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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
If the velocity of a body related to displacement x is given by $$\upsilon = \sqrt {5000 + 24x}$$ m/s, then the acceleration of the body is .................... m/s2.

## Explanation

$$V = \sqrt {5000 + 24x}$$

$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$

Now, $$a = V{{dV} \over {dx}}$$

$$= \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$$

a = 12 m/s2
2

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s2)

## Explanation

Image

When both balls will collied

y1 = y2

$$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$$

$$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$$

0 = 150 $$-$$ 30 t

t = 5 sec

$$\therefore$$ Height at which both balls will collied

$$h = 35t - {1 \over 2} \times 10 \times {t^2}$$

$$= 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$$

h = 50 m
3

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30$$^\circ$$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $$\theta$$ with the line AB should be _________$$^\circ$$, so that the swimmer reaches point B. ## Explanation Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them

$$\theta$$ = 30$$^\circ$$
4

### JEE Main 2021 (Online) 22th July Evening Shift

Numerical
Three particles P, Q and R are moving along the vectors $$\overrightarrow A = \widehat i + \widehat j$$, $$\overrightarrow B = \widehat j + \widehat k$$ and $$\overrightarrow C = - \widehat i + \widehat j$$ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $$\overrightarrow A$$ and $$\overrightarrow B$$. Similarly particle Q is moving normal to the plane which contains vector $$\overrightarrow A$$ and $$\overrightarrow C$$. The angle between the direction of motion of P and Q is $${\cos ^{ - 1}}\left( {{1 \over {\sqrt x }}} \right)$$. Then the value of x is _______________.

## Explanation

$${\widehat n_1}{{\overrightarrow A \times \overrightarrow B } \over {\left| {\overrightarrow A \times \overrightarrow B } \right|}} = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$$

$${\widehat n_2}{{\overrightarrow A \times \overrightarrow C } \over {\left| {\overrightarrow A \times \overrightarrow C } \right|}} = \widehat k$$

$$\cos \theta = {\widehat n_1}.{\widehat n_2} = {1 \over {\sqrt 3 }}$$

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