To lift a load of 30 kgf, Suhas uses a single fixed pulley while Radha uses a single movable pulley. The displacement of efforts in both the cases are equal. In…

Given,

• Load = 30 kgf

(a) As, mechanical advantage (MA) of a machine is given by,

$\text {MA} = \frac{\text {Load}}{\text {Effort}}$

For a single fixed pulley, the mechanical advantage (MA) is 1.

Then,

$$\begin{aligned}\text {MA} = \frac{\text {Load}}{\text {Effort} (\text E_1)} \\[1em] \Rightarrow \text E_1 = \frac{\text {Load}}{\text {MA}} \\[1em] \Rightarrow \text E_1 = \frac{30}{1} \\[1em] \Rightarrow \text E_1 = 30 \text { kgf}\end{aligned}$$

For a single movable pulley, the mechanical advantage (MA) is 2.

Then,

$$\begin{aligned}\text {MA} = \frac{\text {Load}}{\text {Effort} (\text E_2)} \\[1em] \Rightarrow \text E_2 = \frac{\text {Load}}{\text {MA}} \\[1em] \Rightarrow \text E_2 = \frac{30}{2} \\[1em] \Rightarrow \text E_2 = 15 \text { kgf}\end{aligned}$$

Ratio of efforts = $\text E_1 : \text E_2$ = 30 : 15 = 2 : 1

Hence, the ratio of the efforts in the two cases is 2 : 1.

(b) Potential energy is given by,

Potential energy gained = Load × Height raised

If the effort displacement is the same:

• In a fixed pulley, load rises by the same distance as the effort.
• In a movable pulley, load rises by half the distance of the effort.

Let, for the fixed pulley, height raised by the load be h.

Therefore,

Ratio of the potential energies in two cases is given by,

$$\begin{aligned}\text {Ratio of PE} = \frac{\text {PE gained by the load in fixed pulley}}{\text {PE gained by the load in movable pulley}} \\[1em] = \frac {30 \times \text h}{30 \times \frac{\text h}{2}} \\[1em] = \frac {30\text h \times 2}{30 \text h} \\[1em] = \frac {2}{1} \\[1em] = 2 : 1\end{aligned}$$

Hence, the ratio of the potential energies in two cases is 2 : 1.

(c) In an ideal machine, efficiency = 100 % for both pulleys.

Then,

Efficiency ratio = 1 : 1

Hence, the ratio of the efficiencies in the two cases is 1 : 1.