PHYSICS-19-11- 11th (PQRS)

REVIEW TEST-6 Class : XI (PQRS) Time : 100 min Max. Marks : 100 General Remarks: INSTRUCTIONS 1. The question paper contain 20 questions and 24 pages. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. 2. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 3. Use of Calculator, Log table and Mobile is not permitted. 4. Legibility and clarity in answering the question will be appreciated. 5. Put a cross ( × ) on the rough work done by you. 6. Last three pages are Extra pages. You may use them for any unfinished question(s) mentioning the page number with remark "continued on page " Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name For Office Use ……………………………. Total Marks Obtained………………… Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks XI (PQRS) PHYSICS REVIEW TEST-6 Take g = 10 m/s2 where ever required in this paper. Q.1 The graph shown in the figure represents change in temperature of 5 kg of a substance as it absorbs heat at a constant rate of 42 kJ min–1. Find the latent heat of vapourization of the substance. Also find the specific heat of the liquid [Ans. 126 kJ kg–1 , 42/75kJ kg] [Sol. By the graph time taken by 5 kg of substance to fuse by absorbing energy at the rate 42 kJ/min = (30 – 15) = 15 min  Energy absorbed = 15 × 42 kJ = 630 kJ  Latent heat of vapourization = 630 kJ/kg 5 i.e Lf = 126 kJ/kg Ans. (15 10)  42 42 Specific heat of the liquid = 5(125  50) = 75 kJ/kg ] Q.2 A glass flask contains some mercury at room temperature. It is found that at different temperatures the volume of air inside the flask remains the same. If the volume of mercury in the flask is 300 cm3, then find the volume of the flask. [Given that coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8 × 10–4 (°C)–1 and 9 × 10–6 (°C)–1 respectively] [Ans.2000 cm3 ] [Sol. We have Vflask = VHg + Vair As (V)air = 0  (V)flask = (V)Hg  Cavity having same expansion co-efficient for material  Vf f T = VHg Hg T  Vf VHg Hg  f 1.8104 20 = 3 9 106 = 3  Volume of flask = 2000 cm3 ] Q.3 A spaceship of mass m is travelling with speed v0 along +Y-axis suddenly shots out one fourth of its part with speed 2v0 along + X-axis. Find the velocity vector of the remaining part. Also find increase in 4 kinetic energy of the whole system. [Ans. v ˆj – 2 v 5 i , mv ] [Sol. Let mass of spaceship = m Given that u = v0 ˆj According to question 3 0 3 0 6 0 mu = m (2v ) ˆi + 3 mv 4 0 4 i.e. 3 mv = mv ˆj – m v ˆi 4 4 i.e. v = v 0 2 0 ˆj – 2 v ˆi 3 0 3 0 1 3  20 2 1 m 1 5 Increase in KE = (KE) – (KE) = × m    v2 + × 2v2 – mv2 = mv2 ] f i 2 4  3  0 2 4 0 2 0 6 0 Q.4 One mole of an ideal gas is compressed slowly from 0.5 lt to 0.25 lt. During the compression, 35 ×102 J of work is done on the gas and heat is removed to keep the temperature of the gas constant at all times. Find the temperature of the gas. [Take universal gas constant R = 25/3 J mol–1 K–1, e0.7 = 2] [Ans. 600 K] [Sol. Here gas is compressed keeping temperature constant v f  W = –nRT 𝑙n ( isothermal compression) i  35 × 102 = – 25 3 × T 𝑙n (1/2)  T = 600 K ] Q.5 The internal energy of the systemat the point Ais 8000 J and the net heat absorbed by the system along the path ABC is 10 kJ as shown in the figure. Find the internal energy of the system at point C. [Ans.9 kJ ] [Sol. In the given diagram WABC = 1000 × 3 + 1 2 × 3 × 4000 = 9000 J = 9 kJ QABC = 10 kJ  UABC = 1 kJ (increase)  Internal energy at C = 8 kJ + 1 kJ = 9 kJ ] Q.6 A metallic rod of cross-sectional area 9.0 cm2 and length 0.54 m, with the surface insulated to prevent heat loss, has one end immersed in boiling water and the other in ice-water mixture. The heat conducted through the rod melts the ice at the rate of 1 gm for every 35 sec. Find the thermal conductivity of the rod. [Latent heat of fusion of ice = 80 cal/g & J = 4.2 J/cal ] [Ans. 57.6 W m–1 K–1] [Sol. Let the thermal conductivity of rod is 'k' KA (100  0) = Q L t Q t is the power which is melting ice Q  t 1 80 4.2 = 35  9 104  0.54 100 80  4.2 K = 35  K = 57.6 Wm–1K–1 ] Q.7 A force of (3 ˆi 1.5ˆj) N acts on a 5 kg body. The body is at a position of (2 ˆi  3ˆj) m and is travelling at 4 ms–1. The force acts on the body until it is at the position (ˆi  5ˆj) m. Assuming no other force does work on the body, the final speed of the body. [Ans. [Sol. Given ms–1] Mass of the body = 5 kg Force → = 3ˆi  1.5ˆj  a = 3 ˆi  5 3 ˆj 10 Now displacement s = { (ˆi  5ˆj)  (2ˆi  3ˆj) } m = (ˆi  8ˆj) m From Work Energy principle → → 1 W = = m(v2 – u2) F· 2  v = m/s ] Q.8 Consider the thermodynamic cycle shown on PV diagram. The process A  B is isobaric, B  C is isochoric and C A is a straight line process. The following internal energy and heat are given: UAB = + 400 kJ and QBC= – 500 kJ The heat flow in the process QCA is [Ans. –25 kJ] [Sol. For the given thermodynamic cycle, UAB = 400 kJ WBC = QBC = –500 kJ [ isochoric]  UCA = UCB + VBA = 100 kJ Also, 1  (3105  2 105 )  0.5 WCA = –  2  By 1st Law of thermodynamics QcA = –125 kJ + 100 kJ = – 25 kJ ]  J = – 125 kJ [ V is negative]  Q.9 A ball R = 0.1 m radius is suspended in a chamber in vaccum by a metal wire of length L = 2m. The ball is at 300 K and the walls of the chamber are at 0 K. The wire has cross-sectional radius r = 3 mm and thermal conductivity K = 34 Wm–1K–1. The emissitivities of the ball and the walls are 0.1 and 1 respectively. Find the ratio of rate of heat conducted through the wire to the rate of heat radiated by the ball. [Take the value of Stefan constant as 17 × 10–8 W m–2 K–4 ] 3 [Sol. For conduction Q t & for radiation Kr 2 = 𝑙 (1  2 ) 1 [Ans. 40 ] Q = e4R2) (4  4 ) t 1 2 Kr2  𝑙 (1  2 ) 34 32 106  300 1 Required ratio = e(4R 2 )(4  4 )  2 4  0.1(17 / 3) 108  (0.1)2  (300)4  40 ] 1 2 Q.10 Two massless strings of length 5 m hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height 0.45 m as shown in the figure. The collision between two balls is completely elastic. Immediately after the collision, the kinetic energy of ball B is 1 J. Find the velocity of ballAjust after the collision. [Ans. 1 ms–1 to the left] [Sol. From conservation of energy 1 m gh = m 1 v2 + m v2 A 2 A A 2 B B  vA = ±1  e = 1 & mA < mB, vA < 0  vA = – 1 m/s ] Q.11 Two different masses are connected to two light and inextensible strings as shown in the figure. Both masses rotate about a central fixed point with constant angular speed of 10 rad s–1on a smooth horizontal plane. Find T1 the ratio of tensions 2 in the strings. [Ans. 9 ] 8 [Sol. Drawing the FBDs for masses M 1 and M2 T1 – T2 = M1R12 T2 = M2R22 T1  T2  = M1 · R1 = 1 · 1 T2 M2 R 2 4 2 T1 1 9  T2 = 1 + 8 = 8 ] Q.12 A gas having molar mass 25 g is enclosed in a vessel at temperature 300 K. Four of its molecules have speeds 100, 100, 300, 500 m/s. Find the root mean square speed of these four molecules. [Ans. 300 m/s] [Sol. R.M.S. value for the four molecules is = = 100 11 9  25 4 m/s = 300 m/s Ans.] Q.13 A composite cylinder is made of two materials having thermal conductivities K1 and K2 as shown. Temperature of the two flat faces of cylinder are maintained at T1 and T2. For what ratio K1/K2 the heat current through the two materials will be same. Assume steady state and the rod is lagged (insulated from the curved surface). [Sol. We know Q  KA (T ~ T ) [Ans. K1 =3] 2 t l 1 2 since both the cylinders for the arrangement are in parallel, hence heat currents are  Q  K .R 2     1 (T ~ T )  t 1 l 1 2  Q  K [(2R)2  R 2 ] &    2 (T ~ T )  t 2  Q  l 1 2   Q  Given  t   t   1  2 K R 2 K 3R 2  1 (T1 ~ T2) = 2 (T1 ~ T2) K1  K2 l l =3 ] Q.14 A car begins from rest at time t = 0 and then accelerates along a straight track during the interval 0 < t  2s and thereafter with constant velocity as shown in the graph. Acoin is initially at rest on the floor of the car. At t = 1 s, the coin begins to slip and it stops slipping at t = 3s. Find the coefficient of static friction between the floor and the coin. [Sol. Given that graph is parabola having vertex at origin then v = kt2 at (2, 8) we have 8 = k · 4  k = 2 [Ans. 0.4]  v = 2t2  dv dt = 4t the coin slips over floor if a0 = g a 0   = g = 41 10 = 0.4 ] Q.15 Two trains start simultaneously from two towns Aand B located at a distance d from each other on the same track in opposite directions. Both trains move with constant speed v towards each other. Abee is initially sitting on the front of the train starting from town A. As the train pulls out of town A, the scared bee takes off and flies with velocity u > v along the track towards B. As soon as it encounters the other train, it turns back towards A and again scared on reaching first train. It continues flying between the trains with speed u until the end. Find the total distance travelled by the bee before it is crushed by the u two trains. [Ans. 2v d] [Sol. Here relative velocity between trains = 2v  Time of collision = d 2v for this time the bee continuously flies with speed u  Total distance covered = u · d 2v ] Q.16 A body has initial temperature (at t = 0) 52°C and temperature of surroundings is constant at 25°C. At t= 5 min, temperature of body is found to be 43°C. What will be the temperature at t = 15 min? Assume the body cools according to Newton's law of cooling. [Sol. We have d 2 d t1 = –k( –  )   dt = –k  dt 0  1  0  1 27  𝑙n  = kt  k = × 𝑙n s–1      1 5 60 18   Also, for t = 15 min = 900 sec  1  0  3 𝑙n   = k × 15 × 60 = 3 𝑙n  2 27    298 = 27   = 298 + 8 = 306 k 8   = 33°C ] Q.17 A particle moves along a circle of constant radius with radial acceleration changing with time as ar = k tn where k is constant and n > 1. The power developed by the net force on the particle varies with time as [Ans. tn–1] [Sol. Given radial acceleration ar = ktn  v2 = Rktn  v =  n 1 tn/2 dv = n Rk t  dt 2  n 1 dv n  2   Tangential force = m dt = m 2 Rk t   n 1 → → n  2  n/2 mnRk n1  Power developed = F·v  P  tn–1 ] = F × v = m 2 Rk t  · Rk t = 2 t Q.18 A particle of mass m is moving along a straight line under the influence of conservative force such that its velocity v varies with displacement x from a fixed point as v2 =  – x2 where  and  are constants. Considering that fixed point as reference zero for potential energy, find the total energy of the particle at any instant t. [Sol. We have v2 =  – x2  2v dv = –  2x dx dt dt 1 [Ans. 2 m] dv  dt = –x (conservative nature)  Total energy at any point is constant At x = 0 U2 =   K.E. = 1 mv2 = 2 1 2 m P.E. = 0 (given)  Total energy of body at any point = 1 m ] 2 Q.19 A composite heavy rope of two materials is suspended vertically from a high ceiling. The ratios of different quantities for upper to lower rope are: length Lu  1 , cross sectional area Au  2 , density du  2 . Find the ratio Ll 2 Al 1 dl 3 of maximum stress in the two ropes. [Ans. 5/6] [Sol. We have mu = ml Lu Au du Ll Al dl 4 = 6 = 2 3 (mu  ml )g    Ratio of maximum stress =  Au   mu =  1 = 5  1 = 5  m g   l  u 3 2 6  l    ( )  Al  u max (l )max 5   = 6 ] Q.20 A small ball is projected from point P on floor towards a wall as shown. It hits the wall when its velocity is horizontal. Ball reaches point P after on bounce on the floor. If the coefficient of restitution is the same for the two collisions, find its value. [Ans. 1/2] [Sol. We have 2vxvy R = g After first collision v'x = e vx v'y = vy 2 vy , T = g Distance covered before 2nd collision T d1 = v'x · 2 T = e vx · 2 evx vy = g After second collision v''x = v'x v''y = e v'y 2v''x v''y  d2 = g 2e vxe vy = g d1 + d2 = evx vy {1 2e} g R but d1 + d2 = 2 2vxvy = 2g  e{1 + 2e} = 1 ; 2e2 + e – 1 = 0 e = – 1 4 1 8 – 1 3 = 4 Rejecting –ve value e = 1/2 ]