A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12 cm mark, the stick is found to be balanced at 45 cm. The mass of the metre stick is

Question

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12 cm mark, the stick is found to be balanced at 45 cm. The mass of the metre stick is (A) 56 g (B) 66 g (C) 76 g (D) 86 g

Tardigrade Admin · Accepted Answer

Let m be the mass of the metre stick concentrated at C, the 50 cm mark as shown in the figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4e27c5509cfcded831890dfc5b038720-.png" /> In equilibrium, taking moments of forces about C', we get 10 g(45-12) = mg (50-45); 10g × 33 = mg × 5 m = (10 × 33/5) = 66 g

Q. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5 g$ are put one on top of the other at the $12 c m$ mark, the stick is found to be balanced at $45 c m$ . The mass of the metre stick is

$56 g$

$66 g$

$76 g$

$86 g$

Let $m$ be the mass of the metre stick concentrated at $C$ , the $50 c m$ mark as shown in the figure.

In equilibrium, taking moments of forces about $C^{'}$ , we get
$10 g (45 - 12) = m g (50 - 45);$
$10 g \times 33 = m g \times 5$
$m = \frac{10 \times 33}{5} = 66 g$

Q. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12cm mark, the stick is found to be balanced at 45cm. The mass of the metre stick is

56g

66g

76g

86g