Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the wavelength of the sound wave in air (speed of sound = 340 m/s)?
A.0.77 m
B.0.85 m
C.0.90 m
D.1.00 m
Solution
The wavelength λ can be calculated using the formula λ = v/f, where v is the speed of sound and f is the frequency. Thus, λ = 340 m/s / 440 Hz = 0.77 m.
Q. A tuning fork produces a sound wave with a frequency of 440 Hz. What is the wavelength of the sound wave in air, given that the speed of sound in air is approximately 340 m/s?
A.0.77 m
B.0.85 m
C.0.90 m
D.1.00 m
Solution
Wavelength λ is given by the formula λ = v/f. Here, v = 340 m/s and f = 440 Hz. Thus, λ = 340/440 = 0.7727 m, approximately 0.77 m.
Q. A uniform rod of length L and mass M is pivoted at one end and allowed to fall under gravity. What is the angular acceleration of the rod just after it is released? (2019)
A.g/L
B.2g/L
C.3g/L
D.g/2L
Solution
The torque τ = Mg(L/2) and moment of inertia I = (1/3)ML². Using τ = Iα, we find α = 3g/2L.
Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?
A.√(3g/L)
B.√(2g/L)
C.√(g/L)
D.√(4g/L)
Solution
Using conservation of energy, potential energy at the top = rotational kinetic energy at the bottom. mgh = (1/2)Iω^2. For a rod, I = (1/3)ML^2, h = L/2. Solving gives ω = √(3g/L).
Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod when it makes an angle θ with the vertical?
A.√(g/L)(1-cosθ)
B.√(2g/L)(1-cosθ)
C.√(g/L)(1+cosθ)
D.√(2g/L)(1+cosθ)
Solution
Using conservation of energy, the potential energy lost equals the rotational kinetic energy gained. The angular velocity ω can be derived as ω = √(2g/L)(1-cosθ).
Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular speed of the rod just before it hits the ground? (2019)
A.√(3g/L)
B.√(2g/L)
C.√(g/L)
D.√(4g/L)
Solution
Using conservation of energy, potential energy at the top converts to rotational kinetic energy at the bottom. The angular speed ω = √(3g/L).
Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod just before it hits the ground? (2019)
A.√(3g/L)
B.√(2g/L)
C.√(g/L)
D.√(4g/L)
Solution
Using conservation of energy, potential energy at the top is converted to rotational kinetic energy at the bottom. The angular velocity ω can be found using the relation ω = √(3g/L).
Q. A uniformly charged sphere of radius R has a total charge Q. What is the electric field at a point outside the sphere (r > R)?
A.0
B.Q/(4πε₀r²)
C.Q/(4πε₀R²)
D.Q/(4πε₀R)
Solution
For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).
