A hemispherical bowl of radius R=0.1m is rotating about its own axis (which is vertical) with an angular velocity $ω$ . A particle of mass $10^{- 2}$ kg on the friction-less inner surface of the bowl is also rotating with the same w. The particle is at height h from the bottom of the bowl. It is desired to measure ‘g’ using the set-up by measuring h accurately. Assuming that R and $ω$ are known precisely and that the least count in the measurement of h is $10^{- 4} m$ . The minimum possible error (Dg) in the measured value of g is $n \times 10^{- 2} m / s^{2}$ . Then n is (h << R)

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$\frac{1}{2}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} (a) R is the radius of the bowl \\ \sin θ = \frac{r}{R} \\ \Rightarrow r = R \sin θ - - - (1) \\ N \cos θ = m g - - - (2) \\ a n d \\ N \sin θ = m r ω^{2} - - - - (3) \end{array}$

$\begin{array}{l} \Rightarrow N s i n θ = m ω^{2} R s i n θ \\ \Rightarrow N = m R ω^{2} - - - (4) \\ f r o m (2) & (4) \\ ∴ m R ω^{2} \cos θ = m g \end{array}$

$\begin{array}{l} \Rightarrow g = ω^{2} (R - h) = ω^{2} R (1 - \frac{h}{R}) \\ taking 'log' both side \\ \Rightarrow \log (g) = \log ω^{2} R + \log (1 - \frac{h}{R}) \\ \Rightarrow \log (g) = \log ω^{2} R + [- \frac{h}{R} - \frac{h^{2}}{2 R^{2}} - \frac{h^{3}}{3 R^{3}} + \dots] \end{array}$

$\begin{array}{l} D i f f e r e n t i a t i n g w i t h r e s p e c t t o ‘ h ’ \\ \Rightarrow \frac{1}{g} \frac{d g}{d h} = 0 - \frac{1}{R} - \dots [higher order of \frac{h}{R} neglected] \\ \Rightarrow Δ g = \frac{g}{R} \times Δ h = \frac{10}{0.1} \times 10^{- 4} = 10 \times 10^{- 3} = 1 \times 10^{- 2} \end{array}$