JEE Advanced Model Paper 7 with solutions for online practice

We know Fx =

−dUx

dx

and fy =

−dUy

dx

Hence, force on the particle is

→

F

= -(3

^

i

+4

^

j

) N
|F| = 5N = constant ⇒ a =

|F|

m

= 5m∕s2
Hence, the acceleration of the particle is constant, so option (a) is constant
At t = 0 , particle was at rest at (6,4)
From x = 6 +

1

2

axt2 = (6−

3

2

t2) m and y = 4 +

1

2

ayt2 = (4−2t2) m
When the particle crosses the x - axis y = 0 ⇒ t1 = √2s
∴ Displacement during this time, S =

1

2

at2
S =

1

2

× 5 × (√2)2 = 5m
Hence, work done by the force, upto this instant is W = FS = 5 × 5 = 25J
Hence, option (b) is correct
The particle crosses y-axis when x = 0
Hence , 6 -

3

2

t22 = 0 ⇒ t2 = 2 sec
∴ speed of the particle at this instant will be
V = at2 = 5 × 2 = 10 m∕s2
Hence, option (c) is correct
At t = 4 sec , x = 6 -

3

2

(4)2 = -18m
And y = 4 - 2(4)2 = -28m
Hence, option (d) is correct