Physics Test 268

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Physics Test 268

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Question 1
4 / -1

An object is said to be in uniform motion in a straight line if its displacement

A
is decreasing in equal intervals of time

B
is increasing in equal intervals of time

C
is equal in equal intervals of time.

D
is equal in not equal intervals of time.

Solution

Uniform motion in a straight line means the body covers equal displacements in equal intervals of time, implying zero acceleration:

Displacement = constant in equal time intervals
Question 2
4 / -1

For motion in 3 dimensions we need

A
No frame of reference is required ; a set of 3 points will do.

B
Frame of reference consisting of a clock and a Cartesian coordinate system having two mutually perpendicular axes, labeled X-, and Y- axes

C
Frame of reference consisting of a clock and a Cartesian coordinate system having three mutually perpendicular axes, labeled X-, Y-, and Z- axes

D
No frame of reference is required a single point will do.

Solution

Motion is a change in position of an object with time. In order to specify position, we need to use a reference point and a set of axes. It is convenient to choose a rectangular coordinate system consisting of three mutually perpenducular axes, labelled X-, Y-, and Z- axes.

The point of intersection of these three axes is called origin (O) and serves as the reference point. The coordinates (x, y. z) of an object describe the position of the object with respect to this coordinate system.

To measure time, we position a clock in this system. This coordinate system along with a clock constitutes a frame of reference.
Question 3
4 / -1

A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground.

A
The block will not move unless the man also moves

B
The man can move even when the block is stationary

C
If both move, the acceleration of the man is greater than the acceleration of the block

D
option A and option C are correct

Solution

Let the mass of the man be m and that of the block be M, where M > m.

The coefficient of friction between both (man and block) and the ground is μ.

The maximum frictional force on the man = μm_g

The maximum frictional force on the block = μM_g

Since M > m, the block experiences a greater maximum frictional force.

Hence, the man cannot apply a pull on the block greater than μm_gwithout himself starting to move, while the block will start moving only if the applied tension exceeds μM_g.

Because μM_g> μm_g, the man alone cannot make the block move while he remains stationary.

Therefore, the block will not move unless the man also moves.

→ Assertion A is correct.

When both start moving, the same tension T acts on the man and the block.

For the block: T− μM_g= Ma₂

For the man: μm_g− T = ma₁

Since the same tension acts on a smaller mass (the man),

a₁> a₂

Hence, if both move, the acceleration of the man is greater than that of the block.

→ Assertion C is correct.
Question 4
4 / -1

Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms^_2, which of the following statements A, B, C is (are) correct?

[1] The horizontal force acting on the body is 20 N

[2] The weight of the 5 kg mass acts vertically downwards

[3] The net vertical force acting on the body is 30 N

A
1,2,3

B
1,2

C
2, 3

D
1 only
Solution
Given:
- A force of 40 N acts at an angle of 30° to the horizontal.
- The mass of the body is 5 kg.
- The acceleration due to gravity is 10 m/s².
- The surface is smooth, meaning there is no friction.
Analyzing the Statements:

Statement 1: The horizontal force acting on the body is 20 N.

The horizontal component of the force is found using the cosine of 30°: Horizontal force = 40 * cos(30°) cos(30°) = √3 / 2 ≈ 0.866 Horizontal force = 40 * 0.866 ≈ 34.64 N This is not 20 N. So, Statement 1 is incorrect.

Statement 2: The weight of the 5 kg mass acts vertically downwards.

The weight of the body is calculated as: Weight = mass * gravity = 5 * 10 = 50 N This force acts vertically downward. So, Statement 2 is correct.

Statement 3: The net vertical force acting on the body is 30 N.

The vertical component of the force is found using the sine of 30°: Vertical force = 40 * sin(30°) sin(30°) = 1/2 = 0.5 Vertical force = 40 * 0.5 = 20 N

The net vertical force is the difference between the vertical component of the applied force and the weight of the body: Net vertical force = Vertical force - Weight = 20 N - 50 N = -30 N The negative sign indicates the force is acting downward, and the magnitude is 30 N. So, Statement 3 is correct.

Conclusion: The correct statements are Statement 2 and Statement 3.

Correct Answer: C: 2 and 3.
Question 5
4 / -1

A man getting down a running bus, falls forward because-

A
Due to inertia of rest, road is left behind and man reaches forward

B
Due to inertia of motion upper part of body continues to be in motion in forward direction while feet come to rest as soon as they touch the road

C
He leans forward as a matter of habbit

D
Of the combined effect of all the three factors stated in (A), (B) and (C)

Solution

When a man jumps off a moving bus, his upper body continues moving forward due to inertia of motion, while his feet come to rest upon touching the ground. This causes him to fall forward:

Inertia of motion ⟹ Upper body retains velocity
Question 6
4 / -1

Two spheres A and B of masses m₁ and m₂ respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.

A
Same as that of B

B
Opposite to that of B

C
θ = tan^–1 (1/2) to the x-axis

D
θ = tan^–1 (–1/2) to the x-axis

Solution

Here initially sphere A is at rest and B is moving along x axis with velocity V.

After collision, velocity of sphere B becomes V/2 along direction perpendicular to its initial direction i.e. along Y axis.

Sphere A will move with some velocity u at some angle θ as shown in figure.considering law of conservation of momentum along X−axis.

m₂V=m1ucosθ...........(i)

Considering law of conservation of momentum along Y−axis.

m₂v/2=m1usinθ............(ii)

From equation i and ii.

1/2 = tanθ

θ = tan−1(0.5).
Question 7
4 / -1

A ring of mass 200 gram is attached to one end of a light spring of force constant 100 N/m and natural length 10 cm. The ring is constrained to move on a rough wire in the shape of the quarter ellipse of the major axis 24 cm and the minor axis 16 cm with its centre at the origin. The plane of the ellipse is vertical and wire is fixed at points A and B as shown in the figure. Initially, ring is at A with other end of the spring fixed at the origin. If normal reaction of wire on ring at A is zero and ring is given a horizontal velocity of 10 m/s towards right so that it just reaches point B, then select the correct alternative (s) (g = 10 m/s²)

A
Work done by friction is -10 Joule.

B
At B energy stored in spring is more than energy stored at A.

C
Work done by friction is -10.16 Joule.

D
Work done by spring force is positive.

Solution
Question 8
4 / -1

Two spheres of masses m₁and m₂ (m₁>m₂) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

A

B

C

D

Solution

If, r₁ and r₂ are position vectors of the centres of the positional vector of their centre of mass is given by R=(m₁r₁+m₂r₂)/ (m₁ + m₂)

R= (m₁r₂+m₂r₂)/m₁+m₂)

The acceleration of the centre of mass is given by:

A=d²R/dt

= [m₁ d₂r/dt+m₂ d²r/dt² ]/(m₁+m₂)

But, d²r₁/dt² and d²r₂/dt² are the accelerations of masses m1 and m2

Have the same magnitude (m₁-m₂)/(m₁+m₂) g

If we take acceleration of [m₁(m₁-m₂)/ (m₁+m₂)g – m2(m₁-m₂)/ (m₁+m₂)/]/ (m₁+m₂)

On simplifying that we get,

a=[(m₁-m₂)/ (m₁+m₂)]² g
Question 9
4 / -1

Which of the following options are correct,

where i, j and k are unit vectors along the x, y and z axis?

A
i.j = 1 ; j x i =0

B
i.j = 0 ; j x i = -k

C
i.j = 1 ; j x i = -k

D
i.j = 1 ; j x i = k

Solution

Dot product of two different unit vectors is 0 and dot product of two same unit vectors is 1. Cross product of two different unit vectors taken according to right hand thumb rule is the other vector. Cross product of two same unit vectors is 0.
Question 10
4 / -1

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:

A
g / L

B
2g / L

C
2g / 3L

D
3g / 2L

Solution