JEE Main 27-Aug-2021 Shift 2 Solved Paper

According to the given figure, and free body diagrammasses of two bodies m1=2kgm2=8kg ,
Acceleration of mass m1=a
Acceleration of mass m2=‌

a

2

Tension =T
Distance between m2 and ground =20cm
=0.2m
Initial velocity u=0
Equation of motion of 2kg block,
∴‌‌T−2g=2a‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)

Equation of motion of 8kg block, and 8g−2T=8‌

a

2

⇒4g−T=2a‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From Eqs. (i) and (ii),
⇒‌‌T−2g=4g−T
‌ Substituting the value in Eq. (i) we get ‌
‌‌3g−2g=2a⇒g=2a
⇒‌‌a=‌

g

2

=5ms−2
∴‌‌a1=5ms−2‌ and ‌2=‌

5

2

ms−2
‌ Since, ‌s=ut+‌

1

2

at2
∴‌‌‌

20

100

=0+‌

1

2

‌

5

2

×t2⇒t2=‌

20×4

5×100

⇒‌‌t=‌

2

5

=0.4s