You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Displacement of COM:Application

A meter stick...

Question

A meter 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 $12.0 c m$ mark, the stick is found to be balanced at $45.0 c m$ . What is the mass of the meter stick?

Open in App

Solution

Figure shows the forces on the stick
The mass of metre stick is concentrated at $C M = 50 c m$
Consider the torques around $0$ .Since the stick is in rotational equilibrium,
${¯ τ}_{e x t} = m (5 c m) - (10 g) (45 - 12) = 0$
$∴= 10 g \frac{(33 c m)}{(5 c m)} = 66 g m .$

Suggest Corrections

Similar questions

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 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Q. a meter stick is balanced on a knife edge at its center. when two coins , each of mass 5 g are put one on of the other at the 12 cm mark, the stick is to balanced at 45 cm . the mass of the meter stick is

The rock has a mass of 1 kg. What is the mass of the measuring stick if it is balanced by a support force at the one-quarter mark? The centre of mass of the measuring stick is at the one-half mark, one-quarter of the stick's length beyond the pivot.

Q. A metre scale is balanced on a knife-edge at its centre. When a mass of

10

kg is kept at the

12

cm mark, the scale is balanced at

45

cm. What is the mass of the metre scale?

Q. A meter stick is hung from two spring balances

A

and

B

of equal lengths that are located at the

20 c m

and

70 c m

marks of the meter stick. Weights of

2.0 N

are placed at the

10 c m

and

40 c m

marks, while a weight of

1.0 N

is placed at the

90 c m

mark. The weight of the uniform meter stick is

1.5 N

. Determine the scale readings of the two balances

A

and

B

Join BYJU'S Learning Program

A meter 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 12.0cm mark, the stick is found to be balanced at 45.0cm. What is the mass of the meter stick?

Figure shows the forces on the stickThe mass of metre stick is concentrated at CM=50cmConsider the torques around 0.Since the stick is in rotational equilibrium,¯τext=m(5cm)−(10g)(45−12)=0∴=10g(33cm)(5cm)=66gm.

A meter 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 $12.0 c m$ mark, the stick is found to be balanced at $45.0 c m$ . What is the mass of the meter stick?

Figure shows the forces on the stick
The mass of metre stick is concentrated at $C M = 50 c m$
Consider the torques around $0$ .Since the stick is in rotational equilibrium,
${¯ τ}_{e x t} = m (5 c m) - (10 g) (45 - 12) = 0$
$∴= 10 g \frac{(33 c m)}{(5 c m)} = 66 g m .$