Error in weighing due to height difference – Rankers Physics

Acceleration Due to Gravity and its variation: Practice Problem & Solution

Two equal masses m and m are hung from a balance whose scale pans differ in vertical height by ‘h’. The error in weighing in terms of density of the earth \(\rho\) is :
\(\pi G \rho m h\)
\(\frac{1}{3} \pi G \rho m h\)
\(\frac{8}{3} \pi G \rho m h\)
\(\frac{4}{3} \pi G \rho m h\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Acceleration Due to Gravity and its variation. Understanding the underlying formula is key to arriving at the correct answer below:

The error in force is \(\Delta F = m(g_1 - g_2) \approx m \frac{2g}{R} h\). Since \(g = \frac{4}{3} \pi G \rho R\), we have \(\frac{2g}{R} = \frac{8}{3} \pi G \rho\). Therefore, \(\Delta F = \frac{8}{3} \pi G \rho m h\).

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