Question

3.) On a rainy day, a man is running on a hilly terrain in such a way that he is always finding the rain
drops hitting him vertically. It is assumed that rainfall is uniform in the whole terrain from O to E
When the man was at rest on stretch OA, he found the rainfall at an angle of 30° with the vertical.

Match the statements from List I with those in List II and select the correct answer using the code
given below the lists.

*List-1*
(P)The magnitude of velocity of rain observed on stretch AB to the magnitude of velocity of man on stretch BC.

(Q)The magnitude of velocity of rain observed on stretch BC to the magnitude of actual velocity of rain.

(R)The magnitude of velocity of rain observed on stretch DE to the magnitude of velocity of rain observed on stretched CD.

(S)The magnitude of velocity of man on stretch DE to the magnitude of velocity of man on stretch CD

*List 2*
(1) √3/2

(2)2/√3

(3) √3

(4)3/2


*Code:*
Р Q R S

(A) 2 1 4 3
(В) 1 3 2 3
(C) 3 3 1 2
(D) 2 3 3 1​

Answer 1

Answer:

Distance moved / relative speed of the rain = 0.15

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Explanation:

On A rainy day a man is running On a hilly terrain in such a way that he is always finding the rain hitting him vertically .It is assumed that the rainfall is uniformed in the whole terrain from O to E when the man was at rest on stretch OA he found the rainfall at an angle 30 degree with the vertical.

Note: The question is incomplete as to what we have to find. So interpreting the maximum information we can given the situation.

Solution:

Cos 30 = distance moved / relative speed of the rain

 Cos 30 = 0.15

So distance moved / relative speed of the rain = 0.15

Answer 2

The correct answer is option (A).

Given:

It is assumed that rainfall is uniform in the whole terrain from O to E.

The angle made by the rainfall with the vertical at the stretch OA = 30°.

To Find:

We have to match the statements from List I with those in List II and select the correct answer using the given list of codes.

Solution:

(P). The magnitude of velocity of rain observed on stretch AB to the magnitude of velocity of man on stretch BC.

The angle made by the rainfall with the horizontal at the stretch AB = 30°.

∴, It makes an ∠60° with the vertical.

The magnitude of velocity of rain observed on stretch AB is given by,

$V_{rain, observed} =$ Velocity vector of rain - Velocity vector of man $= V_{rain} - V_{man}$

From the given figure, $V_{rain, observed}|_{AB} = V_{rain} . cos$ $30$° - $(V_{man|_{AB} } ) . cos$ $60$°.

Since, the projection of $V_{rain}$ on the horizontal is same as the projection of $V_{man}$ on the horizontal, we get,

$V_{rain} . cos$  $30$° = $(V_{man|_{AB} } ) . sin$ $60$°.

Or, $(V_{man|_{AB} } ) = V_{rain} . \frac{sin 30}{sin 60} = V_{rain} . \frac{1}{\sqrt{3}}$.

$i.e., V_{rain, observed}|_{AB} = V_{rain}. [\frac{\sqrt{3} }{2} - \frac{1}{\sqrt{3} }.\frac{1}{2} ] = V_{rain}. \frac{1 }{\sqrt{3}}$.

Also, $V_{man|_{BC} } = V_{rain} sin 30 = V_{rain}.\frac{1}{2}$

∴, The ratio of the magnitude of velocity of rain observed on stretch AB to the magnitude of velocity of man on stretch BC is $\frac{2}{\sqrt{3} } .$

(Q). The magnitude of velocity of rain observed on stretch BC to the magnitude of actual velocity of rain.

The angle made by the rainfall with the vertical at the stretch OA = 30°.

From the given figure, the ratio of the magnitude of velocity of rain observed on stretch BC to the magnitude of actual velocity of rain is given by,

$\frac{V_{rain, observed}|_{BC}}{V_{rain} } = cos 30 = \frac{\sqrt{3} }{2} .$

(R). The magnitude of velocity of rain observed on stretch DE to the magnitude of velocity of rain observed on stretched CD.

The angle made by the stretch CD to the horizontal = 30°.

The angle made by the stretch DE to the horizontal = 60°.

From the given figure, $V_{rain, observed}|_{CD} = V_{rain} . cos$ $30$° + $(V_{man|_{CD} } ) . cos$ $60$°.

Since the horizontal projection for $V_{rain}$ and $V_{man|_{CD} }$ is the same, we get,

$(V_{man|_{CD} }) sin 60 = V_{rain}. sin 30$

Or, $V_{man|_{CD} } = V_{rain}.\frac{1 }{\sqrt{3} }$

∴, $V_{rain, observed}|_{CD} = V_{rain}. [\frac{\sqrt{3} }{2} + \frac{1}{2\sqrt{3} } ] = V_{rain}. \frac{2 }{\sqrt{3}}$

Now, $V_{rain, observed}|_{DE} = 2.V_{rain} . cos 30 = V_{rain} . \sqrt{3}$

∴, The ratio of the magnitude of velocity of rain observed on stretch DE to the magnitude of velocity of rain observed on stretched CD is $\frac{3}{2 } .$

(S). The magnitude of velocity of man on stretch DE to the magnitude of velocity of man on stretch CD.

Similarly from the figure, we get that the ratio of magnitude of velocity of man on stretch DE to the magnitude of velocity of man on stretch CD is $\sqrt{3}$.

∴, The correct matches between the two lists are: (P - 2), (Q - ), (R - 4), and (S - 3).

Hence, the correct answer from the given list is option (A).

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