PHYSICS-25-02- 11th (PQRS & J) partial marking(1)

PHYSICS FINALTEST Take g = 10 m/s2 wherever required PART -A [6 × 5 = 30] Q.1 The lower end of a long capillary tube of diameter 2 mm is dipped so that its lower end is 10 cm below the surface of water in a beaker. (a) What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? (b) Calculate the capillary rise in the tube if its upper end is left open to atmosphere. The surface tension of water at temperature of the experiment is 7.30 × 10–2 N m–1, 1 atmospheric pressure = 105 Pa, density of water = 1000 kg m–3, angle of contact = 0° [3 + 2] [Ans. (a) 1.01146 × 105 (b) 14.6 mm] [Sol. Excess pressure = 2s/r = 146 Pa Pressure 'h' below the surface = 0 + gh  {1}  Pinside bubble = (P0 + gh) + 2s/r  {1} 7.310–2 = 1 × 105 + 103 × 10 × 0.1 + 2 × = 1 × 105 + 103 + 146 = 1 × 105 + 1146 103 = 1.01146 × 105  {1} 2s 2 7.3102 (b) h = gr = 103 10103 = 14.6 × 10–3 m = 14.6 mm  {2} ] Q.2 One end of a long string of linear mass density 10–2 kg m–1 is connected to an electricallydriven tuning fork of frequency 150 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. Thepulleyendabsorbs all the incomingenergysothat reflectedwaves fromthisendhavenegligible amplitude. At t = 0, the left end (fork end) of the string is at x = 0 has a transverse displacement of 2.5 cm and is moving along positive y-direction. The amplitude of the wave is 5 cm. Write down the transverse displacement y(in cm) as function of x (in m) and t (in sec) that describes the wave on the string. [5] 5sin(300t  x)+  [Sol. At x = 0 and t = 0, y = 2.5 cm and dy/dt is positive   = /6  {1} [Ans. y = =  ]   v = = 900 = 300 m/s  {1} 0.01 frequency = 150 Hz   = (1/2) m  {1} Then equation will be y = Asin2 –  +  = 5sin2150t   +  = 5sin(300t  x)+  {2} ] Q.3 A source of frequency 700 Hz is placed between a man and a wall at the same height of person's ear. Find the velocity v (assume v << c) of the source with which it should approach the wall such that person will detect 3 beats per second. [Take velocity of sound (c) = 350 m/s] [5] [Ans. (3/4) m/s] [Sol. 700 c  c  = 3  {2}  c  v c + v  700 × 2vc = 3c2 – 3v2  {1} as v << c 1400 vc = 3c2  {1}  v = 3c 1400 = 3 350 1400 3 = 4 m/s  {1} ] Q.4 What is the temperature of the steel-copper junction in the steady state of the system shown in the figure. Length of the steel rod = 25 cm, length of the copper rod = 50 cm, temperature of the furnace = 300 °C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50 J s–1 m–1 K–1 and of copper = 400 J s–1 m–1 K–1) [5] [Ans. 100°C] [Sol. k1A1(T1  T) L1 = k2A2 (T  T2 ) L2  {2}  L1   k 2   A2  300 – T =      k     A  (T – 0)  {2}  300 – T = 2T 2   1   1  T = 100°C  {1} ] Q.5 A ring of radius r made of wire of density  is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring as shown in figure. Find the percentage change in tension for two percent increase in angular velocity of rotation. Ignore gravity. [5] [Sol. 2T sin = Ar × 2r2  {2} [Ans. 4 %]    T = Ar22  sin    T  2  {1}   Therefore, the percent change in  for 2% change in  T  T = 2   {1}  %change = 4%  {1} ] Q.6 Two blocks A and B are joined by means of a slacked string passing over a massless pulley as shown in diagram. The system is released from rest and it becomes taut when B falls a distance 0.5 m, then (a) Find the common velocity of two blocks just after string become taut. (b) Find the magnitude of impulse on the pulley by the clamp during the small interval while string becomes taut. [2 + 3] [Ans. (a) 10 m/s, (b) 4 5 N-m/s] 3 3 [Sol. Velocity of B just before the string is taut vB = 2gh = 10 m/s  {1} (a) Common velocity = v  (m + m )v = m v  v = v /3 = 10 m/s  {2} A B B B B 3  10  (b) Magnitude of impulse on A = Magnitude of impulse on B = 1  –  = 3  N-m/s  Impulse on pulley = 4 Impulse on A = 3 N-m/s  {2} ] PART -B [5 × 6 = 30] Q.7 A conical pendulum is composed of a mass m = 100 gm and a massless string of length l = 50 cm. It swings with constant angular velocity 0 = 10 rad/s as shown in the figure. (a) What is the angle, , between the string and the vertical? (b) What is the angular momentum vector with respect to the pivot at the instant shown in the diagram? (c) What is the torque vector with respect to the pivot at the instant shown in the diagram? [2 + 2 + 2] [Ans. (a) 37°, (b) 3 × 10–3 (4ˆi + 3ˆj) N-m/s, (c) 0.3ˆj N-m ] [Sol.(a) Tcos – mg = 0 Tsin = m2r 2r  {1}  2l sin   tan = 0 = 0 g  cos = g g    = cos–1    = 37°  {1} l0  l0  (b)  = (l sin ˆi – l cos kˆ)    {1} & v = l sin 0ˆj L =    2 2 ˆ 2 ˆ 2 ˆ ˆ  r  mv = m(l sin k + l sin cos i) = ml sin (cosi + sin k) = 3 × 10–3 (4ˆi + 3ˆj) N-m-s  {1} (c)   kˆ )  {1} = (lsinˆi  lcoskˆ)  mg(kˆ) = mgl sin  ˆj = 0.3ˆj N-m  {1} ] Q.8 The mass distribution inside the sphere of radius 60 cm is nonuniform, so that the center of mass lies at a distance 30 cm from the geometric center. Suppose that the sphere is attached rigidly to a massless horizontal rod of length 10 cm and negligible volume, oriented along the line formed by the center of mass and the geometric center of the sphere, as shown. The rod is attached on the side of the sphere nearest to the center of mass, and the far end of the rod is attached to a fixed pivot. The entire system is submerged in water and the sphere is held stationary by a vertical string tied to its lowest point. Consider the average density of the sphere = 700 kg/m3 and density of water = 1000 kg/m3. (a) Find the force exerted by water on the sphere. (b) Find the tension T in the string. [2+4] [Sol.(a) Force exerted by liquid on sphere = (b) Sphere is in equilibrium [Ans. (a) 2880 N, (b) 1728  N ] 4 (0.6)3 1000 10 = 2880 N  {2} 3  R + l     = 0  B(R + l) – T (R + l) = W      {2} B  W l + (R / 2)   T =    l + R   l + (R / 2)   T = Vwg – Vsg l + R   {1}   l + (R / 2)  4 3 T = w s  l + R  3 R g    After putting the values, T = 1728  N  {1} ] Q.9 A frog sits on the end of a long board of length L. The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed i.e. relative to ground v that allows the frog to do the trick? The board and the frog have equal masses. [6] [Ans. ] [Sol. Taking v for the plank in ground frame and conserving linear momentum in horizontal direction mv = m(ucos) v = ucos 2u sin  t = g  {2} 2u(u cos + u cos) sin  2u2 sin 2  L = g = g  {2} u =  Minimum u =  {1}  {1} ] Q.10 A heat engine uses an ideal gas ( = 1.40) that undergoes the reversible cycle shown in figure. Obtain the thermodynamic percentage efficiency of the engine. [6] [Ans. 24%] [Sol. Heat supplied (1  2 & 2  3) = nCv9T0 + nCp90T0  {2} Heat rejected (3  4 & 4  1) = nCv90T0 + nCp9T0  {2}  10Cv + Cp    = 1 V +10C  p   {1} = 1 – 10 +  1+10 11.4 = 1 – 15 = 3.6 15 12 = 50  %efficiency = 24%  {1} ] Q.11 A car is moving away from a stationary target at a speed of 10 m/s. A man sitting on the top of the car fires a machine gun that can fire bullets at a constant rate of 3 bullets per second. Assuming that all the bullets hit the target and the speed of the bullets with respect to ground is 80 m/s. [3.5+2.5] (a) Find the number of bullets that hit the target in one minute. (b) Findtheaveragethrustforceexperiencedbythemancarrying gunduetorecoilifmassofeachbulletis 100 gm. [Ans. (a) 160, (b) 27 N] [Sol.(a) Distance between two consecutive bullets = 80 × (1/3) + (10/3) = 30 m  {1.5}  Time difference between two consecutive shot at target = 30/80  {1}  Number of shots per minute = 1  60 = 160  {1} 30 / 80 (b) dm dt = 300 1000 kg/s  {1.5} vrel = 80 – (–10) = 90   F = v dm rel dt = 90 × 3 10 = 27 N  {1} ] PART -C [4 × 7.5 = 30] Q.12 Astudent has a large quantity of a flexible cable. If she cuts a piece of that cable and hangs it vertically, the longest piece that does not break under its own weight is l. The student then cuts another piece and places it on a horizontal smooth table. Asmall part of that cable hangs over the edge so that the cable begins to slide down after being released. How long (L) can this piece be so that it does not break during the slide? [7.5] [Ans. 4l] [Sol. Let x be the varying length of the cable on the horizontal table. The cable's acceleration is provided by the weight of the vertical part only. Considering the whole cable as a system of mass M, we can write M[1 – (x/L)]g = Ma.  {2} The acceleration is then a = (1 – x/L)g. For the horizontal part, the acceleration is provided by the tension at the bending point. Therefore, the force of tension on the horizontal part is xa = x(1 – x/L)g,  {1} where  is the linear mass density. This force has to be less than the maximum tension force gl, i.e. 0 < x2 – xL + Ll. Thus, the discriminant of the quadratic function on the right-hand side has to be negative, leading to L < 4l.  {2.5} ] Q.13 A river has a width d. Afisherman in a boat crosses the river twice. During the first crossing, his goal is to minimize the crossing time. During the second crossing, his goal is to minimize the distance that the boat is carried downstream. In the first case, the crossing time is T0. In the second case, the crossing time is 3T0. What is the speed of the river flow? Find all possible answers. [7.5] 2 2d d [Sol. Case - I If vr < vb Shortest path: Quickest path: [Ans. 3T0 0 d vb sin  = 3T0  {1} d = T0  {1} b & vr – vbcos = 0   sin = 1/3 d  v = v cos = 2 2d =  {1.5} r b T0 3T0 Case - II If vb < vr  d  x =  (vr  vb cos )  {1} vb = d/T0  sin   d 2 2 d2  x = (vr cosec  vb cot ) vr = 3 vr  2  vb  t  for min. x dx = d (v cot  + v cosec) =0 2v 2 = 3d2/T 2 d vb cos = vb/vr vr = d  {1} T0 3T0 =  {1} ] Q.14 A weightless inextensible string which first runs over a fixed weightless pulley D and then coils on a spool B of outer radius R, and inner radius R/2 tightly. The spool rolls without sliding along a horizontal fixed rail, as shown. The total mass of the spool is M. The axis passing through centre O of the spool is perpendicular to the plane of the drawing and the radius of gyration relative to O is (R/2). If the end A of the string is pulled downward with constant acceleration g/2, then find (a) Direction in which centre O moves. (b) Acceleration of centre O. (c) Tension in the string. (d) Direction and magnitude of friction force between spool and rail. [1 + 2.5 + 2 + 2] [Ans. (a) towards left, (b) g/2, (c) 3Mg , (d) 5 Mg (toward left)] [Sol. Equation of motion are MR 2 T × R – f × (R/2) = 2 4 4 ...(1)  {1} T – f = Ma ...(2)  {1} Constraint equation is  {0.5} R a + 2 = 0 ...(3)  {1} On solving Also, – 3Ma T = 2 a + R = g/2  {1} so, a = –g/2  O will move towards left  {1}  T = 3Mg 4  {1} and f = T – Ma = 5 Mg (toward left)  {1} ] 4 Q.15 A block of mass 160 gm is attached to a rigid support by a spring of force constant 104 N/m and it is executing simple harmonic motion with amplitude 10 cm on a horizontal, frictionless table. Abullet of mass 90 gm and speed 100 m/s strikes the block when the block passes the equilibrium (x = 0) point, going in the same direction as the bullet, as shown. The bullet remains embedded in the block. (a) Determine the amplitude and angular frequency of the resulting simple harmonic motion. (b) Write x coordinate (in cm )of combined mass as function of time, considering time of collision as t = 0. [5+2.5] [Ans. (a) 26 cm, 200 rad/s, (b) 26 sin 200 t + ] k [Sol. Velocity of block just before the collision, v0 = M x0 = 25 m/s  {1}  Velocity of combined mass just after collision Mv0 + mv V = M + m = kM  x0 + mv M + m  {2} Amplitude of SHM A = V = = 26 cm  {1} Angular frequency = k M + m = 200 rad/s  {1} Also, at t = 0, x = 0, V = Vmax in –ve x direction   =   {1}  Equation is x = 26 sin (200 t + )  {1.5} ]