A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

$\frac{√hG}{c^\frac{3}{2}}$

According to question,

L ∝ h^p c^q G^r

[M⁰LT⁰] = [ML²T^-1]^p [LT^-1]^q [M^-1L³T^-2]^r

Equating power both sides, we get

p - r = 0        ......(1)

2p + q + 3r = 1        ......(2)

- p - q - 2r = 0        ......(3)

Solving equation (1), (2), (3), we get

p = r = $\frac{1}{2}$ , q = $\frac{-3}{2}$

∴ L = $\frac{√hG}{c^\frac{3}{2}}$