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1

### JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

A
$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$
B
$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$
C
$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$
D
$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$

## Explanation

$${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$$

$${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$$
2

### JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :
A
$$2\sqrt {{k \over p}}$$
B
$$2\sqrt {{p \over k}}$$
C
$$\sqrt {{{2p} \over 2}}$$
D
$$\sqrt {{{2k} \over p}}$$

## Explanation

$${{dp} \over {dt}} = F = kt$$

$$\int_P^{3P} {dP} = \int_0^T {kt\,dt}$$

$$2p = {{K{T^2}} \over 2}$$

$$T = 2\sqrt {{P \over K}}$$
3

### JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be -

A
$${g \over {13l}}$$
B
$${g \over {2l}}$$
C
$${g \over {3l}}$$
D
$${7g \over {3l}}$$

## Explanation

Applying torque equation about point P.

2M0 (2l) $$-$$ 5 M0 gl = I$$\alpha$$

I = 2M0 (2l)2 + 5M0 l2 = 13 M0l2d

$$\therefore$$  $$\alpha$$ = $$-$$ $${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$$ $$\Rightarrow$$ $$\alpha$$ = $$-$$ $${g \over {13\ell }}$$

$$\therefore$$  $$\alpha$$ = $${g \over {13\ell }}$$ anticlockwise
4

### JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - (g = 10 ms–2)
A
30 m
B
40 m
C
20 m
D
10 m

## Explanation

Time taken for the particles to collide,

t = $${f \over {{V_{rel}}}} = {{100} \over {100}} = 1$$ sec

Speed of wood just before collision = gt = 10 m/s

& speed of bullet just before collision v-gt

= 100 $$-$$ 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision -

$$-$$ (0.02) (1v) + (0.02) (9v) = (0.05)v

$$\Rightarrow$$   150 = 5v

$$\Rightarrow$$   v = 30 m/s

Max. height reached by body h = $${{{v^2}} \over {2g}}$$

h = $${{30 \times 30} \over {2 \times 10}}$$ = 45m

$$\therefore$$  Height above tower = 40 m

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