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

REVIEW TEST-1 Class : XI 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 . (Leave the answer in 13 fraction) [4] 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 then find CD in terms of p and q . (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

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more