Answer:
1.6 x 10^-5 T
Explanation:
i = 4 A
r = 0.05 m
The magnetic field due to long wire at a distance r is given by

B = 10^-7 x 2 x 4 / 0.05
B = 1.6 x 10^-5 T
Length of the cord..........................................................
Answer: Option B: 1.3×10⁵ W
Explanation:


Work Done, 
Where s is displacement in the direction of force and F is force.

where, v is the velocity.
It is given that, F = 5.75 × 10³N
v = 22 m/s
P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W
Thus, the correct option is B
Answer:
4v/3
Explanation:
Assume elastic collision by the law of momentum conservation:

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively
Substitute 

Divide both side by
, then multiply by 6 we have



So the final speed of the second car is 4/3 of the first car original speed