PHYSICS KMKK CHALLENGE

A steel wire of diameter 3.3 mm and 40 cm long is attached to a mass of 2.5 kg at  one end and is suspended. The wire elongates by 0.1 mm. Calculate the Young’s Modulus

• 5.0 x 10¹¹ N/m²

• 2.5 x 10¹⁰ N/m²

• 1.15 x 10¹⁰ N/m²

• 1.00 x 10¹¹ N/m²

Answer explanation

Young's Modulus (Y) is calculated using the formula Y = (F/A) / (ΔL/L). Here, F = mg = 2.5 kg * 9.81 m/s², A = π(d/2)², ΔL = 0.1 mm, and L = 40 cm. After calculations, Y = 1.15 x 10¹⁰ N/m², which is the correct answer.

A 50 g object connected to a spring with a spring constant 35 N m^-1 oscillates with amplitude 4 cm on a horizontal frictionless surface. Calculate the total energy of the system.

• 0.014 J

• 0.056 J

• 0.035 J

• 0.028 J

Answer explanation

The total energy in a spring system is given by E = 1/2 k A^2. Here, k = 35 N/m and A = 0.04 m. Thus, E = 1/2 * 35 * (0.04)^2 = 0.028 J. Therefore, the correct answer is 0.028 J.

A progressive wave is represented by equation, y (x, t) = 1200 sin (314t – 0.42x) where x and y are in cm and t is in second. Determine the maximum velocity of the particle.

• 2500 m/s

• 3768 m/s

• 3865 m/s

• 5578 m/s

Answer explanation

The maximum velocity of a particle in a wave is given by v_max = Aω, where A is the amplitude (1200 cm) and ω is the angular frequency (314 rad/s). Thus, v_max = 1200 * 314 / 100 = 3768 m/s.

A mechanical wave propagates at 550 m s^-1 along a string stretched to a tension of 800 N. The string oscillates at fundamental frequency 440 Hz. Calculate the frequency of the second overtone

• 880 Hz

• 6260 Hz

• 1220 Hz

• 1320 Hz

Answer explanation

The second overtone is the third harmonic, which is three times the fundamental frequency. Thus, 3 x 440 Hz = 1320 Hz. Therefore, the frequency of the second overtone is 1320 Hz.

A car is traveling at 25 m s^-1 emits a sound of frequency 1100 Hz approaches a stationary observer. Calculate the apparent frequency of the sound heard by the observer. The speed of sound is 340 m s⁼¹ .

• 1150.7 Hz

• 1250 Hz

• 1280.6 Hz

• 1187.3 Hz

Answer explanation

Using the Doppler effect formula, the apparent frequency f' = f(v + vo)/(v - vs). Here, f = 1100 Hz, v = 340 m/s, vo = 0 (observer stationary), vs = 25 m/s (source approaching). Calculating gives f' = 1187.3 Hz.

The temperature of 3 moles of an ideal diatomic gas is 45.0 °C. The gas is heated and the temperature is increased by 20.0 °C. Calculate the change in the internal energy of the gas.

• 800.5 J

• 1246.5 J

• 1200.0 J

• 1000.0 J

Answer explanation

The change in internal energy (\Delta U) for an ideal diatomic gas is given by \Delta U = nC_v\Delta T. For diatomic gases, C_v = 5/2 R. Here, n = 3 moles, \Delta T = 20 °C. Thus, \Delta U = 3 * (5/2 * 8.314) * 20 = 1246.5 J.

Seven moles of gas of temperature 290 K is expanded isothermally from 1.5 litres to 5.0 litres. Calculate the work done by the gas.

• 4.03 x 10⁴ J

• 2.03 x 10⁴ J

• 3.03 x 10⁴ J

• 1.03 x 10⁴ J

Answer explanation

The work done by the gas during isothermal expansion is calculated using W = nRT ln(Vf/Vi). Substituting n=7, R=8.314 J/(mol K), T=290 K, Vf=5.0 L, and Vi=1.5 L gives W ≈ 2.03 x 10^4 J, which is the correct answer.

Neon is a monoatomic gas and it has a molar mass of 20 g mol^-1 . At 300 K, calculate the translational kinetic energy of a neon atom.

• 8.31 x 10^-21 J

• 5.00 x 10^-22 J

• 6.21 x 10^-21 J

• 4.12 x 10^-22 J

Answer explanation

The translational kinetic energy (KE) of a gas atom is given by KE = (3/2)kT, where k = 1.38 x 10^-23 J/K and T = 300 K. Calculating gives KE = 6.21 x 10^-21 J, which matches the correct answer.

A tank contains nitrogen gas at temperature of 27 °C is heated to 352 °C. Given the molar mass of nitrogen gas is 28 g mol^-1 , calculate the change in root mean square speed (v_rms) of the nitrogen gas molecules.

• 229 ms^-1

• 220 ms^-1

• 129 ms^-1

• 239 ms^-1

Answer explanation

The root mean square speed (v_rms) is calculated using the formula v_rms = sqrt(3RT/M). The change in temperature leads to a change in v_rms, resulting in a final speed of 229 ms^-1, which is the correct answer.

A metal wire with Young’s Modulus 127.0 GPa and 75.0 cm long stretches by 0.035 cm when a load of 8.0 kg is hung on its end. Calculate the diameter of the wire.

• 3.49 x 10^-4 m

• 6.49 x 10^-4 m

• 8.49 x 10^-4 m

• 1.49 x 10^-4 m

Answer explanation

Using the formula for Young's Modulus, E = (F/A) / (ΔL/L), we can rearrange to find the area A. Given the load F = 8.0 kg (78.4 N), length L = 0.75 m, and stretch ΔL = 0.00035 m, we find A = 3.24 x 10^-6 m². The diameter is then 6.49 x 10^-4 m.