Question

Two pans of a balance are 24.1 cm apart. The fulcrum of the balance has been shifted 1.33 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 1

As we know that fulcrum is shifted to one side by 1.33 cm

so here let say left side mass is m1 while on other side mass is m2

so we will have

$m_1g (24.1 - 1.33) = m_2 g(24.1 + 1.33)$

$m_2 = 0.895 m_1$

now in order to find the percentage we have

$percentage = \frac{m_1 - m_2}{m_1}\times 100$

$percentage = (1 - 0.895)\times 100$

$percentage = 10.5%$

so it is 10.5 % less mass

Answer 2

To solve the problem we must know about the bending moment and also know to calculate the percentage.

What is bending moment?

The bending moment is the measure of the bending that can occur due to external forces. It is calculated by multiplying the force with the perpendicular distance between the point and the force.

The shopkeeper is cheating by giving 10.46% fewer things to the customer.

Calculating the momentum about the fulcrum,

$P_1 \times (0.241-0.0133) = P_2 \times (0.241+0.0133)\\\\P_2 = 0.8954P_1$

Now, calculating the percentage change,

$\rm Percentage\ Change = \dfrac{P_2-P_1}{P_1} \times 100 = -0.1046 = -10.46\%$

Thus, the shopkeeper is giving 10.46% fewer things to the customer while the negative sign denotes that the value of P2 is less than before.

Therefore, the shopkeeper is cheating by giving 10.46% fewer things to the customer.

Learn more about Bending moment:

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