A metre scale of weight 50gf can be balanced at 40 cm mass without any weight suspended on it.
(i) lf this ruler is cut at its centre then state which part (0-50 cm or 50-100 cm) of the ruler weight more than 25 gf.
(ii) What minimum weight balanced on this metre ruler can be balance this ruler when it is devoted at its centre.
(i) lf this ruler is cut at its centre then state which part (0-50 cm or 50-100 cm) of the ruler weight more than 25 gf.
(ii) What minimum weight balanced on this metre ruler can be balance this ruler when it is devoted at its centre.
Answers & Comments
Answer:
Hence, the part [ 0 to 50 cm] of the ruler will weigh more 25 gf.
Verified answer
Answer:
(i) When the ruler is cut at its center, both parts will have equal weights since it was balanced at the center without any weights suspended. So, neither part (0-50 cm nor 50-100 cm) will weigh more than 25 gf.
(ii) To find the minimum weight needed to balance the ruler when pivoted at its center, you need to consider the torques. Since the ruler was initially balanced at 40 cm without any weights, when cut at the center, you'd need a weight at a distance that creates a torque equal to the torque exerted by the 50gf weight at 40 cm.
The torque is given by the formula: Torque = Force × Distance
So, the minimum weight (W) needed at the center can be calculated as follows:
\[50 \, \text{gf} \times 40 \, \text{cm} = W \times 20 \, \text{cm}\]
Solving for W, you find the minimum weight needed at the center to balance the ruler.