JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 11)
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with
speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further
increasing the speed of the car to (1 + $$\beta $$)v, this angle changes to 45o. The value of $$\beta $$ is close to :
0.50
0.73
0.37
0.41
Explanation
_6th_September_Evening_Slot_en_11_1.png)
tan 60o = $${{{V_r}} \over V}$$ .....(1)
tan 45o = $${{{V_r}} \over {\left( {1 + \beta } \right)V}}$$ .....(2)
From (i) and (ii), we get
$${{\sqrt 3 } \over 1} = {{{1 \over V}} \over {{1 \over {\left( {1 + \beta } \right)V}}}}$$
$$ \Rightarrow $$ $${\sqrt 3 = \left( {1 + \beta } \right)}$$
$$ \Rightarrow $$ $$\beta $$ = 0.732
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