Choose the correct answer:
1. Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity u at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms–1 [use g = 10 ms–2]
The correct option is(D): 30 m/s.
Let us assume that they meet at t = t0
\(A:80=\frac{1}{2}gt^2_0....(i)\)
\(B:80=u(t_0-2)+\frac{1}{2}+\frac{!}{2}g(t_0-2)^2...(ii)\)
⇒ 80 = 2u + 5(2)2
⇒ u= 30 m/s
A wire of 60 cm length and mass 10 g is suspended by a pair of flexible leads in a magnetic field of 0.60 T as shown in the figure. The magnitude of the current required to remove the tension in the supporting leads is:

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
The velocity with which one object moves with respect to another object is the relative velocity of an object with respect to another. By relative velocity, we can further understand the time rate of change in the relative position of one object with respect to another.
It is generally used to describe the motion of moving boats through water, airplanes in the wind, etc. According to the person as an observer inside the object, we can compute the velocity very easily.
The velocity of the body A – the velocity of the body B = The relative velocity of A with respect to B
V_{AB} = V_{A} – V_{B}
Where,
The relative velocity of the body A with respect to the body B = V_{AB}
The velocity of the body A = V_{A}
The velocity of body B = V_{B}