1

### JEE Main 2019 (Online) 9th January Evening Slot

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point. the rope deviated at an angle of 45o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms$-$2
A
200 N
B
140 N
C
70 N
D
100 N

## Explanation tan 45o = ${F \over {mg}}$

$\therefore$  F = mg

= 10 $\times$ 10

= 100 N
2

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

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
3

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

Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle $\theta$ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is -
A
90o
B
60o
C
30o
D
120o

## Explanation

4F2 + 9F2 + 12F2 cos $\theta$ = R2

4F2 + 36F2 + 24F2 cos $\theta$ = 4R2

4F2 + 36F2 + 24F2 cos $\theta$

= 4(13F2 + 12F2cos$\theta$) = 52F2 + 48F2cos$\theta$

cos $\theta$ = $-$ ${{12{F^2}} \over {24{F^2}}}$ = $-$ ${1 \over 2}$
4

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

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is : [Take g = 10 m/s2] A
${{\sqrt 3 } \over 4}$
B
${1 \over 2}$
C
${{\sqrt 3 } \over 2}$
D
${2 \over 3}$

## Explanation

2 + mg sin30 = $\mu$mg cos30o

10 = mgsin30 + $\mu$ mg cos30o

= 2$\mu$mg cos30 $-$ 2

6 = $\mu$mg cos30

4 = mg sin30

${3 \over 2} = \mu \times \sqrt 3$

$\mu$ = ${{\sqrt 3 } \over 2}$