Physics Test 236

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Physics Test 236

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Question 1
4 / -1

A satellite is in a circular orbit very close to the surface of a planet. At some point it is given an impulse along its direction of motion, causing its velocity to increase η times. It now goes into an elliptical orbit, with the planet at the centre of the ellipse. The maximum possible value of η for this to occur is

A
2

B
√2

C
√2 + 1

D

Solution
Question 2
4 / -1

A conductor lies along the z-axis at -1.5 ≤ z < 1.5 m and carries a fixed current of find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10^–3 s. Assume parallel motion along the x-axis.

A
14.85 W

B
29.7 W

C
1.57 W

D
2.97 W
Solution
Concept:

Power required to move a conductor in a magnetic field:
- When a conductor carrying a current moves through a magnetic field, a force is exerted on it due to the interaction between the current and the magnetic field.
- The power required to move the conductor at constant velocity is equal to the rate of work done against this force.
∴ The correct option is 4
Question 3
4 / -1

N identical cells, each of emf ε and internal resistance r, are joined in series. Out of these, n cells are wrongly connected, i.e. their terminals are connected in reverse of that required for series connection. n < (N/2). Let ε₀ be the emf of the resulting battery and r₀ be its internal resistance.

A
ε₀ = (N – n) ε, r₀ = (N – n) r

B
ε₀ = (N – 2n)ε, r₀ = (N – 2n)r

C
ε₀ = (N – 2n)ε, r₀ = Nr

D
ε₀ = (N – n)ε, r₀ = Nr

Solution

Concept:

In the circuit for every cell that is wrongly connected, the emf decreases by 2ϵ. While internal resistance does not depend on the direction and therefore remains the same for all cells.
Question 4
4 / -1

A skier plans to ski a smooth fixed hemisphere of radius R. He starts from rest from a curved smooth surface of height (R/4). The angle θ at which he leaves the hemisphere is

A
cos^–1 (2/3)

B
cos^–1 (5/√3)

C
cos^–1 (5/6)

D
cos^–1 (5/ 2√3)

Solution

Concept

Energy Conservation:
When the skier starts from a height R/4 above the center of the hemisphere, he has potential energy. As he slides down, his potential energy converts into kinetic energy. At the point where the skier leaves the hemisphere, the total energy (potential + kinetic) remains conserved.

Forces and Centripetal Force:
At the point where the skier leaves the hemisphere, the normal force becomes zero. The component of the gravitational force provides the necessary centripetal force for circular motion.
Question 5
4 / -1

In a region of space, a uniform magnetic field B exists in the y - direction. A proton is fired from the origin, with its initial velocity v making a small angle α with the y-direction in the yz plane. In the subsequent motion of the proton,

A
its x - coordinate can never be positive

B
its x and z-coordinates cannot both be zero at the same time

C
its z-coordinate can never be negative

D
its y-coordinate will be proportional to the square of its time of flight

Solution

Concept:

When a charged particle, such as a proton, moves in a magnetic field, the magnetic force acts perpendicular to both the velocity of the particle and the magnetic field direction. This causes the particle to follow a curved path, resulting in circular or helical motion depending on the initial conditions.

In this case:

Thus, The proton will move in a helical path around the y-axis with a constant velocity component along the y-direction and a circular motion in the xz-plane.

∴ The correct option is 2
Question 6
4 / -1

A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion A is punctured with a pin, the thread

A
becomes concave towards A

B
becomes convex towards A

C
either (1) or (2) depending on the size of A with respect to B

D
remains in the initial position

Solution

Explanation:

When a thread is tied slightly loose to a wire frame and the frame is dipped into a soap solution, the soap film covers the entire frame, including the area inside the loop formed by the thread. The soap film creates surface tension, which acts uniformly across the surface, pulling the thread from all directions and keeping it in equilibrium.

Surface Tension:

Surface tension is a force that acts along the surface of a liquid, trying to minimize the surface area. In a soap film, the surface tension acts uniformly and pulls the thread in all directions.

Puncturing the Film:

When the portion A of the soap film inside the loop of the thread is punctured, the surface tension on that part disappears. The surface tension in the remaining unpunctured parts of the film now pulls the thread towards itself.

As a result, the thread becomes concave towards A due to the unbalanced surface tension forces acting on it after the film is punctured at that point.

∴ The correct option is 1
Question 7
4 / -1

A capacitor is charged and then made to discharge through a resistance. The time constant is τ. In what time will the potential difference across the capacitor decrease by 10%.

A
τ ln (0.1)

B
τ ln (0.9)

C

D

Solution
Question 8
4 / -1

In an npn transistor circuit, the collector current is 10mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?

A
The emitter current will be 8 mA.

B
The emitter current will be 10.53 mA.

C
The base current will be 10.53 mA.

D
The base current will be 2 mA.
Solution
Concept:

Transistor Currents:
- In an npn transistor, the emitter current (I_E) is the sum of the collector current (I_C) and the base current (I_B).
- The relationship can be expressed as: I_E = I_C + I_B.
- Given that 95% of the electrons emitted from the emitter reach the collector, the collector current (I_C) is 95% of the emitter current (I_E).
- Formula: I_C = 0.95 × I_E
Calculation:

Given,

Collector current, I_C = 10 mA

95% of the electrons reach the collector:

I_C = 0.95 × I_E

⇒ 10 mA = 0.95 × I_E

⇒ I_E = 10 mA / 0.95

⇒ I_E ≈ 10.53 mA

∴ The emitter current will be approximately 10.53 mA.
Question 9
4 / -1

A closed coil having 50 turns, area 300 cm², is rotated from a position where its plane makes an angle of 45º with a magnetic field 2.0 Tesla to a position perpendicular to the field in a time of 0.1 sec. The average e.m.f. induced in the coil is

A
0 V

B
15 V

C
11 V

D
none of the above
Solution
Concept:

Electromagnetic Induction:
- When a coil is rotated in a magnetic field, the change in magnetic flux induces an electromotive force (e.m.f.) in the coil.
- The induced e.m.f. is given by e.m.f.
- Magnetic flux through a coil is, ϕ = B A cos(θ) where B is the magnetic field, A is the area, and θ is the angle between the magnetic field and the normal to the coil.
Question 10
4 / -1

If the electric potential of the inner metal sphere is 10 volt & that of the outer shell is 5 volt, then the potential at the centre will be :

A
10 volt

B
5 volt

C
15 volt

D
0
Solution
Concept:

Electric Potential in Conductors:
- In electrostatics, the electric potential inside a conductor is constant throughout its volume.
- This means that the potential at any point inside a conducting sphere is the same as the potential on its surface.
- The electric potential is a scalar quantity and is measured in volts (V).
- In this case, we have an inner metal sphere and an outer conducting shell.
- Potential of the inner sphere = 10 V
- Potential of the outer shell = 5 V
Calculation:

Given,

Potential of the inner sphere, V_inner=10 V

Potential of the outer sphere, V_outer=5 V

The potential at the centre will be = V_inner

V_inner∴ The potential at the center will be 10 V.