JEE Main 16 March 2021 Shift 2 Solved Paper

Given
Force, F=(2

∧

i

+3

∧

j

+5‌

∧

k

)‌N
Mass, m=2kg
Time, t=4s
We know that
Force = mass × acceleration
⇒‌‌F=m×a⇒a=‌

F

m

⇒‌‌a=‌2

∧

i

+3

∧

j

+5

∧ k

2

.....(i)
From second equation of motion,
s=ut+‌

1

2

at2...(ii)
From Eqs. (i) and (ii), we get
s=ut+‌

1

2

‌(2

∧

i

+3

∧

j

+5

∧ k )

2

⋅t2
=0+‌

1

4

(2

∧

i

+3

∧

j

+5

∧

k

)⋅(4)2‌‌[∵u=0 and t=4s]
=0+8

∧

i

+12

∧

j

+20

∧

k

=8

∧

i

+12

∧

j

+20

∧

k

Let s=x

∧

i

+y

∧

j

+z

∧

k

⇒x

∧

i

+y

∧

j

+z

∧

k

=8

∧

i

+12

∧

j

+20

∧

k

.....(iii)
According to question,new coordinates are (8,b,20), it means
x

∧

i

+y

∧

j

+z

∧

k

=8

∧

i

+b

∧

j

+20

∧

k

......(iv)
Comparing Eqs. (iii) and (iv), we get
b=12