$A$ swimmer can swim with speed '$v$' with respect to still water in a river which is flowing with speed '$u$'. There is a float moving with the river. Now the swimmer overtakes the float,gets a lead of '$l$',and returns back to the float. The time taken by the swimmer in this process will be:

  • A
    $\frac{2l}{v}$
  • B
    $\frac{2l}{u}$
  • C
    $\frac{l}{\sqrt{v^2 - u^2}}$
  • D
    $\frac{l}{v + u} + \frac{l}{v - u}$

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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 $60^{\circ}$ from the horizontal. On further increasing the speed of the car to $(1+\beta)v$,this angle changes to $45^{\circ}$. The value of $\beta$ is close to...

Two persons $A$ and $B$ are located in the $X-Y$ plane at the points $(0,0)$ and $(0,10)$ respectively. (The distances are measured in $MKS$ units). At time $t=0$, they start moving simultaneously with velocities $\overrightarrow{v}_A = 2\hat{j} \text{ m/s}$ and $\overrightarrow{v}_B = 2\hat{i} \text{ m/s}$ respectively. The time after which $A$ and $B$ are at their closest distance is:

Match Column-$I$ with Column-$II$ correctly.
Column-$I$ Column-$II$
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$(2)$ Velocity of rain drops relative to a man $(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

$A$ boat is moving with a velocity $3i + 4j$ with respect to the ground. The water in the river is moving with a velocity $-3i - 4j$ with respect to the ground. The relative velocity of the boat with respect to the water is:

$A$ man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \,km/h$. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is:

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