PHYSICS-04-06- 11th (PQRS)

REVIEW TEST-1 Class : XI (PQRS) Time : 90 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 2. Use of Calculator, Log table and Mobile is not permitted. 3. Legibility and clarity in answering the question will be appreciated. 4. Put a cross ( × ) on the rough work done by you. Name : Roll No. Batch Class : XI Invigilator's Full Name For Office Use ……………………………. Total Marks Obtained………………… Part-A Part-B 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-1 Q.1 Answer following short questions: [2 × 6 = 12] → → → → → (i) Two forces and have a resultant . If (in N) = 2 ˆi  3 ˆj and (in N) = 5 ˆi  4 ˆj then find F1 F2 → F2 (in N). F3 F1 F3 → → → [Sol. F1  F2  F3 → → → ˆ 7 ˆj F2  F3  F1 = ( 3 i  ) N Ans.] (ii) Two forces of magnitude 1N and 13 N have resultant of magnitude 6 N, then what is the angle between two forces. [Sol. R2 = A2 + B2 + 2AB cos  6 5 2 = 12 + 132 + 2(1) (13) cos  180 = 170 + 26 cos  10 cos = 26  = cos–1 5 13 Ans.] → → (iii)  2 ˆi  3ˆj  kˆ &  4 ˆi  2ˆj  2kˆ . Find a vector parallel to → → → A but has magnitude twice that of B . [Sol. B = = 2 Aˆ = ˆ → required vector = 2A B = or 2ˆi  3ˆj  6kˆ  Ans. ] (iv) Kinetic energy of a particle moving along elliptical trajectory is given by K = s2 where s is the distance travelled by the particle. Determine dimensions of . [Sol. K = s2 (M L2T2 ) [] = (L2 ) [] = M1 L0 T–2 (M T –2) Ans. ] (v) Given momentum P of a particle is varying with time t as P = a t + t is time, then calculate force on particle as a function of time 't'. where a, b are constant and a  b  [Sol. F = dt =  2 t3 / 2  N Ans. ] (vi) Work done by friction in a certain mechanics problem involving masses m1 & m2 in time t is (m  m )g2t2m 2 1 W = m1 2(m1  m2 ) 1 , here  is coefficient of friction which is a dimensionless constant, g is acceleration due to gravity. Verify answer' s dimensional correctness/incorrectness. [Sol. LHS [w] = M L2 T–2 M2 (LT2 )2 T2 2 –2 RHS M = M L T equation is dimensionally (correct) Ans.] Q.2 The dimensional formula for viscosity of fluids is, [4] =M1L–1T–1 Find how many poise (CGS unit of viscosity) is equal to 1 poiseuille (SI unit of viscosity). [Sol.  = M1 L–1 T–1 1 CGS units = g cm–1 s–1 1 SI units = kg m–1 s–1 = 1000 g (100 cm)–1 s–1 = 10 g cm–1 s–1 Thus, 1 Poiseuilli = 10 poise Ans.] Q.3 Two vectors → A and → B are at an angle of 23°. If →  3ˆi  4 ˆj12 kˆ and → is recorded as 2 ˆi  ˆj  ( ) kˆ where coefficient of kˆ is missing, calculate this coefficient. Given cos23° = A. B [Sol. cos  = → → 12 13 . [4] A B → ˆ ˆ ˆ A  3i  4 j12 k → = ˆ ˆ ˆ B 2 i  j  ( ) k angle between vectors is 23° cos 23 12  6  4 12x 12  13 13 x2  5 = 13 100 + 144 x2 + 240 x = 144 x2 + 144 × 5 620 31 x = 240 = 12 Ans.] Q.4 Three forces of 3N, 2N and 1N act on a particle as shown in the figure. Calculate the (a) net force along the x-axis. (b) net force along the y-axis. (c) single additional force required to keep the body in equilibrium. [4] → [Sol.  3ˆi → ˆ ˆ F2  2[cos 60(i)  sin 60 j] → ˆ ˆ F3  1[cos 60(i)  sin 60 j] → 3 ˆ 3 ˆ FN  2 i  2 j (a) F 3 = N (b) F = 3 (c) → →   0 → [  force required for equilibrium ] x 2 y 2 → → FN F F F  FN →   3 ˆi  ˆj 2 2 N Ans. ] Q.5 Given that x = 120 – 15t – 6t2 + t3 (t > 0), find the time when the velocity is zero. Find the displacement at this instant. [4] [Sol. x = 120 – 15t – 6t2 + t3 v = dx = – 15 – 12 t + 3 t2 = 0 dt solving , we get t = 5, –1 sec. As t > 0 , t = 5 sec. Ans. putting t = 5 in the expression of x, we get x = 20 m Ans. ] Q.6 Answer the following short Questions: [3 × 4 = 12] (i) In the diagram below, the vectors u and v are at right angles to each other. The length of v is d. The horizontal and vertical components of u are a and b respectively. Find the vertical component of v in terms of a, b and d. b [Sol. tan  = a e = d sin (90 – ) = d cos  da e = ] (ii) In the given regular hexagon AB= p and of p and q . BC  q then find CD in terms [Sol. AD  2 BC  2q AB  p BC  q AB  BC  CD  AD p  q  CD  2q CD  q  p Ans. ] (iii) Value of g = 9.8 ms–2 in SI, if unit of length is doubled and unit of time is halved then find numerical value of g in new units. [Sol. g = 9.8 m/s2 x[2m] g' =  1  = 8x m/s2  2 S equating we get 8x = 9.8 4.9 x = 4 = 1.225 m Ans.] (iv) Acceleration of particle moving in straight line can be written as dv dv a = dt = v dx [Sol. from graph . From the given graph find acceleration at x = 20 m. dv  90  50  40 dx 40  20 20 dv dx = 2 v ( at x = 20) = 50 m/s dv a = v dx a = 50 × 2 = 100 m/s2 Ans. ] Q.7 Three forces are acting on a body. →  2 ˆi  3ˆj and it does 8J of work, →  3 ˆi  5 ˆj and it does – 4J on body. Find the displacement of body in form x ˆi  y ˆj . [6] [Sol. → → = 8  2x + 3y = 8 (1) F1.r → → F2 .r = – 4  3x + 5y = – 4 (2) solving (1) and (2) we get y = –32 ; x = 52 r  52 ˆi  32 ˆj Ans. ] Q.8 P = nxyT V0  M g h e nx T , where n is number of moles, P is pressure, T is temperature, V is volume, M is mass, g represents acceleration due to gravity and h is height. Find dimension of x and value of y. [3 + 3] [Sol. [Mgh] = [xnT] (same dimensions because exponents are dimensionless) M L T2L [x] = K mol [x] = M L2 T–2 K–1 mol–1 (K  dimension for temperature) [Pv0] = [nxy T]  M L2 T–2 = mol (M L2 T–2 K–1 mol–1 )y M L2 T–2 = My L 2y T–2y K1–y mol–y compare  y =1 Ans. ] Q.9 A particle moves in such a way that its position vector at any time t is r = t ˆi  1 t 2ˆj  t kˆ 2 . Find as a function of time (a) the velocity, (b) the speed, (c) the acceleration, (d) the magnitude of the acceleration, (e) the magnitude of the component of acceleration along velocity (called tangential acceleration), (f) the magnitude of the component of acceleration perpendicular to velocity (called normal acceleration). [1 + 1 + 1 + 1 + 1 + 2] → ˆ t2 ˆ [Ans. (a) i  tj k , (b) ˆ , (c) j, (d) 1, (e) t / , (f) / ] [Sol. r  t i  2 → j t k →  dr ˆ ˆ ˆ (a) = i  t j k Ans. dt (b) speed → = Ans. → → (c) a  dv  ˆj dt Ans. → (d) a → = 1 Ans. (e) aT = (a.vˆ) vˆ  ˆ (ˆi  t ˆj  kˆ)  (ˆi  t ˆj  kˆ) =   j.     →  t  t(ˆi  t ˆj  kˆ) aT =   vˆ or 2  t2  2  → a T Ans. (t  2) (f) a 2 2  a 2  a 2 t2 a N  1 t2  2 aN = Ans. ] Q.10 The value of acceleration due to gravity g on the surface of a spherical planet is dependent on the mass M and radius of the planet R and G (Universal gravitational constant). (a) find a formula for g in terms of above quantities by dimensional analysis. Given: Force F between two point masses m1 & m2 at distance r is given by G m1m2 r2 (b) If the densities of the moon and the earth are related by M/E = 3/5, and if gM/gE = 1/6, what is RM/RE? [4 + 4] [Sol. (a) g  MaRbGc ; g = kMaRbGc [g]  [M]a [R]b [G]c g = k (M)a Lb [M–1L3 T–2]c M0L1T –2 = Ma–c L b+3c T–2c a = c = 1 ; b = –2 GM g = k R 2 Ans. (b) g = k G  4 R3 3 R 2  g = CGR gm  mRm ge eRe 1  3 6 5      Re  Rm  Re  5 Ans. ]   Q.11 (a) Jerry is standing 20 m away in a direction 37° N of E from plate of swiss cheese. (Treat swiss cheese as origin). Draw position vector of Jerry. (b) Tom is standing somewhere east of swiss cheese. He can finally catch jerry (& end the best ever cartoon series) if he walks in a direction 53° N of E. Draw the displacement vector of Tom. Find initial location of Tom find initial distance between Tom and Jerry. [2 + 6] [Sol. (a) 20[cos 37 ˆi  sin 37 ˆj] → ˆ 12 ˆj rJ = 16 i  Ans. Ans. (b) Unit displacement of Tom = 3 ˆi  4 ˆj position vector of Tom = 5 5 x ˆi x ˆi  y(0.6 ˆi  0.8 ˆj)  16 ˆi 12 ˆj x + 0.6 y = 16 ; 0.8 y = 12 Solving we get y = 15 ; x = 7 Initial location of Tom = 7 m east of plate Initial distance between Tom & Jerry = 15 m Ans. ]