Physics Test 227

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Physics Test 227

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Question 1
4 / -1

Water drops fall at regular intervals from a leaking water tap 20 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. The height of the second drop above the ground at that instant is (take g = 10 m/s²)

A
5 m

B
7.5 m

C
12.5 m

D
15 m

Solution

Concept:

Equation of motion:

These equations are only valid when the acceleration of the body is constant and they move in a straight line. There are three equations of motion:

Where, v = final velocity, u = initial velocity, S = distance traveled by the body, a = acceleration, and t = time taken by the body

Calculation:

Given:

The Water droplets fall with an equal interval of time.

Height of the leaking tab H = 20 m

The third drop is leaving the tap at the instant the first drop touches the ground, therefore the third drop starts after t₁ second.

Water drops fall at regular intervals from a leaking water tap, therefore the second drop starts just midway through the first and third drops i.e t₁/2 seconds.

Let us say while the first drop reached the ground it traveled H height in t₁ seconds:

Therefore third drops starts after 2 seconds and second drop start at 1 second as described earlier.

So distance covered by second drop in one second will be:
Question 2
4 / -1

A point P is moving in uniform circular motion with radius 3m. Let at some instant the acceleration of the point is the position vector is The correct statement is

A

B

C

D

Solution

Concept:

Circular motion is described as a movement of an object while rotating along a circular path.

Circular motion can be either uniform or non-uniform.

During uniform circular motion the angular rate of rotation and speed will be constant while during non-uniform motion the rate of rotation keeps changing.

Solution:

Acceleration,

In a circular motion, we know that the velocity vector is always perpendicular to the acceleration vector.

Acceleration a is perpendicular to velocity v, hence v⋅a=0

We know that the position vector is parallel to the acceleration.

Again, position vector r is parallel to acceleration a, hence r×a=0

The correct answer is option (3).
Question 3
4 / -1

A
π/4

B

C

D
π/2

Solution

Concept Used:-

For a vector when a, b and c are direction ratios of a vector, then the direction cosines of this vector will be,

Also, when the angles made by line segment with the coordinate axis are α,β, and γ, then the direction cosines can be given as,

For x axis,

⇒ l = cosα,

For y axis

⇒ m = cosβ and for z axis,

⇒ n = cosγ

Explanation:-

Now, the given vector is,
Question 4
4 / -1

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is

A
2ω

B
ω

C
ω/2

D
0

Solution

CONCEPT:

The angular momentum is written as;
Question 5
4 / -1

Three blocks of masses 10 kg, 6 kg and 4 kg lie on a frictionless horizontal surface. The masses are connected together by light strings. A string attached to 4 kg block pulls the three blocks horizontally. If the system of blocks is accelerated at 0.5 m/s², the tension in the string connecting the 4 kg and 6 kg blocks is

A
3 N

B
5 N

C
8 N

D
10 N

Solution

Concept:

Three blocks are given and the system of blocks moves with common acceleration.

We have to make individual F.B.D for all the blocks.

Calculation:

Given:

Acceleration of blocks a = 0.5 m/s².

T₁ is the tension between the mass 4 kg and 6 kg.

T₂ is the tension between mass 10 kg and 6 kg.
Question 6
4 / -1

The maximum velocity (in m/s) with which a car driver can safely take a circular turning of radius 100 m along a level road without skidding is (coefficient friction between tyre and road = 0.4):

A
20

B
10

C
30

D
40
Solution
Concept:

Banking of Road:
- It is defined as the phenomenon in which the outer edges are raised for the curved roads above the inner edge to provide the necessary centripetal force to the vehicles so that they take a safe turn.
- The angle at which the vehicle is inclined is defined as the bank angle.
- The inclination happens at the longitudinal and horizontal axis.
Question 7
4 / -1

A man pushes a 100 N box a distance of 6.0 m upward along a frictionless slope that makes an angle of 30° with the horizontal. The force exerted by him is parallel to the slope. If the speed of the box remains constant, the work done by the man is

A
−300 J

B
300 J

C
−600 J

D
600 J

Solution

CONCEPT:

Equation of motion:

A motion in physics can be represented by the form of an equation.

The equation is based on newton's second law of motion. i.e. in the LHS, we write the total force on the object, and in RHS we write the mass×acceleration.

For example, If a body is moving through an inclined plane with an acceleration of 'a' when F force is applied to it, Then the equation of motion will be,

⇒ F ± f₁ ± f₂ + ... = ma (Where, f₁, f₂... can be forces like friction, gravity etc)

Work done:

The dot product of force and displacement is called work done. or in other words, the component of force along the direction of displacement is called the work done.

Mathematically, ⇒ W = F⋅d

EXPLANATION:

As shown in the figure the box is pushed through a distance of 6m.

The weight of the body is mg (acting downwards)

There are two components of the weight mg cosθ and mg sinθ

Where F and mg sinθ acts in the opposite direction.

So, according to the problem

⇒ F - mgsinθ = ma (a = acceleration)

⇒ F - mgsinθ = 0 (velocity is constant, so a = 0)

⇒ F = mg sinθ

Again work done = W = Force×distance travelled = Fd

Given,

mg = 100 N, θ = 30, d = 6 m

So, F = 100×sin30 = 100×½ = 50 N

so, W = 50 × 6 = 300 J

Hence the correct answer is option 2.
Question 8
4 / -1

Statement 1: An astronaut experiences weightlessness in a space satellite.

Statement 2: This because the gravitational force is small at that location in space.

A
Both statement 1 and statement 2 are CORRECT and statement 2 is the CORRECT explanation of the statement 1.

B
Both statement 1 and statement 2 are CORRECT, but statement 2 is NOT THE CORRECT explanation of the statement 1.

C
Statement 1 is CORRECT, but statement 2 is INCORRECT.

D
Statement 1 is INCORRECT, but statement 2 is CORRECT.
Solution
Explanation:
- Weight of an object is the force with which the earth attracts it.
- We are conscious of our own weight when we stand on a surface, since the surface exerts a force opposite to our weight to keep us at rest.
- When an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness
- In a satellite around the earth, every part and parcel of the satellite has an acceleration towards the centre of the earth which is exactly the value of earth’s acceleration due to gravity at that position.
- Thus, in the satellite everything inside it is in a state of free fall.
- This is just as if we were falling towards the earth from a height.
- Thus, in a manned satellite, people inside experience no gravity.
- Gravity for us defines the vertical direction and thus for them there are no horizontal or vertical directions, all directions are the same.
- An astronaut experiences weightlessness in a space satellite.
- This is not because the gravitational force is small at that location in space.
- It is because both the astronaut and the satellite are in “free fall” towards the Earth.
Question 9
4 / -1

A small block of mass 200 g is placed on a horizontal slab at a height of 2 m above the floor. The block is pressed against a horizontal spring fixed at one end to compress the spring through 10.0 cm. Upon releasing, the block moves horizontally till it leaves the spring. Calculate the horizontal distance covered by the block after leaving the slab and just before hitting the ground. The spring constant is 50 N/m.

(Assume g = 10 m/s²).

A
0.99 m

B
0.55 m

C
0.44 m

D
0.33 m

Solution

Concept:

According to Conservation of Energy, the total energy of a system is always conserved.

Total Energy = Kinetic Energy + Potential Energy

Solution:

Given:

Mass m = 200 g = 0.2 kg;

k = 50 N/m; x = 10.0 cm = 0.1 m

s = 2 m and g = 10 m/s^2

Let the block hit the ground with velocity v.

⇒ From energy conservation :
Question 10
4 / -1

The linear momentum of a sphere of mass of 2.0 kg varies with time t as P = 3t² + 4, where P and t are in S.I. units. It follows that the sphere moving with a:

A
Constant acceleration

B
Variable acceleration

C
Constant speed

D
constant velocity
Solution
Concept:
- Momentum: The product of mass and velocity is called the momentum of the body.
Momentum (P) = m × V

Where

m = mass of the body

V = velocity of the body.
- Momentum is the rate of change of momentum with respect to time equals magnitude of force.
F = dP/dt
- Conservation of momentum: Whenever there is no net external force on the system then the total momentum of the system remains constant.
Initial momentum (P₁) = Final momentum (P₂)

m₁v₁ = m₂v₂

Here, mass of two object = m₁& m₂,velocity of two object = v₁ & v₂

Calculation:

Given: Mass = 2 Kg, linear momentum varies with time as P = 3t² + 4.

F = dP/dt

⇒ F = d(3t² + 4)/dt = 6t

We know, force = mass × acceleration

F = 2 × a = 6t

⇒ a = (3t) m/s²

So, a ∝ t

So, The sphere moving with a variable acceleration.