CBSE - MCQ Question Banks (के. मा. शि. बो . -प्रश्नमाला )

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Q. 301  Explain RADAR.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

RADAR stands for radio detection and ranging. Radar is a device to detect the position of aero plane in the sky. Radar works on the principle of reflection of waves called echo. Radio waves are transmitted in all direction and when they hit the target or aero plane, the waves are reflected back. The reflected waves are received by antenna of receiver.  Time gap between the transmission and reception of waves is noted and then by using the formula S = V(T/2) distance of aero plane can be measured.                                       

  


Q. 302 What is the technique used for measuring the large time intervals.  
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

Very large intervals of time are measured by the method of carbon dating or radioactive dating. Carbon dating is used to measure the age of fossils and uranium dating is used to find the age of rock.The age can be calculated by noticing the ratio of number of radioactive atoms, which are decayed to the no. of atoms undecayed.


Q. 303 What type of method is most suitable to measure the time
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

The phenomenon which repeats itself after fixed interval of time is used to measure the time. E.g. the rotation of earth about its own axis. Heartbeat can be used to measure the time but above all the cesium atom clock is most accurate.


Q. 304 What are errors? Explain two types of errors?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

The difference in the true and observed value of any physical quantity is called errors.e.g the velocity of sound in air is 332 m/s. After experiment the value of velocity is found to be 331.3m/s so the difference in the two is called error. The two types of errors are random error and gross error. Random error occurs by chance, its cause is not known. It can be minimized by taking large number of readings and then mean of them. Gross error occurs due to the carelessness of the observer.


Q. 305 Convert 25 Joule into erg.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

By using dimensional formulae we can convert 25 Joule into erg. Joule is the unit of energy of dimensional formulae ML2T-2

                   


Q. 306 Derive the formulae for velocity (v) of water waves that may depend upon their wavelength λ, density of water ∂ and acceleration due to gravity g.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION



Q. 307 If x = a + bt + ct2  where x is in metres and t in seconds, write the units of a, b, c. 
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

   a + bt + ct2 should have unit of distance i.e. metre.    

                   Unit of a = m

                   Unit of bt = m

                   Unit of b= m/t= m/s

                   Unit of ct2 = m

                   Unit of c = m/s2


Q. 308 What do you mean by significant figures? Discuss their rule also.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

Right Answer is:

SOLUTION

 The significant figures are those figures, which are considered in the measurement. The following are the rule to measure the significant figures:  

1: All non zero digits are significant e.g. the digit 3476, there are 4 significant figures.

2: All the zeros between non zero digits are significant e.g. 14005, there are 5 significant figures.                                                 

3: All zeros on the right of the last non-zero digit in the decimal part are significant e.g. 0.00800 has 3 significant figures.                                            

4: All zeros on the right of non-zero digit are NOT significant e.g. 378000 has 3 significant figures.                                           

5: In a number less than one, all zeros to the right of decimal point and to the left of non-zero digit are not significant. E.g. 0.0084 has 2 significant.


Q. 309  A body A moves with a uniform acceleration ‘a’ and zero initial velocity. Another body B starting from the same point moves in the same direction with a constant velocity v. The two bodies will meet after a time


A. 3a/2v.

B. v/2a.

C. v/3a.

D. 2v/a.

Right Answer is: D

SOLUTION

If the two bodies meet after time t, then the 
Distance traveled by body A in time t, SA = 0 + (1/2)at2
Distance traveled by body B in time t, SB = vt Now, SA = SB (1/2) at2 = vt So, t = 2v/a  


Q. 310 The velocity of a particle is v=v0 + bt + ct2 . If its position is x = 0 at t = 0, then its displacement after unit time, t=1 is


A. v0 - b/2 + c

B. v0 + b/2 + 3c

C. v0 + b/2 + c/3

D. v0 + b + c

Right Answer is: C

SOLUTION



Q. 311 A police jeep is chasing with velocity of 45 km/hr and a thief in another jeep is moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is


A. 100 m/s

B. 150 m/s

C. 250 m/s

D. 450 m/s

Right Answer is: B

SOLUTION


Q. 312 A particle under the action of a constant force moves from rest up to 20 seconds. If the distance covered in first 10 seconds is S1 and that covered in next 10 seconds is S2 then


A. S2 = S1

B. S2 = 4S1

C. S2 = 2S1

D.  S2 = 3S1

Right Answer is: D

SOLUTION


Distance covered in first 10 seconds, 
Distance covered in 20 seconds, 
So, the distance covered in second 10 seconds, S2 = 200a - 50a = 150a
So, S2 = 3S1


Q. 313 Two masses m1 and m2 are dropped from heights h1 and h2. The ratio of time taken by them to fall on the Earth’s surface is


A. (h1/h2)1/2

B. (m1/m2)1/2

C. (h2/h1)1/2

D. (m2/m1)1/2

Right Answer is: A

SOLUTION


Q. 314 A car travels the first half of a distance between two places at a speed of 30 km/hr and the second half of the distance at 50 km/hr. The average speed of the car for the whole journey is


A. 42.5 km/hr.

B. 40.0 km/hr.

C. 38.2 km/hr.

D.  37.5 km/hr.

Right Answer is: D

SOLUTION

Time taken to cover first half of the distance =   Time taken to cover next half of the distance =  Total time taken to cover the whole distance =  So, average speed = 


Q. 315 A balloon of total mass 1000 kg floats motion less over the earth’s surface .If 100 kg of sand ballast are thrown over board, the balloon starts to rise with an acceleration of


A. 0.5 m/s2.

B. 1.09 m/s2.

C. 3.8 m/s2.

D. 4.9 m/s2.

Right Answer is: B

SOLUTION

Remaining mass = (1000 - 100)m = 900 m
Force = mg = 100g (as removal of 100 kg mass gives force to the balloon)


Q. 316 The velocity-time curve for a body projected vertically upwards is


A. parabola.

B. hyperbola.

C. straight line.

D. ellipse.

Right Answer is: C

SOLUTION

For upward motion, v = u – gt
So, velocity-time (v-t) graph is a straight line with slope (-g) and an intercept u on velocity axis.


Q. 317 A car has to cover a total distance 60 km. For half of the total time of its journey, its speed is 80 km/h and for rest half of the time its speed becomes 40 km/h. Then, the average speed of the car will be:


A. 180 km/h

B. 120 km/h

C. 80 km/h

D. 60 km/h

Right Answer is: D

SOLUTION


Q. 318 The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s2 .The distance covered by the body in the 5th second of its motion is


A. 15 m

B. 25 m

C. 35 m

D. 50 m

Right Answer is: B

SOLUTION


Q. 319 A person throws balls into air after every second .The next ball is thrown when the velocity of the first ball is zero. The height that the ball attains above his hand is


A. 5 m

B. 6 m

C. 8 m

D. 10 m

Right Answer is: A

SOLUTION


Q. 320  A particle at rest starts moving in a horizontal straight line with uniform acceleration. The ratio of the distance covered during the fourth and the third second is


A. 26/9.

B. 4/3.

C. 7/5.

D.  2.

Right Answer is: C

SOLUTION


Q. 321 The velocity-time graph of a body is shown in the figure. It implies that at point B
 


A. the force is zero.

B. there is a force, which opposes motion.

C. there is a force towards motion.

D.  there is only gravitational motion.

Right Answer is: B

SOLUTION

 Body is being retarded at B, so force opposes the motion.


Q. 322  An elevator in which a man is standing is moving upwards with a speed of 10 m/s . If the man drops a coin from height of 2.45 m, it reaches the floor of the elevator after a time (g = 9.8 m/s2)


A. 2.

B. (2)1/2

C. 1/2.

D. 1/(2)1/2

Right Answer is: D

SOLUTION

Since the elevator is moving with uniform speed so the acceleration of the coin is g. So, the time taken by the coin to reach the floor is, 


Q. 323  Two particles A and B get 4 m closer each second while traveling in opposite direction. They get 0.4 m closer every second while traveling in same direction. The speeds of A and B are respectively


A. 2.2 m/s and 0.4 m/s.

B. 2.2 m/s and 1.8 m/s.

C. 4 m/s and 0.4 m/s.

D.  5 m/s and 0.8 m/s.

Right Answer is: B

SOLUTION

Let the speeds of the two particles be u and v. 
Then u +v=4 and u-v=0.4
From the above two equations we get u=2.2 m/s and v=1.8 m/s


Q. 324  A particle is dropped from the top of the tower and at the same time another one is thrown horizontally from the same point. Their relative velocity is


A. vertical and constant.

B. vertical and variable.

C. horizontal and constant.

D.  horizontal and variable.

Right Answer is: C

SOLUTION

Since, the two particles cover equal vertical distances, their vertical velocities are equal and so there is no relative motion in vertical direction. The horizontal velocity of the particle thrown in vertical direction is zero but the horizontal velocity of the other is not zero. So, there is only a relative velocity in horizontal direction, which remains constant in magnitude.


Q. 325 A train of length 200 m traveling at 30 m/sec overtakes another train of length 300 m traveling at 20 m/sec. The time taken by the first train to pass the second is


A. 60 seconds.

B. 50 seconds.

C. 40 seconds.

D. 30 seconds.

Right Answer is: B

SOLUTION

Relative speed of first train with respect to second train = (30 - 20) = 10 m/s The additional distance traveled by the first train to pass the second train = (300 + 200) = 500.


Q. 326 The acceleration of a particle at time t is given by . Its displacement at time t is


A. .

B. .

C. .

D. .

Right Answer is: A

SOLUTION


Q. 327  A car runs at constant speed on a circular track of radius 100 m taking 62.8 sec on each lap. The average speed and average velocity on each complete lap is


A. velocity 10 m/s, speed 10 m/s.

B. velocity zero, speed zero.

C. velocity zero, speed 10 m/s.

D.  velocity 10 m/s, speed zero .

Right Answer is: C

SOLUTION

Distance traveled in one lap = 2  r = 2  3.14  100
Time for one lap = 62.8 sec
So, average speed = 
Displacement = 0 and so velocity = 0


Q. 328 ABCD is a square of side a. A body starts from A and reaches D through indicated path. The ratio of displacement to distance traveled is
                             


A. 2a.

B. 2.

C. a/2.

D. 1/2.

Right Answer is: D

SOLUTION

Distance traveled =2a, Displacement=a2.


Q. 329 The ratio of displacement to distance is


A. always = 1.

B. always  1.

C. may be  1.

D. always 1.

Right Answer is: D

SOLUTION

Displacement is the shortest distance travelled by a body between the initial point (say A) and the final point (say B). For example 


Q. 330  If an object covers one complete circular track of radius ‘r’, then displacement is


A. zero.

B. 2r.

C. /2r.

D. /r.

Right Answer is: A

SOLUTION

Displacement is the shortest distance travelled by a body between the initial point and the final point. Here, net displacement is zero


Q. 331 If displacement is constant between two points but distance keeps on changing, then


A. velocity is zero.

B. time taken is zero.

C. velocity increases.

D. velocity decreases.

Right Answer is: A

SOLUTION

Displacement is the shortest distance travelled by a body between the initial point (say A) and the final point (say B) and it remains constant always between A and B. However, the distance can change.

Velocity = dx/dt = zero.


Q. 332  The slope of velocity-time graph represents


A. acceleration.

B. distance.

C. displacement.

D. average velocity.

Right Answer is: A

SOLUTION



Q. 333 A man is going up with speed of 50 m/s. After 10 second a second man going same direction with a speed of 70 m/s. Assuming that they travel at constant speeds, the distance covered by first man when second man catches the first, is


A. 200 m.

B. 300 m.

C. 500 m.

D. 1250 m.

Right Answer is: D

SOLUTION

Let first man catch the second man after t second  
70t=50(t+10)
or,    t =25s
Distance covered by first man in 25 s =50x25=1250 m  


Q. 334 Two parallel rail track run north-south. Train A moves due north with a speed of 50 km/h and train B moves due south with a speed of 80 km/h. The velocity of B with respect to A in ms-1, is


A. 36.11 m/s.

B. 34.67 m/s.

C. 30 m / s.

D. 25.45 m/s.

Right Answer is: A

SOLUTION

VBA = VB+ VA  as both are coming opposite directions.    


Q. 335 The velocity-time graph for a uniform motion is


A. parallel to time axis.

B. parallel to velocity axis.

C. diagonal to time axis.

D. diagonal to velocity axis.

Right Answer is: A

SOLUTION

In the uniform motion, the body moves with constant velocity.
Hence, the rate of change of velocity with time is zero i.e. slope of velocity-time graph is zero which corresponds to a straight line parallel to the time axis.


Q. 336 A particle is moving along X-axis. At time t1=2s, its position is x1 = 3m and t2 = 7, its position is x2 = 18m. 
The average velocity of the particle is


A. 3 ms-1.

B. 4 ms-1.

C. 5 ms-1.

D. 6 ms-1.

Right Answer is: A

SOLUTION


Q. 337 The velocity of a body depends on time according to the equation v = 20 + 3 t2. This shows that the body is undergoing


A.  non-uniform acceleration.

B.  uniform acceleration.

C.  uniform retardation.

D.  zero acceleration.

Right Answer is: A

SOLUTION

Acceleration = dv/dt = 3 x 2t = 6t
As it depends upon time t; hence, acceleration is non- uniform.


Q. 338 A body covered a distance of l metre along a semicircular path. The ratio of distance to displacement of the body is


A. .

B. .

C.

D. 1 : 2.

Right Answer is: C

SOLUTION

Let ‘r’ be the radius of the semicircular path.
Circumference = 2r
‘l’ is the semicircular distance travelled.

Diameter is shortest distance travelled between two points in a semicircle.
Magnitude of displacement = Diameter = (2l)/
Distance: Displacement = .


Q. 339 A stone is dropped from a height of 45 m. The distance travelled by it during the last one second of its motion is (g= 10 m/s2)


A. 35 m.

B. 25 m.

C. 12.5 m.

D. 10 m.

Right Answer is: B

SOLUTION

From the 2nd equation of kinematics
 s = ut + ½ at2
 for a vertical downward motion of the body, we have
 45 = 0 + ½ x 10 x t2 or t = 3s
 From
     D3 = u + ½ a (2x 3-1) = 0 + ½ x 10 x 5 = 25 m.


Q. 340 A body starting from rest has an acceleration of 20 m/s2. The distance travelled by it in 8th second is


A. 90 m.

B. 100 m.

C. 115 m.

D. 150 m.

Right Answer is: D

SOLUTION

Given, u =, a = 20 m/s2, D8 =?
 From
     Dn = u + ½ a (2n-1)
 D8 = 0 + ½ x 20 x [(2x8)-1]
         = 10 x 15 = 150m.


Q. 341 A particle under the action of a constant force moves from rest upto 20 seconds. If the distance covered in first 10 seconds is ‘s1’ and that covered in next 10 seconds is ‘s2’, then


A. s1= s2.

B. s2 = 2s1.

C. s2 = 3s1.

D. s2 = 4s1.

Right Answer is: C

SOLUTION


Q. 342 Delhi is at a distance of 150 km from ambala. A sets out from ambala at speed of 20 km/h B sets out at the same time from Delhi at a speed 10 km/h . They will meet each other after


A. 4 hour.

B. 5 hour.

C. 6 hour.

D. 8 hour.

Right Answer is: B

SOLUTION

Distance covered by A = (20 x t) km 
Distance covered by B = (10 x t) km
Total distance = (20 t + 10 t) =150 km
t = 5 hour


Q. 343 The position coordinates of a moving body are given by x=4+10t+15t2. The acceleration is  (Where x is in meters and t is in seconds.)


A. 45 m/s2.

B. 40 m/s2.

C. 35 m/s2.

D. 30 m/s2.

Right Answer is: D

SOLUTION

Velocity, v  = (dx/dt) = 10+30 t,

acceleration, a = dv/dt = d2x / dt2= 30 m/s2.


Q. 344  The graph representing negative acceleration is


A.

B.

C.

D.

Right Answer is: C

SOLUTION


Slope (dv/dt) = -ve value
But dv/dt = acceleration.


Q. 345 A body is dropped from the top of a tower, which falls through 40m during the last two seconds of its fall. The height of tower is (g= 10 m/s2)


A. 20 m.

B. 45 m.

C. 80 m.

D. 150 m.

Right Answer is: B

SOLUTION


Q. 346 Two cars A and B are moving with same speed of 45 km/h along the same direction. If a third car C coming from opposite direction with a speed of 36 km/h meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)


A. 7.25.

B. 6.75.

C. 5.55.

D. 4.75.

Right Answer is: B

SOLUTION

Relative velocity of car C w.r.t. car A or B
 = 45 – (-36) = 81 km/h
 Distance between car A and car B = 81 x (5/60)
                                                         =  6.75 km.


Q. 347 The distance of a particle starting from rest (at t=0) is given by s = 6t2-t3. The time in seconds at which the particle will attain zero velocity again, is


A. 2.

B. 3.

C. 4.

D. 8.

Right Answer is: C

SOLUTION

Given, s = 6t2-t3
velocity,
    v = ds/dt = 12t – 3t2
or v = 12t – 3t2.......... (1)
If velocity is zero, then equ(1) becomes
    0 = 12t – 3t2
Either t =0 s or, t= 4.0s Since time can never be zero; so, t = 4.0 s.


Q. 348 The average speed of a runner making one lap around a 200m circular tarck in 25 seconds is


A. 4 m/s.

B. 8m/s.

C. 16 m/s.

D. 4 m/s.

Right Answer is: B

SOLUTION

Average speed = (total distance)/(total time) = 200/25=8 m/s


Q. 349 A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 seconds .Its average velocity is


A. zero.

B. 2 m/s.

C. 4 m/s.

D. 8 m/s.

Right Answer is: B

SOLUTION

In a circular, the shortest distance travelled in a half revolution is diameter.


Q. 350 A body is dropped from the top of the tower and reaches the ground in 3 seconds. Then, the height of the tower is


A. 98 m.

B. 66 m.

C. 44.1 m.

D. 39.2 m.

Right Answer is: C

SOLUTION

 For a falling body,
 h = ( 1/2 )gt2  h = ( 1/2 ) x 9.8 x 9 = 44.1 m


Q. 351  A body starts from rest and move with constant acceleration. The ratio of distance covered by the body in nth second to that covered in n seconds is


A. 1:n.

B. .

C. .

D. .

Right Answer is: C

SOLUTION


Q. 352 An object is projected upward with a velocity of 80 ms-1. The time after which it will strike the ground is (g= 10 m/s2)


A. 16 s.

B. 17 s.

C. 18 s.

D. 19 s.

Right Answer is: A

SOLUTION

Here s = 0, v= 80 ms-1, a=10 m/s2, t=?
 From the 2nd equation of kinematics
        s = ut + ½ at2
 we have,
       0 = 80t + ½ x (-10) x t2
 or,   t = 16 s.


Q. 353 A heavy and light body of the same size are dropped from the top of a tower in air from the same height. The one reaching the ground first will be


A. the lighter body.

B. the heavier body.

C. both simultaneously at the same point.

D. both simultaneously at the different points.

Right Answer is: B

SOLUTION

Due to more mass, the heavier body will experience more gravitation pull. So, it will be able to overcome the opposing force of air friction easily than the lighter body.


Q. 354 The velocity of a particle is given by, Its acceleration is


A. 4 ms-2.

B. 2 ms-2.

C. -1 ms-2.

D. -2 ms-2.

Right Answer is: D

SOLUTION


Q. 355 A ball is thrown upward. At the highest point the ball can


A. have zero height.

B. have acceleration.

C. be colourless.

D. be dimensionless.

Right Answer is: B

SOLUTION

In motion under gravity, at highest point of its motion velocity is zero but acceleration ‘a’ = ‘g’ in downward direction.


Q. 356 The relative velocity of two bodies having equal velocities is


A. 2v.

B. v.

C. v/2.

D. 0.

Right Answer is: D

SOLUTION


Q. 357 A rifle bullet loses (1/20)th  of its velocity in passing through a plank. The minimum number of planks required to stop the bullet is 


A. 5.

B. 7.

C. 9.

D. 11.

Right Answer is: D

SOLUTION

Let s = thickness of the plank
      a = uniform retardation of the plank          u = initial velocity
The velocity of the bullet after passing through a plank is 
 

So, the number of planks required to stop the bullet completely = 1 +10 = 11.


Q. 358 If a particle is thrown vertically upwards then its velocity so that it covers same distance in 5th and 6th seconds would be


A. 19 m/s.

B. 29 m/s.

C. 39 m/s.

D. 49 m/s.

Right Answer is: D

SOLUTION

The distance covered by the particle will be same in the 5th and 6th second if it reaches at the highest position in the 5th seconds. Using t = 5s, a = -9.8 m/s2, v=0 and u=?
 Now, from the 1st equation of kinematics
      v = u + at
  


Q. 359 A scooter is travelling at a speed of 72 km/h. The brakes are applied so as to produce a uniform retardation of 4 m/s2. The distance scooter covers before stopping is


A. 50 m.

B. 55 m.

C. 60 m.

D. 65 m.

Right Answer is: A

SOLUTION


Q. 360 In one dimensional motion, the average velocity is equal to instantaneous velocity when the body is moving with


A.  non-uniform motion.

B.  uniform motion.

C.  variable speed.

D.  variable speed.

Right Answer is: B

SOLUTION

During uniform motion, a body covers or travels equal displacements in equal intervals of time. Hence, the average and instantaneous velocities have same value as the uniform velocity.


Q. 361 A particle starts from rest. Its acceleration (a) versus time (t) is shown in figure. The maximum speed of the particle will be


A. 45 ms-1.

B. 55 ms-1.

C. 90 ms-1.

D. 110 ms-1.

Right Answer is: C

SOLUTION

  Maximum velocity
 = area under acceleration time graph
 = ½ x 12 x 15
 = 90 ms-1.


Q. 362 The distance ‘x’ covered by a body is directly proportional to the square of time. Therefore, the body travels with


A. uniform acceleration.

B. uniform velocity.

C. non-uniform acceleration.

D. zero velocity.

Right Answer is: A

SOLUTION


Q. 363 A particle shows distance-time graph as shown in the figure. The maximum instantaneous velocity of the particle is around the point is


A. D.

B.
A.

C. B.

D. C.

Right Answer is: D

SOLUTION

 Instantaneous velocity = slope of distance – time graph at the given instant of time.
 As slope is highest at C, so instantaneous velocity
 = ds/dt = max at C.


Q. 364  A ball thrown up is caught by the thrower 8s after start. The height to which the ball has risen is (assuming = 10 m/s2)


A.  8 m.

B.  10 m.

C.  80 m.

D.  90 m.

Right Answer is: C

SOLUTION

As the time of ascent (t1) = time of descent (t2); therefore, as per question,
     t1 + t2 = 8
or, t2 + t2 = 8s
or, t2 = 4s Taking vertical downward motion of the ball from the highest point to the surface of the earth, we have
u =0; a = 10 m/s2; t= 4s, S=?
From the 2nd equation of kinematics,
     S = ut + ½ at2
or, S = 0 + ½ x 10 x 42
        = 80m.


Q. 365  Acceleration is


A.  a dimensionless quantity.

B.  time independent.

C.  a vector quantity.

D.  a scalar quantity.

Right Answer is: C

SOLUTION

Acceleration is defined as the ratio of change in velocity and the corresponding time taken by the object. So, acceleration has magnitude as well as direction like velocity.


Q. 366 In the unit of acceleration, the unit of time appears


A. once.

B. twice.

C. thrice.

D. quadrupled.

Right Answer is: B

SOLUTION

Acceleration is the rate of change of velocity. Velocity is the rate of displacement. Hence, time occurs twice in its unit, i.e., metre per second per second.


Q. 367 The distance covered by a freely falling body during the first second of its motion is


A. 4.9 m.

B. 3.9 m.

C. 2.9 m.

D. zero.

Right Answer is: A

SOLUTION

From the 2nd equation of kinematics
        s = ut + ½ at2
 acceleration, a=g = 9.8 m/s2
 Distance covered,
        s = ½ gt2 = ½ x 9.8 x (1)2 = 4.9 m.


Q. 368 A body starts from rest and slides down an inclined plane (assume to be frictionless) of length 15 m in 15 s. Its speed at the bottom of the plane is


A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Right Answer is: B

SOLUTION


 u= 0; s = 15 m; t= 15s; a=? ; v=?
 From the 2nd equation of kinematics
        s = ut + ½ at2
 or, 15 = 0 + (½ x a x 152)
 or,   a = (2/15) m/s2
 From the 1st equation of kinematics
      v = u + at
 or, v = 0 + [2/15] x 15= 2 m/s.


Q. 369 Explain following velocity time graphs.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

 

                                         

 

(1)The graph shows that body has uniform velocity throughout the motion.

(2)This graph shows that body has some initial velocity and moving with uniform acceleration.

(3)This graph shows that body is moving with some velocity and moving with retardation.


Q. 370 A jet plane is moving with a velocity of 800 km/hr. The gases are ejecting from rear of jet plane with velocity of 1600 km/hr with respect to jet. Find the velocity of gases w.r.t a person on ground.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Consider the directions as according to origin. +ve towards +X axis and negative towards –ve X axis. Jet plane and gases moves in opposite direction. Velocity of jet plane Vg = 60 km/hr with respect to +x-axis

 Vgp = Vgj – Vpj   ( Vgj=vel of gas w.r.t to jet, Vjp=vel of jet w.r.t person )

 Vgp= Vgj+  Vjp

Vgp= 800km/hr(-i)+1600km/hr(i)

     =  1600-800

    =    800km/hr

So, this will be velocity of gases w.r.t person standing on ground.


Q. 371 A particle is moving along X axis the position is given by                                                 x = K + jt2 where K = 8m and j = 4 m/s2 t is time.  Find velocity of particle at t = 0, t= 3 sec.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

 X = k + Jt2

K = 8m, j = 4 m/s2

X = 8 + 4t2

 To get velocity differenciate both sides v =   =  (8 + 4t2)

 v  4 x 2 t  = 8t

 so  = v = velocity

when t = 0, v = 0

when t = 3 sec.  v = 8 X 3

= 24 m/s.

 


Q. 372 Draw the position time graphs for two objects initially occupying different positions but having zero relative velocity.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

The position time graphs for two objects initially occupying different positions but having zero relative velocity are parallel to each other.

 
                                              


Q. 373                                                                       
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

A. Body is at rest.  

B. Body is moving with constant velocity.

C. Body is moving with constant velocity coming back to origin

D. The velocity of body is increasing

E. The body is moving with decreasing velocity.

F. Body first moves with increasing velocity and then velocity 
                    decreases.


Q. 374 Two trains A and B of length 600m each are moving on two parallel tracks with a uniform speed of 60km/hr. in same direction with A ahead of B. B accelerates with 1m/s2 to overtake
A. If after 50sec. the guard of B just brushes past the driver of A, what was the original distance between them?
Right Answer is:

SOLUTION

          uA = 60 km/hr  =  60  ( 5/18 ) m/s = ( 50/3 ) m/s

           S1 = ( 50/3 )  50  = ( 2500/3 ) m

          uB = 60 km/hr = 60  ( 5/18 ) m/s = ( 50/3 ) m/s

          a = 1 m/ s2    t = 50Sec.

          S = ut + ( 1/2 ) at 2

                = ( 50/3 )   50 +  ( 1/2 )  1  50 50

                = ( 6250/3 )  m.   

          The original distance was

                 ( 6250/3 ) - ( 2500/3 ) = ( 3750/3 ) = 1250 m

       


Q. 375 A body is moved from rest along a straight line by a machine delivering constant power. How is the distance moved by the body is related to time?
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION


Q. 376 Define relative velocity. Explain its concept in detail.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Relative velocity of an object A with respect to another object B is the comparison of their velocities. It can also be understood by time rate at which object A changes its positions w.r.t B. Consider two objects A and B moving with velocities V1 and V2  

Displacement of A at t = 0, x01

Displacement of B at t = 0, x02      

 

  

 If X1 and X 2 is the displacement of A and B after time t.

displacement of A in time t 
( X1 – X01) = V1 T             -  (1)

displacement of B in time t

 ( X2 – X02) = V2 t             -  (2)    

Subtract (1)  from (2)

( X2 – X1)  = (X02 + X10)  +(V2 – V1) t  

Let X2 – X1  =  X and X0 = (X02 + X01)

X = X0 + (V2 – V1) t  

Relative velocity of B w.r.t A is

VBA = V2 – V1

When the two velocities are equal V2 = V1 
V2 – V1  = 0

X = X0
The two bodies will remain at constant distance from each other. This can be shown by the graph (1)

                               

         
When two bodies move with unequal velocity the graph will be (2)

In general , we can write

Vab= Va- Vb

If they are moving with respect to a third quantity, then

Vab= Vac- Vbc


Q. 377 Derive equation of motion v = u + at, by graphical method.   
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Consider an object moving along a st. line with uniform acceleration ‘a’.
Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.
Below is the velocity-time graph of the motion which is st . line.


Let ‘u’ be the initial velocity of particle at t = 0, ‘v’ is final velocity after time ‘t .
‘‘A’ represents the velocity of body at time ,t = 0;
‘B’ represents the velocity of body at time, t = t
OA = CD = u
BC = v - u, 
OD = t.
acceleration = slope of graph 

v - u = at,
v = u + at


Q. 378 A body is dropped from the top of a tower, which falls through 40m during the last two seconds of its fall. What is the height of tower ?(g= 10 m/s2)
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION


Q. 379 Derive distance velocity relation by graphical method.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Consider an object moving along a st. line with uniform acceleration ‘a’.
Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.
‘S’ be the distance travelled by the object in time t.
Below is the velocity-time graph of the motion which is st . line.
  
 
                       
In figure,
OA = CM = u.
BM = OE = v.
BC = v-u,
OM = AC = t
Distance traversed in time interval t, is
S = Area of trapezium OABM


Q. 380 Derive the equations of motion v = u + at, s= ut + (1/2)at2, v2 – u2 = 2as by calculus.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Consider an object moving along a st. line with uniform acceleration ‘a’.
Let ‘u’ be the initial velocity of the object at time , t=0
and ‘v’ be the final velocity of the object after time ‘t’.

Acceleration,
a = (dv/dt)

or, dv = a.dt

 
(1) Velocity-time relation:
Let at instant time, t, dv be the change in velocity in time interval dt.
Integrating , within the condition of motion
(i.e. time from 0 to t and velocity from u to v ), we get,
(2) Distance-time relation :
Let at an instant t, dx be the displacement of the object in time interval dt.
Instantaneous velocity ‘v’ is given by
(3)Velocity-displacement relation:
 


Q. 381 Two balls are thrown simultaneously one vertically upwards with speed of 20m/s from the ground and other vertically downwards from height of 40m with same speed and in same line of motion. Determine the point where they will collide.
A. 1 m/s.
B. 2 m/s.
C. 3 m/s.
D. 5 m/s.

Right Answer is:

SOLUTION

Suppose two balls meet at height 'X' above ground after time t sec. from start. 
For upward motion,
u = 20m/s, g= - 9.8m/s2
From 2nd equation of kinematics,
   s = ut +  1/2at2                           
X = 20t - 1/2(9.8 t2)
or,X = 20t – 4.9 t2...........(1)  
For downward motion:           
40 - X = 20t + 4.9 t2..........(2)          
Adding equ (1) & (2) , we get,           
40t = 40, 
t = 1 sec.          
From equ(1)
   X = 20 – 4.9          
or,X = 15.1m.          
Both balls collide after 1 sec at height 15.1 above the ground.


Q. 382 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: A particle is moving in a circular path in a clockwise direction. Then angular velocity will be perpendicular to the linear velocity and angular acceleration.
Reason: The angular velocity will act in a downward direction perpendicular to the plane of rotation.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: C

SOLUTION

According to right-hand thumb rule if fingers point in the direction of rotation then thumb represents the angular velocity vector. Here, the particle is moving in the clockwise direction so angular velocity will act in a downward direction perpendicular to the rotational plane. As in circular motion centripetal acceleration acts towards centre and perpendicular to velocity vector acting in the tangential direction. Both velocity and acceleration are acting in the plane of rotation. So, angular velocity will be perpendicular to both linear velocities as well as centripetal acceleration.


Q. 383 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: When dot product and cross product of any two vectors is zero, then one of the vectors must be null vector.
Reason: A null vector has a zero magnitude.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: D

SOLUTION

Let us assume two vectors a and b and θ is the angle between them. Given, dot product of the vectors is zero, i.e a . b =| a || b |cos θ°= 0 and cross product of vectors is zero, i.e. a × b =| a || b |sin θ° = 0. In this case, if | a |and | b |are not zero, then both sinθ andcosθ must be simultaneously zero.But is not possible so one of the vectorsmust be null vector. [email protected]@[email protected]@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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[email protected][email protected]


Q. 384 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: A river is flowing towards the east at a speed of 4 ms-1. A man on the south bank of the river is capable of swimming at 8 ms-1 in the still water. He can cross this river in the shortest time if he swims by making an angle with the vertical.
Reason: He will cross the river in the shortest time when he swims along the direction of resultant velocity.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: B

SOLUTION

Suppose width of the river is w and swimmer starts swimming with velocity v s making an angle of θ° with the vertical direction.Then effective velocity of the swimmer is v s cos θ. Time taken to cross the river, t = W v s cosθ For time to be minimum, cos θ should have maximum value, i.e. θ = 0°. The swimmer should swim due north. [email protected]@[email protected]@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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[email protected][email protected]


Q. 385 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: A man holds the water hose at 45° so that he can wet the more area of the lawn.
Reason: The water stream will get the greatest range when launched at an angle of 45°.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: C

SOLUTION

In the absence of air resistance, the only force acting on the water stream is the force due to gravity, due to which the water stream follows the parabolic path. The water stream will wet the maximum area when it reaches the maximum distance or range. The range in projectile motion is given by the formula, R= v 2 sin g Given, angle θ = 45°. R max = v 2 sin2( 45 ° ) g = v 2 g [email protected]@[email protected]@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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[email protected][email protected] Thus, the water stream will reach maximum area when launched at 45°.


Q. 386 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The speedometer needle represents the radial acceleration of the body.
Reason: The radial acceleration is a measure of the rate of change in the magnitude of the velocity vector, whereas the tangential acceleration is a measure of the rate of change of the direction of the velocity vector.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: B

SOLUTION

The tangential acceleration is the measure of the rate of change in the magnitude of the velocity vector, thus, always coincides with the speed, and directed along with the tangent drawn at any point on trajectory. The normal acceleration is a measure of the rate of change of the direction of the velocity vector and is directed perpendicular to the tangential acceleration in the direction of the concavity of the curve.


Q. 387 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: When a ball is dropped from the train moving at a constant speed, for a person standing on the ground, the path of the ball is parabola however the motion appears straight line to the person on the train.
Reason: The path of the ball is a straight line as seen from the moving train because both ball and train are moving at the same speed. But, with respect to ground, the path of the ball is parabola as the vertical velocity of the ball changes with time.


A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: C

SOLUTION

The path of the ball depends upon the position of the observer. For an observer inside the train, the path of the ball is vertically downward, as the horizontal velocity of the ball and the train are the same. But, when this is observed from the ground, the path of the ball is a parabola because the vertical velocity of the ball is influenced by the acceleration due to gravity.


Q. 388 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The resultant of three vectors cannot be zero if they are coplanar.
Reason: The three unequal vectors should form a polygon to give zero resultant vector.


A. Assertion is true but reason is false.

B.

 Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

Right Answer is: B

SOLUTION

For getting a zero-resultant vector, the three vectors can be arranged in a polygon. If three vectors are coplanar then their resultant vector can be zero, that is when the resultant of two of them is equal in magnitude and opposite to the direction of the third vector.


Q. 389 When two A and B move towards each other, the distance between them decreases by 10 m/s. If both car moves in the same direction with the same speed, the distance between them increases by 6 m/s. The speeds of the two cars are


A. 8 m/s, 1 m/s.

B. 6 m/s, 4 m/s.

C. 8 m/s, 4 m/s.

D. 4 m/s, 6 m/s.

Right Answer is: A

SOLUTION

Let u and v be the speeds of the two cars
u+v=10, 
When the cars approach each other
u-v=6
2u=16 
or   u=8 m/s 
 2v=10-8=2 m/s or v= 1 m/s.  


Q. 390 A car is going up with speed of 40 m/s. After 10 second a bike going same direction with a speed of 60 m/s .Assuming that they travel at constant speeds, the distance from the turning where the bike catches the car, is


A. 10 s

B. 15 s

C. 20 s

D. 25 s

Right Answer is: C

SOLUTION

Let bike catch the car after t second  60t=40(t+10)
or,    t=20 s


Q. 391 Two train 200 m and 250 m in length are running in same direction with velocities 50 km/h and 32 km/h .The time when they will completely cross each other is


A.   40 s.

B.  50 s.

C.  90 s.

D.  100 s.

Right Answer is: C

SOLUTION

Relative velocity of one train w.r.t. second = 50-32 =18 km/h = 5 ms-1 Total distance traveled = 200 + 250 =450 m, Time taken, t = 450/5 = 90 s.


Q. 392 Two train 100 m and 50 m in length are running in opposite direction with velocities 40 km/h and 20 km/h .The time when they will completely cross each other


A. 4 s.

B. 6 s.

C. 8 s.

D. 9 s.

Right Answer is: D

SOLUTION

 Relative velocity of one train w.r.t. second = 40+20 =60 km/h = 50/3 ms-1  Total distance traveled = 100 + 50 =150 m, Time taken, t = ( 150 x 3) / 50 = 9 s.


Q. 393 A car travels along a straight line with a speed of 20 km/h for first half time and second half time with speed of 10 km/h. The means speed of the car is


A. 15 km/h

B. 12 km/h

C. 10 km/h

D. 5 km/h

Right Answer is: A

SOLUTION

Let  t1= t2 = t/2

The mean speed of car vm = =15 km/h


Q. 394 A train of 300 m length is going towards east direction with a speed of 10 m/s. A parrot fly a speed of 20 m/s towards west direction parallel to the road .The time taken by the parrot to cross the man is


A. 30 s

B. 20 s

C. 15 s

D. 10 s

Right Answer is: D

SOLUTION

Relative velocity of parrot w.r.t. train =10 + 20 = 30 m/s. Time taken by parrot to cross the train = 300 / 30 = 10 s


Q. 395 The slope of velocity-time graph represents


A. speed .

B. velocity.

C. acceleration.

D. time.

Right Answer is: C

SOLUTION

Slope of velocity-time graph represent the rate of change of velocity with time which is acceleration.


Q. 396  Acceleration is negative when


A. velocity is decreasing.

B. velocity is increasing.

C. speed is increasing.

D. velocity is zero.

Right Answer is: C

SOLUTION


Q. 397 The graph representing zero acceleration is


A.

B.

C.

D.

Right Answer is: B

SOLUTION


velocity = dx/dt =constant (straight line)
acceleration, a = dv/dt = 0
i.e., position- time graph is a straight line.  


Q. 398 The acceleration due to gravity is 9.8 ms-2. It means that the acceleration due to gravity is 9.8 ms-2, acting


A. vertically downward.

B. vertically upward.

C. at angle 45.

D. at angle 60.

Right Answer is: A

SOLUTION

A body`s acceleration due to gravity (g = 9.8 ms-2) always acts in vertically downwards direction no matter whether it is rising or falling. When the body rises, the velocity decreases at the rate of 9.8 m/s every second. When the body is falling, its velocity increases at the rate of 9.8 m/s every second.


Q. 399 A runner makes one lap around a 200m circular track in a time of 25 s. The runner`s average speed and average velocity are


A. 0 m/s, 0 m/s.

B. 8 m/s, 0 m/s.

C. 0 m/s, 8 m/s.

D. 4 m/s, 4 m/s.

Right Answer is: B

SOLUTION

Average speed = total distance / total time = 200 / 25 = 8 m/s.
Net displacement = 0.
So, average velocity = 0.


Q. 400  If the velocity of a particle is given by v = , then its acceleration is


A. - 12 ms-2.

B. - 10 ms-2.

C. - 8 ms-2.

D. - 4 ms-2.

Right Answer is: C

SOLUTION


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