QJEEM19Phc9b3: JEE Main 2019 Physics - Class 11 JEEM19Phc9b3 Solution: If surface tension (S), Moment of inertia (I) and Planck's…

We know,

surface tension (S) = ${F \over L}$ = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ L \right]}}$

$\therefore$ [S] = ${\left[ {M{T^{ - 2}}} \right]}$

Moment of inertia (I) = mr²

$\therefore$ [I] = ${\left[ {M{L^2}} \right]}$

Planck's constant (h) = ${E \over f}$ = Et

$\therefore$ [h] = ${\left[ {M{L^2}{T^{ - 1}}} \right]}$

Also linear momentum (p) = mv = ${\left[ {ML{T^{ - 1}}} \right]}$

Now we have to express p in terms of s, I and h.

$\therefore$ Let, [P] = [S^a I^b h^c]

$\Rightarrow$ ${\left[ {ML{T^{ - 1}}} \right]}$ = ${\left[ {M{T^{ - 2}}} \right]}$ ^a ${\left[ {M{L^2}} \right]}$ ^b ${\left[ {M{L^2}{T^{ - 1}}} \right]}$ ^c

$\Rightarrow$ ${\left[ {ML{T^{ - 1}}} \right]}$ = [ M^{a + b + c} L^{2b +2c} T^{- 2a - c} ]

By comparing the dimensions of both sides, we get

a + b + c = 1 .........(1)

2b +2c = 1 ..............(2)

- 2a - c = -1 ...................(3)

By solving those three equations we get,

a = ${1 \over 2}$

b = ${1 \over 2}$

c = 0

$\therefore$ linear momentum [p] = [S^1/2I^1/2h⁰]