College

Two pans of a balance are 39.3 cm apart. The fulcrum of the balance has been shifted 1.46 cm away from the center by a dishonest shopkeeper.

By what percentage is the true weight of the goods being marked up by the shopkeeper? Assume the balance has negligible mass.

Answer in units of %.

Answer :

The true weight of the goods is being marked up 1,936.5% by the shopkeeper.

What is the error in percentage?

To calculate the percentage of increase in weight, we need to calculate the actual increase in weight, then divide it by the true weight and multiply by 100.

The displacement of the fulcrum has caused a torque on the balance, and we can calculate this torque using the formula: torque = force x distance. The force causing the torque is the weight of the goods, and the distance is the displacement of the fulcrum from the center.

Torque = weight x displacement

= weight x (39.3 / 2 + 1.46)

= weight x (20.365)

The balance is in equilibrium, so the sum of torques on the two pans is zero. Therefore, the torque on the pan with the goods is equal in magnitude but opposite in direction to the torque on the other pan.

Torque_goods = -Torque_counterweights

= - (weight_counterweights x 20.365)

= - (weight_goods x 20.365)

Dividing both sides by weight_goods, we get:

20.365 = - weight_goods / weight_counterweights x 20.365

Solving for weight_goods, we find:

weight_goods = weight_counterweights x 20.365 / 20.365

Therefore, the actual weight of the goods is 20.365 times the true weight. The percentage increase in weight is:

percentage increase = (actual weight - true weight) / true weight x 100

= (20.365 - 1) / 1 x 100

= 1,936.5%

So, the true weight of the goods is being marked up by 1,936.5%.

Learn more on fulcrum and balance here: https://brainly.com/question/10100080

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