BPHCT-131 · June 2023 · English

IGNOU BPHCT-131 June 2023 Previous Year Question Paper

MECHANICS

Structured previous year question paper for BPHCT-131, June 2023 session.

Max marks: 50 · Questions: 6

Verified: 24 Aug 2026

BACHELOR OF SCIENCE (B. Sc.)

(BSCG)

Term-End Examination

June, 2023

BPHCT-131 :MECHANTES

Time : 2 Hours Maximum Marks : 50

Note :

(i) Attempt all questions.

(ii) The marks for each question are

indicated against it.

(iii) Symbols have their usual meanings.

(iv) You may use a calculator.

Q1.Answer any five parts : 2 each

(a) Show that for any two vectors p and q : 2 2

(b) State the order and degree of the ODE : (vy +2y'=0. A car of mass 3000 kg is moving in a circular path of radius 10 m at a constant speed of 6 ms!. Calculate the centripetal force on it.

(d) Calculate the work done by the force F = (2.0x -5.0%2) N in moving a particle of mass 2.0 kg from x, =1.0 m_ to

(e) What is the rotational K. E. of a particle of mass 2 kg moving in a circle of radius 1 m with an angular speed of 1 rad s-!?

(f) State Kepler's law of harmonics for planetary motion.

(g) Two tuning forks of frequencies 383 Hz and 389 Hz are sounded simultaneously. Calculate the beat frequency.

(h) Give one example of waves : @ which require a medium for propagation. Gi) which do not require a medium for propagation.

Q2.Answer any two parts :

(a)

(0) Determine torque about the point > nA (1, 0, -1) due to a force F = 31+ j-k being exerted at the point (2, —1, —4). 2 Gi) Determine the velocity and acceleration of a particle with position vector : 3 r (t) = cos? ti + sin? tj + cos 2tk

(b) Show that the following ODE is exact and solve it : 5

(0) Solve the following boundary value problem : 5 y’—4y'+4y = 0 y

(0)=3

Q3.Answer any two parts:

(a) A ship of mass 4x 10° kg is moving at a constant velocity. Its engine generates a forward thrust of 8x10°N. Determine

(6) the upward bouyant force on the ship due to water and

(ii) the resistive force exerted by water on the ship. 5 Take g =10ms~.

(b) Calculate the value of acceleration due to the Earth’s gravity at an altitude of

Q4.0x10?km and at the depth of 10 km. Given gy = 10 ms~. Take Re = 6400 km. 5 (©)

(i) If the force of friction is 20 N, what power is needed to maintain a steady speed of an object at 2.0 ms-! on level ground ? 2

(1) A woman of mass 80 kg and her car are suddenly accelerated from rest to a speed of 6.0 ms“! as a result of rear- end collision. Obtain the impulse on the woman and the average force exerted on her if the duration of the collision is 0.8 s.

Q5.Answer any two parts : 5 each

(a) (४) A girl is sitting on a giant wheel at a distance of 5.0 m from its centre. What is her speed when the wheel is turning at the rate of 1 revolution every 5 s ?

(ii) State the law of conservation of angular momentum. A satellite having moment of inertia of 10000 kgm? is rotating at angular speed of 1 r.p.m. If its moment of inertia is increased to 30000 kgm?, what will its angular speed be (in r.p.m.) ? शक

(b)

(i) What is a central force ? What are the two constants of motion under central conservative forces ? 2+1

(i) Determine the centre of mass and relative coordinates of a system consisting of two particles of masses of 1.5 kg and 2.5 kg placed 3.0 m apart. 2

(c) Two balls of masses 1 kg and 3 kg collide head on in an elastic collision in the equal and opposite velocities v . Determine their final velocities.

Q6.Answer any two parts:

(a)

(i) The amplitude of oscillation of a simple harmonic oscillator is 40 cm. Show that its instantaneous kinetic energy is less than its average kinetic energy when the displacement is 30 cm. 3 Gi) An object undergoes SHM with frequency f =0.45 Hz. The initial displacement is 0.025 m and the initial velocity is 1.5 ms. Calculate the amplitude of the oscillation. 2

(b) The motion of a simple pendulum is described by the differential equation: 5 TX ox = 0 dt? Write the solution of this differential equation for the following set of initial conditions : att =0,x=3cm and & <0,

(c) A travelling wave is given by : y(x,t) = 0.18 sin (10.2¢ - 3.2” +1.8) m where x is measured in metres and ¢ is measured in seconds. Determine the distance by which the origin on the x-axis should be shifted so that the expression for the wave becomes : 5 y (x,t) = 0.18 sin (10.2¢ - 3.25) m or yo oe (vy +2y' =0 F-34+j-k = fas (1,0,-1) # via aa argh stat HATE 2 F(t) = cos? tf + sin? } + cos 2h (3e* -2+ y)dv +(x +e —y)dy =0 yl" —4y" + 4y =0 y

(0)=3 y

(1) =0 g=10ms2 @ 5 (@) Gag 5.0x10? km SK THE 10km K fea @ fF g)=10ms21 Ry = 6400km (giant wheel) % Bah HR A 5.0m [15 ]