To determine the correct pairs of weights that will keep the hydraulic machine balanced, we need to utilize the principle of hydraulic balance. According to this principle, the force applied is proportional to the cross-sectional area of the pistons.
Given:
- The cross-sectional area of the larger piston is twice the area of the smaller piston.
- The weights [tex]\( W_1 \)[/tex] and [tex]\( W_2 \)[/tex] are the weights on the smaller and larger pistons respectively.
For the machine to be balanced, the equilibrium condition in terms of weights and areas should be:
[tex]\[ \frac{W_1}{A_1} = \frac{W_2}{A_2} \][/tex]
Given that [tex]\( A_2 = 2A_1 \)[/tex], substituting [tex]\( A_2 \)[/tex] gives us:
[tex]\[ \frac{W_1}{A_1} = \frac{W_2}{2A_1} \][/tex]
[tex]\[ W_1 = \frac{W_2}{2} \][/tex]
Thus, for balance, [tex]\( W_1 \)[/tex] should be equal to half of [tex]\( W_2 \)[/tex]. So we need pairs where:
[tex]\[ W_1 = 2 \times W_2 \][/tex]
Now, let's evaluate each given pair:
1. (i) [tex]\( W_1 = 4 \, \text{N}, W_2 = 2 \, \text{N} \)[/tex]
- For balance: [tex]\( 4 = 2 \times 2 \)[/tex] which is true.
- Hence, this pair balances the machine.
2. (ii) [tex]\( W_1 = 5 \, \text{N}, W_2 = 10 \, \text{N} \)[/tex]
- For balance: [tex]\( 5 = 2 \times 10 \)[/tex] which is false.
- Hence, this pair does not balance the machine.
3. (iii) [tex]\( W_1 = 2 \, \text{N}, W_2 = 1 \, \text{N} \)[/tex]
- For balance: [tex]\( 2 = 2 \times 1 \)[/tex] which is true.
- Hence, this pair balances the machine.
4. (iv) [tex]\( W_1 = 3 \, \text{N}, W_2 = 1.5 \, \text{N} \)[/tex]
- For balance: [tex]\( 3 = 2 \times 1.5 \)[/tex] which is true.
- Hence, this pair balances the machine.
Therefore, the correct group of weights to keep the machine balanced satisfying the given conditions are:
- [tex]\( W_1 = 4 \, \text{N}, W_2 = 2 \, \text{N} \)[/tex]
- [tex]\( W_1 = 2 \, \text{N}, W_2 = 1 \, \text{N} \)[/tex]
- [tex]\( W_1 = 3 \, \text{N}, W_2 = 1.5 \, \text{N} \)[/tex]
