PHYSICS-04-06- 11th (PQRS) Space

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 Q.1 Answer following short questions: [2 × 6 = 12] (i) Two forces F1 and F2 have a resultant F3 . If F1 (in N) = 2 ˆi  3 ˆj and F3 (in N) = 5 ˆi + 4 ˆj then find F2 (in N). (ii) Two forces of magnitude 1N and 13 N have resultant of magnitude 6 N, then what is the angle between two forces. (iii) A = 2 ˆi + 3ˆj  kˆ & B = 4 ˆi + 2ˆj  2kˆ . Find a vector parallel to A but has magnitude twice that of B . (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 . (v) Given momentum P of a particle is varying with time t as P = at + t is time, then calculate force on particle as a function of time 't'. b where a, b are constant and (vi) Work done by friction in a certain mechanics problem involving masses m1 & m2 in time t is (m  m )g2t 2m 2 1 W = m1 2(m1 1 + m2 ) , here  is coefficient of friction which is a dimensionless constant, g is acceleration due to gravity. Verify answer' s dimensional correctness/incorrectness. 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). Q.3 Two vectors A and B are at an angle of 23°. If A = 3ˆi + 4 ˆj +12kˆ and B is recorded as 2 ˆi + ˆj + ( ) kˆ where coefficient of kˆ is missing, calculate this coefficient. Given cos23° = 12 . [4] 13 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. 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] 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. (ii) In the given regular hexagon AB= p and BC = q of p and q . then find CD in terms (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. (iv) Acceleration of particle moving in straight line can be written as dv dv a = dt = v dx . From the given graph find acceleration at x = 20 m. Q.7 Three forces are acting on a body. F1 = 2 ˆi + 3ˆj and it does 8J of work, F2 = 3ˆi + 5 ˆj and it does – 4J on body. Find the displacement of body in form x ˆi + y ˆj . [6] 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] Q.9 A particle moves in such a way that its position vector at any time t is r = t ˆi + 1 t2ˆj + 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] 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] 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] EXTRA PAGE