Physics Test 245

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Physics Test 245

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Question 1
4 / -1

Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heights of P and Q from the Earth surface are h_P, and h_Q, respectively, where The accelerations of P and Q due to Earth's gravity are g_P and g_Q. respectively. If , what is the value of h_Q?

A

B

C

D

Solution
Question 2
4 / -1

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is:

A
4 V

B
6 V

C
8 V

D
10 V

Solution
Question 3
4 / -1

At any instant of time t, the displacement of any particle is given by 2t – 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):

A
10

B
5

C
7

D
6

Solution

Formulas Involved

Power (P): The rate at which work is done or energy is transferred. SI unit: Watt (W), Dimensional Formula: [M¹L²T^-3]

Instantaneous power is given by:

P = F × v

Force (F): An interaction that causes an object to change its velocity. SI unit: Newton (N), Dimensional Formula: [M¹L¹T^-2]

Velocity (v): The rate of change of displacement with respect to time. SI unit: meter per second (m/s), Dimensional Formula: [L¹T^-1]
Question 4
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
Solution

Formula involved:

The formula for completely inelastic collision is highlighted as follows:

m_Av₁ + m_Bv_B = (m_A + m_B)v₂

where:
- m_A = Mass of body A (kg)
- v₁ = Initial velocity of body A (m/s)
- m_B = Mass of body B (kg)
- v_B = Initial velocity of body B (m/s)
- v₂ = Final velocity of the system after collision (m/s)
Calculation

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 5
4 / -1

10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is:

A
10

B
100

C
1

D
1000

Solution

Calculation:

The increase in current when switching from the series to the parallel configuration is the ratio n:
n= I _parallel/ I _series

I _series = E/10R

I _parallel = E/(R/10) = 10E/R

n = 100

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

Light travels a distance x in time t₁ in air and 10x in time t₂ in another denser medium. What is the critical angle for this medium?

A

B

C

D

Solution

Calculation:

Speed of light in air = V₁ = x/t₁

Speed of light in medium = V₂ = 10x/t₂

Sin θ = V₂/V₁

Sinθ = 10x/t₂× t₁/x

θ = Sin^-1(10t₁/t₂)

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

An ac source is connected to a capacitor C. Due to decrease in its operating frequency:

A
capacitive reactance decreases.

B
displacement current increases.

C
displacement current decreases.

D
capacitive reactance remains constant

Solution
Question 8
4 / -1

The temperature of a gas is -50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

A
669°C

B
3295°C

C
3097 K

D
223 K

Solution

Calculation:

The root-mean-square (rms) speed of gas molecules is given by the formula:

v_rms = √(3RT/m)

Where:

v_rms is the root-mean-square speed

R is the gas constant

T is the absolute temperature in Kelvin

m is the mass of one molecule of the gas

Initial temperature (T₁) is -50ºC, which is 223 K (since T (K) = T (ºC) + 273)

T = (-50) + 273 = 223 k

We need to find the final temperature (T₂) such that the rms speed increases by 3 times.

Given that v_rms1,= 3 × v_rms1 + v_rms1,we can write:

v_rms1 = 4_vrms1

√(3RT₂/m) = 4√(3RT₁/m)

T₂ = 16 × 223 = 3568 K

Converting back to Celsius: T₂ = 3568 - 273 = 3295ºC.
Question 9
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
Question 10
4 / -1

An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 10⁵ N C^-1. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.

A
8 mC

B
6 mC

C
4 mC

D
2 mC
Solution
Concept:

Electric Dipole:
- Electric Dipole: A pair of equal and opposite charges separated by a distance. The dipole moment (p) is given by p = q × d, where q is the charge and d is the separation distance.
- Torque on Dipole: When placed in an electric field (E), the torque (τ) experienced by the dipole is given by τ = p × E sinθ, where θ is the angle between the dipole moment and the electric field.
- SI Unit: The SI unit of torque is Newton-meter (N·m), and the SI unit of electric field is Newton per Coulomb (N/C).
- Dimensional Formula: The dimensional formula for torque is [M L² T²], and for electric field is [M L T³ A¹].
- Important Formula: τ = pEsinθ and p = q × d.
Calculation:

Given,

Electric field intensity, E = 2 × 10⁵ N/C

Torque, τ = 4 N·m

Angle, θ = 30°

Dipole length, d = 2 cm = 0.02 m

We know, T = q × d × E × sin30

⇒ 4 = q × d × E × sin30

⇒ 4 = q × 0.02 × 2 × 10⁵ × (1/2)

q = 2 × 10^-3

q = 2mC

∴ The correct option is 4)

4 = q × 0.02 × 2 × 10⁵ × 0.5

Hence, the magnitude of charge on the dipole is 2 mC.