PHYSICS. A balloon is moving horizontally in air with speed of 5 m/s towards north. A car is moving with 5 m/s towards east. If a person sitting inside the car sees the balloon, the velocity of the balloon as seen from the car is (a) m/s towards NW (b) m/s towards SE (c) m/s towards NE (d) m/s towards SW Solution.(A) Let us take X-axis along east-west and Y-axis along north and south. = and = 5i To calculate the balloon s velocity as seen from the car, we have to view the situation by placing ourselves inside the car. The equation: gives an idea about how to do this. For calculation relative to can, we will impose the reversed velocity of car on the balloon velocity The net effect these gives the velocity of the balloon as seen from the car. Hence the velocity of the balloon as seen from the car is m/s towards NW.. Six particles are located at the vertices of a Hexagon of side a. They all start moving simultaneously with a constant speed v but move in such a way that the first particle is continuously headed for the second, the second for the third and the third for the fourth and so on. When will the particles meet?
(a) a/v (b) a/v (c) a/v (d) a/3v Solution. (A) As the particles are moving with equal speeds, symmetry leads us to conclude that the hexagon joining the particles A, B, C,D,E,F always remains hexagonal. The size of the hexagon diminishes tending towards the centre. The velocity at which A approaches B (the component of the velocity of A relative to B along AB) is always v/. cos. This velocity is the approach at the rate which the separation between A and B decreases with time. As this rate is constant, the time after which the separation decreases from a to zero is simply. 3. A ball is thrown with a velocity of at an angle of 45 with the horizontal. It just clears two vertical poles of height 9 cm each. The separation between the poles (take g = 9.8 ms - ) (a) 6 m (b) 8 m (c) m (d) m Solution:.(B) Let us first calculate the time t after which the ball is at the top of the poles. During this time interval:
sin Alternative method: Using equation of trajectory for y =.9 m, we should get the values of ( ) cos has roots On simplification, the equation reduces to; 4. A stone is thrown with a velocity of 9.6 m/s at an angle of 3 above horizontal from the top of a building 4.7 m high. The distance of the landing point of the stone from the building is (approx) (take g = 9.8 ms - ) (a) 5 m (b) 6 m (c) 7 m (d) 8 m Solution : (A) 3
Consider the interval from O to C Along vertical direction: For time of flight : -4.7=9.8 t+/ (9.8) t, Stone lands at C after 3 seconds. Distance travelled from O to C; the horizontal displacement= Sx cos Distance of C from the building = AC = 5.9 m ~ (5 m) 5. A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) 3 t (b) 4 t (c) t (d) t 3 4 Solution: (A) 4
dv P. dt Power P F. v m v vdv dt m v P or t C when t, v C= m or v Pt dx Pt or m dt m P dx t dt or x t m 3 6. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as a k v where k is a constant, k. At the initial moment the velocity of particle is v What time will it take to cover that distance? (a) v t= k (b) v t= k (c) v t= k (d) 3 v t= k Solution: (A) dv dt dv dx k v k v dx dt dv v k v vdv k. dx dx v x 3 / 3 / v kx or Distance x v o and 3 3k dv dv k vdt k dt v or v kt or - v v or t = k v kt v t 7. Two projectiles, one fired from the surface of the earth with speed 5 m/s and the other fired from the surface of a planet with initial speed 3 m/s, trace identical trajectories. Neglecting air resistance, the value of acceleration due to gravity on the planet will be if g = m/s on earth (a) 5.9m/s (b) 3.6 ms/s (c)6.3 m/s (d) 8.5 m/s 5
Solution: (B) y gx xtan u cos Trajectory is identical so g g u u g 5 9 8 g m / s 3.6 m / s 5 8. Ratio of minimum kinetic energies of two projectiles of same mass is 4 :. The ratio of the maximum height attained by them is also 4 :. The ratio of their ranges would be (a) 6 : (b) 4 : (c) 8 : (d) : Solution : (B) Use the formula for range and compare. 9. If the time period of a drop of liquid of density d, radius r, vibrating under / surface tension s is given by the formula t ( d a r b s c ) and if a =, c = -, then b is (a) (b) (c) 3 (d) 4 Solution:. (C) T = (M L -3 ) a/ L b/ (ML T - ) c/ M L T = M a/ + c/ L -3a/+b/ T -c 3a b - 3 a + b = b = 3 a b = 3 b = 3 6
. A physical quantity P is given by P = maximum percentage error in P is A C 3 4 B D 3. The quantity which brings in the (a) A (b) B (c) C (d) D Solution : (C) Quantity C has maximum power. So it brings maximum error in P.. For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is (a)3 o (b)45 o (c)6 o (d)9 o Solution : (A) Maximum range up the inclined plane ( Rmax ) up u g ( sin) Maximum range down the inclined plane ( Rmax ) down u g ( sin) and according to problem : u u 3 g( sin) g( sin) By solving = 3 o. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first (a) The faster one (b) Depends on their mass (c) The slower one (d) Both will reach simultaneously Solution :.(D) 3. A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball fall (a) Outside the car (b) In the car ahead of the person (c) In the car to the side of the person (d) Exactly in the hand which threw it up Solution : (d) Because the horizontal component of velocity are same for both car and ball so they cover equal horizontal distances in given time interval. 7
4. A mass M is suspended by a rope from a rigid support at P as shown in the figure. Another rope is tied at the end Q, and it is pulled horizontally with a force F. If the rope PQ makes angle with the vertical then the tension in the string P PQ is (a) (b) (c) (d) F sin F / sin F cos F / cos Solution: (b) From the figure v Q M F For horizontal equilibrium T sin F T F sin T sin T T cos mg F 5. A cricket ball of mass 5 gm is moving with a velocity of m/s and is hit by a bat so that the ball is turned back with a velocity of m/s. The force of blow acts for.s on the ball. The average force exerted by the bat on the ball is Solution : (a) (a) 48 N (b) 6 N (c) 5 N (d) 4 N v m / s and v m / s [because direction is reversed] m 5 gm. 5kg, t. sec Force exerted by the bat on the ball Newton m[ v v].5[ ( )] F = 48 t. 6. You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface (a) By jumping (c) By rolling your body on the surface (b) By splitting or sneezing (d) By running on the plane Solution : (b) By doing so we can get push in backward direction in accordance with Newton s third law of motion. 8
7. Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter mass will be (a) 5 N (b) 4 N (c) N (d) None of the above Solution : (c) Let m kg, m 4kg and F 5N (given) 6 Force on the lighter mass = m m F m 4 5 6 4 N 8. Three masses of 5 kg. kg and 5 kg are suspended vertically as shown in the fig. If the string attached to the support breaks and the system falls freely, what will be the tension in the string between kg and 5 kg masses. Take g ms. It is assumed that the string remains tight during the motion 5N 6 kg 4 kg 5kg kg (a) 3 N (b) 5 N (c) 5 N (d) Zero Solution : (d) In the condition of free fall, tension becomes zero. 5kg 9. A sphere is accelerated upwards with the help of a cord whose breaking strength is five times its weight. The maximum acceleration with which the sphere can move up without cord breaking is (a) 4g (b) 3g (c) g (d) g Solution : (a) Tension in the cord = m(g + a) and breaking strength = 5 mg For critical condition mg a5mg a 4g This is the maximum acceleration with which the sphere can move up with cord breaking. 9
. Two masses m and m are attached to a string which passes over a frictionless smooth pulley. When m kg, m 6 kg, the acceleration of masses is Solution : (c) (a) m / s (b) 5m / s (c).5 m / s (d) m / s m m 6 a g m m 6.5 m / s kg m m 6 kg. The acceleration of block B in the figure will be m g (a) ( 4m m ) m g (b) (4m m ) (c) ( m m g 4m m g (d) ( m m ) ) Solution : (a) When the block m moves downward with acceleration a, the acceleration of mass m will be a because it covers double distance in the same time in comparison to m. Let T is the tension in the string. By drawing the free body diagram of A and B T m a i m g T m a ii by solving (i) and (ii) a m g 4m m. A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a. What will be the angle of inclination with vertical (a) tan ( a / g) (b) tan ( g / a ) (c) cos ( a / g ) (d) cos ( g / a) Solution : (a) From the figure a tan g tan a / g A m a m m (a) m B T g m T m g a m a
Velocity (m/sec) Velocity (m/sec) CLASS XI 3. The correct statement from the following is (a) A body having zero velocity will not necessarily have zero acceleration (b) A body having zero velocity will necessarily have zero acceleration (c) A body having uniform speed can have only uniform acceleration (d) A body having non-uniform velocity will have zero acceleration Solution : A When body is thrown upward and at the highest point velocity becomes zero but acceleration remains g. 4. A particle moves along a straight line such that its displacement at any time t is given by t t meter. The displacement when the acceleration becomes zero is (a) meter (b) meter (c)3 meter (d) meter Solution : A. Differentiate displacement twice for getting time when acceleration becomes zero and put in displacement expression 5. A rocket is projected vertically upwards, whose velocity-time graph is shown in fig. The maximum height reached by the rocket is (a) km (b) km (c) km (d) 6 km Solution : D. Area under v-t graph gives displacement. A C B 4 6 8 4 Time (sec) 6. A lift is going up. The variation in the speed of the lift is as given in the graph. What is height to which the lift takes the passenger (a) 3.6 m (c) 36. m (b) 8.8 m 3.6 (d) Cannot be calculated from the above graph Time (sec) Solution : C. Area under v t graph gives displacement. Here displacement of the lift is equal to height. 7. The potential energy of a particle varies with distance x from a fixed origin as A x U, where A and B are dimensional constants then dimensional formula x B for AB is (a)ml 7/ T / (b) ML T (c) 9 / M L T (d) ML 3/ 3 T
Solution : B 8. From the top of a tower, a particle is thrown vertically down wards with a velocity of m/s. The ratio of the distances covered by it in the 3 rd and nd seconds of the motion is (Take g = m/s ) (a) 5: 7 (b) 7 : 5 (c) 3 : 6 (d) 6 : 3 Solution : B. Use second equation of motion. 9. A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of m each will then be Solution : C (a) All equal, being equal to lgsecond (b)in the ratio of square roots of the integers,,, (c)in the ratio of the difference in the square roots of the integersi.e., ( ), ( ), (d)in the ratio of the reciprocal of the square roots of the integers i.e.,,, 3. The equation of motion of a projectile is y x x. Given that g = ms, what is the range of the projectile (a).4 m (b)6 m (c) 3.6 m (d)36. m Solution : B. Compare with trajectory equation of parabolic motion