Set the drifts suffered by boat in increasing order
(a) d=1000 m, `V_(R) = 2m//s, V_(b) = 4 m//s`
( b) `d=500 m, V_(R) =1 m//s, V_(b) = 6 m//s`
( c) `d=1000 m, V_(R) = 6 m//s, V_(b) = 6 m//s`
(`d to` width of river `V_(R) to` velocity of river `V_(b) to` velocity of botat). The boat moves perpendicular to width of the river

Options

A. a,b,c
B. b,c,a
C. b,a,c
D. c,b,a

Answer

To find the drift suffered by the boat in each case, we can use the following formula:

\[ \text{Drift} = \frac{d \cdot V_R}{V_b} \]

where:
- \( d \) is the width of the river,
- \( V_R \) is the velocity of the river,
- \( V_b \) is the velocity of the boat.

We will calculate the drift for each scenario provided in the question.

### Step 1: Calculate Drift for Each Case

**Case (a):**
- \( d = 1000 \, m \)
- \( V_R = 2 \, m/s \)
- \( V_b = 4 \, m/s \)

Using the formula:
\[ \text{Drift}_a = \frac{1000 \cdot 2}{4} = \frac{2000}{4} = 500 \, m \]

---

**Case (b):**
- \( d = 500 \, m \)
- \( V_R = 1 \, m/s \)
- \( V_b = 6 \, m/s \)

Using the formula:
\[ \text{Drift}_b = \frac{500 \cdot 1}{6} = \frac{500}{6} \approx 83.33 \, m \]

---

**Case (c):**
- \( d = 1000 \, m \)
- \( V_R = 6 \, m/s \)
- \( V_b = 6 \, m/s \)

Using the formula:
\[ \text{Drift}_c = \frac{1000 \cdot 6}{6} = \frac{6000}{6} = 1000 \, m \]

### Step 2: List the Drifts

Now we have the drifts for each case:
- Drift (a) = 500 m
- Drift (b) ≈ 83.33 m
- Drift (c) = 1000 m

### Step 3: Set the Drifts in Increasing Order

Now we can arrange the drifts in increasing order:
1. Drift (b) ≈ 83.33 m
2. Drift (a) = 500 m
3. Drift (c) = 1000 m

### Final Answer

The drifts suffered by the boat in increasing order are:
1. Case (b) ≈ 83.33 m
2. Case (a) = 500 m
3. Case (c) = 1000 m

---