De-bye is the unit of electric dipole moment.

y = a sin ( ωt + kx + π / 2) = a cos (ωt + kx).

Mechanical waves always transfer energy from one point to another point in a medium without any physical transport of the material between these points.

It’s a fact.

Work = force × displacement = mass × acceleration × displacement = [M][LT - 2][L] = [ML2T - 2] = [M x LyTz]... x = 1, y = 2, z  = - 2.

By definition , β = (∆IC / ∆IB) at constant VCE, therefore , ∆IB = ∆IC / β = 50/10 = 5mA.

Change of momentum of one bullet = m(v - u) = 0.02Χ[50 - ( - 20)] = 1.4 kg ms-1Average force = rate of change of momentum of 300 bullets = 300 Χ 1.4 = 420 N

Kinetic energy of body is K = p2/2m Where p is the momentum and m is the mass of a body respectively ... k proportional to p2 When the momentum of a body is increased by 25%, its momentum will become P' = p + (25/100)p = (125/100)p = (5/4)p ... k'/k = p' 2/p2 = (5/4)2 = 25/16 or k' = (25/16)k Percentage increase in the kinetic energy of the body = (k ' -k)/k = {(25/16)k - k}/k= (9/16) Χ 100 = 56%

Electric intensity at centre O, due to small element dl of the charged ring dE = λ dl / 4πє0r2 = λ(a dӨ) / 4πє0r2.  As is clear from figure, horizontal components of dE will cancel out in pairs and vertical components will add. π π ... E = ∫dE sinӨ = ∫ λ / 4πє0r (sinӨ dӨ) 0 0 π = λ / 4πє0r [ - cos Ө] o dl dE E = λ / 4πє0r ( - cosπ + cos0) = λ / 2π є0r.

Two electric lines of force never intersect, because in that case at the point of intersection, there will be two tangents and hence, there would be two values of the electric field at that point, which is impossible. So, this is a correct option.

Let T be the tension of a rope. When man climbs up with acceleration a, ... T - m1g = m1a …(1) When a monkey climbs down with same acceleration a ... m2g - T = m2a ….(2) Adding (i) and (ii), we get (m2 - m1)g = (m1 + m2)a Or (1 - m1 / m2)g = (m1 / m2 + 1)a Or (1 - 2 / 3)g = (2 / 3 + 1)a Or (1 / 3)g = (5 / 3)a Or, a = g / 5. Clearly, it is not equal to 5g, hence it is a wrong option.

The acceleration of the system is The acceleration of the system is a = (m1 - m2)g / (m1 + m2) = g / 10 (m1 - m2) / (m1 + m2) = 1 / 10 10m1 - 10m2 = m1 + m2 9m1 = 11m2 m1 / m2 = 11 / 9

Thrust, F = u(dm/dt) = 4 Χ 10⁴ Χ 40 = 1.6 Χ 10⁶N

P = mgh / t or m = (p Χ t) / (g Χ h) m = (1000W Χ 60s) / (10 ms-2 Χ15 m) = 400 kg = 400 litre.

The time period of simple pendulum is given by T = 2π√(L / g). On the moon, g is much smaller as compared to g on earth. Therefore, T increases.