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 :
1. \(\pi G \rho m h\)
2. \(\frac{1}{3} \pi G \rho m h\)
3. \(\frac{8}{3} \pi G \rho m h\)
4. \(\frac{4}{3} \pi G \rho m h\)
View Answer
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, \(\Delta F = \frac{8}{3} \pi G \rho m h\).
The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. A man jumps to a height of $2\text{ m}$ on the surface of $A$. What is the height of jump by the same person on the planet $B$:
(2003)
1. $\frac{2}{9}\text{ m}$
2. $18\text{ m}$
3. $6\text{ m}$
4. $\frac{2}{3}\text{ m}$
View Answer
The muscular work done in jumping is the same, so the potential energy gained is constant: $m g_A h_A = m g_B h_B$. Substituting $g_A = 9 g_B$ and $h_A = 2\text{ m}$, we get $9 g_B \times 2 = g_B \times h_B \implies h_B = 18\text{ m}$.