4) Water vapour condenses on the surface of water drops while they are falling at rate 2 milligram. $$ / \mathrm{sec} $$, the mass of one falling water drop was 0.2 gm . at a moment , then pura mass of the falling water drop after $$ 1 \frac{1}{4} $$ minutes from this moment $$ =\ldots $$ $$ \begin{array}{llll} ext { (a) } 3.50 & ext { (b) } 35 & ext { (c) } 3.5 & ext { grams }\end{array} $$

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4) Water vapour condenses on the surface of water drops while they are falling at rate 2 milligram. $$ / \mathrm{sec} $$, the mass of one falling water drop was 0.2 gm . at a moment , then pura mass of the falling water drop after $$ 1 \frac{1}{4} $$ minutes from this moment $$ =\ldots $$ $$ \begin{array}{llll}	ext { (a) } 3.50 & 	ext { (b) } 35 & 	ext { (c) } 3.5 & 	ext { grams }\end{array} $$

UpStudy AI · Accepted Answer

To find the mass of the falling water drop after $$ 1 \frac{1}{4} $$ minutes, we need to follow these steps:

1. **Convert the time from minutes to seconds**:
   - $$ 1 \frac{1}{4} $$ minutes is equal to $$ 1.25 $$ minutes.
   - Since there are $$ 60 $$ seconds in a minute, we convert this to seconds:
   $$
   1.25 \text{ minutes} = 1.25 \times 60 \text{ seconds}
   $$

2. **Calculate the total time in seconds**:
   $$
   1.25 \times 60 = 75 \text{ seconds}
   $$

3. **Determine the total mass of water vapor condensed**:
   - The rate of condensation is $$ 2 $$ milligrams per second.
   - Over $$ 75 $$ seconds, the total mass condensed is:
   $$
   2 \text{ mg/sec} \times 75 \text{ sec} = 150 \text{ mg}
   $$

4. **Convert the mass of condensed water vapor to grams**:
   - Since $$ 1000 $$ mg is equal to $$ 1 $$ gram:
   $$
   150 \text{ mg} = \frac{150}{1000} \text{ grams} = 0.15 \text{ grams}
   $$

5. **Calculate the new mass of the falling water drop**:
   - The initial mass of the water drop is $$ 0.2 $$ grams.
   - After $$ 75 $$ seconds, the new mass will be:
   $$
   0.2 \text{ grams} + 0.15 \text{ grams} = 0.35 \text{ grams}
   $$

Now, let's summarize the calculations and find the final answer.

The mass of the falling water drop after $$ 1 \frac{1}{4} $$ minutes is $$ 0.35 $$ grams, which can be expressed as $$ 3.5 $$ grams when considering the options provided.

Thus, the answer is $$ \text{(c) } 3.5 \text{ grams} $$.
After $$ 1 \frac{1}{4} $$ minutes, the mass of the falling water drop is $$ 3.5 $$ grams.

Water vapour condenses on the surface of water drops while they are falling at rate
2 milligram. , the mass of one falling water drop was 0.2 gm . at a moment , then pura
mass of the falling water drop after minutes from this moment

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Water vapour condenses on the surface of water drops while they are falling at rate 2 milligram. , the mass of one falling water drop was 0.2 gm . at a moment , then pura mass of the falling water drop after minutes from this moment