Physics Test 299

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Physics Test 299

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Weekly Quiz Competition

Question 1
4 / -1

A mass of 1 kg is suspended from a spring of force constant 400 N, executing S.H.M. total energy of the body is 2J, then maximum acceleration of the spring will be:

A
4 m/s²

B
40 m/s²

C
200 m/s²

D
400 m/s²

Solution
Question 2
4 / -1

A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision, the subsequent motion of the combined body will be :

A
in the same circular orbit of radius R

B
such that it escapes to infinity

C
in an elliptical orbit

D
none of these

Solution

It is given that the satellite and the meteorite collide completely in-elastically. Hence its momentum will be conserved but its kinetic energy will not be conserved and will move together as a single body. The collision of the meteorite and the satellite should be taken at an angle of 90, since the satellite is moving in a circular orbit and the meteorite is moving towards the earth.

Let us consider the Initial Momentum of the satellite and the meteorite to be P = Mv

where M is the mass of the meteorite as well as the satellite and v is the orbital velocity through which the satellite revolves around the earth and is the same for the meteorite too.

In the above diagram, P₁ is the momentum of the satellite orbiting the earth and P₂ is the momentum of the meteorite moving towards the earth. Their resultant is given by P by the laws of vector addition.

Initially, the satellite was orbiting around the earth with an orbital velocity of v but when the collision happened the meteorite and the satellite started moving with velocity, V = 0.707v, which signifies that the velocity decreased.

Since the Velocity after the collision is less than of orbital velocity. Hence it does not follow a circular path and travels in an elliptical orbit.
Question 3
4 / -1

A mass of 100g strikes the wall with speed 5m/s at an angle as shown in figure and it rebounds with the same speed. If the contact time is 2 × 10⁻³sec, what is the force applied on the mass by the wall.

A

B
250N to right

C

D
250N to left

Solution

The negative sign indicates the force acts in the opposite direction.
Question 4
4 / -1

The magnitude and direction of the current in the following circuit is

A
0.2 A from B to A through E

B
0.5 A from A to B through E

C

D
1.5 A from B to A through E

Solution

Calculation:

Using, V = IR

I = V/R

I = (10-5)/(7+2+1)

I = 5/10

I = 0.5 A to B through E

∴ The correct option is 2)
Question 5
4 / -1

When the light of frequency 2v₀ (where v₀ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5v₀, the maximum velocity of electrons emitted from the same plate is v₂. The ratio of v₁ to v₂ is

A
1 ∶ 2

B
1 ∶ 4

C
2 ∶ 1

D
4 ∶ 1

Solution
Question 6
4 / -1

Two identical capacitors C₁ and C₂ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C₁ using battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy ?

A
50%

B
25%

C
75%

D
0%

Solution
Question 7
4 / -1

Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁ : v₂is

A
1 : 2

B
2 : 1

C
4 : 1

D
1 : 4

Solution

Given data:

Mass of body A (m_A) = m

Initial velocity of body A (v₁) = v₁

Mass of body B (m_B) = m

Initial velocity of body B (v_B) = 0

Final velocity of the system (v₂) = v₂

m_Av₁ + m_Bv_B = (m_A + m_B)v₂

⇒ mv₁ + m×0 = (m + m)v₂

⇒ mv₁ = 2mv₂

⇒ v₁ = 2v₂

∴ The ratio v₁ : v₂ is 2:1.
Question 8
4 / -1

Given below are two statements :

Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (IdI) of a current carrying conductor only.

Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q.

In light of above statements choose the most appropriate answer from the options given below:

A
Statement I is incorrect and Statement II is correct

B
Both Statement I and Statement II are correct

C
Both Statement I and Statement II are incorrect

D
Statement I is correct and Statement II is incorrect

Solution

Statement I - Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (IdI) of a current carrying conductor only which is written as;

Therefore this statement is true.

Statement II - In statement II is given that Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q. The Biot- Savart's law is the equation of the magnetic field produced by the current carrying conductor whereas in coulomb's law we find the force of attraction and repulsion between the two masses. Biot-Savart's law is not analogous to Coulomb's inverse square law of charge q.

Therefore, this statement is false.

Hence, option 4) is the correct answer.
Question 9
4 / -1

A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which ball can be moved?

A
14 m/s

B
3 m/s

C
3.92 m/s

D
5 m/s

Solution

Given :

Mass of the ball m = 0.25kg

Length of the string (radius of the circular motion) l = 1.96m

Maximum tension in the string T = 25N

Let the maximum speed with which the ball can be moved be v.

The tension provides the necessary centripetal force required for the ball to move in circular path.
Question 10
4 / -1

The ratio of total kinetic energy to the rotational kinetic energy of a solid sphere rolling on a horizontal surface is-

A
2 : 3

B
2 : 7

C
5 : 2

D
7 : 2
Solution
- The rolling motion is a combination of translation and rotation.
- The rolling kinetic energy of a sphere is the sum of the translational kinetic energy of its centre of mass plus the rotational kinetic energy about the centre of mass.