Question

A uniform metre scale is in equilibrium position.Calculate the mass of the ruler​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

A uniform metre scale is in equilibrium position.Calculate the mass of the ruler

$\huge\underline{\underline{\bf \green{Solution-}}}$

50 is the mid-point of uniform scale

Let mass of rod be M

Sum of clockwise movement ➝

$\implies{\bf \blue{Mass×Distance\:of\: weight\:from\:pivot}}$

$\implies{\sf 40 × (30-5)}$

$\implies{\sf 40×25}$ ➝ (eq. 1)

Sum of anticlockwise movement ➝

$\implies{\bf \blue{Mass\:of\:rod×Distance\:of\: weight\:from\:rod}}$

$\implies{\sf M × (50-30)}$

$\implies{\sf 20×M}$ ➝ (eq. 2)

$\large\underline{\underline{\rm From\: Equation \: 1 \: and \: 2-}}$

$\implies{\sf 25×40=20×M}$

$\implies{\sf 1000=20M}$

$\implies{\sf M = \dfrac{1000}{20}}$

$\implies{\bf \red{M = 50\:g }}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Mass of rod is ${\bf \red{50\:g}}$