Test: Newton's laws of Motion (June 4) - NEET MCQ

• It is usually seen that during rainy days we tend to slip because rainwater acts as a lubricant between our feet and the ground.  

• This lubrication (rainwater) converts dry friction into fluid friction. 

• That is, the friction between feet and ground reduces, making us slip.

Therefore, due to less friction, we slip on a rainy day.

When the bus moves, the passenger's body comes into a state of motion but when the bus stops the lower part of the body which is in contact with the floor comes into a state of rest whereas the upper part of the body still remains in the state of motion and because of this the upper part of the body falls in the forward direction. This is due to the presence of inertia. 

Hence the correct answer is C.

Given: 

Speed of the particle after a time 2T, 

Planck's Constant (h)  = 6.626176 x 10-34  m2 kg/s
So, Unit of planck constant=  m² kg/s

Dimensions =M L² T ⁻¹   ________ (1)

Angular momentum l = mvr

Where, m-mass

v-velocity

r-radius

Dimensions of angular momentum  =  M L T ⁻¹ L   = M  L²  T ⁻¹  _______________ (2)
From (1) and (2). 

Planck's constant and angular momentum have the same dimensions.

Net external force F = dp/dt, so using dimensional analysis, unit of momentum should be the same as unit of force times unit of time which is Newton second.

1N = kg m /sec
1 dyne = gram cm/sec
Thus 1N in CGS system = 1000 gram 100 cm/sec
That is 1N = 100000 dyne

The sixth coin is under the weight of four coins baove it. Hence,
Reaction of the 6^th coin on the 7^th coin = Force on the 6^th coin due to 7^th coin
= 4mg

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Here, mass of the stone, m = 1 kg
As the stone is lying on the floor of the train, its acceleration is same as that of train.
∴ Force acting on the stone F = ma = (1kg)(1 m s^-2)
= 1 N

Here, m = 2 × 10⁴kg
Initial acceleration, a = 5ms^-2
Initial thrust = upthrust required to impart acceleration a + upthrust required to overcome gravitational pull
∴ F  = m(a + g) = (2 × 10⁴kg)(5 + 10)ms⁻²
= 3 × 10⁵N