A straw `6 cm` long floats on water. The water film on one side has surface tension of `50 dyn//cm`. On the other slide, camphor reduces the surface tension to `40 dyn//cm`. The resultant force acting on the straw is

Options

A. `(50xx6-40xx6)dyn`
B. `10 dyn`
C. `(50/6-40/6)dyn`
D. `90dyn`

Explanation

To find the resultant force acting on the straw floating on water, we can follow these steps:

### Step 1: Identify the given data
- Length of the straw (L) = 6 cm
- Surface tension on one side (T1) = 50 dyn/cm
- Surface tension on the other side (T2) = 40 dyn/cm

### Step 2: Calculate the forces due to surface tension
The force due to surface tension can be calculated using the formula:
\[ F = T \times L \]

#### Force on the side with water:
\[ F_1 = T_1 \times L = 50 \, \text{dyn/cm} \times 6 \, \text{cm} \]
\[ F_1 = 300 \, \text{dyn} \]

#### Force on the side with camphor:
\[ F_2 = T_2 \times L = 40 \, \text{dyn/cm} \times 6 \, \text{cm} \]
\[ F_2 = 240 \, \text{dyn} \]

### Step 3: Determine the net force acting on the straw
Since the forces act in opposite directions, the resultant force (F_net) can be calculated as:
\[ F_{\text{net}} = F_1 - F_2 \]
\[ F_{\text{net}} = 300 \, \text{dyn} - 240 \, \text{dyn} \]
\[ F_{\text{net}} = 60 \, \text{dyn} \]

### Final Answer:
The resultant force acting on the straw is **60 dyn**.

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