A particle moves with a velocity 5i-3j+6k m/s under the influence of a constant force F= 10i+10j+20k . The instantaneous power applied to the particle is

Q:A particle moves with a velocity m/s under the influence of a constant force . The instantaneous power applied to the particle is:
(A) 200 J/s
(B) 40 J/s
(C) 140 J/s
(D) 170 J/s

Solution : P = F.v
= (10i + 10j + 20k ).(5i – 3j + 6k )
= 50 – 30 + 120
= 140 J/s

A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is μ, the power delivered by the external agent after a time t from the beginning is equal to

Q:A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is μ, the power delivered by the external agent after a time t from the beginning is equal to
(A) ma2t
(B) μmgat
(C) μm(a+μg)gt
(D ) m(a+μg)at
Sol: Instantaneous power delivered = P = F.v = F v
where, F – f = ma
 F = f + ma
 P = (f + ma) v
Put f = μmg
 P = ( μmg + ma)v = m(a + μg).at

Two point charges +q and -q are held fixed at (-d, 0) and (d, 0) respectively of a (x, y) coordinate system

Q:Two point charges +q and -q are held fixed at (-d, 0) and (d, 0) respectively of a (x, y) coordinate system, then
(A)the electric field at all points on the x-axis has the same direction.
(B) E at all points on the Y-axis is along î.
(C)No Work has to be done in bringing a test charge from infinity to the origin.
(D)the dipole moment is 2qd directed along î.

Solution : E at all points on the y-axis is along . Electric potential at origin is zero so no work is done in bringing a test charge from infinity to origin.
 Hence (C) is correct.

A neutron, a proton and an electron are placed in a uniform electric field

Q: A neutron, a proton and an electron are placed in a uniform electric field.
(A) The forces acting on them will be equal.
(B) Their accelerations may be equal.
(C) Magnitude of acceleration may be equal.
(D) Magnitude of acceleration will be different.

Solution :Force on neutron = 0 (as charge = 0)
Force on proton = qE
Force on electron = qE
Masses of proton and electron are different.
Hence their accelerations will be different.
 Hence (D) is correct.

Two bodies are charged by rubbing one against the other. During the process, one becomes positively charged while the other becomes negatively charged

Q: Two bodies are charged by rubbing one against the other. During the process, one becomes positively charged while the other becomes negatively  charged. Then

(A) mass of each body remains unchanged.
(B) mass of each body changes marginally.
(C) mass of each body changes slightly and hence the total mass.
(D) mass of each body changes slightly but the total mass remains the same.

Solution: The transfer of electrons from one body to the other results is development of charges.

Hence, no. of electrons given by one body = No. of electrons obtained by the other.
Mass of negatively charged body slightly increases while mass of positively charged body slightly increases.
whereas the total mass of the system remains the same.
Hence (D) is correct.