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Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A bird weight 2 kg and is inside a cage of 1kg. If it starts flying then the weight of the bird and cage assembly is

  1. 1 kg

  2. 2 kg

  3. 3 kg

  4. 4 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since cage is at rest no pseudo force will be acting on bird hence weight of bird will be $2kg$ in air, therefore weight of bird and cage assembly will be  equal to 2kg and 1kg i.e equal to $3kg$.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be $a$. If the bird flies about inside the cage, the reading of the spring balance is $b$. Which of the following is true?

  1. $a=b$
  2. $a>b$
  3. $a$

  4. Cannot be obtained
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reading shown on the spring balance is weight of bird + weight of cage. When bird starts flying normal force exerted by bird on the case become zero. When bird flies his weight is carried by air but case in made of wire so no additional force of weight of bird act on the cage so reading on the cage is less when bird is flying.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

If the position of two like parallel forces shifted by one-fourth of the distance between the forces when the two forces are interchanged. The ratio of the two forces is:

  1. $1:2$
  2. $2:3$
  3. $3:4$
  4. $3:5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let forces be F1 and F2 at distance d. The resultant acts at x = F2*d / (F1+F2). If forces are swapped, new position x' = F1*d / (F1+F2). The shift is |x - x'| = d * |F1-F2| / (F1+F2) = d/4. Thus, 4|F1-F2| = F1+F2. If F1 > F2, 4F1 - 4F2 = F1 + F2 => 3F1 = 5F2 => F1/F2 = 5/3. If F2 > F1, 4F2 - 4F1 = F1 + F2 => 3F2 = 5F1 => F1/F2 = 3/5. Option D matches 3:5.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A particle is observed from two frames $S _{1}$ and $S _{2}$. The frame $S _{2}$ moves with respect to $S _{1}$ with an acceleration $a$. Let $F _{1}$ and $F _{2}$ be the pseudo forces on the particle when seen from $S _{1}$ and $S _{2}$ respectively. Which of the following are not possible?

  1. $F _{1} = 0, F _{2}\neq 0$
  2. $F _{1} \neq 0, F _{2} = 0$
  3. $F _{1} \neq 0, F _{2}\neq 0$
  4. $F _{1} = 0, F _{2} = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pseudo force is defined as F_p = -m * a_frame. If a_frame = 0 (inertial frame), then F_p = 0. If the frame is non-inertial (a != 0), the pseudo force is non-zero. It is impossible for a frame to be both inertial and non-inertial simultaneously, so F1=0 and F2=0 is impossible if S2 is accelerating relative to S1.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A cyclist riding with a speed of $27kmph$. As he approaches a circular turn on the road of radius $80m$, he applies brakes and reduces his speed at the constant rate of $0.50m/s$ every second. The net acceleration of the cyclist on the circular turn is 

  1. $0.5 \quad m/s^2$
  2. $0.86\quad m/s^2$
  3. $0.56 \quad m/s^2$
  4. $1 \quad m/s^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ v = 27 Km/hr$ 

$  = \cfrac{27 \times 1000}{3600} = 7.5 m/sec$
Centripetal acceleration = $ \cfrac{(7.5)^2}{80} = 0.7 m/s^2$
Tangential acceleration = $ -0.5m/s^2$ 
$ \therefore a _{net} = \sqrt{(0.7)^2 + (0.5)^2} = 0.86m/s^2$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

The backside of a truck is open and a box of 40kg is placed 5m away from the rear end.The coefficient of friction of the box with the surface of the truck is 0.15.The truck starts from rest with $2m/s^2$ acceleration.Calculate the distance covered by the truck when the box falls off

  1. 20m

  2. 30m

  3. 40m

  4. 50m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The box experiences a pseudo-force F = m*a = 40 * 2 = 80N. The friction force is f = mu * m * g = 0.15 * 40 * 10 = 60N. The net force on the box is F_net = F - f = 80 - 60 = 20N. Acceleration of the box relative to the truck is a_rel = F_net / m = 20 / 40 = 0.5 m/s^2. Time to fall off: s = 0.5 * a_rel * t^2 => 5 = 0.5 * 0.5 * t^2 => 5 = 0.25 * t^2 => t^2 = 20 => t = sqrt(20). Distance covered by truck: S = 0.5 * a_truck * t^2 = 0.5 * 2 * 20 = 20m.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A jet water issues from a nozzel with a velocity of $20 m/s$ and it impinges normally on a flat plate moving away from it at $10 m/s$. If the cross-sectional area of the jet is $0.02 m^2$ and the density of water is taken as $1000 kg/m^3$, then the force developed on the plate will be

  1. $ 10 N$
  2. $ 100 N$
  3. $1000 N$
  4. $2000 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A jet water issues from a nozzel with a velocity$=20m/s$

It impinges normally on a flat plate moving away from it$=10m/s$
Cross-sectional area of the jet $=0.02m^2$
The density of water is taken$=1000kg/m^3$
The force developed on the plate will be
Hydrostatic force o the bottom surface of the tank will be
$F _b=\rho g(l\times b)\times h\F _b\rho g\times(1\times2)\times 2\ F _b=4\rho g(lbh)\rightarrow(1)$
Hydrostatic force on vertical surface will be
$F _v=\rho g(l\times h)\times\cfrac{h}{2}\F _v=\rho g(2\times 2)\cfrac{lh^2}{2}\F _v=2\rho g(lh^2)\rightarrow(2)$
Ratio will be
$\cfrac{F _b}{F _V}=\cfrac{4\rho g(lbh)}{2\rho g(lh^2)}\ \cfrac{F _b}{F _V}=1\F _b=1\times10\ \quad=10N$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A man drops an apple in the lift. He finds that the apple remains stationary and does not fall. The lift is:

  1. Going down with constant speed

  2. Going up with constant speed

  3. Going down with constant acceleration

  4. Going up with constant acceleration
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the apple is dropped, it is under free-fall meaning that the force of gravity is acting on it. With respect to the person inside the lift, the apple seems not to be falling Hence, the man and the lift must also be falling with the action of acceleration due to gravity i.e, a constant acceleration.

option - C is correct.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

When a lift is going up with uniform acceleration, the apparent weight of a person is $W _{1}$
When the lift is stationary, his apparent weight is $W _{2}$
When the lift falls freely his apparent weight is $W _{3}$
When the lift is going down with uniform acceleration which is less than the acceleration due to gravity, his apparent weight is $W _{4}$
The increasing order of these four weights is

  1. $W _{1},W _{3},W _{2},W _{4}$
  2. $W _{3},W _{4},W _{2},W _{1}$
  3. $W _{3},W _{2},W _{4},W _{1}$
  4. $W _{2},W _{3},W _{4},W _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. $W _{1}>mg$
When lift is stationary, $W _{2}=mg$
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. $W _{3}=0$
When lift accelerates down, pseudo force acts upwards,  hence it decreases apparent weight $W _{4}<mg$, but also the acceleration is less than g, therefore $W _{4}=m(g-a)>0$

Hence, $W _{3}<W _{4}<W _{2}<W _{1}$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west and B moves from west to east. Which train exerts greater force on the track?

  1. A

  2. B

  3. they will exert equal force

  4. The mass and the speed of each train must be known to reach a conclusion.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Working in the frame of universe:

$mg-N=m{ \omega  }^{ 2 }R$
$N=mg-m{ \omega  }^{ 2 }R$
Earth moves from west to east.
Now, since the train B moves from west to east, its $\omega$ is greater than A. 
Hence, ${ N } _{ B }<{ N } _{ A }$