Question

A 20g mass is suspended from a meter rod at 15cm mark. The meter rod is balanced on the 40cm mark
on a horizontal wedge. Find the weight of the meter rod
A) 5ON
C) 0.5N
B) 5N
D) 0.05N​

Answer 1

Answer:-

$\red{\bigstar}$ The weight of the meter rod

$\large\leadsto\boxed{\rm\purple{5 \: N}}$

• Given:-

A 20g mass is suspended from a meter rod at 15 cm mark.

The meter rod is balanced at 40cm mark on a horizontal wedge.

• To Find:-

The weight of meter rod.

• Solution:-

$\pink{\bigstar}$ Counter clockwise torque:-

Counter clockwise torque = $\sf{0.20 \times 10 \times (40 - 15)}$

→ $\sf{2 \times 25}$

→ $\sf{50}$

The weight of the rod will be at the center of the rod.

$\pink{\bigstar}$ Clockwise torque:-

Clockwise torque = $\sf{Weight \times (50 - 40)}$

→ $\sf{Weight \times 10}$

$\pink{\bigstar}$ Counter clockwise torque = Clockwise torque

→ $\sf{Weight \times 10 = 50}$

→ $\sf{Weight = \dfrac{50}{10}}$

→ $\bf\green{Weight = 5 N}$

Therefore, the weight of meter rod is 5N.