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
Answer & explanation
Correct answer: option 1
According to question,
L ∝ hp cq Gr
[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 = $\frac{√hG}{c^\frac{3}{2}}$