Question

plzz help its urgent...plzz tomorrow is exam nd explain it too.

a uniform meter rod of weight 50 gf is balanced at 40 cm mark , a weight of 100 gf is suspended at 5 cm mark. where must be a weight of 80gf should be placed to balance the meter rod.

Answer 1

Acordng to the principle of equilibrium

Anticlockwise = Clockwise moment moment

=> {50 × 10} + {80 × (40 - x)} = 100 × 35

=> 500 + 3200 - 80x = 3500.

=> 3700 - 8x = 3500

=> 8x = 200

=> x = 25 cm

ANS  = weight of 80gf should be placed at 25 cm mark balance the meter rod. if it is helpful plz mark brainly

Answer 2

therefore x= 25 SEE It helps you