Light Lite-6

Classical Mechanics Level pending

A car of length 3 m moves at a constant velocity 36 km/hr. A man wants to take a photo of side view of the car. The size of the image of the car is 1.5 cm long. The time of exposure needed for the car to get a clear picture, if the image should not move more than 0.1 mm, is :
Light Lite 5

1 \times 10^{-3}

1.5 \times 10^{-3}

2.8 \times 10^{-3}

2 \times 10^{-3}

3 \times 10^{-3}

1 solution

Aryan Sanghi
Jul 18, 2020

Now, the scale factor for conversion from Reality to photo is

$1.5cm \text{ in photo}\equiv 3m \text{ in reality}$

$\text{Now, time of exposure } = \text{ Time taken to cover 0.1mm in photo}$ .

$1.5cm \equiv 3m$ $15mm \equiv 3m$ $0.1mm \equiv \frac{3}{150}m = \frac{1}{50}m$

Now, $\text{speed of car }= 36km/h = 10m/s$

So, $\text{time of exposure } = \frac{distance}{speed}$

$t = \frac{\frac{1}{50}}{10}$

$\color{#3D99F6}{\boxed{t = 2 × 10^{-3}s}}$

It's nice solution.

A Former Brilliant Member - 10 months, 4 weeks ago

@Aryan Sanghi ,i was also trying for same method,but i forget to see that length of image is in centimetres,😅 .

A Former Brilliant Member - 10 months, 4 weeks ago