All questions
- A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position Given : T = 4 s, x = A/3 The displacement of a particle starting into SHM from the mean position is x =A sin ωt = A sin(2π/T)t ∴ A sin(2π/T)... 1
- The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. Given : m = 2 kg, E = 40 J The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is enti... 1
- Potential energy of a particle performing linear S.H.M is 0.1 π2 x2 joule. If mass of the particle is 20 g, Given : PE = 0.1π²x^(2 )J, m = 20 g = 2 X 10⁻² kg PE = ½ mw²x² = ½m(4π²η²)x^(2 )= 0.1π²x² ∴ 2mη² = 0.1 ∴ η² = 0.1/[2(2 X 10⁻²)]= 2.5 ∴ The f... 1
- A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Given : A=4 cm = 4 x 10⁻²m, η = 5Hz, t= 1/30 s ω = 2πη = 2π x 5 = 10π rad/s Therefore, the magnitude of the acceleration, |a| = ω²x = ω²A si... 1
- In SI units, the differential equation of an S.H.M. is \(\frac{d^2x}{dt^2}=-36x\) . Find its frequency and period. Given : d²x/dt²=-36x Comparing this equation with the general equation, We get, ω^(2 )= 36, ∴ ω = 6 rad/s ω = 2πη The frequency, η = ω/2π =... 1
- At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed Given : v = 1/2v_max , 2A=10 cm A=5 cm v = ω√(A²-x²) and v_max = ωA Since v = 1/2v_max ω√(A²-x²)=ωA/2 ∴ A²-x²=A²/4 ∴ x²=A²-A²/4 = 3A²/4 ∴ x=... 1
- Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M. Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of... 1
- State the laws of simple pendulum. The period of a simple pendulum at a given place is T = 2π√(L/g) where L is the length of the simple pendulum and g is the acceleration due... 2
- Obtain the expression for the period of a simple pendulum performing S.H.M. An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string,... 1
- Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M The general expression for the displacement of a particle in SHM at time t is x = A sin(ωt + α) ….. (1) where A and ω is a constant in a par... 2
- Define linear simple harmonic motion. Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always direc... 3
- The graph shows variation of displacement of a particle performing S.H.M. with time t. (B) The force is maximum at time 3T/4. 1
- Two identical springs of constant k are connected, first in series and then in parallel B) 1:2 The ratio of their frequencies of vertical oscillations will be in a ratio 1:2 1
- The length of second's pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be A) 1/6 m Its length on the surface of moon should be 1/6 m 2
- A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + p/ D) 18 J Maximum kinetic energy of the body is 18J 2
- A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. (A) √(3)/2A 1
- A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles. Given : a = fracΔV_max2=6-2/2×10⁻³m³ = 2 x 10⁻³ m³, b = fracΔP_max2=11-1/2×10⁵Pa = 5 x 10^(5 )Pa Cycles = 25 The work done in one cycle, oin... 1
- An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. Given : T_f = 2T_i, monatomic gas ϒ = 5/3 P_iV_i^ϒ = P_fV_f^ϒ in an adiabatic process Now, PV= nRT ∴ V = nRT/P ∴ V_(i )= nRT_i/P_i and V_(f... 1
- An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment Given : W = 2000J , isothermal process In this case, the change in the internal energy of the gas, ΔU=0, as the gas is taken through an isot... 1
- A Carnot refrigerator operates between 250ºK and 300ºK. Calculate its coefficient of performance. Given : T_C = 250 K, T_H = 300 K K = T_C/T_H-T_C=250/300-250=250/50 = 5 this is the coefficient of performance of the refrigerator. 1
- Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC Given : η = 75% = 0.75, T_H = (273 + 727) K = 1000 K η = 1−T_C/T_H ∴ T_C/T_H= 1− η T_C = T_H (1 − η ) = 1000(1 - 0.75) = 250 K = (250 − 273)... 1
- A system releases 125 kJ of heat while 104 kJ of work is done on the system. Calculate the change in internal energy. Given: Q = −125kJ, W= −104kJ ΔU = Q − W= −125−(−104) = (−125 + 104) = −21kJ. Negative sign shows that decrease in energy. This is the change... 1
- A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm Data: P = 1 atm = 1.013 x 10⁵ Pa, V₁ =5 litres =5 x 10⁻³ m³, V₂ = 10 Litres = 10 x 10⁻³ m³, Q = 400J. The work done by the system (gas in th... 1
- A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stat... 2