ⇒ x = y
3x + 2y + z = 0 ⇒ 5x + z = 0 ....(i)
–2x – y – z = 1 ⇒ 3x + z = –1 ...(ii)
on solving (i) & (ii) x = y =

1

2

, z = -

5

2

t α √

Gh

c5