1

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

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms$-$2) A
32 N
B
18 N
C
23 N
D
25 N

## Explanation At equilibrium,

N = mgcos45o    . . . . . . (1)

3 + mgsin45o = P + $\mu$N

$\Rightarrow$   3 + ${{mg} \over {\sqrt 2 }}$ = P + 0.6 $\left( {{{mg} \over {\sqrt 2 }}} \right)$

$\Rightarrow$   3 + ${{100} \over {\sqrt 2 }}$ = P + 0.6 $\times$ ${{100} \over {\sqrt 2 }}$

$\Rightarrow$    P = 3 + ${{40} \over {\sqrt 2 }}$

$\Rightarrow$   P = 3 + 20 ${\sqrt 2 }$

$\Rightarrow$   P = 31.28 $\simeq$ 32 N
2

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

A heavy ball of mass M is suspendeed from the ceiling of a car by a light string of mass m (m < < M). When the car is at rest, the speed of transverse waves in the string is 60 ms$-$1. When the car has acceleration a, the wave-speed increases to 60.5 ms$-$1. The value of a, in terms of gravitational acceleration g, is closest to :
A
${g \over {30}}$
B
${g \over 5}$
C
${g \over 10}$
D
${g \over 20}$

## Explanation Resultant force on the ball of mass M when car is moving with a acceleration a is ,

Fnet = $\sqrt {{{\left( {mg} \right)}^2} + {{\left( {Ma} \right)}^2}}$

= $M\sqrt {{g^2} + {a^2}}$

$\therefore$   T = M$\sqrt {{g^2} + {a^2}}$

We know,

Velocity, V = $\sqrt {{T \over \mu }}$

When Car is at rest then,

60 = $\sqrt {{{Mg} \over \mu }}$   . . . . (1)

and when is moving then

60.5 = $\sqrt {{{M\sqrt {{g^2} + {a^2}} } \over \mu }}$    . . . . (2)

By dividing (2) by (1) we get,

${{60.5} \over {60}} = \sqrt {{{\sqrt {{g^2} + {a^2}} } \over g}}$

$\Rightarrow$   $\left( {1 + {{0.5} \over {60}}} \right)$ = ${\left( {{{{g^2} + {a^2}} \over {{g^2}}}} \right)^{{1 \over 4}}}$

$\Rightarrow$   ${{{g^2} + {a^2}} \over {{g^2}}}$ = ${\left( {1 + {{0.5} \over {60}}} \right)^4}$

$\Rightarrow$   ${{{g^2} + {a^2}} \over {{g^2}}}$ = 1 + 4 $\times$ ${{{0.5} \over {60}}}$ [Using Binomial approx]

$\Rightarrow$   ${{{g^2} + {a^2}} \over {{g^2}}}$ = 1 + ${1 \over {30}}$

$\Rightarrow$   1 + ${{{a^2}} \over {{g^2}}}$ = 1 + ${1 \over {30}}$

$\Rightarrow$   ${a \over g}$ = ${1 \over {\sqrt {30} }}$

$\Rightarrow$   a = ${g \over {\sqrt {30} }}$

$\therefore$   Closest answer, a = ${g \over 5}$
3

### 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
4

### 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